MACHINE PROOFS IN GEOMETRY Automated Production of Readable Proofs for Geom.etry TheoreJlS
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Series on Applied Mathematics Volume 6
MACHINE PROOFS IN GEOMETRY Automated Production of Readable Proofs for Geometry Theorems Shang-Ching Chou Department of Computer Science The Wichita State University
Xiao-Shan Gao Institute of Systems Science A cademia Sinica, Beijing
Jing-Zhong Zhang Chendu Institute of Computer Application Academia Sinica, Chendu
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Library of Congress Cataloging-in-Publication Data Chou, Shang-Ching, 1946Machine proofs in geometry: automated production of readable proofs for geometry theorems I Shang-Ching Chou, Xian-Shan Gao, Jing -Zhong Zhang. p. cm. -- (Series on applied mathematics; vol. 6) Includes bibliographical references and index. ISBN 9810215843 1. Axioms--Data processing. 2. Automatic theorem proving. I. Gao, Xian-shan . II. Chang, Ching-chung, 1936- III. Title. IV. Series. QA481.C48 1994 94-5809 516'.OO285'51--dc20 CIP
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Foreword Polya named Descartes' Rules for the Direction of the Mind his favorite book on how to think. In the Rules, Descartes speculates on the possible existence of a powerful, secret method known to the ancients for doing geometry: But my opinion is that these writers then with a sort oflow cunning, deplorable indeed, suppressed this knowledge. Possibly they acted just as many inventors are known to have done in the case of their discoveries, i.e., they feared that their method being so easy and simple would become cheapened on being divulged, and they preferred to exhibit in its place certain barren truths, deductively demonstrated with show enough of ingenuity, as the results of their art, in order to win from us our admiration for these achievements, rather than to disclose to us that method itself which would have wholly annulled the admiration accorded. Descartes, that master of both method and geometry, would have been amazed to see the publication of this volume by Chou, Gao, and Zhang, who here present a method for proving extremely difficult theorems in geometry, a method so simple and efficient it can be carried out by high school students or computers. In fact, the method has been implemented in a computer program which can prove hundreds of difficult theorems and moreover can produce simple, elegant proofs. In my view, the publication of this book is the single most important event in automated reasoning since Slagle and Moses first implemented programs for symbolic integration. All too often, research in automated reasoning is concerned with technical questions of marginal interest to the mathematics community at large. Various strategies for efficient substitution and propositional or equational reasoning have dominated the field of automated reasoning, and the mechanical proving of difficult theorems has been a rare event. Mechanical searches for the proofs of difficult theorems are usually guided extensively by the 'user' . Almost never do we find exhibited a computer program that can routinely treat hard problems in any area of mathematics, but in this book we do! The skeptical reader is urged to flip through the 400 difficult theorems in Chapter 6, all mechanically proved, and try his hand. The exceptional simplicity of the mechanically generated proofs presented for many of these theorems illustrate that the authors' method is not some algorithm of mere "in principle," proof-theoretic relevance, such as Tarski's method. No, by applying the simple method fonnulated by the authors, the reader may also quickly become an expert VII
viii
Forward
at geometry proving. The stunning beauty of these proofs is enough to rivet the reader's attention into learning the method by heart. The key to the method presented here is a collection of powerful, high level theorems, such as the Co-side and Co-angle Theorems. This method can be contrasted with the earlier Wu method, which also proved astonishingly difficult theorems in geometry, but with low-level, mind-numbing polynomial manipulations involving far too many terms to be carried out by the human hand. Instead, using high level theorems, the Chou-Gao-Zhang method employs such extremely simple strategies as the systematic elimination of points in the order introduced to produce proofs of stunning brevity and beauty. Although theorems are generally regarded as the most important of mathematical results by the mathematics community, it is methods, i.e., constructions and algorithms, that have the most practical significance. For example, the reduction of arithmetical computation to a mechanical activity is the root of the computing industry. Whenever an area of mathematics can be advanced from being an unwieldy body of theorems to a unified method, we can expect serious practical consequences. For example, the reduction of parts of the calculus to tables of integrals and transforms was crucial to the emergence of modem engineering. Although arithmetic calculation and even elementary parts of analysis have reached the point that computers are both faster at them and more trustworthy than people, the impact of the mechanization of geometry has been less palpable. I believe this book will be a milestone in the inevitable endowment of computers with as much geometric as arithmetic prowess. Yet I greet the publication of this volume with a tinge of regret. Students of artificial intelligence and of automated reasoning often suffer from having their achievements disregarded by society at large precisely because, as Descartes observed, any simple, ingenious invention once revealed, seems first obvious and then negligible. Progress in the automation of mathematics is inherently dependent upon the slow, deep work: of first rate mathematicians, and yet this fundamentally important line of work: receives negligible societal scientific support in comparison with those who build bigger bombs, longer molecules, or those multi-million-line quagmires called 'systems' . Is it possible that keeping such a work of genius as this book secret would be the better strategy for increased research support, using the text as the secret basis for generating several papers and research proposals around each of the 400 mechanically proved theorems, never revealing the full power of the method?
Robert S. Boyer December, 1993
Preface This book reports a recent major advance in automated theorem proving in geometry which should be of interest to both geometry experts and computer scientists. TIle authors have developed a method and implemented a computer program which, for the first time, produces short, readable, and elegant proofs for hundreds of geometry theorems. Modem computer technology and science make it possible to produce proofs of theorems automatically. However, in practice computer theorem proving is a very difficult task. Historically, geometry theorem proving on computers began in earnest in the fifties with the work of Gelemtner, J. R. Hanson, and D. W. Loveland [102]. This work and most of the subsequent work [128, 142, 82] were synthetic, i.e., researchers focused on the automation of the traditional proof method. The main problem of this approach was controlling the search space, or equivalently, guiding the program toward the right deductions. Despite initial success, this approach did not make much progress in proving relatively difficult theorems. On the other hand, in the 1930s, A. Tarski, introduced a quantifier elimination method based on the algebraic approach [34] to prove theorems in elementary geometry. Tarski's method was later improved and redesigned by A. Seidenberg [145], G. Collins [83] and others. In particular, Collins' cylindrical decomposition algorithm is the first Tarski type algorithm which has been implemented on a computer. Solutions of several nontrivial problems of elementary geometry and algebra have been obtained using the implementation [45,113] . A breakthrough in the use of the algebraic method came with the work of Wen-Tsiin Wu, who introduced an algebraic method which, for the first time, was used to prove hundreds of geometry theorems automatically [162, 36]. Many difficult theorems whose traditional proofs need an enormous amount of human intelligence, such as Feuerbach's theorem, Morley's trisector theorem, etc., can be proved by computer programs based on Wu's method within seconds. In Chou's earlier book [12], there is a collection of 512 geometry theorems proved by a computer program based on Wu's method. However, if one wishes to look at the proofs produced using Wu's method, helshe will find tedious and formidable computations of polynomials. The polynomials involved in the proofs can contain hundreds of terms with more than a dozen variables. Because of this, producing short, readable proofs remains a prominent challenge to researchers in the field of automated theorem proving. Recently, the authors developed a method which can produce short and readable proofs
ix
Preface
x
for hundreds of geometric statements in plane and solid geometries [193, 71 , 72, 73]. The starting point of this method is the mechanization of the area method, one of the oldest and most effective methods in plane geometry. One of the most important theorems in geometry, the Pythagorean theorem, was first proved using the area method. But the area method has generally been considered just some sort of special trick for solving geometry problems. J. Zhang recognized the generality of the method and developed it into a systematic method to solve geometry problems [40,41,42]. By a team effort, we have further developed this systematic method into a mechanical one and implemented a prover which has been used to produce elegant proofs for hundreds of geometry theorems. This book contains 478 geometry problems solved entirely automatically by our prover, including machine proofs of 280 theorems printed in full. The area method is a combination of the synthetic and algebraic approaches. In the machine proofs, we still use polynomial computation; but we use geometric invariants like areas and Pythagoras differences instead of coordinates of points as the basic quantities, and the geometric meaning for each step of the proof is clear. Another important feature of the area method is that the machine proofs produced by the method/program are generally very short; the formulas in the proofs usually have only afew terms, and hence are readable by people. The method is complete for constructive geometry theorems, i.e., those statements whose diagrams can be drawn using a ruler and a pair of compasses only. The area is used to deal with geometry relations like incidence and parallelism. Another basic geometry quantity in our method is the Pythagoras difference, which is used to deal with geometry relations like perpendicular and congruence of line segments. Besides the area and Pythagoras difference, we also use other geometry quantities, such as the full-angle, the volume, the vector, and the complex number. The reason we use more geometry quantities is that for each geometry quantity, there are certain geometry theorems which can be proved easily using this quantity.
Another aspect of automated geometry theorem proving relates to the difficulty of learning and teaching geometry. About two thousand years ago, an Egyptian king asked Euclid whether there was an easier way to learn geometry. Euclid's reply was, ''There is no royal road to geometry." Of course, the difficulty here was not the basic geometry concepts, such as points, lines, angles, triangles, lengths, areas, etc. The difficulty has been with many other fascinating facts (theorems) and how to use logical reasoning to justify (prove) these theorems based on only a few basic facts (axioms) that are so obvious that they can be taken for granted. One can draw dozens of triangles with three medians and find the fact that the three medians of a triangle intersect at the same points. However, the empirical observation is only a justification of formation of a conjecture whose correctness must be proved by other means - logical reasoning. 1 The
prover is available via ftp at emcity.cs.twsu.edu: pub/geometIy.
xi Unlike algebra, in which most problems can be solved according to some systematic method or algori~ a human proof requires a different set of tricks for each geometry theorem, making it difficult for students to get started with a proof. In the two thousand years since the time of the Egyptian king, people have continued to wish for an easy way to learn geometry. In response to this difficulty and in working with middle school students, J. Z. Zhang has established a new geometry axiom system based on the notion of area [40, 41 , 42]. Using his new system, Zhang has made a great effort in promoting a new reform in high school geometry education in China. 1be successful implementation of his method has led to its use in Chinese geometry textbooks for teachers' colleges. In addition, the area method has been used in recent years to train students of Chinese teams for participation in the International Mathematical Olympiads. One of the goals of this book is to make learning and teaching of geometry easy. The machine proofs generated have a shape that a student of mathematics could learn to design with pencil and paper. By reading the machine produced proofs in this book, many readers might be able to use the mechanical method to prove difficult geometry theorems themselves.
This book consists of six chapters. 1be first chapter is about the basic concepts of geometry, area and ratio of lengths. Then we introduce the new method, the area method for proving geometry theorems. This chapter can stand alone as a supplement to textbooks of high school or college geometry. We present this chapter at an elementary level with many interesting examples solved by the new method, with the intention of attracting many readers at various levels, from high school students to university professors. It is also our hope that people will be able to prove many difficult geometry theorems using the method introduced in this chapter. Beginning with Chapter 2, we formalize or mechanize the method by describing the method in an algorithmic way. Those who are interested in geometry only will have a clearer idea about this mechanical procedure of proving geometry theorems. Those who wish to write their own computer programs will be able to produce short and readable proofs of difficult geometry theorems. Experts will be able to find further extensions, developments and improvements in this new direction. Only when more and more people are participating in projects of this kind will real advances in geometry education be possible. One of our goals in writing this book is to encourage such research and advances. The last chapter, Chapter 6, is a collection of 400 theorems proved by our computer program, including machine proofs of 205 theorems printed in full. Most of this chapter was generated mechanically, including machine proofs in TsX typesetting form. This chapter is an integral and important part of the book, because it alone shows the power of our mechanical method and computer program. Even reading the proofs produced by our computer program is enjoyable.
Preface
xii
Funding from the National Science Foundation gave us a unique opportunity to form a very strong research team at Wichita State University and to make this book possible. Also the Chinese National Science Foundation provided further support for Gao and Zhang. The authors wish to thank many people for their support of our research project. Dr. K. Abdali of National Science Foundation constantly encouraged Chou and recommended further funding to invite the third author, J. Zhang, to the U.S. This turned out to be the crucial step for this exciting new development and publication of this book. The encouragements from R. S. Boyer, L. Wos, W. T. Wu, W. W. Bledsoe, and J S. Moore when they heard about our early work were invaluable. They also thank Frank Hwang for his encouragement.
In Wichita, Mary Edgington has also played a key role in promoting the project. The authors also thank Patsy P. Sheffield and L. Yang for their careful reading of the manuscript and helpful suggestions, and thank Tom Wallis, the computer system manager, for providing an excellent computing environment. Chou used a part of this work in his graduate course "Symbolic and Algebraic Computation;" the manuscript of this book was also read by many students in his class. In particular, we wish to thank Joe G. Moore for his proof-reading and Nirmala Navaneethan for her suggestions. The support and patience of the authors' families were also very important for the success of this project.
S. C. Chou X. S.Gao J. Z. Zhang December, 1993
Table of Contents Foreword
vii
Preface
ix
Part I. The Theory of Machine Proof 1 Geometry Preliminaries
1 1 3 5 8
......
1.1
Introduction
1.2
Directed Line Segments
1.3
Areas and Signed Areas
1.4
The Co-side Theorem
..... .. .
1.8
Trigonometric Functions . .
16 20 26 33
1.9
Circles
... . ..
38
1.5
Parallels
1.6
The Co-angle Theorem
1.7
Pythagoras Differences
1.10 Full-Angles
44 49
. . .
Summary of Chapter 1 .
2 The Area Method 2.1 2.2
2.3
2.4
51
Traditional Proofs Versus Machine Proofs .
51
Signed Areas of Oriented Triangles
54 54
2.2.1
The Axioms . . . . . . . . . . . .
2.2.2
Basic Propositions . . . . . . .. .
57
The Hilbert Intersection Point Statements .
59 60 63
2.3.1
Description of the Statements .
2.3.2
The Predicate Form . . . . . . . . . . . . . . .. .
64
2.4.1
Eliminating Points from Areas .
2.4.2
Eliminating Points from Length Ratios
65 66
The Area Method
xiii
CoBteBta
xiv
2.5
2.6
2.7
2.4.3
Free Points and Area Coordinates
2.4.4
Working Examples
. .. .. .
71
More Elimination Techniques . . . . .
76 76 79
2.5.1
Refined Elimination Techniques
2.5.2
The 1\vo-line Configuration
Area Method and Affine Geometry
. .
. ...
2.6.1
Affine Plane Geometry
2.6.2
Area Method and Affine Geometry
Applications . . ..
........ .
2.7.1
Formula Derivation . . . . . .
2.7.2
Existence of n3 Configurations
2.7.3
Transversals for Polygons
3 Machine Proof in Plane Geometry
3.2
3.4
3.5
3.6
. . . . . .
3.1.1
Pythagoras Difference and Perpendicular
3.1.2
Pythagoras Difference and Parallel
3.1.3
Pythagoras Difference and Area . . .. .
3.2.1
3.3
101
The Pythagoras Difference
Constructive Geometry Statements
.. . .. . .
Linear Constructive Geometry Statements
3.2.2
A Minimal Set of Constructions
3.2.3
The Predicate Form
Machine Proof for Class CL . . . . . . 3.3.1
The Algorithm .. . .. . . . .
3.3.2
Refined Elimination Techniques.
The Ratio Constructions . . . . . . . 3.4.1
More Ratio Constructions ..
3.4.2
Mechanization of Full-Angles
Area Coordinates
81 82 83 86 86 90 93 99
Summary of Chapter 2 . . . . . . . . . . .
3.1
68
.. .. . . . . . .
3.5.1
Area Coordinate Systems ..
3.5.2
Area Coordinates and Special Points of Triangles
Trigonometric Functions and Co-Circle Points . 3.6.1
The Co-circle Theorems .. .
3.6.2
Eliminating Co-Circle Points . . . . .
101 101 103 105 107 107 109 111 112 112 117 119 120 125 130 130 132 138 138 140
xv
CoDteDts
3.7
Machine Prooffor Qass C . . . . . . . . . . . . . . .
144
3.7.1 3.7.2
Eliminating Points from Geometry Quantities . Pseudo Divisions and Triangular Forms . . . .
145 147
3.7.3
Machine Prooffor Qass C . . . . . . . . . . .
3.8
Geometry Information Bases and Machine Proofs Based on Full-Angles 3.8.1 Building the Geometry Information Base . . . . . . 3.8.2 Machine Proof Based on the Geometry Information Base . Summary of Chapter 3 . . . . . . . . . . . . . . . . . .
4 Machine Proof in Solid Geometry 4.1 The Signed Volume . .. 4.1.1 Co-face Theorem . . . . . 4.1.2 Volumes and Parallels . . .
150 154 154 159 165 167 167 169 171
4.1.3 Volumes and Affine Geometry of Dimension Three Constructive Geometry Statements . . . . 4.2.1 Constructive Geometry Statements 4.2.2 Constructive Configurations . . .
173 176 177 180
4.3
Machine Proof for Class SH . . . . . . . . 4.3.1 Eliminating Points from Volumes . 4.3.2 Eliminating Points from Area Ratios 4.3.3 Eliminating Points from Length Ratios 4.3.4 Free Points and Volume Coordinates 4.3.5 Working Examples . . . . . . . . . .
4.4
Pythagoras Differences in Space . . . . . . . .
183 183 186 188 192 194 197
4.2
4.5
4.6
4.4.1
Pythagoras Difference and Perpendicularity .
4.4.2
Pythagoras Difference and Volume
The Volume Method . . . . 4.5.1 The Algorithm . . .
203 203
4.5.2 Working Examples Volume Coordinate System
206 211
Summary of Chapter 4 . . . . . .
5 Vectors and Machine Proofs 5.1
197 200
Metric Vector Spaces of Dimension Three . 5.1.1 Inner Products and Metric Vector Space . 5.1.2
Exterior Products in Metric Vector Space
214 217 217 219 223
Contents
xvi
5.2
The Solid Metric Geometry . . . . . . . . 5.2.1 Inner Products and Cross Products . 5.2.2 Constructive Geometry Statements 5.3 Machine Proof by Vector Calculation .. . 5.3.1 Eliminating Points From Vectors . 5.3.2 Eliminating Points from Inner and Exterior Products 5.3.3 The Algorithm . . . .. . . . . . . . . . . 5.4 Machine Proof in Metric Plane Geometries . . . . . . . 5.4.1 Vector Approach for Euclidean Plane Geometry . 5.4.2 Machine Proof in Minkowskian Plane Geometry 5.5 Machine Proof Using Complex Numbers Summary of Chapter 5 . . . . . . . . . . .
225 227 230 231 231 234 236 239 240 243 246 253
Part II. Topics From Geometry A Collection of 400 Mechanically Proved Theorems
6 Topics From Geometry 6.1 Notation Convention . . 6.2 Geometry of Incidence . 6.2.1 Menelaus'Theorem 6.2.2 Ceva's Theorem . . 6.2.3 The Cross-ratio and Harmonic Points 6.2.4 Pappus' Theorem and Desargues' Theorem 6.2.5 Miscellaneous. . . .. 6.3 Triangles . . . . .. . . . . . . . . 6.3.1 Medians and Centroids .. 6.3.2 Altitudes and Orthocenters 6.3.3 The Circumcircle .. 6.3.4 The Euler Line. . . . . . . 6.3.5 The Nine-Point Circle . . . 6.3.6 Incircles and the Excircles . 6.3.7 Intercept Triangles . . 6.3.8 Equilateral Triangles . 6.3.9 Pedal Triangles 6.3.10 Miscellaneous .. . .
255 255 259 259 262 266 270 274 281 281 293 303 321 323 327 341 346 348 350
XVll
6.4
Quadrilaterals . . .. . . . . . . 6.4.1 General Quadrilaterals . . 6.4.2 Complete QUadrilaterals . 6.4.3 Parallelograms . . . . 6.4.4 Squares . . . . . . . . . 6.4.5 Cyclic Quadrilaterals . . . . 6.4.6 Orthodiagonal Quadrilaterals 6.4.7 The Butterfly Theorems 6.5 Circles . . . . . . . . . . . . . . . . . 6.5.1 Chords, Secants, and Tangents . 6.5.2 Intersectional Circles 6.5.3 The Inversion . . . 6.5.4 Orthogonal Circles . . . 6.5.5 The Simson Line .. . . 6.5.6 The Pascal Configuration . . . 6.5.7 Cantor's Theorems 6.6 A Summary
357 357 365 369 380 387
393 396 400 400 415 421 426 429 435 439 440
Bibliograpby
445
List of Symbols
455
Index
457
Chapter
1
Geometry Preliminaries In this chapter, we will provide geometry background for the rest of this book. The presentation is informal and the prerequisite in geometry is minimal. Anyone with a semester of high school geometry can read and understand this chapter. We will present a new systematic proof method, the area method, which can be used to solve numerous geometry problems at various levels of difficulty, ranging from problems in high school textbooks to those in mathematics competitions. Those who are mainly interested in machine proofs may skip this chapter and start from Chapter 2 directly.
1.1
Introduction
Geometry, like other sciences, is concerned with the laws in a specific domain. For geometry this domain is space. However, reasoning plays a much more important role in geometry and in other branches of mathematics than in other sciences. There are two kinds of reasoning: inductive reasoning and deductive reasoning. Necessity and curiosity have at all times caused people to investigate phenomena and to find the laws governing the physical universe. Drawing three medians of a triangle, one finds that the three medians intersect at the same point. By making repeated experiments, one can come to the conclusion that the three medians of any triangle always intersect at the same point. This kind of inductive reasoning, which is fundamental in experimental sciences, is also very important for observing new facts and laws in mathematics. However, in geometry or mathematics, inductive reasoning generally cannot serve as the justification of the correctness of the observed fact. To justify the observed fact, we have to give a proof of the fact. Here the terminology "proof' has a distinctive meaning. Based on already proved facts (theorems), we need to use logical or deductive reasoning to derive this new fact about three medians. Once we prove a fact in this way, we call it a theorem. The truth of the fact is thus beyond doubt. Here we prove a theorem "based on already proved facts." However, the proof of these "already proved facts" is based on other .. already proved facts." In the final analysis, we have to choose a few basic facts whose
1
2
Chapter 1. Geometry Preliminaries
correctness is so evident in our everyday life that we can take them for granted without any proofs. For example, the fact that for any two distinct points there is one and only one line passing through them is so evident that we can use it without proof. Such kinds of facts are very basic and are called axioms or postulates. The selection of axioms is by no means evident. For example, Euclid's fifth postulate, "from any point not on a line, there is one and only one line passing through the point and parallel to the given line," is a very evident fact for many people. However, for over two thousand years, mathematicians tried to prove this fact using other basic postulates. The failure of these attempts finally led to the Non-Euclidean geometries and a revolution in mathematics. There is a special branch of mathematics, foundations of geometry, which is exclusively concerned with the related topics. Generally, the traditional proof method in geometry proceeds as follows: first we establish (prove) various basic propositions or lemmas, e.g., the theorems of congruence of two triangles. Then we use these basic propositions to prove new theorems. We can enlarge the set of basic propositions by including some of these newly proved theorems. In the traditional methods used by Euclid and many other geometers since then, theorems of congruence and similarity of triangles are the very basic tools. Although the method based on congruence and similarity leads to elegant proofs for many geometry theorems, it also has weaknesses: (I) In a diagram of a geometry statement to be proved, rarely do there exist congruent or similar triangles. In order to use the propositions on congruent or similar triangles, one has to construct auxiliary lines. As we know, adding auxiliary lines is one of the most difficult and tricky steps in the proofs of geometry theorems. This also leads to a changeable and uncertain strategy in the effort of finding a proof. (2) In propositions on congruent or similar triangles, there is asymmetry between the hypothesis and conclusion. For example, in order to prove the congruence of the segments AB and XY, we need to construct (or find) 6.ABC and 6.XYZ and to prove the congruence of the two triangles. Generally, this in tum requires us to prove the congruence of three pairs of geometry elements (segments or angles). In order to prove one identity, we need to find three other identities! As a consequence, it is very difficult to find an effective method for solving geometry problems based on congruence and similarity. In this book, we will use the area of triangles as the basic tool for solving geometry problems. The traditional area method is one of the oldest and most effective methods in plane geometry. One of the most important theorems in geometry, the Pythagorean theorem, was first proved using the area method. But the area method has generally been considered as a set of special tricks for solving geometry problems. 1. Z. Zhang has been studying the area method since 1975. He has recognized its generality and has developed it
3
1.1 Directed Line Segments
into a systematic method for solving geometry problems [40. 41 . 42].2 A large number of geometry problems at various levels of difficulty. ranging from basic propositions in high school textbooks to those in mathematics competitions. have been provided with elegant proofs using the area method. This chapter is a summary of the basic facts about the area method together with many geometry theorems solved by this method. Another feature of the area method is that the deduction in the method is achieved mainly by algebraic computation. This makes the area method ready for mechanization. which is the main theme of this book.
1.2
Directed Line Segments
For most of the book. we are concerned with plane geometry. Thus our starting point is a plane. which we sometimes refer to as the Euclidean plane. The basic geometry objects on a plane are points and lines. We use capital English letters A . B. C • ... to denote points on the Euclidean plane. For two distinct points A and B . there is one and only one line I that passes through points A and B. We use AB or BA to denote this line. In addition. we can give a line one of the two directions and talk about directed lines. Thus directed line AB has the direction from point A to point B. whereas the directed line BA has the opposite direction from point B to A. Two points A and B on a directed line determine a directed line segment whose length
AB is positive if the direction from A to B is the same as the direction of the line and negative if the direction from A to B is in the opposite direction. Thus
AB= -BA
(1.1)
and AB
= 0 if and only if A = B .
Let A . B. p . and Q be four points on the same line such that A#- B . Then the ratio of PQ and AB is meaningful; let it be t. We have
PQ AB
=
= t,OT PQ = tAB .
If the two directed line segments AB and PQ have the same direction then t ~ 0; if they have opposite directions then t ~ O. If P or Q is not on line AB. then we cannot compare or do arithmetic operations between AB and PQ. because the sign of AB depends on the direction of line AB which has nothing to do with the direction of line PQ. 2This systematic method bas been used to train students of Chinese teams for participating in the International Mathematical Olympiad in solving geometry problems.
Chapter 1. Geometry PreIimlnaries
4
If point P is on line AB then AB
(l.2)
= AP + P B or AP AB
PB
+ AB = l.
We call ~ and !:.!Z. the position ratios or the position coordinates of point P with respect AB AB th. . to AB. It is clear that for two real numbers s and t such that s + t = I, ere IS a uruque point P on AB which satisfies
AP AB
=t,
PB AB
=5 .
In particular, the statement that point 0 is the midpoint of segment AB means 1~
= ~: =
I
2'
Two distinct points always determine a line. However, three points are generally not on the same line. If they are, we say that the three points are collinear. In fact, the most fascinating facts about many elegant geometry theorems discovered over the past two thousand years have been that no matter how you draw a certain geometry figure, the three particular points in the figure are always collinear. Connected with each theorem, there is the collinear line named after the mathematician who discovered it, for example, Pappus' line, Euler's line, Gauss' line, Pascal's line, Simson's line, etc.
c. Example 1.1 (Pappus' Theorem) Let points A, B and C be on one line, and AI, BI and C I be on another line. Let ABI meet AlB in P, ACI meet AIC in Q, and BC I meet BI C in S. Show that P, Q, and S are collinear.
B
C
Figure 1-1
Please draw a few diagrams for this geometric configuration on a piece of paper. Note that the three intersection points P, Q and S are always collinear. If you have our software package, you can see this fact more vividly. First, the program helps you to draw one diagram in a few seconds on the computer screen. Then you can use the mouse to move any of the user-chosen points, e.g .• point C I . The diagram is continuously changed on the screen while you can see that the three moving points p. Q. and S are always on the same line. This observation convinces almost everyone that this statement is always true. However, in order to justify the truth of a fact in mathematics. empirical observations are not enough. We need a proof of the truth of the fact using the mathematical reasoning. The proof of Pappus' theorem (or of many other geometry theorems) is by no means easy. especially to high school students. The main objective of this book is to present a new systematic proof method. We believe that the serious reader can learn this method easily. Once you know this method and work with a few examples. you can prove a difficult theorem in a few minutes. For a proof of Pappus' theorem, see page 14.
1.3
5
Areas and Signed Areas
Exercises 1.2 2 1. Let A , B , and C be three collinear points. Show that AB2 + BC (Use (1.1) and (1.2). Do the same for the following exercises.)
= E + 2AB · CB.
2. Let A , B , C, and D be four collinear points. Then AB· CD + AC · DB + AD · BC
= O.
3. Four collinear points A , B , C, and D are ~Ied a harmonic sequence if ~g = -~~ . Show that four collinear points A , B , C, and D fonn a hannonic sequence if and only if AB AB 2
C'B+DB'= .
4. Show that four collinear points A, B , C, and D fonn a hannonic sequence if and only if OC · OD = <JA2 where 0 is the midpoint of AB.
1.3
Areas and Signed Areas
The next geometric object is the triangle. As we Icnow, three non-collinear points A , B , and C fonn a triangle which is denoted by 6.ABC. The area of 6.ABC is denoted by \l ABC. The reader can safely assume that \l ABC is ~h . BC where h is the altitude of the triangle on the side BC . In our area method, however, we do not consider this a basic fact. Instead, we use other simple facts about the area as basic propositions. If A, B , and C are collinear, 6.ABC is called degenerate and \l ABC is defined to be zero. For any four points A , B , C , and P in the same plane, we can fonn four triangles 6.ABC, 6.P AB, 6.P BC, and 6.P AC and the areas of the four triangles satisfy seven different relations, depending on the relative position of the points A, B , C and P.
.. B
... p •
'
Figure 1-2
If point P is inside the triangle ABC as shown in Figure 1-2, we have
\l ABC
= \l P AB + \l P BC + \l PC A.
Figure 1-3
Chapter 1. Geometry PrelimInaries
6 For the three cases shown in Figure 1-3, we have
'VABC = 'VPAB + 'VPBC - 'VPCA 'VABC = 'VPAB - 'VPBC + 'V PCA 'VABC = - 'V PAB + 'VPBC + 'V PCA
If ABC P is convex. If AB PC is convex. If AP BC is convex.
/~
~ ~,;~ .:___ -- - - - -- -- --
B
Lj,\
A
----- - -- ___ : : : ; '. p
Figure 1-4
For the three cases shown in Figure 1-4, we have
'VABC 'V ABC 'V ABC
= 'VPAB =-
=-
If C is in the interior of 6.P AB. If A is in the interior of 6.P BC. If B is in the interior of 6.P AC.
'VPBC - 'VPCA 'V P AB + 'V P BC - 'V PC A 'V P AB - 'V P BC + 'V PC A
We see that the above relations among the areas of triangles formed from the four points are very complicated. However, if we introduce the signed area of an oriented triangle, we can greatly simplify the relations of these areas; the seven relations can be reduced to just one equality. A triangle ABC has two orientations: if A-B-C is counterclockwise, triangle ABC has the positive orientation; otherwise triangle ABC has the negative orientation. Thus 6.ABC, 6.BCA, and 6.CAB have the same orientation, whereas 6.ACB, 6.CBA, and 6.BAC have the opposite orientation. The signed area of an oriented triangle ABC, denoted by S ABC, has the same absolute value as 'V ABC and is positive if the orientation of triangle ABC is positive; otherwise S ABC is negative. We thus have SABC
=
SBCA
=
S C AB
=
-SACB
=
-SBAC
=
-SC BA.
Now the seven equations for 'V ABC, 'V P AB, 'V P BC, and 'V PC A can be summarized as just one equation: regardless of the position among points A , B , C, and P, we always have
(1.3)
ReTTUlrk for Advanced Readers. Based 00 analytic (coordinate) geometry, the proof of (1.3) is more strict and straightforward, and we do oot have to discuss each of the seveo cases separately. Suppose we have a coordioate system 00 the plane and the coordinates for all four points. For example, the coordinates of the points A, B and C are (a" , ay), (b" , bll ), and (c", cll ) , respectively. Twice the signed area SABC of the
1.3
7
Areas and Signed Areas
biangle ABC is the polynomial (a r - br )(b y of (1 .3) is a straighuOIward computation.
cy) - (a y - by)(br - cr) . Then checking the validity
Funhermore. we do not even need to use the brute force computation of polynomials to prove (1 .3). Since 0 .3) is valid wheo P is inside the biangle ABC and (1.3) is equivalent to a polynomial equation. that equation must be true. Since the truth of the equation is independent of the relative position of the four points. 0 .3) is valid in all cases. With this argument in mind. the understanding of many other identities in this book or in geometry is much easier. For many geometry statements or theorems of equality
type. the order
relation (inside. between. etc.) is irrelevanl By the above argument based on polynomials. if the statement is true in one case. then it is true in all cases regardless of the relative order position of the points involved. There are many geometry theorems in which the order relation is essential. Proving of these theorems is beyond the scope of our current computer program based on the area method. One of the important examples is "A biangle with two equal internal angle bisectors is an isosceles biangle." In this chapter we will prove
many such theorems using the area method. However. the proofs are informal in the sense that we use some facts other than the basic propositions of the area method used in our computer program. End of Remark.
Similarly, we can also define oriented quadrilaterals. Given four points A, B , C, and D, we define an oriented quadrilateral ABCD according to the point orientation AB-C-D . Thus BCDA, CDAB, and DABC denote the same oriented quadrilateral as ABCD because their point orientation is the same. Four points can form six (4!/4) different orientations, hence six different oriented quadrilaterals: ABC D, ADC B, AC B D, AD BC, ACDB, and ABDC, as shown in Figure 1-5.
Figure 1-5
Now we can define the area of an oriented quadrilateral ABC D to be
In order to justify that SABC D is well-defined, we need to prove that for the above definition, we have SABCD
= SBCDA = SCDAB = SDABC
which is a direct consequence of (1.3).
The definition of the signed area here can be generalized to an oriented n-polygon with any n > 4.
Exercises 1.3 I. Prove the following properties of the areas of quadrilaterals.
Chapter 1. Geometry Prelimiuaries
8 SABCD
=
SBCDA
=
SCDAB
-SADCB, SABCD
= SABC -
SADC
=
=
SDABC
= SBCD -
SBAD,
-SDCBA
=
-SCBAD
=
-SBADC =
and
SABBC =SABCC =SAABC =SABCA =SABC.
2. For any five points
A, B,
e, P, and Q in the same plane, we have SPAQB + SPBQC =
SPAQC.
1.4
The Co-side Theorem
The following is the first basic proposition of the area method.
Proposition 1.4 Let A , B, and C be three distinct collinear points, and P a point not on line AB. Then we have ~p!~ = ~~, or if Be = tAB then SPBC = tSPAB.
c
B
A
Figure 1-6
Proposition 1.4 is obvious if we use the area formulas: 'V P Be
= ~h . Be, 'V P AB =
~h . AB where h is the distance from point P to line AB. Notice that the areas involved
here are signed areas of oriented triangles. Two triangles with a common side are said to be a pair of co-side triangles. From Proposition 1.4 we can easily infer the co-side theorem, which is the most important proposition of the area method. By using this theorem alone, we can prove many difficult theorems easily.
A~""'~ A
M
B
Figure 1-7
A
!sL
A
B
Proposition I.S (The Co-side Theorem) Let M be the intersection of the lines PQ and Q i= M. Then we have ~s = QPMM . QAB
M
AB
and
Figure 1-7 shows four possible cases of the co-side theorem. Here we give two proofs of Proposition 1.5, which are valid for all the four cases.
lA
9
The Co-side Theorem
Proof 1. Let N be a point on the line AB such that M N SPAB SQ A B
P
~
.1'2
roo) .
_
5qAS -
~
~
= AB. By Proposition 1.4
PM
SPMN
= SQMN = QM '
5904 .. _
5PA .. • 5qA .. . 5qAS -
AB
PM
I
AM
AM . QM . AB
PM = QM'
We mentioned early that in the diagrams of geometry theorems, rarely are there similar or congruent triangles. But there are plenty of co-side triangles. For instance, in each of the four diagrams in Figure 1-7 there are 18 pairs of co-side triangles! Propositions 1.4 and 1.5 are two basic propositions. We will expand the set of basic propositions and, based on them, introduce our area proofmethod. Using the area method to prove theorems, no skillful trick is needed : we need only to follow systematic or prescribed steps to reach the completion of a proof. We will use several non-trivial examples to illustrate how to use the basic propositions to prove theorems. Example 1.6 Let 6ABC be a triangle and P be any point in the plane (inside or outside of the triangle). Let D be the intersection of lines AP and C B, i.e., D = AP nCB. Also let E = BPnAC. and F = Cpn AB. Show that ~g + ~~ + ~~ = 1.
A
F
A
B
F
A
F
B
Figure \-8
The hypotheses of most geometry theorems can be stated in a constructive way: beginning with some arbitrarily chosen points, lines, and circles, we introduce new points by taking arbitrary points on or taking intersections of the lines and circles; from the constructed points, new lines and circles can be formed; we now can introduce new points from these new lines and circles. etc; finally. a figure is formed consisting of points, lines, and circles. Such kinds of geometry theorems are called theorems of constructive type. Example 1.6 is a theorem of constructive type. Beginning with the arbitrarily chosen (free) points A, B, C and P, we introduce new points D, E, and F by constructing
Chapter 1. Geometry PreUrninaries
10
intersections of lines AP and BC, lines BP and AC, and lines CP and AB. Our aim is to eliminate the constructed points from the left-hand side of the conclusion
in the reverse order of the points in which they are introduced until all points in the expression are free points. Then the expression is equal to or is easily proved to be equal to the right-hand side of the conclusion. Proof. By using the co-side theorem three times we can eliminate points D,E, and F respectively: PE SPGA PD SPBG = = - -, AD = SABG' BE SABG By (l.3) on page 6, PD AD
+ PE + PF = SPBG + SPGA + SPAB = SABG = 1. BE
CF
SABG
SABG
If not using the signed area, we would have to discuss several (seven) cases when P is inside or outside the triangle ABC. Figure 1-8 shows three possible cases of the example. The use of the signed area makes the proof concise and more strict. The reader will see this advantage throughout this book in other examples. At this point, we need to mention that the non-degenerate conditions which are necessary for a geometry statement to be true are not stated explicitly in the example. Here triangle ABC must be non-degenerate, i.e., SABG i= 0, as we generally implicitly assume in geometry textbooks. But this is not enough. We need each of the intersection points D, E and F to be normal, i.e., there is one and only one intersection point. This imposes conditions on the point P . For detailed discussion of non-degenerate conditions, see Chapter 2.
Example 1.7 (Ceva's Theorem) The same hypotheses (constructions) as Example 1.6. Show that AF BD CE = ·= · ==1 FB DC EA . Proof. Our aim is still to eliminate the constructed points F, E and D from the left-hand side of the conclusion. Using the co-side theorem three times, we can eliminate E, F, and D AF BD CE SAPG SBPA SGPB =·= · ==-- · --·--=1 FB DC EA SBGP SGAP SABP .
The above proof for Ceva's theorem can be extended immediately to prove Ceva's theorem for an arbitrary (2m + 1)-polygon with m ~ 1.
lA
The Co-side 1'beorem
11
Example 1.8 (Ceva's Theorem for a (2m + 1)-polygon) Let a point 0 and an arbitrary polygon V1··.v2m+1 be given. Let P; be the intersection ofline OV; and the side V;+m V;+m+l' Then 2m+1 - -
II
=1
V;+kP;
;=1 P;V;+k+1
where the subscripts are understood to be mod 2m + 1. Proof By the co-side theorem,
~ P.;1T ";+m+l
_ -
SOV;V;tm
SOV;tmtl V; '
i
= 1, ... ,2m+ 1.
Multiplying the above equations together and noticing that the (i + m )-th element in the denominator SovHm)tmtl VHm = SOV;VHm is just the i-th element in the numerator, we prove the result. I
B
A
F
A
B
F
Figure 1-9
Example 1.9 (Menelaus' Theorem) F, D , and E are three points on sides AB, BC, and CA of.!! triangle ABC respectively. Show that E, F, and D are collinear if and only if ~ . !JJl . ~ = -1 FB DC EA •
Proof A transversal may meet two sides of a triangle and the third side produced, or all three sides produced (Figure 1-9). The following proof is valid for both cases. If E, F, and D are collinear, by the co-side theorem,
AF FB Then
SAEF
= - SBEF'
BD DC
= - SCEF '
CE EA
SCEF
= - SAEF '
eFB . !lJl. . ~ = -1 . DC EA
Conversely, let E , F , and D be points on AC, AB, and BC such that :~. ~g. ~! = -1 . Let EF meet BC in H. Then we need only to show D H. By what we just proved AF BH CE BD· FB . Hc . EA = - 1. A s a consequence Bii Hc -- Dc' I.e., D -- H . I
=
We can extend the Menelaus' theorem to m-sided polygons. Example 1.10 (Menelaus' Theorem for an m-polygon) Let Vt 1, ... , min p;. Then
A line XY meets V;V;+l> i
=
IT
V;P;
;=1 P;V;+l
= (-lr.
... Vm be an m-polygon.
Chapter 1. Geometry Preliminaries
12
Proof. By the co-side theorem,
Multiplying the m equations together, we obtain the result.
We can use the area method to solve many geometry problems other than theorem proving.
Example 1.11 Let ABC be a triangle, D and E be two points on the lines AC and AB such that CD = uAD and AE = vBE. Let P be the intersection of BD and CEo Express ~~ in terms ofu and v. Solution. We need to eliminate the constructed points P, E, and D successively. By the co-side theorem we have3 PD PB
SDCE
= SBCE·
Figure 1-10
Now the right-hand side of the above equality is free of the point P . We still need to eliminate the points E and D from it. By the co-side theorem again, we have: SDCE
SEDC
SEAC
SBCE
= SEAC
. SEBC
DC
EA
-uAD vEB uAD· EB
= AC · EB = AD -
Let us look at the special case when u side AC and AB respectively. Then
uv
= (u -
1)"
= v = -1, i.e., D and E are the midpoints of the
The following theorem follows from this fact immediately.
Example 1.12 (The Centroid Theorem) The three medians of a triangle meet in a point, and each median is trisected by this point. Example 1.13 The same hypotheses as Example 1.11. Express ~s in terms of u and v. ABO
3Here we mention triangle DeE. The side DE of the triangle is not in the figure. Our method will add those lcinds of auxiliary lines automatically.
1.4 The Co-sIde Theorem
13
Solution. By (1.3) on page 6 and the co-side theorem, we have SABC SPBC
SAPC
SABP
SPBC
AE
AD
1
= SPBC + SPBC + SPBC = EB + DC + 1 = -v - ~ + 1 =
S Therefore, i!£BJ:4 s ABC
u - 1- UV u
= -IU -' -uv u-
Example 1.14 Take three points D , E, and F on sides AG. AB. and BG of the triangle ABC such that ~ AD = u • ~ BE = v • !!L CF = W . Let R = AF n BD. P = BD nEG, and R Q = AF n CEoExpress sSP9 in terms ofu. ABC V . andw.
B
c
F
Figure 1-11
Solution. By (1.3) on page 6 and Example 1.13,
=
SPQR
SABC -
SPBC -
SQCA -
SRAB
w
v )SABC 1 - uv w - 1 - wu v-I - wv (1 + wuv)2 S (1 - u + uv)(1 - v + wv)(1 - w + wu) ABC · U
(1 -
U -
Remark 1.IS
1. In Example 1.14. ifu
= v = w = -~ then we have ~:~~ = t.
2. As a consequence of Example 1.14, we have proved a stronger version of the Ceva theorem: lines AF, BD. and C E are concurrent if and only if S PQR = O. i.e., if and
only ifuvw
+ 1 = 0 or
Example 1.16 4 Let L be the intersection of AB and CD, K the intersection of AD and BC, F the intersection of BD and KL. and G the intersection of AC and K L. Then
LF KF
LG
= GK '
L
F
K
G
Figure 1-12
4This theorem is sometimes referred as the basic principle of the projective geometry. It can be used to define the hannonic sequence geometrically. For more details see page 365.
14
Chapter 1. Geometry PreIlmlaarfes
Proof 1. By the co-side theorem, we can eliminate point F: ~~ to eliminate points K and L: LB
SLBD LF Then KF
= ABSABD =
=~
SBA C D '
SBCDSABD ,SKBD SBCAD
= ~~:~ . We need further
KB
= CB SCBD =
S'mu'1 ar1y, we can prove that GK LG
SBADSCBD S BACD
LF = ~ = 'K"F' SBACD
Proof 2. The following proof is shorter. LF KF
=
SLBD SKBD
SLBD SKBL
DA
LC
SDAC SLAC
SLAC
LG
= SKBL SKBD = AK' DC = SAKC SDAC = SAKC = GK '
Example 1.17 (Pappus' Theorem) Let points A, Band C be on one line, and Ai> BI and C l be on another line. Let P = ABI n AlB, Q = ACI n AlC, and S = BCI n BIG. Show that P, Q, and S are collinear. Proof We generally convert a problem of collinearity into a ratio problem in the following way. Let Zl = PQnBCl and Z2 = PQnBlC. We need only to prove that Zl = Z2' This is equivalent to
(1)
Figure 1-13
We can eliminate points Z2 and Zl from (1) using the co-side theorem. PZ1 Q Z 2 SPBC. SOCB. QZl . PZ2 SOBC, . SPCB, . (2) Now we want to eliminate points Q and P from the right-hand side of (2) using the following identities which can be easily obtained using the co-side theorem.
=
1.4 The Co-sIde Theorem
15
The following extension of the co-side theorem is also very useful.
~
Proposition 1.18 Let R be a point on line PQ. Then for any two points A and B
A
B
Figure 1-14
Proof By the co-side theorem ~s QAP
SRAB = SPAB = SPAB
= _pPQR, ~ss = !.!l. QPB PQ By (1.3) on page 6,
+ SRAP + SRPB
= SPAB
PR
+ PQ(SAPQ -
SBPQ)
PR
+ PQ(SAPB + SABQ)
PR (1 - PQ)SPAB PR = PQSQAB
PR
+ PQSQAB
RQ
+ PQSPAB.
Example 1.19 (0.116, 4, 7) The circumdiameters AP, BQ, C R of a triangle ABC meet the sides BC, CA, AB in the points K. L, M . Show that (KP/AK) + (LQ/BL) +
(MR/CM)
= 1.
Proof By the co-side theorem.
RM SABR QL CM SABC BL
-SACQ PK SBCP SABC AK SABC
= = - - , = = - - - , ==--. Note that 0 is the midpoint of AP. BQ. and CR. Using Proposition 1.18. we can eliminate points R, Q, and P . Figure 1-15 SABR =2SABO - SABC, SACQ =2SAco + SABC. SBCP =2SBco - SABC . Thus
RM MC
+ QL + PK = -(2SBCO + 2SAoc + 2SABO LB
KA
SABC
3SABC) = SABC = 1. SABC
Remark 1.20 In summary, we see that to prove a geometry theorem using the area method systematically, we generally follow three steps: first, formulate the geometry theorem in
Chapter 1. Geometry Preliminaries
16
a constructive way and state the conclusion of the theorem as an expression in areas and ratios of directed line segments; second, based on the basic propositions about areas, try to eliminate the points from the conclusion in the reverse order in which the points are introduced; finally state whether the conclusion is true or not.
Exercises 1.21 I. There are 48 pairs of co-side triangles in each diagram of Figure 1-8. Try to count how many co-side triangles are there in Figure 1-12. 2. If A , B , C, and D are on the same line, then for points P and Q such that SPCQD '" O. we have SPAQB/SPCQD = AB/CD. 3. The same hypotheses as Example 1.6. Show that 1~
+ ~~ + g; = 2.
4. Let ABC D be a quadrilateral and 0 a point. Let E. F, G, and H be the intersections of lines AO, BO, CO, and DO with the corresponding diagonals BD, AC. BD, and AC al Show that HcFAwGi'i AHCFBE7JG_1 of the quad n'l ater. - . 5. (Lesening's Theorem) Continuing from Example 1.17, let L 1 , L 2 , L3 be the intersections of lines OP and CCl> lines OQ and BBl> and lines OS and AA 1 • Show that LI, L 2 , L3 are collinear.
1.5
Parallels
Definition 1.22 Let AB and CD be two non-degenerate lines. If AB and CD do not have any common point, we say that AB is parallel to CD. We use the notation AB II CD to denote the fact that A , B, C, and D satisfy one of the following conditions: (1) AB and CD are parallel; (2) A = B or C = D; or (3) A. B. C and D are on the same line. A parallelogram is an oriented quadrilateral ABC D such that AB II CD. BC II AD. and no three vertices of it are on the same line. Let ABC D be a parallelogram. It is clear that line AB and line DC have the same direction. Let ABC D be a parallelogram and P, Q two points on CD. We define PQ PQ AB = DC to be the ratio of two parallel line segments. We have the following very useful and simple result which is also basic to our method.
Proposition 1.23 Let A. B, C. and D be four points. AB = SABD or SADBC = O.
II
CD if and only if SABe
1.5
17
Parallels
If we assume the area for a triangle to be ah/2. then Proposition 1.23 is obvious. However. we can prove it using Proposition 1.4 (see page 57 for details). With Proposition 1.23. we can prove many other theorems easily.
Example 1.24 Let 0 be the intersection of the two diagonals AC and BD of a parallelogram ABCD. Show that AO = OC, or ~=1 DC .
l>
A
B
Figure 1-16
The construction of the figure proceeds as follows. First we have three arbitrarily chosen points A. B and C . Then we take a point D such that AB = DC. Finally. we take the intersection of two lines AC and BD to obtain the point O. Thus we need to eliminate 0 from the expression ~g . then eliminate the point D from the new expression. Here is the proof.
Proof AD_~
Oc -
-
by the co-side theorem
SB C D
-~ SBCA
(since AB
II CD and AD II BC, SABD = SABC, SBCD = SBCA.)
=~=1 SABC .
Example 1.25 Three parallel lines cut two lines at A. B. C, and X, Y. Z respectively. Show that AB XY CB = ZY ·
Figure 1-17
This proposition is considered a very basic property of parallel lines in high school geometry. Here we can prove it very elegantly .
Proof By the co-side theorem and Proposition 1.23. we have
Example 1.26 Lines AB and CD are parallel. Let P = ACn BD andQ = ADn BC. Line PQ intersects line AB at M . Show that M is the midpoint of AB. i.e.• AM = M B .
B
Figure 1-18
Chapter 1. Geometry Prellmlnaries
18 Proof 1. By the co-side theorem. we can eliminate M: theorem again. we can eliminate Q:
Now by Proposition 1.23 • ~ MB
=~ = SPCB
SecatSaAQ SPC DtSCBD
'it~
=
~=~~. By
the co-side
=1
Proof 2. The following proof is shorter. AM MB
SPAQ _ SPAQ SQAB _ PD GA _ SCPD SCDA _ SCDA - 1 - SQAB . SPQB - DB' PG - SCDB . SCPD - SCDB - .
= SPQB
Example 1.27 (Pascalian Axiom 5 ) Let A. B and G be three points on one line. and P . Q. and R be three points on another line. If AQ II RB and BP II QG then AP II RG. Figure 1-19
Proof. We need to prove SRAP 1.23 we have SRAP
=
= SCAP . Since AQ II RB and BP II QG. by Proposition
SRAQ SBAP
+ SAPQ = SBAQ + SAPQ = SBAPQ + SBPQ = SBAP + SBPC = SCAP .
Example 1.28 (Desargues' Axiom) SAA Io SBB I • and SGG I are three distinct lines. If AB II AlBI and AG II AlGI then BG II BlG I .
Proof. We need to show SB,BC = SC,BC. Noting that AB II AlBI and AG II AlGI. by the co-side theorem we can eliminate Bl and Gl
Hence SB,BC
=
Figure 1·20
SC,BC .
SThis result is a special case of Pappus' theorem. It was referred to as Pascal's theorem by Hilbert in [24] and was used as an axiom in [5].
J.5
19
Parallels
Example 1.29 Let AI . B I. C I • and DI be poinlson the sides CD. DA. AB. and BC ofa parallelogram ABCD such thatCA.jCD = DBI/DA = AC.jAB = BD.jBC = 1/3. Show that the area ofthe quadrilateralformed by the lines AAI BBl. CCI • and DDI is one thirteenth of the area of parallelogram ABCD.
c
Figure 1-21
Proof By the co-side theorem, SABC D SABA,
Then SA, B,C,D,
= SABCD -
SABA, - SBCB, - SCDC. - SDAD,
SABC D
=1 _
SABCD
12 13
=~ 13
Proposition 1.30 Let ABC D be a parallelogram. P and Q be two points. Then SAPQ + S C PQ = SBPQ + SDPQ or SPAQB = SPDQC.
A
B
Figure 1-22
Proof Let 0 be the intersection of AC and BD. Since 0 is the midpoint of AC. by Proposition 1.18. SAPQ + SCPQ = 2SoPQ. For the same reason. SBPQ + SDPQ = 2SoPQ .
We have proved the first fonnula. The second fonnula is just another fonn of the first one.1 Exercises 1.31 I. In Example 1.28. let AAI II BBI AC II AIC I then BC II BICI .
II
CCI are three parallel lines. If AB
II
AIBI and
2. Let I be a line passing through the vertex of M of a parallelogram M N PQ and intersecting the lines N p. PQ. and NQ at points R. S. and T. Show that IjMR+ IjMS = IjMT. 3. The diagonals of a parallelogram and those of its inscribed parallelogram are concurrent. 4. Use the same notations as Example 1.29. If CA.jCD BD JBC = T then SA,B,C,D, = _r_'-. I
SABCD
r'-2r+2
= DBdDA = ACdAB =
20
Chapter 1. Geometry Preliminaries
5. Let P, Q be the midpoints of the diagonals of a trapezoid ABC D. Then PQ is half of the difference of the two parallel sides of ABC D. 6. A line parallel to the base of trapezoid ABC D meets its two sides and two diagonals at H, G, F, and E . Show that EF = GH.
The Co-angle Theorem
1.6
In this section, we will discuss more basic theorems, the co-angle theorems, in the area method. However we have not incorporated these theorems into our computer program. An angle is the figure consisting of a point 0 and two rays emanating from O. The symbol for angle is L. We assume the value of the angle is ~ 0 and ~ 180°. In a triangle ABC, we use LA, LB, and LC to represent the three inner angles of the triangle at vertices A, B, and C respectively. Then we have the following formula for the area of a triangle ABC (for details see Section 1.8 below.)
(1.4)
VABC
If LABC triangles.
= ~AB ' BCsin(LB) = ~AC. CBsin(LC) = ~AB' ACsin(LA).
= LXY Z or LABC + LXY Z = 180°. we say /::,.ABC and /::,.XY Z are co-angle
Proposition 1.32 (The Co-angle Theorem) If LABC
=
LXY Z or LABC
we have gABC _ I;7XYZ -
+ LXY Z =
180°,
AB·BC Xy.yz·
Figure 1-23
Proof 1. This is a consequence of (1.4). Proof2. The following proof uses the co-side theorem only. Without loss of generality, we assume that B = Y and Z is on line BC (Figure 1-23). If LABC = LXY Z, X is on line AB. We have VABC VXYZ If LABC + LXY Z
VABC
VABZ
BC
AB
AB · BC
= VABZ' VXYZ = BZ ' XY = Xy· YZ'
= 180°, the result can be proved similarly.
The co-angle theorem, though obvious, can be used to prove nontrivial geometry theorems easily.
Example 1.33 In triangle ABC, if LB Prooif.. B y the co-angIe theorem, 1 --
= LC then AB = AC.
gABC I;7ACB -
AB·BC Ac.BC -
AB AC '
21
1.6 The Co-lUIgie Theorem
Example 1.34 If ABC D is a parallelogram. then AB
= CD.
Proof. Since ABC D is a parallelogram, we have LC AB \lBCD = \lACD. By the co-angle theorem, we have
=
LAC D and \l ABC
=
1 _ \l ABC _ AC · AB _ AB
- \lACD - AC· CD - CD·
I
Example 1.35 F is a point on side BC of ~ABC such that AF is the bisector of LBAC. Then AB _ FB AC -
Fe ·
Proof. By the co-side and co-angle theorems,
FB FC
\lFAB
FA · AB
Example 1.36 In triangles ABC and XY Z.
if LA = LX, LB = LY then ~~
Proof. From the hypotheses, we also have LC
\lABC \lXYZ
AB
= \lFAC = FA · AC = AC ·
AB · AC XY·XZ
= LZ.
= ~~ = ~~.
By the co-angle theorem,
AB·BC XY·YZ
AC · CB XZ·ZY·
The result follows from the above fonnula immediately. Two triangles are said to be similar if their corresponding angles are equal. The above example implies that the corresponding sides of two similar triangles are proportional.
Example 1.37 In triangle ABC. AB = AC. AB.l.AC. and M is the midpoint of AB. The perpendicular from A to C M meets BC in P. Show that PC = 2P B. Proof. Note that LP AC = LAM C and LP AB LAC M . By the co-side and co-angle theorems,
PC PB
= =
\lPAC \lPAC \lMAC \lPAB = \lMAC \lPAB PA·AC AC · MC MA · MC PA · AB AC . AC = AB· AB = AB MA · AB MA·AB MA
= 2.
=
~
A
M
B
Figure 1-24
Example 1.38 AM is the median of triangle ABC. D, E are points on AB, AC such that AD = AE. DE and AM meet in N. Show that ~~ = 1~ ·
Chapter 1. Geometry PreUmlnaries
22
From BM = MC, we have \lABM \l AC M_Now by the co-angle theorem,
Proof
DN NE
\lADN \lANE AD-AN AB -AM
\lADN \lACM \lABM \lANE AC-AM AD-AC - AN -AE = AB -AE
AC
= AB -
Lt1 B
M
C
Figure 1-25
Example 1.39 Four rays passing through a point 0 meet two lines sequentially in A, B, C, D and P, Q, R , S _Show that ~~:~g = ~~:~_ Proof By the co-side and co-angle theorem
=
=
AB-CD·PS-QR AD·BC PQ-RS AB CD PS QR - -- -- - AD BC PQ RS \lOAB _\lOCD _\lOPS _\lOQS \lOAD \lOBC \lOPQ \lORS \lOAB \lOCD \lOPS \lOQR \lOPQ - \lORS - \lOAD - \lOBC OA·OB-OC -OD-OP-OS-OQ-OR OP · OQ -OR -OS · OA -OD·OB-OC = 1.
Figure 1-26
Example 1.40 AM is the median of triangle ABC_ A line meets the rays AB,AC, and AM at P, Q, and N respectively_ Show that ~~ = ~(~g + ~~)_ Proof By the co-angle theorem,
~
AP -AQ \lAPQ \lAPN \lAQN = --=--+-AB -AC \lABC \lABC \lABC B M \lAPN \lAQN 2 \l ABM + 2 \l ACM Figure 1-27 AP -AN AQ -AN 1 AP AQ AN = 2AB· AM + 2AC . AM = 2{ AB + AC) AM
C
Multiplying the two sides of the above formula by ~~:1::18, we obtain the result.
23
1.6 The Co-angle Theorem
Example 1.41 Take two points M, N on the sides AB, AC of triangle ABC such tlw.t LMCB = LNBC = LA/2 . Then BM = CN.
B__._ _ _ _----''''C
Figure 1-28
Proof Since LBMC = LA + LACB - LA/2 and LCNB = LA + LABC - LA/2, we have LBMC + LCNB = LA + LACB + LABC = 180°. Hence in niangles BMC and CNB, we have LMCB LNBC, LBMC + LCNB 180°. 8ythe co-angle theorem,
=
MC · BC NB·BC i.e., 1
=
VBMC
= VCNB =
BM · MC CN · NB
= BM/CN.
Example 1.42 (Pythagorean Theorem6 ) In triangle ABC, let BC = a,CA = b,AB = c, and LAC B = 90°. Show tlw.t a 2 + b2 = 2.
A
D
B
Figure 1-29
Proof Let the altitude on side AB be CD = h. Then VACD + VBCD = VABC, i.e., ~~~g + ~!~g = 1. Since LACD = LABC, LBCD = LCAB, by the co-angle theorem we have bh/{ae) + ah/{be) = 1. Also note that ab = 2 V ABC = eh, we have h = able. Substituting this into the above equation, we prove the result. I
Proposition 1.43 (The Co-angle Inequality) If LABC 1800 then
2ABC 'i7XYZ
>
> LXY Z and LABC + LXY Z <
AB·BC Xy ·yz ·
Proof I . This is a consequence of (1.4) on page 20 and the property of the sine function. I Proof 2. The following proof does not use the nigonomenic functions. Draw an isosceles niangle PQR such that QP = PR and LQPR = LABC - LXYZ. Produce QR to S such that LRPS = LXYZ. Then we have LQPS = LABC and vQPS > VRPS. By the co-angle theorem
eIn Chinese literature, Ibis theorem is called the Gou-Gu theorem and is attributed to Shang-Gao (1100 B.C.). This celebrated theorem is one of the most important theorems in the whole realm of geometry. Tbere are about 370 proofs for Ibis theorem in [28].
Chapter 1. Geometry Preliminaries
24
\lABC \lQPS \lRPS \lQPS \lRPS\lXYZ AB . BC \lQPS Rp· PS QP · PS \lRPS Xy · YZ
\lABC \lXYZ
>
AB · BC \lQPS XY · YZ\lRPS AB·BC Xy . YZ ' I
(QP
= RP) s
R
Q
Figure 1-30
Corollary 1.44 1. 1'LABC > LXY Z and LABC + LXY Z > 180° then 'J
2• (The Converse of the Co-angle Theorem) /£ 'J LXY Z or LABC + LXY Z = 180°. Example 1.45 In triangle ABC,
if LB >
P roo),£. By the co-angle 'mequ al'lty, 1 --
LC then AC
'jZABC \7ACB
>
'jZABC
\7XYZ
'jZABC
\7XYZ
=
<
AB·BC
XY·YZ·
AB·B C
XY ·YZ
then LABC
=
> AB.
AB·BC AC.BC -
..
AB AC '
Corollary 1.46 1. Among the segments from a point to any point on a line, the perpendicular is the shortest. I 2. The area of any quadrilateral is less than or equal to half of the products of its two diagonals. Example 1.47 If AB 2: AC and P is a point between B and C then AB > AP.
Proof Since AB 2: AC, we have LACB 2: LABC. Then LAPB = LACB + LCAP > LABC = LABP. By the co-angle inequality, we have AB > AP. I
B
P
C
Figure 1-31
Example 1.48 The sum of any two sides of a triangle is larger than the third one.
1.6
25
'The c.ucJe Theorem
Proof Produce BC to D such that CD = AC. Then LBDA = LCDA = LCAD = LBAD - LBAC < LBAD. In triangle ABD, by Example 1.45 we have AB < BD = BC + CD = BC + AC.
c
B
D
Figure 1-32
7 Take three points M, K, and L on the three sides AB, BC, and CA of triangle ABC respectively. Show that at least one of\JAML, \JBMK, and \JCKL is less that ~ \J ABC.
Example 1.49
Proof Let AM = r · AB, BK numbers less that 1. Also BM By the co-angle theorem,
=
=
= s · BC, and CL = t · AC. Then r, s , and t are positive = (1- r)AB,CK = (1- s)BC, and AL = (1 - t)AC.
~
\JAML \JBMK \JCKL \J ABC · \J ABC . \J ABC AM CK CL . -AL . _BM . -BK . .AB AC AB BC BC AC r · (1 - t) · (1 - r)· s· (1 - s) · t r(1 - r)· s(1 - s)· t(1 - t) ~ (1/4)· (1/4)· (1/4).
As a consequence, one 0 f
VAML VBMK 'V ABC' 'V ABC '
an d
CKL V 'V ABC
A
M
B
Figure 1-33
mustbe Iess than 1/4. A
Example 1.50 (Erdos' Inequality) P is a point inside the triangle ABC. Let the distances between the point P and the lines BC,CA, and AB be x,y, and z respectively. Show that P A + P B + PC ~ 2(x+y+z).
M
B
Figure 1-34
Proof Take points N and M on AB and AC such that AM = AB,AN = AC. Then MN = BCand z . AC + y. AB
= z . AN + y . AM = 2( \J APN + \J APM) ~ P A . M N = P A . BC
+ y . (AB/BC) ~ PA. Similarly x · (AB/AC) + z· (BC/AC) ~ PB; X · (AC/AB) + y. (BC/AB)
i.e., z· (AC/BC)
~ PC.
Adding the three equations together, we have
AB AC BC AB BC AC PA+PB+PC ~ x·(Tc+A"B)+Y(AB + BC)+z · (AC + BC) ~ 2(x+y+z). I 7This is a problem from the 1966lntematiooal Mathematical Olympiad.
26
Chapter 1. Geometry PrelimInaries
Example 1.51 (The Sleiner-Lehmus Theorems) In triangle ABC,
angles Band C are equal then AB
if the bisectors for
= AC.
Proof Without loss of generality, we assume AB ~ AC. Then LACB ~ ABC. Let I be the intersection of BD and CE o Then LDCI ~ EBI. Take a point P between DI such that LPCI = LEBI. We need only to show that P = D . In triangles PCI and EBI, LPCI = LEBI, LPIC = LEIB. Then LCPI = LBEI . In triangles PBCand EBC, LCPB = LBEC, LPCB
A
B
~
Figure 1-35
C
LEBC. By the co-angle theorem
and the co-angle inequality,
PC . P B BE · CE
= _'V_P_B_C > -::p-=c,.....-::B-=C 'VEBC - BE · BC'
i.e., ~~ ~ 1. Therefore PB ~ CE = DB. Since P is on DB, we have P
= D.
Exercises 1.52 I . We mentioned earlier that rarely are there congruent and similar triangles in the diagram of a geometry theorem, but there are many co-side triangles. There are also many coangle triangles in any figure. Try to count the co-angle triangles in Figures 1-23, 1-24, and 1-25. This is why the co-angle theorem works well for many geometry theorems. 2. On the two sides AB and AC two squares ABDE and ACFG are erected externally. Show that 'V ABC = 'V AEG. 3. In triangle ABC, the bisector of the external angle (at vertex A) meets BC in D . Show that ~~ = ~g . 4. In triangle ABC, the bisectors of the inner and external angles (at vertex A) meet Be in D and E . Show that ~g = ~~ .
1.7
Pythagoras Differences
To solve geometry problems involving perpendiculars and congruence of line segments, we need to introduce a new geometry quantity: the Pythagoras difference. To do that, we first introduce the concept of co-area of triangles.
8For interesting extensions of this theorem, see (37).
1.7
27
Pythagoras Dll1'ereoa5
On side AB of a triangle ABC, a square ABPQ is erected such that S ABC and S AB PQ have the same sign (Figure 1-36). The co-area C BAC is a real number such that
Q--------"p
A I<'--------""B
Figure 1-36
Similarly CABC is equal to 'ilBPC or - 'il BPC according to whether LB is acute or obtuse. Generally speaking, C BAC, C ABC, and C AC B are different. But we have CBAC = CCAB , CABC = CCBA, and CACB = CBCA'
Proposition 1.53 For triangle ABC, we have C ABC
+ CBAC = AB2/2.
Proof As in Figure 1-36, if both LA and LB are acute then
CABC + CBAC
= 'ilBPC + 'ilACQ = 'ilABPQ/2 = AB2/2.
If LA is obtuse and LB is acute then
CABC
+ C BAC = 'ilBPC -
'ilACQ
= 'ilABPQ/2 = AB2/2.
If LA is acute and LB is obtuse then
C ABC
+ C BAC = -
'il BPC + 'ilACQ
= 'ilABPQ/2 = AB2/2.
By Proposition 1.53,
CABC + CBAC
AB2
CA 2
BC2
= -2-,CBCA + CABC = -2-'CBAC + CBCA = -2- '
From the above equations. we have
CABC C BAC C ACB
(AB2 + BC 2 - AC 2)/4, (AB 2 + AC 2 - BC 2)/4 , (AC 2 + BC 2 - AB2)/4.
Definition 1.54 We call AB2 + BC 2 - AC2 the Pythagoras difference of triangle ABC with regard to B. and denote it by P ABC . i.e..
PABC = 4CABC = AB2
Proposition 1.55 (Pythagorean Theorem)
1. PABC = 0 if and only if LABC = 90°.
+ BC 2 -
AC2.
28
Chapter 1. Geometry PrelImInaries
2. PABC > 0 if and only if LABC < 90°. 3. PABC < 0 if and only if LABC > 90°. Proof As shown in Figure 1-36, it is clear that V BPC = 0 if and only if LABC = 90°. The second and third cases come from the definition of the co-areas directly. I
Proposition 1.56 (The Pythagoras Difference Theorem) 1. LABC
If LABC '" 90°, we have
= LXY Z if and only if ;~e; = ~~. ~~ ;
2. LABC + LXY Z = 180° if and only if ;~e; = - ~~.~~ . Proof This proposition is a consequence of the definition of Pythagoras difference and the converse of the co-angle theorem on page 24. I
Example 1.57 In triangle ABC, let BC = a, C A m of the median C M in terms of a, b, and c.
= b, and AB = c. Express the length
Solution. Without loss of generality, let us assume that LA is not a right angle. Applying Proposition 1.56 to triangles BAC and MAC , we have
PMAC MA · CA 1 --= AB·CA -2
(AB
= 2MA) .
A
M
B
Figure 1-37
Hence PBAC
= 2PMAC . i.e., b2 + c2 -
a2
= 2(b2 + c2 /4 -
Example 1.58 In triangle ABC, let BC = a, CA of the altitude CD in terms of a, b, and c.
= b, and AB = c. Express the length h
Solution. Without loss of generality, we assume that LA is acute. Applying Proposition 1.56 to triangles DAC and BAC, we have
PDAC / PBAC
= (AD · b)/(b· c) = AD/c.
m 2 ) . Now it is clear that
L1
A
D
Figure 1-38
B
1.7
29
Pythagoras Dltrerence5
Let X Thus
= AD. BythePythagoreantheorem.b2 -h2 = x2. ThenPDAc = x 2+b2_h 2 = 2x2. x/c =
i.e .• x
= (b 2 + ~ -
P DAC / P BAC
a 2 )/(2c). Since x 2
= 2x2/(b2 + ~
= b2 -
Example 1.59 In triangle ABC, let BC the bisector C F in terms of a, b. and c.
-
2 a ),
h 2 • we obtain the final result:
= a, C A = b, and AB = c. Express the length of
Solution. We assume that LA '" 90°. Applying Proposition 1.56 to triangles BAC and F AC P FAC / P BAC
= (AF · b)/b · c = AF/c.
F
A
B
Figure 1-39
= b/a; hence AF = (bc)/(a + b) . Let x = CF . We have AF/c = P FAC / PBAC = (AF2 + b2 - x 2)/W + ~ - a2). Substituting AF = bc/(a + b) into the above equation. we obtain By Example 1.36. AF/BF
x
2
= ab((a+W -
~)/(a+W ·
I
We define the Pythagoras difference for an oriented qruu:lrilateral ABC D as follows.
It is easy to derive the following properties for the Pythagoras differences of quadrilaterals.
3.
= P CDAB = P BADC = PDCBA. P ABCD = - PBCDA = -PDABC = -PADCB = -PCBAD . P ABCD = PBAC - P DAC = P ABD - P CBD = P DCA - P BCA = PCDB -
5.
PABAC
= 0, PABCB = O.
6.
PAPBQ
+ PBPcQ = P APCQ . (The additivity of the Pythagoras difference.)
I.
2.
PABCD
Proposition 1.60
If A, B, and C are collinear. we have PABC
= 2BA· BC.
P ADB .
Chapter 1. Geometry Preliminaries
30 Proof Since AC
= AB -
CB, we have
~oc=d+d-~=~+~-~-C~=2M . OC. Definition 1.61 We use the notation AB .lC D to denote that four points A , B, C, and D satisfy one of the following conditions: line AB is perpendicular to line CD; A = B ; or C=D. The following result gives a criterion for AB .lC D using the Pythagoras difference.
Proposition 1.62 AC .lBD if and only if PABCD = PABD - PCBD = O. Proof Let M and N be the feet of the perpendiculars from points A and C to line B D . Then by the Pythagorean theorem,
PABD
B~~~________~D
Figure 1-40
=
AB2 + BD2 - AD2 AM2 + BM2 + BD2 _ AM2 _ MD2
= =
BM2 + BD2 - (BD - BM? 2BM ·BD .
Similarly PCBD = 2BN . BD. We have PABD = PCBD if and only if M B = N B , i.e., M = N which is equivalent to AC.lBD. I
Proposition 1.63 Let P and Q be the feet of the perpendiculars from points A and C to BD. Then PABCD = 2QP · BD. Proof By Propositions 1.62 and 1.60,
PABD - PCBD = PPBD - PQBD 2Bp· BD - 2BQ· BD = 2QP · BD . Proposition 1.64 Let ABC D be a parallelogram. Then for two points P and Q, we have PPAQB = PPDQC. Proof This is a consequence of Proposition 1.63.
Proposition 1.65 Let D be the foot drawn from point P to a line AB (A have
AD PPAB = = - -, DB PPBA
t= B).
Then we
1.7
Pythago... J)jlfe~Dces
31
Proof. By Proposition 1.62,
The result is clear now.
Proposition 1.66 Let AB and PQ be two nonperperulicular lines and Y be the intersection of line PQ and the line passing through A and perperulicular to AB (Figure 1-41). Then PY
PPAB
QY
P QAB
==--, Q
y
Proof. We need only to show the first equation. Let Pl and Ql be the orthogonal projections from P and Q to line AB respectively. By Proposition 1.62 ~ = pp, .u =~'AB , PQAB
PQ,AB
AQ, ·AB
AQ,
A
~
=~=t::i.
QY '
~
B
Figure 1-41
Example 1.67 (The Orthocenter Theorem) Show that the three altitudes of a triangle are concurrent. A
Proof. Let the two altitudes AF and BE of triangle ABC meet in H . We need only to prove C H l.AB, i.e .• P ACH = P BCH . By Proposition 1.62, P ACH = P ACB
= P BCA = P BCH .
I
c
F
B
Figure 1-42
Example 1.68 (Orthocenter-dual) Let ABCO be a quadrilateral. From point 0 perperuliculars to OA , OB , and OC are drawn which meet BC, CA, and AB in D, E, and F respectively. Show that D , E , and F are collinear. Proof. Let DE and AB meet in Z. We need only to show Z By Proposition 1.66, we can eliminate F : ~~ eliminate Z: ~ AZ
=
EC
SBDE
§..s..DLs •
ADB
= ::~~.
= F.
To eliminate E, by Proposition 1.66 we have P
COB = ACSBDA = -p,--SBDA. COAB
i.e., ~~
= ~~ .
By the co-side theorem, we can
32
Chapter 1. Geometry Preliminaries
Then P COAB P BOC
P COAB
P AOC . SBDA P AOB
PAOBSACD
= PAOC
SACD
P AOB
.
BD CD
= P AOC
. PAO B
P AOB
PAoc
= l.
Example 1.69 On the two sides AB and AC of triangle ABC, two squares ABDE and ACFG are drawn externally. Show that BGl.CE.
Proof
B
P BCGE
=
Figure 1-44
= P BCGA + P BAGE
P BCAA
(Additivity) (Additivity)
+ P ACGA + P BAAE + P AAGE
= - PBAC -
(AGl.AC, ABl.AE) (LBAC + LGAE = 180°) I
PGAE
=0
A
Example 1.70 In triangle ABC, take a point J on the altitudes AD. Lines BJ and CJ meet AC and AB in N and M respectively. Show that LM DA = LADN.
Proof To prove LM DA = LADN, by the co-angle theorem and Proposition 1.56 we need only to show PMDA/SMDA
Figure 1-45
= PADN/SADN .
S
_AMS -~S M DA - AB BDA - SAJBC BDA
the co-side theorem;
S
-~S -~S ADN - AC ADC - SABCJ ADC
the co-side theorem;
P ADN PM DA
= NACC P ADA = = MABB P ADA =
S ..2JJ.!
Proposition 1.65; Proposition 1.65;
Then P ADN SMDA
SBDA SAJC
P MDA SADN
SADC SABJ
---- = ---- =
BD DC DC BD
== =
1
.
Exercises 1.71 1. In Example 1.69, show that BG
= C E.
2. In Example 1.69, let M be the midpoint of BC. Show that AM l.EG and EG
= 2AM.
3. The sum of the squares of the diagonals of a parallelogram is equal to sum of the squares of the four sides of the given parallelogram. (Use Example 1.57.)
33
J.8 TrtCoaomebic Functio...
4. In the quadrilateral ABC D, if LABC and BD then !:JuIJ. = CP AP . PSCD
1.8
= LC DA = 90° and P is the intersection of AC
Trigonometric Functions
We have discussed two major geometric quantities: the area and the Pythagoras difference. Area is a well-known concept and has been used since the time of Euclid. On the other hand, the Pythagoras difference is unfamiliar to most readers. In this section, we will introduce the trigonometric functions and use them to represent areas and Pythagoras differences.
c The sine and cosine functions can be defined in the usual way. Let ABC be a triangle with LB = 90°. Then sin( LA)
= ~~,
cos(LA)
AB
= AC. A
We can also use the area to define trigonometric functions.
B
Figure 1-46
Dermition 1.72 sin(LA) is twice the signed area of a triangle ABC such that AB AC = 1.
=
It is easy to see that the two definitions are consistent. The following properties of the trigonometric functions can be derived from the definition directly. l. sin(O)
= sin(1800) = 0; sin(900) = l.
2. sin(LX)
= sin(180° -
3. sin(LA)
= sin{LB) if and only if LA = LB or LA + LB = 180°.
LX).
4. Let us assume that LA + LB ~ 180°. Then A < B implies sin{LA) < sin{LB), and vise versa. (This is a consequence of Proposition 1.43.)
Proposition 1.73 SABe
= ~AB · AC · sin(LA) .
Proof Take two points X and Y on AB and AC respectively such that AX = AY = l. By the co-angle theorem
\lABC = \lAXY
AB · AC = \lAXY · AB· AC AX·AY
1 . = -AB· AC· sm(LA).
2
34
Chapter 1. Geometry Preliminaries
Applying the above proposition to the three angles of triangle ABC respectively, we have
'VABC
= ~bcsin(LA) = ~acsin(LB) = ~absin(LC) .
2 As a direct consequence, we have
2
2
Proposition 1.74 (The Sine Law) In triangle ABC, let BC
Then sin(LA)ja
= a, CA = b, and AB = c.
= sin(LB)jb = sin(LC)jc.
The following is a very useful property of trigonometric functions.
A\
Proposition 1.7S (The Visual Angle Theorem)
Emanating from P, there are three rays P A, P B, and PC such that LAPC = Q , LC P B = (3, and LAPB = "( = Q + (3 < 180°. Then A,B, andC are collinear if and only if
A
C
B
Figure 1-47
sin(L"()j PC
= sin(LQ)j PB + sin(L(3)j PA.
Proof If points A , B, and C are collinear then we have 'V P AB PA· PB· sin(L"()
= 'V P AC + 'V PC B, i.e.,
= PA· PC· sin(LQ) + PB · PC· sin(L(3).
We obtain the result by dividing P A . P B . PC from both sides of the above formula. Conversely, from the above formula, we have 'V P AB = 'V PAC + 'V PCB which implies 'V ABC = I 'V P AB - 'V PAC - 'V PC BI = 0, i.e., A, B , and C are collinear. I Proposition 1.75 has many applications.
Example 1.76 In triangle ABC, LACB = 120°; C E is the bisector of angle C. Show that 1 j C E =
IjCA
+ 1jCB.
Proof By Proposition 1.75,
~
A
E
B
Figure 1-48
Since sin(1200) = sin(180° - 120°) = sin(600), we obtain the result by dividing sin(600) I from both sides of the above equation.
35
1.8 Tri&onometric Functions
/1\
Example 1.77 In 1M right triangle ABC, let BC = a , AC = b, and CD = h be tM altitude on tM hypotenuse. Show that 1/h2 = 1/a2 + 1/b2 .
A
Proof. Let 0-
D
= LACD, {J = LBCD. By Proposition 1.75 sin(o- + {J)/h
B
Figure 1-49
= sin(o-)/a + sin({J)/b. = 1, sin(0-) = sin(LCBD) =
We have sin(o- + {J) = sin(LACB) h/a, sin({J) sin( LCAD) = h/b. Substituting these into the above equation, we obtain the result. I
=
Definition 1.78 Let LA be an angle. TM cosine of LA is defined asfollows. _ { sin(90° - LA) . ( LA - 90° ) - sm
cos (LA) -
if LA ~ 90°; if. LA> 90°.
Proposition 1.79 (The Cosine Law) In triangle ABC, we have PABC
Proof. In Figure 1-36, if LB CABC
~
= 2AB . C B . cos( LB) .
90°
= ~AB. BC· sin(90° -
= ~AB. BC· cos(LB) .
LB)
Thus PABC = 4CABC = 2AB · BC · cos(LB). Other cases can be proved similarly.
I
Example 1.80 Let 0- and {J both be acute angles. Show that sin(o-
+ {J)
= sin(o-)cos({J)
+ sin({J)cos(o-).
Proof. In Proposition 1.75, let us assume that PC l..AB, PC Then
sin(o-
= h, P A = b, and P B = a.
+ {J) = (h/a) sin(o-) + (h/b) sin({J) = sin(o-) cos({J) + sin({J) cos(o-).
I
It is easy to show that the angles in the preceding example are not necessarily acute. In Example 1.80, let 0- + {J = 90°. We have the well-known formula: sin(0-)2 Example 1.81 sin(300)
+ cos(o-?
= 1.
= ~ ; sin(45°) = v'2/2;sin(600) = ../3/2.
36
Chapter 1. Geometry Preliminaries
Proof. In Proposition 1.80, setting (t
= f3 = 30°, we have
sin(600)
= 2 sin(300) cos(300).
Since cos(300) = sin(90° - 30°) = sin(600), we have sin(300) = ~ . Similarly setting (t = 30°, f3 = 60°, we have sin(600) = J3/2; setting (t = f3 = 45°, we have sine 45°) = .,fi/2.1 So far, the angles are always positive. To represent the signed areas using trigonometric functions. we need to introduce the oriented angle. which is also denoted by L. Definition 1.82 The oriented angle ~ABC is a real number such that (1) the absolute value onABC is the same as the ordinary angle LABC. and (2) ~ABC has the same sign with SABC. It is clear that for any angle (t, we have -180° < (t ::; 180°. The arithmetic for oriented angles is understood to be mod 360° and between -180° and 180°, Definition 1.83 We extend the definition of the sine and cosine functions to the oriented angles as follows. Let ~(t be a negative angle. Then sin(~(t)
=-
sine -~(t),
cos(~(t)
= cos( -~(t).
With the above definition, the properties for these two functions proved before are still valid. In particular. we have Proposition 1.84
SABC
= ~AB . BC . sin(~ABC), PABC = 2AB· BC . cos(~ABC).
Example 1.85 (The Herron-Qin Formula9 ) In triangle ABC. 2
16SABC
~
= 4AB
==2
. CB -
2
P ABC '
Proof. By Proposition 1.84, sin(~ABC) = ~~~Zg, cos(~ABC) sinUABC)2 + cos(~ABC)2 = 1, we have
Thus
2 16SABC
-2-2
= 4AB . CB -
= 2:~Rffc' Since
2 P ABC '
9Qin (1247 A.D.) is an ancient Chinese mathematician who discovered the area fonnula for triangles in exact fonn of this proposition [169).
37
1.8 Triconometric Functions
y
Definition 1.86 The oriented angle between two directed lines PQ and AB. denoted by 4(PQ, AB). is defined as follows. Take points O , X , and Y such that OYQP and OX BA are parallelograms. Then we define 4(PQ , AB) = tXOY.
---,.::------?x
A
Figure 1-50
B
With the above definition. it is easy to check the following properties for the oriented angles. 1. 4(PQ,AB)
= -t(AB,PQ) .
2. 4(PQ,AB)
= 180
0
+4(QP,AB)
= 180
0
+t(PQ , BA).
3. 4(PQ,AB) =4(QP,BA) . 4. 4(PQ,AB)+4(AB,CD)=t(PQ,CD). We can now represent the signed area and the Pythagoras difference of quadrilaterals using trigonometric functions.
Proposition 1.87 SABCD
= ~AC . BD . sin(t(AC, BD».
Proof Take a point X such that AXDB is a parallelogram. Then AX = BD. By Proposition 1.30. SABCD = SAACX = SXAC = ~XA. AC· sin(tXAC) = ~AC· BD · sin(4(AC, BD». I
Proposition 1.88 PABCD
= 2AC . BD . cos(t(AC, DB»
.
Proof Take a point X such that CXDB is a parallelogram. Then CX = BD. By Proposition 1.64. P ABCD = -PCBAD = -PCCAX = PXCA = 2XA· AC · cos(tXCA) = 2AC· BD . cos(t(AC, DB». I
Example 1.89 (The Herron-Qin Fonnula for Quadrilaterals) ForaquadrilateralABCD. we have 16S~BCD = BD2 - plBcD '
4E .
Proof By Propositions 1.87 and 1.88. sin(t(AC, BD» = 2%tlfg, cos(t(AC, DB» = ~,,
4S~BCD AC2. BD2 Thus 16S~BCD
....,-=2
= 4AC
-2
2
P~BCD
+ 4AC2. BD2 = 1.
. BD - PABCD '
Chapter 1. Geometry PreUmhl8lies
38 Exercises 1.90 1. Use Example 1.80 to prove the following fonnulas sin(a - 13) cos(a + 13 ) cos(a - 13) 2. Let tan( a)
= =
sine a) cos(f3) - cos( a) sin(f3) sine a) sin(f3) - cos( a) cos(f3) sine a) sin(f3) + cos( a) cos(f3)
= ;~:~:l . Show that
tan (a
tan( a) + tan(f3) tan () a tan () 13 ,
+ 13 ) = 1 -
13 tan( a -
tan( a) - tan(f3)
) = 1 + tan( a) tan(f3)"
3. Prove the following version of the Herron-Qin fonnula for quadrilaterals.
= (p - a)(p - b)(p - c)(p - d) - a · b · c· d· cos( LB +2 LD )2 where a = AB,b = BC, c = CD , d = DA, and P = (a + b + c + d)/2. (Hint. 2 SABC D
have S~BC D
We
= (SABC + SACD)2. Then use Proposition 1.84.)
4. Continue from the above exercise. If ABCD is convex and cyclic then S~BCD = (p - a)(p - b)(p - c)(p - d) . 5. Continue from the above exercise. If ABCD also has an inscribed circle then S~BCD = abed.
1.9
Circles
In this section, we will discuss another important geometric object: the circle. Points on the same circle are called co-circular or cyclic.
A
x Figure 1-51
Figure I-52
Let J be a fixed reference point on the circle, I the antipodal of J. and X a point on the tangent of the circle at point J such that S/Jx > 0 (Figure I-51). For any point A on
the circle, ( A is defined to be the oriented angle # A J X . i.e.. the (oriented) inscribed angle corresponding to the arc JA. Then for two points A and B on the circle, # B - bA = ( BJ A (Figure 1-51). We now define the oriented chord on the circle. The absolute value of the oriented chord is A B while its sign is the same as the sign of the oriented angle 4 B JA, or equivalently the same as thesgn of SBJA.In Figure 1-51, since SBJA> 0. we have TB> 0. The oriented chord J A is always positive. Proposition 1.91 With the above notations, we have
TB=
a=
d sin(( B J A ) = d sin@B - ( A ) dsin(4AJX)
where d is the diameter of the circle. Proof: The signs of the two sides of the equation are equal. Thus we need only to check the absolute values of both sides of the equation. Since L BA J = L BI J or L B A J L BI J = 180". we have
+
BJ BJ AB = sin(LBJA) = sin(LBJA)= sin(LBJA) . IJ. sin LBAJ a n L BI J P r o p o s i t i ~1.92 LeGhe diameter of the circumcircle of triangle ABC be d. Then SABC= AB BC . ACI(2d). Proof: As shown in Figure 1-52, we have LABC = LCJA or LABC = 180" - LCJA. By Propositions 1.73 and 1.91,
1 1 vABC = - A B . BCl sin(LABC)I = -AB . BC 2 2
1 . I sin(LCJA)I = -AB . BC . C A . 2d
We still need to check whether the signs of both sides of the formula are the same. At first, it is easy to see that when interchanging two vertices of the triangle, the signs of both sides of the equation will change. Therefore, we need only to check a particular position for the three vertices A, B, and C , e.g.. the case shown in F@re 1-52. In this case, we have 2 0.1 S A B ~0,SCJB2 O,SBJA 0,and S C J2~0.Hence AB 2 0,FC2 0,and
>
>
---
Proposition 1.93 (The Co-circle Theorem) If the circumcircles of triangles ABC and X Y Z are the same or equal then = -.
&
Proof: This is a direct consequence of Proposition 1.92.
I
40
Chapter 1. Geometry Prelim/Daries
Example 1.94 (Ptolemy's The~m) Let A,l!.; C, and D be four points on the same circle. Then AB· CD + AD· BC = AC· BD. Proof. We chose A to be the reference point. By Proposition 1.91, AB· CD+AD· BC-AC·BD d2(sin(4B) sin(4D - 4C) + sin(4D) sin(4C - 4B) - sin(4C) sin(4D - 4B)) O.
I
Example 1.95 (Brahmagupta's Formula) Let A, B, C, and D befour points on the same circle. Then 2 SABCD = (p - AB)(p - BC)(p - CD)(p - AD) where p = !(AB + BC + CD
+ AD).
Proof. By Example 1.89 and Ptolemy's theorem -2
-2
4AC . BD -
=
2 P ABCD
4(AB · CD + AD· BC? - P~BCD ((AB + CD)2 - (AD - BC)2)((AD + BC)2 - (AB - CD)2) 16(p - AB)(p - BC)(p - CD)(p - AD). I E
Example 1.96 A, B, C, and D are four co-circle points. For any point E on the same circle. lines DE and C E meet AB in F and G. Show that ~~ . ~g is independent of E . Proof. By the co-side theorem and the co-circle theorem, AF BG BP'AG =
=
Figure 1-53 SBDESACE
Ab·J5E·Eit·fiG·8E·EiJ
Ab · fiG
BD·DE · EB·AC·CE · EA
BD·AC
which is independent of E.
Example 1.97 (Pascal's Theorem) Let A. B. C . Alo B I. andC I be six points on a circle. Let P = ABI n AlB. Q = AC I n AIC. and S = BC I n BIC. Show that p . Q. and S are collinear. Figure 1·54
1.9
Orda
41
Proof Note that the points P, Q, and R in this example are constructed in the same way as in Example 1.17 on page 14. Let ZI = PQ n BCI and Z2 = PQ n BIC. We need only to show G _ PZI . QZ2_ - QZI PZ2 - 1.
Figure I-55
Figure I-56
Example 1.98 (The General Butterfly Theorem) A, B, C, D, E, and F are six co-circle points. Lines CD and EF meet AB in M and N. Lines CF and DE meet AB in G and H. Show that ~g ~Z = ~~. (Figure 1-55)
.
Proof By the co-side theorem
By the co-side theorem again, SMCF SABC SFDE SAFBE - = MC = ---, - = =NFEE = ---. SDCF DC SADBC SNDE SABE '="
Then G ately.
= SaGrSARcSIIQ6SAf". SACPSABESPDESBDC
In Example 1.98, when G butterfly theorem:
Now G
=H
= 1 follows from the co-circle theorem immediI
becomes the midpoint of AB, we obtain the ordinary
Example 1.99 (The Butterfly Theorem) C, D, E, and F are four points on circle 0 (Figure 1-56). G is the intersection of DE and CF. Through G draw a line perpendicular to OG, meeting CD in M and EF in N . Show that G is the midpoint of M N.
42
Chapter 1. Geometry Preliminaries
In the above examples, no Pythagoras differences occur in the proof. To deal with Pythagoras differences, we need to develop some new tools. We still assume that there is a fixed reference point J on the circle whose diameter is d.
c
Figure 1-57
Proposition 1.100 Let d be the diameter of the circumcircle of triangle ABC. Then PABC = 2AB· CBcos(~CJA). Proof By the cosine theorem for oriented angles, IPABC I = 2AB . BCI cos( LABC)I. As shown in Figure I-57, we have LABC = LCJA or LABC = 180° - LCJA. Then IPABCI = 2AB· BC'I cos(LCJA)I .
We still need to check whether the signs of both sides of the equation are equal. At first, when interchanging the position of A and C, the signs of both sides of the equation will not change. Therefore, we need only to consider the following two cases: J is on the arc AC or on the arc AB. In the first case, we have AB 2: O,CB ~ O. Since~ABC+~CJA = 180°, PABC and cos(t C J A) always have opposite signs. Therefore the proposition is true in this case. In the second case, we have AB ~ 0, CB ~ O. By the inscribe angle theorem, ~C J A = ~C BA. Thus PABC and cos(tC J A) always have the same sign. I Proposition 1.101 (Co-circle Theorem for Pythagoras Differences) Ifthecircumcircles of triangles ABC and XY Z are the same and PXY Z '" O. then PABC PXYZ
AB. fiG cos(tAJC) = Xy · Y Z cos(tX J Z)
Proof This is a direct consequence of Proposition 1.100.
I
Example 1.102 (Simson's Theorem) Let D be a point on the circumcircle of triangle ABC. From D three perpendiculars are drawn to the three sides BC. AC, and AB of triangle ABC. Let E. F. and G be the three feet respectively. Show that E. F and G are collinear. Figure 1-58
43
1.9 Circles
Proof By Menelaus' theorem (Example 1.9 on page 11). we need only to show
AG BE CF · ==-1 GB EC FA .
G== · =
By Propositions 1.65 and 1.101
G
= =
PABOPOACPOCB BA · ADcos(JBJD)AC. CDcos(JAJD)i5B . BCcos(JDJC) AB· BDcos(JAJD)DA · ACcos(JDJC)DC · CBcos(JDJB) -1
Example 1.103 Let VI.·· ·, Vm • and P be m + 1 co-circle points. Pj the orthogonal projections from P to V;V;+\. i = 1, ...• m. Show that
where the subscripts are understood to be mod m.
Proof By Propositions 1.65 and 1.101
V;P; _ PPV;V,±1 _ pv; . V;V;+1 cos(JPJV;+1) Pj V;+l - PPVH1V, - PV;+\· V;+1V;cos(JPJV;) '
i
= 1, ... ,m.
Multiplying the m equations together, we obtain the result.
Exercises 1.104
1. A , B, C , and D are four co-circle points. AB and CD meet in P. Show that PA · PB
=
PC · PD. 2. Show that the diameter of the circumcircle of triangle ABC is equal to the product of two sides dividing the altitude on the third side of the given triangle. 3. Prove Example 1.99 directly. 4. In Example 1.102, if D is an arbitrary point. Show that D0 2
= A02(1 -
~ss ). ABC
5. Continue from the above exercise. Show that D is on a fixed circle with 0 as its center if and only if SEFG is a constant.
44
1.10
Chapter 1. Geometry PrelimInaries
Full-Angles
To define the concept of angles. we need the concept of rays. As a consequence. we need to compare the order of points on a line. In algebraic language. this means that we need to use inequalities to describe angles. In this section. we introduce a new kind of angle. the description of which does not require inequalities. Angles of this kind will be used to simplifying the proofs of many geometry theorems. Definition 1.105 A full-angle consists of an ordered pair of lines I and m and is denoted by L[l, mI. Two full-angles L[l. m] and L[u, v] are equal if there exists a rotation K such that K(l) II u and K(m) II v. If A, Band C, D are distinct points on I and m respectively. then L[I, m] is also denoted by L[AB, CD], L[BA, CD], L[AB, DC]. L[AB, mI. and L[l, DC]. For three points A, B, and C. let L[ABC] L[AB, Be].
=
Definition 1.106 If ll..m, L[l, m] is said to be a right full-angle and is denoted by L[l].1f L[I, m] is said to be aflatfull-angle and is denoted by L[O].
III m,
To give a criterion for the equality of two full-angles. we need to introduce the tangent function for full-angles. Definition 1.107 The tangent function for the full-angle L[PQ, AB] is defined to be
tan(L[PQ,AB])
= sin(~(PQ,AB)). cos(~(PQ,
AB))
We need to check that this definition is well-defined. That is when interchanging P, Q or
A, B, :~:( (~Q,~!» does not change. This comes from the following equations sin(~(PQ,AB))
= -sin(~(PQ.BA)) = -sin(~(QP,AB)).
cos(~(PQ,AB))
= -cos(~(PQ,BA)) = -cos(~(QP,AB)).
As a consequence. we see that the sine and cosine functions for a full-angle are meaningless. Proposition 1.108 L[AB, PQ] ~(AB,PQ) -~(XY,UV)
= L[XY, UV] if and only if~(AB, PQ) = ~(XY, UV) or
= 180°.
Proof Without loss of generality. we may assume AB II XY. Then L[AB,PQ] = L[XY,UV] if and only if PQ II UV. i.e .• if and only if ~(AB,PQ) = ~(XY,UV) or ~(AB, PQ) - ~(XY, UV) = 180°. I
45
1.10 FulI-ADeles
Proposition 1.109 tan{L[AB,PQ])
= 4pSAPB9. A9BP
Proof This is a consequence of Propositions 1.87 and 1.88.
Proposition 1.110 L[AB, PQ] = L[XY, UV] ifand only iftan(L[AB, PQ])
= tan(L[XY, UV]).
=
L[XY, UV], by Proposition 1.108 we have ~(AB, PQ) = 180°. It is clear that tan(L[AB,PQ]) tan( L[XY, UV]) for both cases. If tan( L[AB, PQ]) = tan( L[XY, UV]) then we have
Proof
If L[AB, PQ]
~(XY,UV) or ~(AB,PQ) - ~(XY,UV)
sin(~(AB,PQ))
= =
sin(~(XY,UV))
cos{~(AB, PQ))
= cos(~(XY, UV)) ' if ~(AB, PQ) = ~(XY, UV)
The above equation holds if and only or = 180°. i.e .• L[AB. PQ] = L[XY, UV] by Proposition 1.108.
~(AB,
PQ) -
~(XY, UV)
Example 1.111 Let ABG be a triangle such that AB = AG. Then L[ABG] = L[BGA] . Conversely, if L[ABG] = L[BGA] and A, B, and G are not collinear then AB = AG. Proof If AB
= AG, we have
tan(L[ABG])
= 4SABC = 4SBCA = 4SBCA = tan(L[BGA]). PABC
BG2
PBCA
Then L[ABG] = L[BGA]. Conversely. if L[ABG] = L[BGA] and SABC '" 0, by the definition of the tangent function we have PABC = PBCA • i.e., AB2 = AG2 In the above example. we do not need to say that the corresponding "inner" angles of an isosceles triangles are equal. To describe the "inner" angle. we need inequalities. Example 1.112 Let I, m, and t be three lines. Then
III m if and only if L[l, t]
= L[m, t].
Notice that in the above criterion for parallel, we do not need to mention that the angles should be the "corresponding angles," the exact description of which needs inequalities. Example 1.113 (The Inscribed Angle Theorem) L[AB,BG] = L[AD,DG].
If A, B, G, and D are cyclic points then
If using angle in the usual sense. we need two conditions: LABG = ~ADG or LABG LADG = 180° and to distinguish these two cases, we need inequalities.
+
The proofs for the above two examples will be given later. From these examples. we see that the concept of full-angles makes many geometry relations concise. We will see later that this will lead to short proofs for many geometry theorems.
46
Chapter 1. Geometry PrelImInaries
Definition 1.114 Let I, m, u, and v be Jour lines. Let K be a rotation such that K(l) We define L[u, v] + L[l, m] = L[u, K(m}] .
II v.
It is easy to check that the addition of full-angles is associative and commutative. The main properties of the full-angles are summarized as follows.
Ql L[u, v] = L[O] if and only if U Q2 L[u, v]
II v.
= L[l] if and only if u.lv.
Q3 L[l] + L[l] = L[O].
+ L[O] = L[u, v]. L[u, v] + L[l, m] = L[l, m] + L[u, v].
Q4 L[u, v] Q5
Q6 L[u, v]
+ (L[l, m] + L[s, tJ) = (L[u , v] + L[l, mJ) + L[s, t].
Q7 L[u, s]
+ L[s, v] = L[u, v].
Q8 if L[u, v] = L[O] then for any line t we have L[u, t] we have L[u, t] = L[v, t] then L[u, v] = O.
= L[v, t] . Conversely, if for a line t
Q8 is Example 1.112 which can be derived from QI-Q7 as follows. If L[u, v]
= 0, we have
L[u, t] = L[u, v] + L[v, t] (Q7) = L[O] + L[v, t] the hypothesis = L[v, t] + L[O] (Q5) (Q4). = L[v,t] Conversely, if L[u, t]
= L[v, t] L[u, v]
= L[u, t] + L[t, v] = L[v, t] + L[t, v] = L[v, v]
= L[O]
(Q7) the hypothesis (Q7) (Ql).
The following properties of the full-angles are also often used. Q9 If AB = AC, we have L[AB,BC] = L[BC,AC]. Conversely, if L[AB,BC] = L[BC, AC] then AB = AC or A, B, and C are collinear.
QI0 Points A, B, C, and D are on the same circle or on same line if and only if L[AB, BC] = L[AD, DC].
Qll If AB is the diameter of the circumcircle of triangle ABC then L[AC, BC]
= L[l]
47
1.10 FulI-ADeles
Q12 If a is the circumcenter ofttiangle ABC then LIBa , aC]
= 2LIAB, AC] .
Q9 is Example 1.111. For the proofs of QlO. Qll. and Q1l2. see Example 3.52 on page 128. Example 1.11S Two circles a and Q meet in two points A and B . A line passing through A meets circles a and Q in C and E. A line passing through B meets circles a and Q in D and F . Show that CD liEF. Figure I-59 shows five possible cases for this example. The following proof based on full-angles is valid for aU cases. If we do not use full-angles. we must give different proofs for different figures. F
E
Proof.
Figure 1-59
LlDC, FE]
= LlDC,DB] + LIDB,FEj
(byQ7)
= LlAC, AB) + LIFB,FEj
(by QIO and Q8 since A, B, C, D are cyclic and DE BF)
= LlAE, AB] + LIAB, AE] (by Q8 and QIO since E E AC and A, B, F, E are cyclic) (by Q7)
= LIAE,AE] = LlO]. (by Q2)
A~_ _ _ _+~
H
A
Figure 1-60
48
Chapter t.
Geometry PreUminarles
Example 1.116 In triangle ABC. two altitudes AD and BE meet in H. G is the foot of the perpendicular from point H to AB. Show that L[DG, GH] = L[HG, GE).
Proof. L[DG, GH] + L[GE, HGJ = L[DB,BH] + L[AE,HA] = L[BC,BE] + L[AC,AD] = L[BC, AC] + L[AC, BE] + L[AC, BC] + L[BC, AD] = L[BC,AC] + L[l] + L[AC,BC] + L[l] = L[BC, BC] + L[O]
(QlO;B,D ,G.H ;A,E,G,H
cyclic.)
(Q8;DEBC ;BEEH;EEAC;HEAD.)
(Q7) (ACl.BE and BAl.AD) (Q7,Q3)
= L[O] .
As in Figure 1-60, if both LBAC and LABC are acute angles. we have LDGH = LEGH; if one of them is obtuse. then LDGH + LEGH = 180°. Thus, if we do not use full-angles, we must give two proofs for the two cases.
Example 1.117 (The Nine Point Circle) Let the midpoints of the sides AB, BC, and C A of 6ABC be L , M, and N, and AD the altitude on BC. Show that L , M, N, and D are on the same circle. Proof. We need to show L[LM,MD] L[LN,ND] or equivalently, L[LM,MD] L[ND,LN] = L[O]. L[LM,MD)
= +
+ L[ND,LN]
= L[AC,BC] + L[ND,LN] = L[AC,BC] + L[ND,BC] + L[BC,LN] = L[AC, BC] + L[BC, AC] + L[O] = L[AC,AC]
= L[O].
c Figure 1-61
(Q8; LM II AC; M, D, B, C are collinear) (Q7) (Q9,Ql; ND = NC; BC II LN)
Example 1.118 The circumcenter of triangle ABC is O. AD is the altitude on side BC. Show that LOAD = ILC - LBI. Proof. Let M be the midpoint of BC. MO and AB meet in E . We need only to show that L[AD,AO) = L[CE,AC]. Figure 1-62
L[AD,AO] + L[AC,CE] L [AC,AO] + L[AC,BC] + L[BC,CE] = /[AD,BC] + L[CO,AC]+ L[BE,BC] = L[1] L[CO,MO] + L[MO,AC]+ L[BA,BC] = L[1]+ L[AC,BA] + L[MO,AC] + L[BA,BC] = L [ 1 ] + L[MO,BC] = L[1]+ L[1]= L[O]. I = L[AD,AC]
+
+
(Q7) (47.49 (BE = CE)) (Q7; E E B A ; A D I B C ) (Q12) (47)
Summary of Chapter 1 Signed areas and Pythagoras differences are used to describe some basic geometry relations: collinearity, parallelism, perpendicularity, congruence of line segments. and congruence of full-angles. 1. Three points A, B, and C are collinear if and only if SABC= 0. 2. PQ
il AB if and only if SPAQB= SpAB- SQAB= SBPQ- SAPQ= 0.
3. P Q l A B if and only if PPAQB= PPAB- PQAB= PBPQ- PAP* = 0. 4. L[AB,PQ] = L[XY,UV]if and only if
=
E.
r The following results are powerful tools for solving difficult geometry problems.
1. (The Co-side Theorem) Let M be the intersection of the lines AB and PQ and Q # M. T h e n w e h a v e e =
g.
2. (The Co-angle Theorem) Let X , Y and Z be three non-collinear points. Then LABC = LXYZ or LABC LXYZ = 180" if and only if = 3. (The Pythagoras Difference Theorem) Let LABC # 90". Then LABC = LXYZ if and only if = -; LABC + LXYZ = 180" if and only if
m.
+
EAEC Pxyz
&
AB.BC XY.YZ'
4. (The Visual Angle Theorem) Emanating from P, there are three rays PA, PB, and PC such that LAPC = a. LCPB = P, and LAPB = 7 = a P c 180". Then A, B, and C are collinear if and only if sin(L7)lPC = sin(La)/PB sin(LP)/PA.
+
+
r The concept of oriented angles is introduced to represent the signed areasand Pythago-
ras differences.
50
Chapter 1. Geometry Preliminaries
PABC SABCD P ABCD
= 2 · AB . BC . cos(jABC) = ~. AC· BD . sin(j(AC, BD)) = 2 · AC . BD . cos(j(AC, DB))
• The Herron-Qin fonnulas give connections between the signed area and the Pythagoras difference.
16S~BC 16S~BcD
4AF· CB2 = 4~· Bd -
plBc plBcD ·
Chapter
2
The Area Method In this chapter, we will present the area method in the narrow sense, that is, as a method of mechanical theorem proving for constructive geometry statements only involving two geometry relations : collinearity and parallel. In other words, we are dealing with constructive geometry statements in affine geometry.
2.1
Traditional Proofs Versus Machine Proofs
We start this section with a comment on the traditional Euclidean proof method from [2] . One of the main defects in the traditional Euclidean proof is its almost complete disregard of such notions as the two sides of a line and the interior of an angle. Without clarification of these ideas, absurd consequences result. The following example shows that this defect may occur even in very simple proofs.
Example 2.1 Let ABC D be a parallelogram (i.e. AB II CD. BC II AD). E be the intersection of the diagonals AC and BD. Show that AE = CEo
l><J'
A
B
Figure 2-1
The traditional proof of this theorem is first to prove 6.ACB 9:! 6.CAD (hence AB = CD), then to prove 6.AEB 9:! 6.CED (hence AE = CE). In proving the congruence of these uiangles, we have repeatedly used the fact LCAB = LACD. This fact is quite evident because the two angles are the alternative angles with respect to parallels AB and CD. However, here we have implicitly assumed the "uivial fact" that points D and B are on opposite sides of line AC. The last fact is harder to prove than the original statement. (Please try it!) 0:;,
52
Chapter 2. The Area Method
This extremely simple example reveals the difficulty in implementing a powerful and sound geometry theorem prover based on congruence and similarity of triangles. Of course, one may develop an interactive prover so that the user can input some trivial facts such as the one in the previous paragraph. These facts can be stored by the program in a data base. Then one would face a much more severe problem of the consistency ofproofs. Example 2.2 Every triangle is isosceles. Let ABC be a triangle as shown in Figure 2-2. We want to prove CA = CB.
Proof Let D be the intersection of the perpendicular bisector of AB and the internal bisector of angle ACB. Let DE 1. AC and DF 1. CB . It is easy to see that !:::.CDE ~ !:::.CDF and !:::.ADE ~ !:::.BDF. Hence CE+EA = CF+FB, i.e., CA = CB. B
Try to solve this paradox.
Figure 2-2
Another defect in the traditional Euclidean proofs is the lack of non-degenerate conditions . Each geometry theorem is valid only under some auxiliary conditions which are not stated explicitly in the theorem. For instance, in Example 2.1 we need to assume that A , B , and C are not collinear. We call such kinds of conditions the non-degenerate conditions of the theorem, which may become complicated for difficult theorems. Without explicitly stated non-degenerate conditions, the traditional proofs of geometry theorems are generally not strict. First, in each step of the proof, we use some lines or triangles which are implicitly assumed to be in normal positions, i.e., each line is uniquely determined and each triangle does not degenerate to a line. However, in a machine proof these conditions should be explicitly stated or justified. Second, during a proof, we need to use other known theorems; but in the statement of the cited theorems the nondegenerate conditions are usually not given explicitly. Hence the correctness of the use of these known theorems is not fully justified. Partially due to these defects, it is very difficult to incorporate the traditional Euclidean proof methods into a computer program so that skillful proofs can be automatically produced by computers. Researchers have been studying automated generation of traditional proofs using computer programs since the work by H. Gelernter, J. R. Hanson, and D. W. Loveland [102] in the early 60s. In spite of the enormous amount of research and great improvements [43,82,105, 140,128,142], the successes in this direction have been limited in the sense that no program has been developed which can prove non-trivial geometry theorems efficiently. On the other hand, A. Tarski, introduced a decision procedure for what he called elementary geometry, based on the algebraic method in the 1930s [34]. Tarski's quantifier
2.1
Tradldonal Proofs Versus Machine Proofs
53
elimination method was later improved and redesigned by A. Seidenberg [145], G. Collins [83] and others. In particular, Collins' cylindrical decomposition algorithm is the first Tarslci type algorithm which has been implemented on a computer. Solutions of several nontrivial problems of elementary geometry and algebra have been obtained using the implementation [45,46, 113]. Meanwhile, Wen-Tsiin Wu introduced a highly successful algebraic method of mechanical geometry theorem proving [162]. Inspired by Wu's work, many researchers have developed efficient computer programs for proving geometry theorems [36, 12, 54, 91 , 119, 125, 132, 191 , 154]. Nearly one thousand theorems from Euclidean geometry, nonEuclidean geometry, differential geometry, and mechanics have been proved by various provers based on Wu's method and its variants [12,68,69, 135,155]. Many hard theorems whose traditional proofs need an enormous amount of human intelligence, such as Feuerbach's theorem, Morley's trisector theorem etc., can be proved by computer programs based on algebraic methods within seconds. In addition, Wu first recognized the importance of the non-degenerate conditions in mechanical geometry theorem proving. However, algebraic methods can only tell whether a statement is true or not. If we want to know the proofs, we usually have to look at tedious computations of polynomials in the coordinates of the related points. So the goal of automatically producing readable proofs for geometry theorems has not been achieved. The goal of this book is to present a method which can produce short and readable proofs for geometry statements efficiently. The starting point of this book is the mechanization of a special case of the area method discussed in Chapter 1. Our machine proof method has the following advantages. • Auxiliary points and lines will be added automatically if they are needed. • Sufficient non-degenerate conditions can be generated automatically. • The proof produced according to the method is independent of the diagram. A key fact behind the success of our method (or Wu's algebraic method) is that the validity of most elementary geometry theorems involving equalities only is independent of the relative order positions of the points involved. Such geometry theorems belong to unordered geometry. In unordered geometry, the proofs of these theorems can be very simple. However, the ordinary proofs of these theorems involve the order relation (among points and lines), hence are not only complicated, but also not strict, see for example, Example 2.1. Here we list several other examples. In the statement of Ceva's theorem (Example 1.7 on page 10) we usually assume that P is inside the triangle ABC. But this restriction on P is not necessary: the statement is true regardless of whether P is inside or outside triangle ABC. The proof produced by the area method is valid for all cases. Also see Examples 1.9 on page 11, 1.115 on page 47, and 1.116 on page 48.
54
Chapter 2. The Area Method
N
Figure 2-3
For the Butterfly theorem (Example 1.99 on page 41), as shown in Figure 2-3, we have three different diagrams, which are often treated as different theorems in geometry textbooks. The proof for this theorem based on the area method is valid for all three cases.
Remark 2.3 1. For a strict proof of Example 2.1 using the area, see Example 1.24 on page 17. To prove it using congruence triangles, you may start from points A, E , and C to construct point D as follows: E is the midpoint of AC and D is the symmetry of E with respect to E . 2. The problem in the "proof' of Example 2.2 is that we use a wrong diagram. The correct diagram for Example 2.2 is Figure 2-4.
2.2
B
Figure 2-4
Signed Areas of Oriented Triangles
We will fonnally define two geometry quantities: the ratio of the signed lengths of directed parallel line segments and the signed areas of oriented triangles. Properties of these two quantities will serve as the basis of our area method. Those who are mainly interested in machine proofs may skip the next subsection and read Subsection 2.2.2 directly.
2.2.1
The Axioms
We use capital English letters with or without subscripts to denote points. The only basic geometry relation is a trinary relation collinear, i.e., three points A, E, and C are collinear. The exact meaning of collinear is given by Axioms A.I-A.6.
2.1
55
SIgned Areas 01 Orlenlecl Trlancla
Let R be the field of the real numbers. 10
Axiom A.I For three collinear points P, A, and B such that A:/; B, ~= is an element in R and satisfies
AP AB
PA
PA
AP
= - AB = BA = - BA and ~; = 0 iff (abbr. if and only if) P = A.
~= is called the ratio of the directed segments AP and AB. Let r also write AP = r AB.
= ~=. We sometimes
Axiom A.2 Let A and B be two distinct points. For r E R. there exists a unique point P which is collinear with A and B and satisfies (1) ~= = rand (2) ~= + ~: = 1. Three points A, B, and C determine an oriented triangle ABC. We use the order of the vertices of a triangle to represent its orientation. Thus triangles 6.ABC, 6.BCA, and 6.CAB have the opposite orientation, whereas 6.ACB, 6.CBA, and 6.BAC have the same orientation, i.e., a triangle has two orientations.
The signed area of an oriented triangle ABC, denoted by which satisfies the following four basic properties.
Axiom A.3 SABC = SCAB = SBCA = -SBAC = three non-collinear points. we have SABC :/; O.
-SCBA
=
SABC,
is an element in R
-SACB . l f A,
Axiom A.4 There exist at least three points A, B. and C such that SABC
:/;
B. and Care O.
Axiom A.S For any four points A. B. C. and D. we have
Axioms A.4 and A.5 are called the dimension axioms. Axiom AA ensures that not all the points are collinear. Axiom A.5 ensures that all the points are in one plane. As a consequence of Axiom A.5, we can define the signed area of oriented quadrilaterals. The signed area of an oriented quadrilateral ABC D is defined to be
By Axioms A.3 and A.5, it is clear that SABCD SABCD SABCD
= SABD = =
SBCDA
SCBD;
=
-SADCB
SCDAB
=
=
SDABC;
-SDCBA
=
-SCBAD
= -SBADC ·
IOThe use of real numbers here is not essential. We can use a number field with any characteristic for this purpose. See Section 2.6 for details.
Chapter 1. The Area Method
56
Axiom A.6 Let A. B. and C be three collinear points such that AB point p. we have SPAB = .xSPAC .
= .xAC. Then for any
Axiom A.6 is the most important property of the area We will see in the next section that most of the interesting and nontrivial properties of area come from this axiom. It is convenient to extend the notion of collinearity to be a geometry relation among any set of points: one or two points are always collinear, and a set of points are collinear if any three points in it are collinear. We can thus introduce a new geometry object: the line. Definition 2.4 A line is a maximal set of collinear points. Let 1 be a line and PEL. Then we say that P is on line I (instead of in line I). Proposition 2.5 Three points A, B. and C are collinear iff SABC
= O.
=
Proof If S ABC 0, then by Axiom A.3 A, B, and C are collinear. Now let us assume that A, B, and C are collinear. If A = C, since CC = 2CC = 0, by Axiom A.6 we have SACC = 2S ACC =O. If A", C and.x = ~~, by Axiom A.6, SABC = .xSACC =O. Corollary 2.6 Two distinct points A and B determine a unique line AB which is the set of all points P satisfying S ABP = O.
Proof Let C, D, and E be three distinct points on line AB. We need to show that C, D, and E are collinear, i.e., SCDE = O. By Proposition 2.5
Then A, D, and E are collinear. Similarly C, D, and A are collinear. By Proposition 2.5
DE
SCDE
= DA . SCDA = O.
In what follows, when speaking about a line AB, we always assume that A '" B. A point P on line AB is determined uniquely by ~~ or ~:. We thus call
AP
Xp
= AB'
PB yp= AB
the position ratio or position coordinates of the point P with respect to AB. It is clear that xp+yp=l.
1.1 Signed Areas or Oriented Triangles
2.2.2
57
Basic Propositions
The basic propositions presented in this section are the basis of our area method. We first extend Axiom A.6 to the following convenient form.
Proposition 2.7 If points C and D are on line AB and P is any point not on line AB
~
(Figure 2-5), then SPCD SPAB
CD = AB ·
A
SPAB
8
C
Figure 2-5
Proof Without loss of generality. let us assume C SPCD
D
.... ..... .
SPCD
= SPCA
SPCA • SPAB
# A . Then
CD CA . AB
= CA
CD
= AB·
Proposition 2.8 (The Co-side Theorem) Let M be the intersection of two lines AB and PQ. Then PM SPAB QM SQAB PM SPAB = = - - ; -=- = - - ; -=- = - -. QM SQAB PQ SPAQB PQ SPAQB Proof See Proposition 1.5 on page 8.
Proposition 2.9 Let R be a point on line PQ. Then for any two points A and B
Proof See Proposition 1.18 on page 15.
Similar to Chapter 1. we use the notation AB
Q satisfy one of the following conditions: (1) A
II
PQ to denote the fact that A, B, P, and
= B or P = Q; (2) A. B. P and Q are on
the same line; or (3) line AB and line PQ do not have a common point.
Proposition 2.10 PQ
II AB iff SPAB = SQAB. i.e.• iff SPAQB = O.
Proof If SPAB # SQAB. it is clear that A # B. P # Q. and A, B, P, and Q are not collinear. Let 0 be a point on line PQ such that PpOQ = ~sS • Thus !2fi = - SSQ.u • By PQ PIIQB PIIQB
Proposition 2.9. SOAB = ~gSQAB + ~SPAB = O. By Proposition 2.5. point 0 is also on line AB. i.e .• AB is not parallel to line PQ Conversely. if PQ It' AB then A # B.
58
Chapter 2. The Area Method
P '" Q and lines AB and PQ intersection at a unique point O. By Proposition 2.8, g~ = ~~~: = 1. Thus P = Q which is a contradiction. 1 A parallelogram is a quadrilateral ABCD such that AB II CD, BC II AD, and no three vertices of it are on the same line. Let ABC D be a paral1elogram and P, Q be two points on C D . We define the ratio of two parallel line segments as follows
In our machine proofs, auxiliary parallelograms are often added automatically and the following two propositions are used frequently. Proposition 2.11 Let ABC D be a parallelogram. Then for two points P and Q, we have
Proof See Proposition 1.30 on page 19.1
Proposition 2.12 Let ABC D be a parallelogram and P be any point. Then
Proof By Proposition 2.11 ,
Figure 2-6
So far, we do not mention the existence of parallel lines. The following statement shows that Euclid's parallel postulate is a consequence of our axioms. Example 2.13 (Euclid's Parallel Axiom) Passing through a point not on a line 1, there exists a unique line which is parallel to I. Proof Let P be a point not on the line AB. By Axiom A.2, we can take points 0 and Qsuch that 0 is the midpoint of P A and Q is the symmetric point of B with 0 , i.e., QO = OB. By the co-side theorem SQA B = 2SOAB = SPAB. By Proposition 2.10, PQ II AB. To show the uniqueness, let T be another point such that TP II AB. By Proposition 2.12, STPQ = STAB - SPAB = 0, i.e., T is on line PQ. Example 2.14 If PR
II
AC and QS
II BD then
;;~~~
= ~~. ~.
1.3
59
Hilbert Intersection Point Statements
Proof Let X , Y be points such that PR SPQRS
=
SPBRY
=
Similarly, we have SPBRD
SpBY -
= AX, QS = BY. By Proposition 2.11,
SRBY
=
BY =(SPBD -
BD
SRBD)
=
QS =SPBRD .
BD
= ~SABCD .
Example 2.1S If line PQ is parallel to line AB then ~ PQ
Proof Let R be a point such that AR
= &AiLs AQP
= PQ. By Propositions 2.7 and 2.11,
AB AB SPAB SPAB = = = = - - = - -. I PQ AR SPAR SPAQ
2.3
The Hilbert Intersection Point Statements
In Chapter VI of Hilbert's classic book, "Foundations of Geometry", he introduced a class of geometry statements which he called the "pure point of intersection theorems." According to Hilbert, every "pure point of intersection theorem" can be put in the following fonn: Choose an arbitrary set of a finite number of points and lines. Then draw in a prescribed manner any parallels to some of these lines. Choose any points on some of the lines and draw any lines through some of these points. Then, if connecting lines, points of intersection and parallels are constructed through the points existing already in the prescribed manner, a definite set of finitely many lines is eventually reached, about which the theorem asserts that they either pass through the same point or are parallel. Hilbert also gave a mechanical proving method for statements of this type. His result can be summarized as follows. Theorem 62 in [24]. Every pure point of intersection theorem that holds in affine geometry takes, through the construction of suitable auxiliary points and lines, the fonn of a combination of finite number of Pascal's configurations. By Pascal's configuration, he meant the diagram of Example 1.27 on page 18. The above result called Hilbert's Mechanization theorem by Wu [164], is the first mechanical theorem proving method for a class of geometry statements, i.e., class C B in this book.
60
Chapter 1. The Area Method
Hilbert's mechanization theorem works as follows. First, we prove a theorem using algebraic calculations (see [164] and Chapter 3 of [36]). Since each arithmetic operation Of numbers, e.g. a + b = b + a, a * b = b * a, can be represented by Pascal configurations, the algebraic proof can thus be converted into a series of Pascal configurations. But the geometric proofs produced in this way are expected to be very long and cumbersome, and as far as we know no single theorem has been proved in this way. The aim of this chapter is to provide an efficient method of producing short and readable proofs for the Hilbert pure point of intersection statements.
2.3.1
Description of the Statements
To describe the Hilbert intersection point statements precisely, we need the concepts of geometry quantities and constructions. In this chapter, by a geometric quantity we mean • the ratio of two oriented segments on one line or on two parallel lines, or • the signed area of an oriented triangle or an oriented quadrilateral. A construction is used to introduce a new point from some known points. We need the following constructions.
Cl Take arbitrary points Yl> ... , Ym on the plane. Y; are free points, i.e., Y; can move freely on the plane. C2 Take a point Y on line PQ. Point Y is a semi-free point, i.e. point Y can move freely on the line PQ . To make sure that point Y can be taken properly, we will introduce a nondegenerate (abbr. ndg) condition P =I Q, Le., the line PQ is well defined. C3 Take a point Y on line PQ such that PY = >'PQ where >. can be a rational number, a rational expression in geometry quantities, or a variable. Notice that>. is the position ratio of point Y with regard to PQ. If >. is a fixed quantity then Y is a fixed point; if >. is a variable then Y is a semifree point. The ndg conditions for this construction are P =I Q and >. is meaningful, Le., its denominator does not vanish.
C4 Take the intersection Y of line PQ and line UV . Point Y is a fixed point. The ndg condition is that P =I Q, U =I V, and the lines PQ and UV have one and only one common point, Le., PQ !YUV. C5 Take a point Yon the line passing through point R and parallel to line PQ. Here Y is a semi-free point. The ndg condition is P =I Q.
2.3
Hilbert IDlersection Point Statements
61
C6 Take a point Y on the line passing through R and parallel to line PQ such that RY = >'PQ, where >. can be a rational number, a rational expression in geometry quantities, or a variable. If >. is a fixed quantity then Y is a fixed point; if >. is a variable then Y is a sernifree point. The ndg conditions are the same as those of C3.
C7 Take the intersection Y of line UV and the line passing through R and parallel to line PQ. Point Y is a fixed point. The ndg condition is that PQ It' UV.
C8 Take the intersection Y of the line passing through point R and parallel to PQ and the line passing through point W and parallel to line UV. Point Y is a fixed point. The ndg condition is that PQ It' UV . Point Y in each of the above constructions is said to be introduced by that construction. We need to show that the above constructions are always possible. That is the introduced points do exist. CI and C2 are trivial. The existence of the point Y in C3 comes from Axiom A.2. C5 and C6 come from Example 2.13 and C3. By Example 2.13, C7 and C8 can be reduced to C4. For C4, since PQ It' UV,line PQ and UV have a unique common point. Definition 2.16 A Hilbert intersection point statement can be represented by a list
where 1. Each construction C j introduces a new point from the points which are introduced by the previous C j , j = I , . .. i-I; and
2. G = (El ' E 2) where El and E2 are polynomials in some geometric quantities about the points introduced by the constructions Cj and El = E2 is the conclusion of S . The ndg condition of S is the set of ndg conditions of Cj and the condition that the denominators of the length ratios in El and E2 are not zero. The set of all the Hilbert intersection point statements is denoted by CR. As indicated by the definition, the ndg conditions of a statement in CD can be generated automatically. Take Ceva's theorem (Example 1.7 on page 10) as an example.
Example 2.17 (Ceva's Theorem) We describe the statement in thefollowing constructive way. Take four arbitrary points A. B. C and P
Chapter 2. The Area Method
62
Take the intersection D of BG and AP. Take the intersection E of AG and BP. Take the intersection F of AB and G P . .~ .~ Show that ..i£ FB DC EA
= 1.
According to the definition, the ndg conditions for Ceva's Theorem are
BG
lY AP;AG lY BP ; AB
lYGP; F
=I B; D =I G; E =I A,
i.e., point P can not be on the three sides of l::.ABG and the three dotted lines in Figure 2-7.
~..~~ .... :/
FE
···· · · ....
---.!'------- --- ---- -- -- ----- ---- ------ ----- ------".".-. 1: --:
A
Figure 2-7
You may wonder that the condition "A, B , and G not collinear" is not in the ndg conditions. Indeed, when A , B, and G are three different points (this comes from the ndg condition) on the same line, Ceva's theorem is still true (now F = G, D = A , and E = B) and the proofs based on the area method is still valid in this case. The ndg conditions produced according to our method guarantee that we can produce a proof for the statement. Certainly, we can avoid this seemingly unpleasant fact by introducing a new construction, TRIANGLE, which introduces three noncollinear points. But theoretically, this is not necessary. Also the ndg conditions are not unique for a geometry statement: they depend on the constructive description of the statements. For instance, Ceva's theorem can be described as follows . Take three arbitrary points A, B, G. Take a point E on line AG. Take a point F on line AB. Take the intersection P of BE and G F . take the intersection D of BG and AP AF BD CE 1 Show th at FB • DC . EA = . Now the ndg conditions of Ceva's theorem are A AP, B =I F , G =I D, and A =I P.
=I
G, A
=I
B , BE
lY Ge, BG lY
The ndg conditions generated according to Definition 2.16 are sufficient, i.e., if a geometric statement is true in the usual sense then it must be true strictly under these ndg
l.J
63
HUbert Intersection Point Statements
conditions. For more details, see Algorithm 2.32 on page 70.
2.3.2
The Predicate Form
A Hilbert intersection point statement can be transfonned into predicate form. We first introduce some basic predicates.
1. (POINT P): P is a point in the plane. 2. (COLL PI, P2 , P3 ) : points plo P2 , and P3 are on the same line. It is equivalent to Sp,P,P3
= o.
Each construction is equivalent to the conjunction of several predicates. C2 Take a point Y on line PQ. The predicate form is (COLL Y P Q) and P C3 Take a point Y on line PQ such that P A Q), >.. = ~~, and P =I Q.
=I Q.
= >"PQ. The predicate form is (COLL Y P
C4 Take the intersection Y of line PQ and line UV . The predicate form is (COLL Y P Q), (COLL Y U V), and ...,(PARA U V P Q). C5 Take a point on the line passing through point R and parallel to line PQ. The predicate form is (PARA Y R P Q) and P =I Q. C6 Take a point Y on the line passing through R and parallel to line PQ such that RY = >"PQ. The predicate form is (PARA Y R P Q), >.. = ~~, and P =I Q. C7 Take the intersection Y of line UV and the line passing through R and parallel to line PQ. The predicate form is (COLL Y U V), (PARA Y R P Q), and ...,(PARA P Q U V). C8 Take the intersection Y of the line passing through point R and parallel to PQ and the line passing through point W and parallel to line UV. The predicate form is (PARA Y R P Q), (PARA Y W U V), and ...,(PARA P Q U V) . The predicate form of each construction C has two parts: the equation part E( C) and the ndg condition ...,D( C) . Now a constructive statement S following predicate form
= (C.. · ·· ,C.. (E,F))
can be transformed into the
64
Chapter 1. The Area Method
where Pi is the point introduced by Ci . lt is clear that the predicate form of a statement depends on how we describe the statement constructively. For the first constructive description of Ceva's theorem (Example 2.17 on page 61), its predicate form is
VA , B,C,P,E, F , D(HYP ~ CONC) where
HYP=
(COLL DB C) 1\ (COLL D A 0) 1\ -.(PARA B C A 0) 1\ (COLL E A C) 1\ (COLL E B 0) 1\ -.(PARA A C B 0) 1\ (COLL F A B) 1\ (COLL F C D) 1\ -.(PARA ABC 0) 1\ B=/=FI\D=/=CI\A=/=E
CONC=
( AF . BD . CE FB DC EA
= 1).
Exercises 2.18 1. We define a new predicate (CONC PI P2 P3 P4 Ps P6 ) which means that the lines PI P2 , P3 P4 • and PS P6 are concurrent. Use an equation in areas to represent this predicate. 2. Construction C3 is to take a point on a line with position ratio A. Show that if point Y is introduced by one of the eight constructions then Y can also be introduced by constructions Cl and C3 . The reason we use more constructions is that we want to describe geometry statements using fewer constructions, and as a consequence to obtain short proofs for the statements.
2.4
The Area Method
Before presenting the method, let us re-examine the proof of Ceva's theorem (Example 1.7 on page 10). By describing Ceva's theorem constructively (Example 2.17 on page 61), we can introduce an order among the points naturally: A, B, C, P, D, E, and F, i.e., the order according to which the points are introduced. The proof is actually to eliminate the points from the conclusions according to the reverse order: F, E, D, P, C, B, and A. We thus have the proof: Eliminate point F. Eliminate point E . Eliminate point D . Then
AF BD CE SACPSBCPSABP . == =1 FB DC EA SBC PSACPSABP . Thus the key step of the method is to eliminate points from geometry quantities. We will show how this is done in the following three subsections.
= .=
65
lA 1be Area Method
2.4.1
Eliminating Points from Areas
We first show that construction C2 is a special case of construction C3. This is because taking an arbitrary point Y on line UV is equivalent to taking a point Y on UV such that U A = >.UV for an indeterminate >.. Similarly, construction C5 is a special case of construction C6: taking an arbitrary point on the line passing through point W and parallel to UV is equivalent to taking a point Y such that W A = >.UV for an indeterminate >.. We will discuss Cl in Section 2.4.3. Thus we need only to consider five constructions C3, C4, C6, C7, and Cg. Lemma 2.19 Point Y is introduced by construction C3, i.e., Y satisfies PY eliminate point Y from SABY, we have SABY
= >'SABQ + (1 -
= >'PQ. To
>')SABP .
Proof This is Proposition 2.9.
Lemma 2.20 Point Y is introduced by construction C4, i.e., Y point Y from SABY, we have SABY
= PQ n UV. To eliminate
1
= -S--(SPUVSABQ + SQVUSABP) . PUQV
Proof By Proposition 2.9, we have PY
SABY
YQ
= PQSABQ + PQSABP .
S By the co-side theorem, we have pPQY = ..fU!1JJL.. . Substituting this into the SpUQV ' ~PQ = sSqvu pUQV previous equation, we obtain the conclusion. Since PQ It' UV, we have SPUQv :f. o. I
Lemma 2.21 Point Y is introduced by construction C6, i.e., Y satisfies RY eliminate point Y from SABY, we have SABY
Proof We take a point S such that RS By Lemma 2.19,
= SABR + >'SAPBQ = PQ. A
B
Figure 2-8
By Proposition 2.11, we have SABS = SABR
= >'PQ. To
+ SABQ -
SABP = SABR
+ SAPBQ .
Substituting this into the previous formula. we obtain the conclusion.
66
Chapter 1. The Area Method
Lemma 2.22 Point Y is introduced by construction C7, i.e., Y is the intersection of line UV and the line passing through R and parallel to line PQ. To eliminate point Y from S ABY, we have
Proof. Take a point S such that RS SABY
= PQ. By Lemma 2.20, we have
= - S1- - . (SUSRSABV + SVRSSABU) .
(1)
RUSV
We also have SRUSV
=
by Proposition 2.11 . by Proposition 2.12. by Proposition 2.12.
SpuQv
SUSR
=
SUQP - SRQP
=
SPUQR
SVSR
=
SVQP - SRPQ
=
SPRQv
Substituting these into (1), we obtain the conclusion. Lemma 2.23 Let point Y be introduced by construction CB. To eliminate point Y from SABY, we have
= -SPWQR S - - · SAUBV + SABW . . PUQV
SABY
Proof. By Lemma 2.21, 2.26 below.
2.4.2
S ABY
=
S ABW
+ ~~ S AU BV.
Now the lemma comes from Lemma
I
Eliminating Points from Length Ratios
ex
Lemma 2.24 Let point Y be introduced by construction C3. To eliminate Y from ~~, we have
if A
D
E PQ;
Y
otherwise. Proof. If A E PQ then ~~
-
-
P
Q
Figure 2-9
.M:.+a
.4l:.H
= AWY = P\jg = Pb Pg
that AS = CD. Then Y is the intersection of PQ 2.8 and 2.11,
Pg and AS
.
Otherwise, take a point S such
and
AS
II CD. By Propositions
1.4
67
The Area Method
Lemma 2.25 Let point Y be introduced by construction C4. To eliminate point Y from AY Co' we ha ve
AY = {~if A is not on UV '§.UD V CD ~ otherwise SCPDq
=
Proof The proof is the same as the second case of Lemma 2.24.1
Lemma 2.26 Let point Y be introduced by construction C6. Then we have
~~ ~ {
4.11.+r .!:!L.~
Pq SAPRq S C PDq
if AERY.
y
T P
if A ¢ RY.
Figure 2-10
Proof The first case is obvious. For the second case, take points T and S such that ~ and ~;
=1
= 1. By the co-side theorem,
Lemma 2.27 Let Y be introduced by construction C7. To eliminate point Y from G we have
AY CD
= {{cA~';V ~ SCPDq
= ~~ ,
if A is not on UV ifAisonUV
Proof If A is not on UV then the proof is the same as the second case of Lemma 2.24. If A is on UV, then the proof is the same as the second case of Lemma 2.26. Since PQ II' UV. we have S C PDQ '" O. 1 Lemma 2.28 Let point Y be introduced by construction CB. To eliminate point Y from G ~~, we have
=
_AY _
CD -
{ ;CAPR 9 PDq
if AY is not parallel to PQ ~S otherwise. CUD V
Proof The proof is the same as the proof of the second case of Lemma 2.26.
68
2.4.3
Chapter 2. The Area Method
Free Points and Area Coordinates
In Subsections 2.4.1 and 2.4.2, we present methods of eliminating fixed or semi-free points from geometry quantities. For a geometry statement S (C}, C2 , • • • , Ck , (E, F)), we can use these lemmas to eliminate all the nonfree points introduced by Cj . Now the new E and F are rational expressions in indeterminates, areas and Pythagoras differences of free points. These geometric quantities are generally not independent, e.g. for any four points A, B, C, D we have
=
SABC
= SABD + SADC + SDBC·
In order to reduce E and F to expressions of independent variables, we introduce the concept of area coordinates. Definition 2.29 Let A. O. U. and V be four points such tlult 0, U, and V are not collinear. The area coordinates of A with respect to OUV are SOUA XA=--, Souv
SOAV YA=-S ' OUV
SAUV ZA=--· Souv
It is clear tlult XA + YA + ZA = 1. Since XA, YA, and ZA are not independent. we also call XA, YA the area coordinates ofQ with respect to OUV.
Proposition 2.30 The points in the plane are in a one to one correspondence with the triples (x, y, z) satisfying x + y + Z = 1.
o
u Figure 2-11
Proof. Let 0, U, and V be three non-collinear points. Then for each point A, its area coordinates satisfy XA +YA +ZA = I . Conversely, for any x, y, and Z such that x+y+z = 1 we will find a point A whose area coordinates are x, y, and z. If Z = 1, take a point A such that ~~ = x . Then by Lemma 2.21, XA = ~~~: = x, YA = -x = Y, and ZA = 1. Hz =F I,
= l~z; take a point A on OT such that ~ = z . By the = ~~ = z. By the co-side theorem again, we have
take a point T on UV such that ~~ co-side theorem, ZA
= ~~~V XA
= Sou A = (1 _ z) SOUT = X SOU V = x. Souv
Similarly, YA
Souv
SOU V
= y.
The f~llowing lemma reduces any area to an expression of area coordinates with respect to three given reference points.
2.4
69
The Area Method
Lemma 2.31 Let O. V. and V be three noncollinear points. Then for points A. B . and Y. we have 1 SOUA SOYA 1 SABY
=Souv
SOUB
SOVB
1
SOUY
SOVY
1
------".y
u
u
v
Figure 2-12
Proof Since by Axiom A.5 SABY = SOAB + SOBY - SOAY, we need only to compute SOBY and SOAY. Let W be the intersection of VV and OY. Then by Lemma 2.20 we have SOBW
1 = -S--(SOBV . SOUY + SOBU . SOYv) . OUYV
By Proposition 2.8, we have ~ss OBW SOBY
= ~s . Thus OUV
1
= - 8 . (SOBV . SOUY + SOBU . SOYv).
(1)
OUV
If OY
II
VV, (l) can be proved as follows . By Example 2.15, ~~
= ~~~;.
By Lemma
2.21 S OBY
OY 8 SOUY (S = VV· OUBV = - 8 OBV ouv
+
8)
=
OUB
SOBV . SOUY S
+ SOBU . SOYV ouv
.
Now we have proved that (1) is true under the condition that 0, V, V are not collinear. Similarly, we have SOAY
=
1 - S . (SOAV . SOUY ouv
1 SOAB
- S . (SOAV ' SOUB OUV
Substituting (1) and the above formulas into the conclusion.
SABY
+ SOAU . SOYv) ; + SOAU' SOBV) .
= SOAB + SOBY -
SOAY,
we obtain
I
Use the same notations asinLemma2.31,letxA = ~~~:'YA = ~~~:;XB = ~~~:'YB xY = !i.w.Ll:.. ,YY = .&ua. . Then the formula in Lemma 2.31 becomes s ouv s ouv ouv
~s ;
SABY
= Souv
XA
YA
1
XB
YB
1
Xy
YY
1
=
Chapter 2. The Area Method
70
which is quite similar to the fonnula of the area of a triangle in tenns of the Cartesian coordinates of its three vertices.
Algorithm 2.32 (AFFINE)
INPUT: S = (C I , C2 , . •• ,Ck, (E,F)) is a statement in Ca.
a UTPUT:
The algorithm tells whether S is true or not, and if it is true, produces a proof
for S . S1. For i = k, ' . . , 1, do S2, S3, S4 and finally do S5.
S2. Check whether the ndg conditions of C j are satisfied. The ndg conditions of a statement have two fonns : A i- Band PQ It' UV . For the first case, we check whether ~~ = 0 where X, Yare two arbitrary points on AB. For the second case, we check whether S puv = SQuv. If the ndg condition of a geometry statement is not satisfied, the statement is trivially true. The algorithm terminates. S3. Let G I ,' .. ,Gs be the geometric quantities occurring in E and F. For j S4.
= 1"" , s do
S4. Let H j be the result obtained by eliminating the point introduced by construction Cj from G j using Lemmas 2.24-2.31 and replace G j by H j in E and F to obtain the new E and F. S5. Finally, E and F are expressions of free parameters. If E is identical to F, S is true under the ndg conditions. Otherwise S is false in the Euclidean plane geometry.
Proof of the correctness. If E = F then it is clear that S is true. Notice that all the elimination lemmas in this section have the property that after applying a lemma to a geometry quantity that has geometric meaning (i.e., its denominator is not zero), the expression obupned also has geometric meaning under the ndg conditions of this statement. Therefore all the geometric quantities occurring in the proof have geometric meaning. The geometric quantities in E and F are all free parameters, i.e., in the geometric configuration of S they can take arbitrary values. Since E i- F , by Proposition 2.33 below we can take some concrete values for these quantities such that when replacing these quantities by the corresponding values in E and F, we obtain two different numbers. In other words, we obtain a counter example for S. I
Proposition 2.33 Let P be a nonzero polynomial of indeterminates Xl, ' .• ,Xn with real numbers as coefficients. Show that we can find rational numbers el, '" ,en such that P( el, ' .. ,en) i- O.
1.4 The Area Method
71
The proof is left as an exercise. For the complexity of the algorithm, let m and n be the number of free and non-free points in a statement respectively. To eliminate each non-free point, we need to apply the lemmas in Subsections 2.4.1 and 2.4.2 to each geometry quantity involving this point once. Also note that each lemma will replace a geometric quantity by a rational expression with degree less than or equal to two. Then if the conclusion of the geometry statement is of degree d, the expression obtained after eliminating the n non-free points is at most degree 2n d. Also note that after eliminating the m free points using Lemma 2.31, each quantity will be replaced by an expression of degree two. Then the final result is at most degree 2d2 n = d2 n +l. This simple exponential complexity of the algorithm seems discouraging. But we will see that on the contrary this method can produce short proofs for almost all statements in e H very efficiently. One reason is that during the proof, the common factors of E and F can be removed. This simple trick alone usually reduces the sizes of the polynomials occurring in a proof drastically. Also the algorithm still has much room for improvement in order to obtain short proofs, as shown in Section 2.5 below. Exercises 2.34 1. Let 0, U, and V be three noncollinear points. Then for points A, B, and Y, we have
1
SABY
SOUA
So VA
SUVA
= S--
SOUB
SOVB
SUVB
OUV
SOUY
Sovy
SUVY
2. Show that each polynomial P{ x) E R[x] of degree d has at most d different roots. Use this result to prove Proposition 2.33. 3. Let P(x) = x d + ad_Ixd- 1 + ... + ao be a polynomial, and m = max{l, E1=llad). Then for any r > m we have P{ r) ::f. O. Use this result to prove Proposition 2.33. 4. Prove the Examples in Sections 1.2 and 1.3 using Algorithm 2.32.
2.4.4
Working Examples
Before going further, we want to explain a little bit about the meaning of the axioms, propositions, lemmas, and algorithms in this book. Since the goal of this book is to provide a method of proving theorems, the algorithms are our final goals. The input to an algorithm is a geometry statement. The output of the algorithm is a proof or a disproof of the statement. The algorithms use the lemmas to eliminate points from geometry quantities. In the proofs of the lemmas, only the basic propositions are used. Finally, the basic propositions are consequences of the axioms. We can thus divide the proofs produced by Algorithm 2.32 into three levels:
72
Chapter 1. The Area Method
1. a proof is at the first level or lemma level if in that proof we only use the lemmas; 2. a proof is at the second level or proposition level if in that proof we not only give the result obtained by applying the lemmas but also the process of how the results are obtained by using the basic propositions; 3. a proof is at the third level or axiom level if in that proof we only use the axioms.
Theoretically, proofs at all levels can be produced automatically. But only proofs at the first or second level are relatively short. If we limit ourselves to the six axioms only, the proofs produced according to our algorithms are generally very long. Also, it is not reasonable to limit oneself to axioms only. The proofs of the geometry statements in Chapter 1 are all at the proposition level and most of the proofs given in this chapter are at the lemma level. Algorithm 2.32 has been implemented as a prover on a computer. At the present time, this prover can only produce proofs at the lemma level. In what follows, when speaking about a machine proof, we mean the proof (in LaTeX form) produced by this prover. For instance, the machine proof for Ceva's theorem (Example 1.7) is as follows. The machine proof
g!:.~
-!z£ . !..Il.~ AE
f..
CD
BF
BF- SBOP
( SloO P) O E BD -SB O P •
AE' ·co
-
!!..
SRGe,SAGe BD SB O p· ( -SABP) • CD
-
simpli fy -
~
B5
S A BP •
CD
Q
SA BP ,S Ace
-
SA B p· SA O P
The eliminants
'CE E 2.B.c..e.... AE'=-SABP BD D
§.aa.e.
Cij= SloOP
sim~ifY 1
In the proof, a ~ b means that b is the result obtained by eliminating point P from plify b · 0 b· a; a sim= means that b IS talned by canceling some common f actors firom the denominator and numerator of a; "eliminants" are the results obtained by eliminating points from separate geometry quantities. The prover can also give the ndg conditions and the predicate form of the geometry statement. We use a sequence of consecutive equations to represent a proof. Some might argue that this proof looks different from the usual form of proofs. It is actually very easy to rewrite a proof in consecutive equations as the usual form. For instance, the above machine proof of Ceva's theorem is essentially the same as the proof of Ceva's theorem on page 10. It is clear that the proofs produced according to Algorithm 2.32 depend on how we describe a geometry statement constructively. For the same statement, some descriptions will lead to long proofs while other descriptions will lead to short ones. Both the way of introducing points and the way of formulating the conclusions will affect the output. We
1.4
73
The Area Method
use some examples to show some heuristic rules in specifying the statement in constructive fonn which may lead to short proofs.
Example 2.35 (Menelaus' Theorem) A transversal meets the sides AB, BC. and CA of a triangle ABC in F, D, and E. Slww that :~ . ~g ~! = -1.
.
First describe the statement in the constructive way. Take three arbitrary points A, B , C . Take a point D on line BC . Take a point E on line AC. Take the intersection F of line DE and line AB. Show that ~ . !J.!J. .~ FB DC EA
The ndg conditions are C"# B. A"# C. DE
~!:.~
~ . JUl.4l:.
BF- SBDE
AE CD SF
~ . ~ . JUl -SBDE
AE
-
Ae Alko . IJ.Il. (-SABD . ~+SABD) . ~ CD
. im1!!i!y ~. -
g
CD
( ~-l) ·(-SACD) · ~
E
=
II' BA, B"# F, C"# D, and A"# E. The eliminants
The machine proof
t;,
Figure 2-13
= -1 .
S ABD
SBDE~
-
(~-1)-SABD)
SADE~
-
(SACD · ~)
-E~-l ~=~ AE
~
IJ.Il.
AC
CD
SABDgSABC · 1J.Il. BC
~.(SABC . ¥c;.;:"SABC) SABC · ~ ·(~-l)
.img1i!y
SACDg(lJ.Il.-l).SABC BC
IJ.Il.gL CD
lJ.Il.-l BC
The above proof produced according to our algorithm is not the simplest one. By describing the example constructively as follows, we can obtain a much shorter proof. Take arbitrary points A , B , C, X, Y . D is the intersection of BC and XY. E is the intersection of AC and XY .
F is the intersection of AB and XY. Show that ~ . !J.!J.. ~ =-1 . FB DC EA
74
Chapter:z. The Area Method
The eliminants
The machine proof
4L!::.~ SF- SSXY
~.!!£.M. AE CD SF
f::.~.~ ..!!,g
~!.~ AE- SAXY
.§..
Sexy,SAXY • BD
t.!..2.p.§..a.Ja..
-
SSXy ,SAXY
CD- SCXY
SSXY
AE CD
CD
sim~ifY ~. IJJ,
SSXY
!2. -
CD
Scxy ·Snxy SSXy,SCXY
si~ifY I
Example 2.36 (Gauss-line Theorem) Let Ao, AI. A 2• and Aa be four points on a plane. X be the intersection of AIA2 and AoAa. and Y be the intersection of AoAl and A2Aa. Let MJ, M 2• and Ma be the midpoints of AIAa. AoA2. and XY respectively. Then MI. M 2• and Ma are collinear.
The constructive description Take arbitrary points Ao, AI, A 2 , and Aa. X
= AoAa n A I A 2.
Y = A2Aa n AIAo . MI is the midpoint of AIAa. M2 is the midpoint of AoA2' M3 is the midpoint of XY. Show that SM,M.M, = 0 Here is the machine proof.
y
Figure 2-14
SM 1 M 2 M 3
~ tSYMIM2+tSXMIM2 ~ (t)·(tSA.YM, +tSA.XM, +tSAoYM, +tSAoXM,) ~ (t ).(-tSA.A,X+tSA,A.y-tSAoA, y-tSAoA,X)
~ (- k)(S~OA'A, A, SA.A,X +S~oA.A, A, SAOA, X+SAOA.A,A, SA, A.A, SAOA, A. -SAOA:zAIA3SAOA2A3SAOA)A3)
simplify n
=
(-i ),(SAoA.A, A, ,SA.A,X+SAoA.A, A, ,SAOA,X+SA,A.A, ,SAoA,A, -SAOA,A, ,SAoA,A,)
= (- i ).( -SAoA,A, A, ·SAoA, A,A2 ,SA,A2A, ,SAoA,A, +SAOA,A,A, ,SAOA,A.A2 ·SAOA, A. ,SAOA,A, + S!OAJA3A.2 ·SAIA:zA3 · SAOAIA2 -S!OAJ A 3 A 2 . SAOA2A3 ,SAOA1A,)
8im2!ify
(k ),(SA oA2A,A, ,SA,A2 A, ,SAOA2A, -SAOA2A,A. ,SAoA,A, ,SAoA,A,SAOA) A3A:z . SA t A2A3 . SAOA} A2 +SAOA) A3A:z ' S AOA2A3 .SAOA) A3)
= W '(O) simplify = n
0
Here a ~ b means b is the numerator of a. To show a numerator is zero.
= 0, we need only to show its
2.4
75
The Area Method
If we describe the statement as follows, then we can obtain a shorter proof. Take arbitrary points A o, Aj, A 2, A 3. X AoA3 n A 1 A2. Y A2A3 n AIAo. MI is the midpoint of A 1 A 3. M2 is the midpoint of AoA2' M3 is the midpoint of XY. Z = M2MI n XY . Showthat ~
=
=
YM.
The machine proof (~)/(XZM3) YM3
YZM
~
'1J -
YM.
W,SYM,M SXM,M. ·(-.)
i
-(tsAaYM,+tsAgYM,) fSA.XM, +tSAoXM,
~ -
r
SAoA, AzA,
-
SAOA)A2A]
XZM, z~. SXM,Ma YZM. - SYM,M.
~"!,. -
3
Z~3 -SYM,Ma . ~ SXM,M.
The eliminants
YM.
(I)
SXM,M. "!,'i(SA.XM,+SAoXM,) SYM,M. ~i(SA.YM, +SAOYM,) SAoXM,
= - 2I( SAOA,X )
1.1, 1.1,
SA.XM, = -
21 ( SA.A.X )
SAOYM, "!,I -i(SAOA.Y) SA.YM,
M'I( = 2 SA,A.Y )
SA2A,X
+5"'OAI x =5"'OAI A2A,
sim~i!y 1
So for a "nice" expression of the conclusion of a statement, we can obtain a short proof. The idea is to express the conclusion as a = b such that a and b are symmetric in some sense. For example, if we want to prove three points P, Q, and R are collinear, and if one of the three points, say R, is on (LINE E F) where E and F are points in the statement, then we usually introduce a new point N by (INTER N (LINE P Q) (LINE E F» and prove the equivalent conclusion N = R or ~~ = ~Z. According to our experience with many examples, the proof for this new conclusion is shorter than the proof for SPQR = O. The following example shows that this rule is also true for the length ratios.
Example 2.37 A line parallel to the base of trapezoid ABC D meet its two sides and two diagonals at H , G, F, and E . Show that EF = GH. We may describe the statements as follows. Take arbitrary points A, B, C. Take a point D such that DC II AB. Take a point E on line BC. Take the intersection H of line AD
Figure 2-15
Chapter 1. The Area Method
76
and the line passing through E and parallel AB. F=BDnEH. G= ACnEF. EF _ HG Prove that AB - -AB· The eliminants
The machine proof
Il!i.~~
EF
--4a... _llJJ.
AB-SABC
u:.!.~
AB
g 2A=-. -SAC H
E..
AB- SABD
EF
AB
SROf;" , SABC
-
-SACWSABD
!!..
-SBDE ·SABq-(-S A BD) (-SACD·SABE)·SABD
-
simJ!!.ify _E
SBDE~
-SRD6' SAB C SACD·SABE
- ( -SBC D · B B eE).SABC
-
SACD·SABC ·
-
(SBCD·~)
SACDg -
(SABC·~)
SBCDg -
(SABC · ~)
:~
si mE!. i! Y §..B.aJJ. -
SACD
D -SAB C·
=:
CD 6 8
-SABC '~~ sim!!1i!y 1
By changing the conclusion to ~~
2.5
= 1, the proof will become much longer.
More Elimination Techniques
The eleven lemmas (2.19-2.31) provide a complete method of eliminating points from geometry quantities. But if only these lemmas are used, the proofs for some geometry theorems are still too long to be readable. In order to produce short and readable proofs for geometry theorems, we will provide more elimination techniques.
2.5.1
Refined Elimination Techniques
Lemmas 2.19-2.31 only give the elimination result in the general case. In some special cases, the elimination results could be much simpler, e.g., see Excises 2.38, 2.39, and 2.40 below. By only using these refined elimination results, we can obtain much shorter proofs for many geometry theorems.
z.s
77
More EUminatloD Tecbalques
Exercise 2.38 Prove the following version of Lemma 2.20. Let Y eliminate point Y from SABY , we have
= PQ n UV.
To
if AB II UV; ifAB II PQ; if u, V, A are collinear;
SABU SABP SUBV ,SAPQ sUPvQ SAUV,SBPQ sUPvQ SAPO ,SBUV sPUQv SPBQ,SAUV sPUQv
if U, V, B are collinear; if P, Q, A are collinear;
sU!VQ(SUPQ' SABV -
sP~Qv(Spuv '
SABQ
S VPQ' SABU)
+ SQvu ' SABP)
if P, Q, B are collinear; ifU or V is on AB; otherwise.
Exercise 2.39 Prove the following version of Lemma 2.22. Let Y be the intersection of the line UV and the line passing through R and parallel to line PQ. To eliminate point Y from SABY, we have
if AB II UV; if AB II PQ; if u, V, A are collinear;
SABU SABR SUBV,SAPRQ SUPVQ SAUV ,SBPRO SUPVQ SAUV ,SBORP sPUQv SBUVSAPRO SPUQV
if U, V, B are collinear;
II PQ; if BY II PQ;
ifAY
-S_i_(SPUQR • SABV pUQV
+ SPRQv • SABU)
otherwise.
Exercise 2.40 Prove the following version of Lemma 2.23. Let Y be the intersection of the line passing through point R and parallel to PQ and the line passing through point W and parallel to line UV. We have SABW SABR SAUWV,SBORP SPUQ V SBUWV ,SAPRO SPUQV SBVWU,SAPRO SUPVQ SAUWV ,SBPRO SUPCQ SABY
= SSPWOR . SAUBV + SABW pUQV
if AB if AB if AY
UV; PQ; PQ;
if BY
PQ;
if AY
UV;
if BY
UV;
otherwise.
To use these elimination methods, we have first to decide whether three points are collinear, or whether two lines are parallel. We can use Algorithm 2.32 to do so. But this process is too time consuming. A faster way is to collect all the obvious collinear and
78
Chapter 2. The Area Method
parallel relations from the constructions and use them as criteria. For instance, in Ceva's theorem (Example 2.17 on page 61) we can find the following collinear point sets easily: {A , B,F}, {A,C, E}, {B,C, D}, {A,D,P}, {B,E,P}, and {C, F,P}. Once we obtain all the lines and parallels for a geometry statement, we can use them to simplify some geometry quantities. For instance,
0 Sp,P3 P• Sp,P2 P3P•
=
Sp,P,P.
{
Sp,P,P3 Sp,P3 P,
if P I P3 II P2 P4 ; if PI, P2 , P3 are collinear; if P2 , P3 , P4 are collinear; if P b P3 , P4 are collinear; if PI, P2 , P4 are collinear.
Example 2.41 Let A , B, and P be three noncollinear points, and C be a point on line PA The line passing through C and parallel to AB intersects P B at D. Q is the intersection of AD and BC. M is the intersection of AB and PQ. Show that M is the midpoint of AB. The example can be described in the following constructive way. Take three arbitrary points A, B, and P. Take a point C on line AP. D is the intersection of line B P and the line passing through C and parallel to AB. Q is the intersection of lines AD and BC. M is the intersection of lines AB and PQ. Show that ~~ = -1. The ndg conditions: A
Figure 2-16
i= P, P If. AB, AD IY BC, and AB IY PQ.
The machine proof
-,w BM !1. -SAP9 -
,w~SAP9 BM-SBPQ
S
SBPQ
!:l
BPQ
-SAPp ·SABq-{ -SABPC)
-
The eliminants
(-SBPC ·SABP)·SABPC
S
QSRPC·SAAQ SABPC
OSAPP -SABG APQ
SABPC
sim!!li/y -5 A PooS A BG -
Q -
SBPC ·SABP
SABP!!,SABC
SAPe ,SABe -SBPC ·SABC
sim!!:.i/y I
Our prover first collects the collinear point sets:
{M,P,Q}; {Q,A, D}; {Q,B,C}; {M,A,B}; {D,B,P}; {C,A,P}; and the parallel lines: DC
II MAB.
l.S
79
More Elimination Techniques
To eliminate D from SABD. we use the first case of Exercise 2.39: S A BD = SABC. To eliminate D from S A PD . since p . B . and D are collinear we can use the fourth case of Exercise 2.39: S
APD -
Since p . A . and C are collinear.
2.5.2
SABPSPAC B SBAPB -
SPAC B
=
SP AC
S -
PACB
+ SPC B = SBPC .
The Two-line Configuration
Another commonly used technique is the two-line configuration. This trick works only when there exist at least five free or semi-free points and these points are on two lines 11 and 12 . If II II 12 • let a he the oriented distance from line II to line 12. If II 11'12 let (3 = sin(NI , /2 )) and 0 he the intersection of II and 1211 . Let II and 12 he two lines satisfying the above conditions. Then we have the following elimination procedure. Case 1. G
= S A B C.
o laBC 21 --aBC I 2::.:::. -{30A · BC 2 1 --"2{30A · BC Case 2. G = ~~ . We have G = (~~) into the ratio of two quantities. Case 3. G
if A , B , C are collinear if II 11/2 • A E Ii> and B, C if II 11/2 • A E 12 • and B, C if II 11'/ 2 • A E Ii> and B , C if II 11'1 2 • A E 12 • and B , C
E 12 Eh E 12 Ell.
= AB/CD. i.e.• we break one geometry quantity
= AB if II and 12 intersects at 0 if II 1112 • A E 11 and BEll if II 11/2 • A E 12 and B E 12
where 0 1 and O 2 are fixed points on II and 12 respectively. For the constructive description of Menelaus' theorem on page 73. we have the following machine proof using the two-line trick. 11 We can avoid the concept of the distance between two parallel lines and the sine function. Our intention here is to write the proving process in as shon a fonn as possible. The reader may tty to develop methods for tbe two-line configuration using tbe basic propositions only.
80
Chapter 2. The Area Method
The eliminants
~ . !!.2. . 4i:.
~~.fi.B. . ~
4.f:.~~ BF- SBDE SBDE= - !(CE.BD·fJ)
21~es
SADE= -
AE CD
BF
-SBDE
AE CD
CE.BD""i;-CD.AW:m
1(-2 CD ·AE·fJ)
(-C E·BD·fJ)·(2) ·C D·AE simJ!!.ijy
Example 2.42 (Pappus' Theorem) Let points A. B and C be on one line. and AI. BI and C 1 be on another line. Let P = ABI n AlB. Q = ACI n AIC. and S = BCI n BIG. Show that p. Q. and S are collinear.
The input to the program. Take arbitrary points A, At. B, B I . Take a point C on line AB. Take a point Cion line AlBI • p=AIBnABI . Q=AClnA1C. S=B1CnBCI . T= BlCnPQ . Prove that lbd =~ CS CT' The machine proof
(W)/(W) J;,
SCPQ . ~ SB,PQ cs
.§.. (-SBB]C] ),scPQ -
SB,PQ ,( -SBCC,)
g
SSB,C"SA,CP ,SACC"SAA,C,C
SB,C,P,SAA,C) ,SBCC,,(-SAA,C,C) siTnJ!!.ify SBS, 0, , SA,OP ,SAGO, SB,C, P,SAA, C,SBCC, P
SaB,c, ,SA] BO ,SAA IS, ,SACO, ,SAA I S, B
= ( SA,BB"SAB,C,) ,SAA,C ,SBCC, '(-SAA,B,B) simJ!!.ify BBS, 0,
-
,SA) BO'SAA , 8, ,SACC t
SA, BB, ,SABIC, ,SAA, C,SBCC,
21~es B;C";·QB·fJ·( -BC.OAd)·A;B;'·QA·(J·( -AG'·oc.·(J)·«2))4 ( A, B, ·OB·fJ)·B, C,·OA·(J·AC ·OA, ·fJ·( -BC·OC,·fJ)·«2))4
simglijy
Ct
Figure 2-17
The eliminants ~.ISB]PQ
CT - SCPQ ~~SBB]C] cs - SBCC, S Q-SB]c]rSAA]c B,PQ SAA,C,C S g SA]CP ,SACC] CPQ- -SAA,C,C S
P-SA,BB"SAB,Ct
B,C,P SAA,B,B S !SA] BC ,SAA]B] A,CP- -SAA,B,B SBCC,= -
!(BC.OC, ·fJ)
SAA,C=!(AC.OA,·fJ) SAB,C, =!(B,C,·QA·fJ) SA,BB,= SACC,= -
!(A,B, .OB'(J) !(AC.OC"(J)
SAA,B, =!(A,B,.OA·fJ) SA,BC= -
!(BC.OA"(J)
SBB,C, =!(B,C,·QB·fJ)
1.6 Am..., Geometry
2.6
81
Area Method and Affine Geometry
We shall first discuss briefly the relationship between geometry and algebra, beginning with passages from E. Artin's book "Geometric Algebra", [5] : We are all familiar with analytic geometry where a point in a plane is described by a pair (x, y) of real numbers, a straight line by a linear, a conic by a quadratic equation. Analytic geometry enables us to reduce any elementary geometric problem to a mere algebraic one. The intersection of a straight line and a circle suggests, however, enlarging the system by introducing a new plane whose points are pairs of complex numbers. An obvious generalization of this procedure is the following. Let k be a given field ; construct a plane whose "points" are the pairs (x , y) of elements of k and define lines by linear equations .... A much more fascinating problem is, however, the converse. Given a plane geometry whose objects are the elements of two sets, the set of points and the set of lines; assume that certain axioms of geometric nature are true. Is it possible to find a field k such that the points of our geometry can be described by coordinates from k and the lines by linear equations? These passages suggest that there are two approaches to defining geometry. The Algebraic Approach. Starting from a number system £ (usually fields), we can define geometry objects and relations between those objects in the Cartesian product £n (or £n / E" in projective geometry). In modem geometry, especially in algebraic geometry, this approach indisputably prevails. If we take this approach, then there are only a few differences between algebra and geometry; geometry can be regarded as a part of algebra. However, the second approach suggested by Artin is more attractive from the point of view of traditional proofs of geometry theorems. The Geometric Approach. By this approach we mean the one that was used by Euclid and Hilbert. In the Euclid-Hilbert system, number systems are developed as parts of the geometry. For each model of a theory ofgeometry, we can prove the existence of a number system (usually a field) inherent to that geometry. This field is called the field associated with that geometry. That geometry then can be represented as the Cartesian product of its associated field. Though beautiful and elegant, the Euclid-Hilbert approach is on the other hand a heavy burden to develop. The axiom systems which we have adopted in this chapter is a mixture of the above approaches. First we take the number systems for granted. On the other hand we use a geometric language instead of an algebraic one. This system is a modification of an axiom system developed by J.Z. Zhang for the purpose of geometry education [40] . It has the advantage of providing simple but also general methods of solving geometric problems, a virtue the algebraic and the geometric approaches do not possess.
82
Chapter 2. The Area Method
2.6.1
Affine Plane Geometry
Affine geometry is the study of incidence and parallelism. There are two kinds of geometric objects: points and lines. The only basic relation in this geometry is that of incidence, i.e., a point A is on a line I, or equivalently, a line I passes through (contains) a point A. Two lines which do not have a point in common are called parallel lines. The following is a group of axioms of affine plane geometry [5]. Axiom H.l. Given two distinct points P and Q. there exists a unique line passing through both PandQ. Axiom H.2. Given a line I and a point P not on I. there exists one and only one line m such that P lies on m and such that m is parallel to 1. Axiom H.3. There exist three distinct points A. B, G such that G does not lie on the line passing through A and B. Axiom H.4 (Desargues' Axiom). Let II. 12• 13 be distinct lines which are either parallel or meet in a point S . Let A, Al be points on II, B, BI points on 12 and G, GI points on 13 which are distinct from S if our lines meet. We assume line AB II AIBI and BG II BI GI . Show that AG II AlGI. Axiom H.S (Pascalian Axiom). Let I and II be two distinct lines, and A, B, G and AI> B I . GI be distinct points on I and II, respectively. If BGI II BIG and ABI II AlB, then AGIIiAIG. A geometry in which all the above five axioms hold is called an affine geometry.
The above is a geometric approach for defining affine geometry. Now let us start at the other end and give a definition of affine geometry based on the algebraic approach. Let [ be a field. From [ we can construct a structure n as follows. Let
i= {(a,b,c) la,b,cE
[, a:;i:Oorb:;i:O}.
We define a relation ,..., in i as: (a,b,c)"'" (a' , b' , d) if and only if there is ak E [suchthat k :;i: 0 and (a, b, c) = (ka', kb', kd). It is easy to see that,..., is an equivalence relation. Let if ,..., (the set of all equivalence classes of i) be denoted by L . Define Inl to be [2 U L. An element p in Inl is a point if and only if p E [2 (i.e., p = (x, y), x, y E E); an element I in Inl is a line if and only if 1 E L. A point p = (x, y) is on a line I = (a, b, c) if and only if ax + by + c = O. Two lines II = (a, b, c) and 12 = (a' , b', d) are parallel if there exists a k E [ and k :;i: 0 such that a = ka', b = kb'. It is easy to check the following theorem. Theorem 2.43 Axioms H.I-H.5 are valid in the structure n.
Proof It can be easily checked that the five axioms are valid in n. H4 and HS, particularly, can be proved automatically using Wu's method (Example 121 and Example 346 in [12]).1
83
2.6 Affine Geometry
The converse of the above theorem is a much deeper result. Theorem 2.44 Every geometry G of the theory H.l-H.5 is isomorphic to a structure with some field E.
n
The key step of the proof is to introduce the segment arithmetic and hence to introduce the field E inherent to G. The field E. uniquely determined by geometry G up to isomorphism. is called the field associated with geometry G . Desargues' axiom makes it possible to introduce a division ring E and Pascalian axiom makes E a commutative field . Each algebraic rule of operation (e.g .• associativity of addition) corresponds to a geometry theorem. The process of introducing number systems in this way is the core of the EuclidHilbert approach. For details. see [24). [5). and [36).
2.6.2
Area Method and Affine Geometry
Suppose that the number field E in the six axioms A.I-A.6 is not the real number field R but an arbitrary field . We shall show that these six axioms define an affine geometry. Theorem 2.45 Show that all the fi ve Axioms H.l-H.5 are consequences ofAxioms A.l -A.6. Proof Axiom H.I follows from Corollary 2.6. Axiom H.3 is a consequence of Axioms A.3 and A.4. For Axiom H.2. see Example 2.13 on page 58. H.4 and H.5 can be proved automatically by our prover. For their proofs. see the following examples.
Example 2.46 (Desargues' Axiom) SAA I • SBB I • and SCCI are three distinct lines. AB II AIBI and AC II AICI then BC II BICI.
Take arbitrary points S. A. B. and C . Take a point Al on line SA. Take the intersection BI of line SB and the line passing through Al and parallel to AB. Take the intersection C I of line SC and the line passing through Al and parallel to AC. Prove that SB, BC = SC,BC .
S~~____~~~4-__~8_,
Figure 2-18
If
84
Chapter 2. The Area Method
The eliminants
The machine proof SBGB] SBGG,
SBGG,
01 SAGA) ,SSBC SSAG
G
SBGB] ,SSAG
S
~ SABA] ,SSBG
-
SAGA, , SSBG
~
SABA, ,SSBC , SSAC
-
SAGA, ,SSBG,SSAB
simJ!!ify A,
=
=
BCB, -
SSAB
SABA] ,SSAG SAGA, ,SSAB
(-SSAB,~+SSAB)·SSAG ~
( -SSAG· ~+SSAG)·SSAB
simJ!!ify
The ndg conditions are S i= A, S, A, B and S, A, C are not collinear, which are consequences of the hypotheses of the statement.
Example 2.47 (Pascalian Axiom) Let A, B and C be three points on one line, and AI, B I, C I be three points on another line. If ABI II AlB and ACI II AIC then BCI II BIC.
The constructive description. Take arbitrary points A, B, and AI. Take a point C on line AB. Take a point BI such that BIA II BAl . Take the intersection CI of line AIBI and the line passing through A and parallel to CAl. Prove that SBGB, = SG,GB, .
A
B
c
Figure 2-19
The eliminants
The machine proof SBGB] SCBIC l
Qj -
SBCBI,SAlCB)
-SAA,B,G ,SA,GB,
simJ!!ify
SBGB] -SAA1B1C
!!!
-SBA -SBA
G'~
'~ 1
G'~ BA]
simgl/f Y I
The ndg conditions are A ofH.5.
i= B, B i= At. and AIBI !yCA I which are all in the statements
Now we have the converse theorem.
85
2.6 Amne Geometry
Theorem 2.48 In the affine geometry associated with any field ratios and areas such that Axioms A.l -A.6 are valid.
e, we
can define length
= (Xi, Yi), i = 1,·· · , 4, be four points on a line I such that P3 t= P4 . Then Pl P2 {%, -%, if Y3 = Y4 · P3 P4 = ~=~ otherwise. Let Pi = (X" Yi), i = 1, 2, 3, be three points. Then define
Proof Let Pi
Xl Yl 1 Sp,P,p, = k
X2
Y2
X3
Y3
1 1
where k is any nonzero element in e. Axioms A.I-A.6 can be verified by direct calculation. I
Now it is clear that Algorithm 2.32 is for the constructive statements not only in Euclidean geometry but also in affine geometry associated with any field, even finite fields . In other words, the area method works also for finite geometries. For examples related to various fields, finite or infinite, see Subsection 2.7.2. The completeness of Algorithm 2.32 is based on Proposition 2.33. For an arbitrary field
e, we have.
Proposition 2.49 Let e be an infinite field and P a nonzero polynomial of indeterminates
Xl, . . . , Xn with coefficients in e. Show that we can find elements el, . . . , en in e such that P(e l , "" en) t= O. Proof We prove the result by induction on n . If n = 1, let P(xt} be of degree d. Then P( x d has at most d different roots. Since is infinite in any d + 1 distinct elements of there exists one which is not a root of P(XI)' Suppose that the result is true for n - 1. We write P as follows
e
e
If s = 0, we need do nothing. If s > 0, by the induction hypotheses there are elements el, .. , en- l in e such that a.( el , ... , en- d t= O. Let Q(xn) = P( el , ..., en- I, Xn) t= O. Now the result can be proved similarly to the case n
= 1.
I
e
If is a finite field, then the above result is false and we do not know whether there exist efficient algorithms to check the existence of such elements. Obviously, a slow algorithm exists: we can check all possible elements in since is finite.
en
e
Note that the area is not an invariant in the affine geometry. But due to the following fact the ratios of areas are invariants.
86
Chapter 2. The Area Method
Exercise 2.50 Let M be a 2 x 2 matrix, F'i E £2 , i Then
= 1, 2,3.
Let Q,
= P,M, i = 1, 2,3.
SQ,Q,Q3 = IMISp,P,P3
where
IMI is the determinant of M .
So we can use ratios of areas instead of areas as geometry quantities. Also it is worth mentioning that in the proofs of all the examples in this and the preceding chapters, the areas always occur in the form of ratios. This is not a coincidence. Let C(TI ,··· ,Td, al , ' " , a.) = 0 be the conclusion of a geometry theorem where the T, are length ratios and the are areas of triangles. Let M = >..J be the multiplication of an indeterminate .x and the unit matrix I . After transfonning each point P in the plane to PM, C = 0 is still valid. By Exercise 2.50, C = (TI,"" Td, .x2al,· ·· ' .x2a.) = O. Therefore, if £ is an infinite field, each homogeneous component of P in the variables aI, . .. , a. must be zero, i.e., without loss of generality we can assume P is homogeneous in the area variables. That is C can be expressed as a polynomial of the ratio of lengths and the ratio of areas.
a,
Applications
2.7
Besides theorem proving, the area method can be used to solve other geometry problems such as deriving unknown formulas automatically. In this section, we will show the application of the area method in three geometry topics: the formula derivation, the existence of n 3 configurations, and the transversal problems.
Formula Derivation
2.7.1
Algorithm 2.32 can be used to derive unknown formulas . We use a simple example to illustrate how this works.
Example 2.51 Let L, M , and N be the midpoints of the sides AB, BC, and C A of triangle ABC respectively. Find the area of triangle LMN. Solution. Since N is the midpoint of AC, by Proposition 2.9, SLMN=HSCLM + SALM). ) SCLM=2I (S BCL ) . Then SLMN= lSBCL-1S,tCL B Ythe CO-SI'de theorem, SALM= - 2I (S ACL, 2 2· --~ 4 .
I
Example 2.52 Let Aio B I , and C I be points on the sides BC, CA, and AB of a triangle ABC such that BAt/AIC = Tio CBt/BIA = T2. and ACt/CIB = Ta. Show that SA]B) C )
_
SAB C
-
rJ rz rJ±l
(TI+l)(r,+l)(r3+l) '
2.7
87
AppUaodons
Constructive description Take arbitrary points A. B. and C. Take a point Al such that ~ = TI . Take a point BI such that ~BB ,A
= T2.
Take a point C I such that ~ Compute SA,B,C, '
= T3 .
B
c
A,
Figure 2-20
SABC
The machine derivation.
The eliminants
SA , B,C, SABC
S
Cl S8"') 8 , ·,-,+5"".4,8,
=
A,B,C,
~ SSA,B"rl+SAA,B,
-SACA] SAA,B, .. +1
SAB c ·( r3+ 1) ~ -SACA, ·,.,-5.4C"', +SABA1 · r, ·r~+SABA1 · rl · f"]
B]
SBA]"]
SABc ·(r3+1)-(,.,+l)'
SABA] ''''
= .. +1
A,~
.imJ!lify - (SA CA, -SABA, .., .,.,) SABC ·(.,+I)·(,.,+l)
SABA, =
1l -( -SABC · r3 ·,.,·r~-SABc ·r3 ·,., ·r,-SABc · r,-SABc) -
.,+1
'b
SABC ,(.,+l)·(,.,+l) ·(r,+l)'
A,
SACA,
=
r,+1
=..5..uu;. r,+1
(r3+1)-(,.,+I)-(r,+I)
Remark. As a consequence of Example 2.52. we "discover" Menelaus' theorem: AI, B I , and C I are collinear iff TI T2T3 = -1.
Example 2.53 Let AI. Blo C I • DI be points on the sides CD. DA. AB. BC 0/ a parallelogram ABCD such that CAt/CD = DBt/DA = ACt/AB = BDt/BC = T. Let A 2 B 2 C2D 2 be the quadrilateral/ormed by the lines AAI BBl. CCI • DD I . Compute SABA, SABCD
a1Ui
SAaBa CaDa SABCD
The constructive description Take arbitrary points A. B . and C. Take a point D such that ~~ = 1. Take a point Al such that ~
= T.
Take a point BI such that ~ A2 = AAI n BBl. Compute SABAa •
=
SABCD
"""'---'7."",---.....",c
T.
Figure 2-21
88
Chapter 1. The Area Method
The eliminants
The machine derivation.
~
ABA:;!: -
SABB,·SABA1
SABAIB}
SABeD,SABA tB I
~
~ SADAI .r-SADAI +SABA 1 (r-1)-SABD)
SADA, ~(r-I).SACD
- (r-I) ,SABD,SABA! SA8CD ,( SADA, .r-SADA, +SA8A,)
SABA, ~ SABD·r-SABC·r+SA8C
~
-(r-I) ,SA8D ,(SA8D ·r-S A8c·r+SA8 C) SA BCD,(SACD ·r'-2SACD ·r+SA CD+SABD ·r-SABc ·r+SA8C)
Q
-(r-I) ,(SA 8c)' (2S A 8 C),(SA8c·r'-2S A Bc·r+ 2SA8C)
simJ!!.i fy
SABAIB}
SA88, ~ -
(-SA8D·r+SABD),SABA! SABCD ,(SADA,·r-SADA ! +SABA,)
s imJ!!.ify
-
~ SA88! 'SA 8A!
S
SA8A, SA8CD
SACDflsABC
- (r-I) (2)·(r' - 2r+2)
Thus
SABA, SA8 CD -
I r 2(r'-2r+2)"
To compute SABG D -
SA,8,C,D" SABCD •
SABA. -
we have
SBGB. -
SGDG. -
SDAD.
l-r
(1 - 4·
=
2(r2 _
2r + 2))SABGD
r2 r2 _ 2r + 2SABGD
Example 2.54 Let E. F . H . and G be points on sides AB. CD. AD. and BC such that AE = DC DF AH = BC BG = r2 · Le t EFandHG meet In . I . Compute EF EI and Hc' HI AB = rl and AD c Constructive description Take arbitrary points A. It-C. D . Take a point E such that = rl. Take a point F such that ~ = rl' Take a point H such that ~ = r2 . Take a point G such that = r2 . 1= EFrtHG. HI Compute 7Ji '
1!
:g
H[L~L----" G A
B
E
Figure 2-22
The machine proof for this example is a little long. The following proof is the modification of the machine proof. !!J. CJ
-~
-
SCEF
_ -
r"SDEF+(l-r,)·S,<\EF r"SCEF+(I-r,) ,SBEF
89
1.7 Applicadons _ -
... .r"SpEc +(l-r, )·r" S6BF ... .( I-r, )·SoEC+(I- ... ).( I-rJ) ,S6BF
• •mJ?!iJy -!:L-
I-r,
Example 2.55 The sides AB and DC of a quadrilateral are cut into 2n + 1 equal segments by points PI , ... , P 2n and Q I , ... , Q2n respectively. Show tlult
(2) If sides BC and AD are cut into 2m + 1 equal segments by points R I ,'" , R2m and SI , ... , S2m respectively. then the area of the quadrilateral formed by the lines PnQn. P n +1Qn+1 . RmSm.
and Rm+1Sm+1 is
(2n+1)(2m+1)SABCD'
= =
Figure 2-23 shows the case n m 2. Note that in the following machine proof for (1), we use some different names for points P n , P n +1, Qn+i> Qn . We also use a trick: point Qn+1 is introduced two times (Q and V) so that different elimination methods will be used to eliminate Qn+1 in different cases. Constructive description Take arbitrary points A, B , C , D .
= 2n~I' Take a point U such that ~~ = 2n~I ' Take a point Q such that ~ = ;"-;'11' Take a point V such that ~~ = 2:+1' Take a point Y such that ~~ = 2n~I' Take a point X such that ~~
Compute
The eliminants y-S6Bq SXQY= 2n+1 v~
Sxuv= 2n+1 5 Q SARo -ntSABC"ntSABC 6BQ 2n+1 S x SaGP "n±Sdep'n±SAon OOX 2n+1 SABOO=SAOO+SABO SBOO=S600- S ABO+SABO
(S6XY+SUXY) . S6BCO
The machine proof -(SxqY+sxuv) S6BOO ~ -(2Sxuv ·n +SXUV-SABq) SABoo ·(2n+1) }:: -( - 2Soox ·n-S oox-2SABq·n-S6Bq) S6Boo ·(2n+I)' . imJ?!ify SOOX+S6Bq SABOO ·(2n+1)
!1. -
2Scnr ·n+Scnr+SARp·n+SARc·n+SARG S6Boo ·(2n+1)'
s, s. J--I~-+--+--t-...,
Figure 2-23
II.
90 !S.. -
Chapter 2. The Area Method 25 BCp · n2 ±SRCQ ·n±2S Acp·n 2 t3S Ac o -niS Acn±2SA Rp·n 2 tS A R Q·n±2SA 80-n 2 +35 A Be ·niS A 8C S ABCD ·(2n+l)3
simJ?!.i fy -
SaGP·ntS AC Q'ntSA c n±SA so·ntS A Be ·niS .!BG SABCD·(2n+l)'
are.!!...-co 25 AGn -n±SACn±2SA Be -n±SA Be
-
simElify
(SACD+SABc)·(2n+l)' 1 2n+l
By Example 2.54. PnQn and Pn+lQn+l are cut into 2m+ 1 equal segments by RiSi , i = 1, ... , 2m respectively. Now (2) comes from (1) directly. I For more examples of formula derivation. see Examples 1.11. 1.12. 1.13. and the examples in the next subsection.
2.7.2
Existence of n3 Configurations
We define a plane configuration as a system of p points and I lines arranged in a plane in such a way that every point of the system is incident with a fixed number A of lines of the system and every straight line of the system is incident with a fixed number 'IT of points of the system. We characterize such a configuration by the symbol (P>., I,,). For example. the triangle forms the configuration of (3 2 , 32 ). The four numbers p, I, A, and 'IT may not be chosen arbitrarily. For. by the conditions we have stipulated, Ap straight lines of the system. in all. pass through the p points; however. every line is counted 'IT times because it passes through 'IT points; thus the number of lines I is equal to ~. Therefore. for every configuration (p>.,I,,). we have AP = 'ITt. We only discuss those configurations in which the number of points is equal to the number of lines. i.e .• for which p = I. Then it follows from the relation AP = 'lT1. that A = 'IT. The symbol for such a configuration is always of the form (p>.,p>.). We shall introduce the more concise notation (P>. ) for such a configuration. We shall further limit the number A. A = 1 yields only the trivial configuration consisting of a point and a line passing through it. The case A = 2 is realized by the closed polygons in the plane. On the other hand. the case A = 3 includes the most important configurations in projective geometry. the Fano configuration. Desargues' configuration. and Pappus' configuration. In this case the number of points. P. must be at least seven. For through any given point of the configuration there pass three lines. on each of which there must be two further points of the configuration. If in a geometry. there exist p points and lines consisting of a (P3) configuration. we say that the configuration (P3) can be realized in that geometry. As an application of the area method. we obtain the sufficient and necessary conditions for the existence of the (73 ). (8 3 ). and (9 3 ) configurations. For more complicated configurations. see [150] . Example 2.56 There exists only one (7 3 ) configuration and this configuration can only be
2.7 ApplkadoDS
91
realized in the geometry whose associated field
= 1,· · · ,7.
Proof Let the seven points be Pi, i consists of the lines :
PI PI P2 P4 Pa Ps
c is of characteristic 2. (Fano plane)
P2 P4 P6
P2 Ps P7
The only possible (7a) configuration
Pa Pa PI P4 Ps P6 P7 P6 P7
Consider the following geometry problem. Take arbitrary points PI , P2 , P4 . Take a point Pa on line PI P2 • Take a point P s on line PI P4 . P6 = P2 P4 n PaPs . P7 = P2 PSn PaP4 . Compute S P, P, Pr .
P,
Figure 2-24
Using Algorithm 2.32, we have
The (7 a) configuration exists iff Sp,p,Pr
= 0, i.e.
PIPS PIPa PIPa PIPS (2)· ( = -1) · ( = -1) · = . Sp,FV. · = PI P4
PI P2
If ~ - 1 = 0, we have P4
P,
P,
= Ps.
If
** -
PI P2 1
PI P4
= 0.
= 0, we have P2 = Pa.
If
**
= 0,
we have PI = Pa. If Sp,P,P. = 0, we have PI, P2 , and P4 are collinear. If ~ = 0, we have PI = Ps . All the above cases will lead to degenerate configurations. Then the (7a) configuration exists iff 2 = 0, i.e., the associate field of the geometry is of characteristic 2 .• Configuration (8 a) also has only one possible table:
PI P2 Ps
PI PI P4 P6 Pa P7
P2 P2 Pa Pa P4 Pa P7 P4 Ps Ps P6 Pa P7 Pa P6
Theorem 2.57 The (8a) configuration only exists in the geometry such that to c, the field associated with the geometry.
Proof Consider the following geometry problem. Take arbitrary points PI, P2 , P4 .
A
belongs
92
Chapter 1. The Area Metbod
** **
= TI. Take a point P6 such that ~ = T2. Take a point Ps such that = T3. P7 = P2 Ps n P I P6 • P3 = P 2 P6 n PsPs . Take a point Ps such that
Compute
SP3PrP, .
Using Algorithm 2.32. (T~ (T2 T2
+ T2TI
-
2T2
SP3PrP,
+ TITI -
(T3T2TI -
Then
SP3PrP,
T2TI
is found to be
TI + 1) - T3TI(T2 - 2TI + 1) + T?)Sp,P2J>,(TI + T2 - 1)(T3(T2TI - T2 - TI + 1) + TI)
-
1)T2
= 0 iff
T~(T2T2
+ T2TI
-
2T2
+ TITI -
TI
+ 1) -
T3TI(T2 -
2TI
+ 1) + Ti = 0
has solutions for T3 . The discriminant of the quadratic equation is therefore the result. I
-3(T2 -
1)2T~. and
Contrary to the (7 3) and (83) configurations which do not exist in the Euclidean plane. the case n = 9 gives rise to three essentially different configurations. all of which can be realized in the Euclidean plane. The first (93) configuration is that related to the Pappus theorem. a machine proof of which can be found in Example 2.42 on page 80. Example 2.58 Prove the existence of the (9 3 ) configuration as shown in Figure 2-25. Proof. Consider the following geometry problem.
******
Take arbitrary points Plo P3 • and Ps. Take a point P7 such that = TI . Take a point Ps such that Take a point P9 such that
Take a point P2 such that ~ P4 = P 1 P9 n P 2 P s . P6 = P3 P s n P2 P9 • Compute S P, Po Pr . Using Algorithm 2.32. (T4(T3 T 2
+ T3 T I -
(T4 T 2 T I -
T4
2T3 -
T2TI -
T2
T3 -
= T3 . = T4. Figure 2-25
is found to be
SP,PoPr
+ T3 T 2 -
= T2.
+ 2) - 2T3T2 + 2T3 + 2T2 + 1)(T4T3TI - T4T3 + T4 -
T2
2)TIT2(1 T3T2
TJ)T3T4Sl')J>,Ps
+ T3 + T2 -
Then SP,PoPr = 0 iff or equivalently. iff T4
= (2T3 T2 -
2T3 -
2T2
+ 2)/(T3T2 + T3TI -
2T3 -
T2TI -
T2
+ 2)
I
1)
1.7
93
Applk:adoDS
Example 2.59 Show the existence of the (9 3 ) configuration as shown in Figure 2-26. Proof Consider the following geometry problem. Take arbitrary points Plo ~and P7 . Take a point P3 such that ~ = rl . Take a point P6 such that ~ = r2 . Take a point Ps such that ~ = r3 . Take a point Pg such that ~ = r 4 . Ps = PIPS n P3 P9 • P2 = P4 PSn P6 P9 • Compute SP.P.p, . Using the program, we have SP,I\P,
11 = h =
r4(r2 - rd - r2rl
dl d2
r4(r3(r2 - rd
= -kh-Sp,p.Pz d • .d.
Figure 2-26
where
+ rl
r4(r~r~ - r~r2rl + r~r~ - r~rl + r3r2rl - r3r2 - 2r3r~ + 2r3rl -r~r~ + r~rl + 2r3r~ - 2r3rl - r~ + rl r4(r3(r2 - rl)
+ rl - 1) - r3rl(r2 + 1) + r2rl + rt} - r3rl(r2 + 1) - rl.
+ r~ -
rl)
- rl
Using the program again, we may check that 11 = 0 (i.e., r4 = (-r2rl + rl)/(rl - r2) implies P3 , P6 , and Pg are collinear, which is a degenerate case. If h = 0, that is,
then S P, p. p,
= 0, and we obtain a realization for the configuration shown in Figure 2-26.1
Remark 2.60 From the above two examples and Example 2.42, the three 93 configurations can be realized rationally, i.e., they can be realized in the geometry associated with the field of rational numbers.
2.7.3
Transversals for Polygons
First, Example 1.8 on page 11 can be further generalized to the following form. Notice that in these theorems involving m points, the subscripts are understood to be mod m . Theorem 2.61 (Ceva's Theorem for an m-polygon) Let Vi ... Vm be an m-polygon , and
o a point. Let P; be the intersection of line OV; and the side V;+k V;+k+l' Then C(m, k) = I1~1 p'~eP; = 1 iJfm is an odd number and k = m;l. ,Yi+'+l
94
Chapter 2. The Area Method
Proof By the co-side theorem,
~ _ Pi Vi+k+ 1 -
Sov;v;+.
i
SOV;+H . V;'
= 1, ... , m.
Multiply the above equations together. We have that C( m, k) = 1 iff the elements in the numerator are the same as the elements in the denominator. Let us assume that the i-th element in the numerator is the same as the j-th element in the denominator, i.e., SOY. V;+. = SOI';+.+. Vj ' Then i = j
+ k + 1 mod (m) ;
+k =
i
j mod (m).
The above two equations have solutions for i and j iff 2k
+ 1 = 0 mod (m).
The only nontrivial solution of the above equation is 2k + 1 = m which proves the theorem. I
By polygrams, we mean the figures formed by the diagonals of polygons. The Menelaus and Ceva type theorems are about the transversals for the sides of polygons. We will discuss some results involving the transversals of polygrams which were discovered by B. Grunbaum and G. C. Shephard using numerical searching ([107]). Using the area method, we can not only prove these results easily but also strengthen some of them. Example 2.62 Let ABC D be a quadrilateral and 0 a point. Let E, F , G, and H be the intersections of lines AO, BO, CO, and DO with the corresponding diagonals of the qu adrI'latera ISh . ow thaAHCFBEDG_l t HcFAEDGB - .
Constructive description Take arbitrary points A, B, C , D , and O. E=BDnAO . F=ACnBO . G = BDnCO. H=ACnDO. Prove that ~ . ~ . tlli . DG = 1 FA
ED
HC
GB
.
The eliminants
The machine proof
4lI.!!~
gg . ~ . M . l!ML B e AF DE CH
tt:
g
CH-SCDO
~£~
~ . !Mi. . f.l. . U. SCDO
Be- SBCO
Be AF DE
t;;L!. =..§..a.ca. AF- SABO
(-SeDo) ,S,1DO • s;;!. ,,M SeDo ,SBeo AF DE
u.~~ DE- SADO
simg1i!y ~. t;;L .U SBeo
K:
AF DE
-(-SBCO) ,SADO • U. SBeo ,SABO DE
D
simg1i!y ~. U. SABO
DE
!i
(-SABO)·S ,1DO
-
SABO '(
SADOl
8img1i!y 1
A
B
Figure 2-27
Example 2.62 is a special case of the following result.
95
1.7 Applbdoas
Theorem 2.63 Let an arbitrary polygon VI " .Ym and a point 0 be given, together with a positive integer k such that 1 ~ k ~ 'f. Let P;,Ie be the intersection of line OVi and line Vi -Ie Vi+/c. Then
Proof By the co-side theorem
Vi+/cP;,Ie _ =;<=,=;=;== -
SOV;V;tl
P i,/eVi-Ie
SOV; _.v;
. _
,I -
1
, .. ,
m.
Multiplying the m equations together and noticing that the elements in the numerator and I in the denominator are the same, we prove the result. The following more intricate extension of Ceva's theorem contains the above example = k.
as the special case j
Theorem 2.64 Let an arbitrary polygon VI .. .Ym and a point 0 be given, and integers j and k that satisfy 1 ~ j ~ n - 2, 1 ~ k ~ n - 2 and j + k ~ n - 1. Let Pi,/e,i denote the intersection point of the line Vi+/e Vi-i and the line OVi. We put C( m , ),. k) -- rrm Vi+/cPi,/e,i . i=1 Pi,/e,. Vi-i Then C(m,j, k)
= (-l)mC(n, j , m
-
k)
= C(';,leJ) '
Proof By the co-side theorem
C( m,),. k ,I.) -- Vi+/eP;,/eJ -_ SOV;V;tl S . P;,/eJ Vi-i OV;_jV;
(1)
Replacing k by m - k in (1). we have
' k ') Vi+m-/ep.,m-leJ = C( m) m- ,I = , , Pi,m-leJ Vi- i
SOV;V;+ ... _. SOV;_ j V;
=
SOV;_.v;
.
(2)
SOV;_jV;
Interchanging k and j in (1), we have
C( m, k , ) ·, ·) 1 From (1) and (2), it is clear that C(m , j , k) have C(m,j, k)
= SSOV;V;+j .
(3)
OV;_.V;
= (_1)mC(n,j, m -
k). From (1) and (3), we
= C(';,leJ) '
Example 2.6~s show~in Figure 2-28, ABC D E - PQ RST is a pentagram. Then ~!l!Hl!.~~ = e~~!i!.~ = 1. PD S8 QE T C RA
TD RB PE SC QA
Chapter 1. The Area Melhod
96 To prove £. DR!!.!:. ES ~ PD S8 QETC RA
= I, we formulate the problem as follows.
Constructive description Take arbitrary points A, B, C, D, and E. p=ADnBE. Q =ACnBE. R= BDnAC. S=BDnCE. T = ADnCE. Show that
B
A
Figure 2-28
..u;JZ.!!.llu££.=~a~~.!!.4 ADDBBEECCA ADDBBEECCA
The elirninants
The machine proof -1
.
~.2.!:..g . ~.;1!!.
~!..§..M;..rL
U.'!'u!' .££' .ll ..1l;
CE- SACDE
CE BD AC BE AD
BE AD OE SD AC
1:. ....,....",==--...::S"'A..C'"'E'-·(L-...!S.,A¥C'"'D~E,.,)<-._ _
•
¥l,%.(-SACD)' ffr~ .SACDE
U . ~ . ££.. u CE BD AC BE
~!~ AD-SACDE
~~..§.aJ::.E.... BD-SBCDE
U~~ CE--SBCDE
,u!!hBJL AC- SABCD
~!!~ BD--SABCD
~g~ BE--SABCE
!f:
SRO" ,SACO,SACE , SASC D
•
~,%.SACD.SBCE,SABD.(-SABCD) sim~ify
SSD/"SAm:
-~.*.SBCE . SABD
g
AC BE
lJ1:.
-SACE , %,SBCE oSABD ,SABCE
BE
%oSABD
-
££. ../l,&,
SBDE,(-SBCE),SACE,(-SABCE) •
sim~ifY ~.
f.
•
££. .U AC BE
~ g =.§..a.c.E. AC- SABCE
w;,!.~ AD- SABDE
u,!.hBJL BE- SABDE
II BE
-SBDd-SABD) ,SABDE (-SBDE),SABD,(-SABDE)
si~ifY
The second result in Example 2.65 is equivalent to the following statement.
1.7
97
Appllcadou
Example 2.66 (Theorem of Pratt.Kasapi) Let ABCDE be a pentagon, AIBI II AC, BICI II BD, CID I II CE, DIEI II AD, EIAI II EB. Show that AlB, BIC · CID , DIE , EIA =BBl ' CCI . DDI ' EEl' AA I , The constructive description, Take arbitrary points A. B. C, D. and E , Take the intersection Al of the line passing through passing through A and parallel to E B . Take the intersection BI of the line passing through passing through B and parallel to AC. Take the intersection C I of the line passing through passing through C and parallel to B D . Take the intersection DI of the line passing through passing through D and parallel to C E , Take the intersection EI of the line passing through passing through E and parallel to AD, Show that ~~~~~ = 1. BBt eC DOl EEl AA)
figure 2-29
B and parallel to CA and the line C and parallel to BD and the line D and parallel to CE and the line E and parallel to AD and the line A and parallel to BE and the line
I
The eliminants
The machine proof
~~.§...w.E..
-~ , ~.~.!!i. , ~ EEl
DD1
AA, -
ce) BB) AA.)
~ -(-SBED,H-S,WE) ,~ , ~,!!i. (-SADA,H-SABE)
DD, CC,
SADAI ,SABE' (
.im!!ii/"
EE, -
BB,
l2! -( -SABDE) ,SCDE '( -SA DC, ) ,SADE
CCI
SADA"SABE ,SACDE ,SCDE ' ( -SBCDE)
-
SABDE ,SBCD ' SCEB, ,
!!i.
SADA, ,SABE ,SBCDE
BB,
SADA, ,SABE ,SBCDE,SBCD ' (
aim!!!i/" -
11 -
SABCD)
SADE
~
SADAI ,SABE ,SABCD
SARQE"Scng SACDE
~~-SCEB, CCI -
!!i.
SCDE
~ S'Qpg*Saon
S
ADCI -
BB,
-SBCDE
!!i. '!.! -SBDA, BBI -
S
SBCD
!J SaeDE -SARa CEB, -
5
~ ADA, -
-SABDE ,SABC ,SBDA,
-SABDE ,SABC ·S ,
.im!!p/"
DD, -
5
BEDI -
~ SABDE ,SBCD ' ( -SBCDE ,SABC)-( -SBDA,) -
BBI
SABDE ,SCDE 'SADC, • ~ , !!i. SADA, ,SABE ,SACDE CC, BB,
~ SABDE ,SCDE ,SACDE,SBCD '( -SCEB,) ,
.im!!!i/"
SABE
~l!!SADC,
• ~ . !!i.
SADE),SACDE
SADA,
~~SBED,
5
SABCD SA8QB *SABC
-SABCE
~ S,!BCp *SA8t:; BDA, -
SABCE
Chapter 1. The Area Method
98
Examples 2.65 and 2.66 are special cases of the following general results. Note that the proofs of these general results are a natural extension of the machine proofs for the two examples. Let VI .. .vrn be a polygram and 1 ~ d ~ T' 1 ~ j ~ the intersections of lines ViVi+d and lines Vi+i Vi+i+d. i intersection of line Vi-i Vi-i+d and line ViVi+d. Let
T( m, d,J·)
= rrrn p. ~ . tr' i=1
Theorem 2.67 T(m, d,j) • d+ 2j • 2d+ j
We denote by PdJ,i Then Pd,i,i-i is the
·) - rrrn ViPd,i,i-i S( m, d ,J . i=1 PdJ,i Vi+d
dd,i-j v i+d
The result in Example 2.65 is T(5, 2, 1)
T integers. = 1, ... , m.
= S(5 , 2,1) = 1. In general. we have
= 1 iff one of the following cases are true
= m; = m.
Proof By the co-side theorem. ViPd,i ,i Pd,j,i-j Vi+d
= ViPd,j,i. ViVi+d
~
=
Pd,i,i-j Vi+d
SV;VHi\';+i+J
S\';-iV;\';-i+J\';+J.
SV;\';+i\';+J\';+i+d S\';-i\';-i+d\';+J
Multiplying the m equations together. we see that T( d, j, m) = 1 iff the areas of triangles and the areas of quadrilaterals in the numerator are the same as the ones in the denominator respectively. Let us assume that the i-th area in the numerator is the same as the xth area in the denominator. i.e. S\';\';+i\';+i+4 = ±SV'_iV'_i+4 V.+4. Then the point sets {Vi, Vi+i> Vi+i+d} and {V:t-i, V:t-j+d, V:t+d} should be the same. Considering the order, there are six possible matches. One of the matches is i +j = x
+ d;
i +j
+d =
x - j;
i= x- j
+d
where the = is understood to be mod(m). Then it is easy to see that the three equations are true for all i and x iff 2d + j = 0 mod(m). Since 1 ~ d ~ T' 1 ~ j ~ T' the only possible solution is 2d + j = m. The other five cases can be treated similarly and d + 2j = m is the only nontrivial solution. Theorem 2.68 S( m, d, j) • d+2j • 2d
= m;
= j;
= 1 iff one of the following cases are true
2.7
99
AppUcaIioas
• 2j = d.
Proof By the co-side theorem, ViPdJ.i-i _ ViPdJ .i- j PdJ.iVi+d - ViVi+d
v.v.:;:;; _ PdJ.iVi+d -
SV;V;-;V;-iH SV;+;V,V;+i+4 VH4 SV'+iV;+i+4V;+4 SV,V;_,V,+dV;_i+4 ·
Now, we can prove it in a similar way as Theorem 2.67. At the present time, our prover can not deal with geometry statements about m-polygons for an arbitrary number m . But our prover can prove these results for any concrete m, e.g. quadrilaterals. pentagons. etc.
Summary of Chapter 2 • The fonowing basic propositions are the deductive basis of the area method.
1. If points C and D are on line AB and P is any point not on line AB, then &=L _ CD SPAB
-
AB·
2. (The co-side theorem) Let M be the intersection of two non parallel lines AB and PQ and M 1= Q . Then
PM QM
SPAB SQAB
==--;
PM PQ
SPAB QM S PAQB PQ
SQAB S PAQB
~=--;~=-- .
3. Let R be a point on line PQ. Then 5MB
PR
RQ
= PQSQAB + PQSPAB .
4. PQ II AB iff SPAQB = SPAB - SQAB = SBPQ - SAPQ = o. 5. Let ABC D be a parallelogram, P and Q be two points. Then SAPQ
+ SCPQ =
SBPQ
+ SDPQ or SPAQB
= SPDQC .
6. Let ABC D be a parallelogram and P be any point . Then
• The Hilbert intersection point statements are geometry statements whose hypotheses can be described constructively and whose conclusions can be represented by polynomial equations in two geometry quantities: the ratio of collinear or parallel segments and the signed areas of triangles or quadrilaterals.
100
Chapter 1. The Area Melhod
• The area method can efficiently produce short and readable proofs for the Hilbert intersection point statements. The proving process is to eliminate points from geometry quantities using Lemmas 2.19 - 2.31. • The area method works for constructive statements in the affine geometry associated with any field. • The area method is used to solve the following geometry problems: deriving unknown geometry formulas, finding the necessary and sufficient conditions for the existence of n3 configurations, and proving theorems about the transversals for arbitrary polygons.
Chapter
3
Machine Proof in Plane Geometry In Chapter 2. we presented an automated theorem proving method for constructive statements involving collinearity and parallelism. In this chapter. we will discuss constructive statements involving perpendicular lines and circles. The key tool for dealing with perpendicularity is the Pythagoras difference. which is essentially the same as the inner product. Therefore. the method presented in this chapter is actually for constructive statements in metric geometry.
3.1
The Pythagoras Difference
For three points A, B , and C. the Pythagoras difference PABe is defined as PABe
~
= AB
=='.!
~
+CB -AC .
Note that in the above definition. we use a new geometry object: the square distance between two points A and B. i.e .• Alf. For four points A , B , C , and D. we define PABeD
~
= AB
==2 + =='.! CD - BC - DA . ~
For basic properties of the Pythagoras difference. see Section 1.7.
3.1.1
Pythagoras Difference and Perpendicular
Besides collinear and parallel considered in Chapter 2. we now have a new basic geometry relation: line I is perpendicular to line I'. denoted by 11.1'. Following are some basic properties of the perpendicularity.
I. If 11..1' then 1'1..1. 2. Let P be a point and I a line. Then there exists a unique line I' which passes through point P and is perpendicular to line I.
101
102
Chapter 3.
Machine Proof in Plane Geometry
3. If two distinct lines [' and [" are both perpendicular to line I then I' is parallel to line [".
4. (Pythagorean Theorem) ABl.BC iffE
= AB2 + BC 2, i.e., iff PABC = O.
For the proof of the Pythagorean theorem, see Example 1.42 on page 23 and Proposition 1.55 on page 27. But here, we take Pythagorean theorem as a basic property of the Pythagoras difference. Other propositions in this subsection can be derived from it. For four points A , B, C, and D , the notation ABl.CD implies that one of the following conditions is true: A = B . or C = D, or the line AB is perpendicular to line CD . Proposition 3.1 ABl.CD iff P ACD =PBCD or P ACBD
= O.
Proof See Proposition 1.62 on page 30. The above generalized Pythagorean proposition is one of the most useful tools in our mechanical theorem proving method. Proposition 3.2 Let D be the foot of the perpendicular drawn from point P to a line AB. Then we have AD PPAB DB = PPBA'
Proof See Proposition 1.65 on page 30. Proposition 3.3 Let AB and PQ be two nonperpendicular lines and Y be the intersection of line PQ and the line passing through A and perpendicular to AB. Then PY QY
=
PPAB PY = PQAB PQ
= --,
PPAB QY PQAB =- -, = = ---. PPAQB PQ PPAQB
Proof See Proposition 1.66 on page 31. Proposition 3.4 Let A, B, and C be three different collinear points. Then for any point P if PPAC i- 0 we have ~;!~ = ~~ . p
Proof Let Q be the orthogonal projection from P to line AB. By Proposition 3.1, PPAB = PQAB = 2AQ · AB; PPAC = PQAC = 2AQ . AC. Now the result is clear.•
A
Q B Figure 3-1
c
3.1
The Pythagoras Ditfereoc:e
103
Q
Proposition 3.5 Let R be a point on line PQ with position ratios Tl = ~~ , T2 = ~ with respect to PQ. Then for any points A and B ,
'r1:J
we have
B
Figure 3-2
Proof We first assume -2
RA RB
-2
Then PRAB
-2 -2 2 = TIQA +T2PA -TIT2P Q =2 -2 -2 = TIQB + T2PB - T)T2PQ .
= -RA2 +AB
....,.--=I!
-=2
-RB
(I) (2)
-2....,.--=1! =2 =-:-2....,.--=1!-=2 = T)(QA +AB -QB )+T2(PA +AB -PB ) =
T)PQAB + T2 PPAB . The second one can be proved similarly. To prove (I). let us first notice that by Proposition 3.4.
Then
3.1.2
Pythagoras Difference and Parallel
Proposition 3.6 For a parallelogram ABCD, we have i.e., PABC = -PBAD .
E + BD2 = 2AB2 + 2BC2•
Proof Let 0 be the intersection of the diagonals AC and BD. By Proposition 3.5,
E = 4m = 40AF + lAd -
~BD2)
= 2AF + 2AD2 -
BD2.
The following proposition shows how the Pythagoras difference changes under a parallel translation.
Proposition 3.7 Let ABC D be a parallelogram. Then for any points P and Q, we have
PAPQ + PCPQ PPAQ + PPCQ
=
PBPQ PPBQ
+ PDPQ or PAPBQ = PDPCQ + PPDQ + 2PBAD
104
Chapter 3. Machine Proof In Plane Geomelly
Proof. Let 0 be the intersection of AC and B D . By the first equation of Proposition 3.5, = PAPQ + P CPQ = P BPQ + PDPQ . By the second equation of Proposition 3.5,
2POPQ
1
1
= PPAQ + PPCQ - 2PACA = PPBQ + PPDQ - 2PBDB . We need only to show that 2PBAD = ~(PACA - P BDB ) which is a consequence of Propo2Pp OQ
sition 3.6.
I
Proposition 3.8 Let ABC D be a parallelogram aru1 P be any point. Then PPAB
=
PPDC - PADC
P APB
=
PPDAC
PAPA - PPDAC
Proof. By Proposition 3.7, PPAB = PPAC - PPAD = PCADP = PPDAC = PPDC - PADe . For the second equation, P APB = PAPA + PAPC - P APD = PAPA + PCPDA = PAPA I
PPDAC·
Proposition 3.9 If PR
II
AC
aru1 QS
II
P
BD
Proof. Let X and Y be points such that PR 3.4,
= PPBRY = PPBY -
PPQRS
Similarly we have
PPBRD
then p~:;~
-
-
= ~~ . ~.
= AX, QS = BY.
By Propositions 3.7 and
QS
PRBY
= BD(PPBD -
QS
P RBD )
= BDPPBRD.
= ~~PABCD'
Proposition 3.10 Let AB
II CD. Then CABD
Proof. By Proposition 3.9,
AB CD
= ~. 2CD
= CD AB . CD = ~ = E.4.u.B.a. . CD PCCDD 2CiJ
Exercises 3.11 I . Let
ABC D
be a parallelogram. Then
1. PABC
2. 3.
P ABD
= PADC = -PBAD = -PBCD. = PBDC; PCBD = P ADB .
P ADB - PADC
= 2AD2.
2. Let ABC D be a parallelogram and 0 be the intersection of its diagonals. For any point P, show that -2
-==2
+ PPBC + PPCD + PPDA = 2(AB + BC ). 2 2. PAPD + PBPC - P CPD - P DPD = 2(AB - BC ). 3. P APD + PBPC + P CPD + P DPD = BW. 1. PPAB
~
J.l
105
1be Pythagoras Dirreruce
3.1.3
Pythagoras Difference and Area
In this subsection, we will prove the Herron-Qin formula which connects the area and the Pythagoras differences of a triangle.
Definition 3.12 Let F be the foot of the perpendicular drawn from point R to line PQ. The signed distance from R to PQ. denoted by hR,PQ. is a real number which has the same sign as SRPQ andlhR,PQI = IRFI. Proposition 3.13 For any two triangles ABC and RPQ. let hA Sho w tha t ~-~ IBClh... - IPQlhR · Proof Without loss of generality, we assume that points B, C , P, and Q are on the same line. As in Figure 3-3, let RF be the altitude of triangle RPQ and M be a point on RF such that AM II BC . Then SABC = 5MBC. By Propositions 2.7 and 2.8,
SABC SAPQ
BC PQ
= hA,BC.
hR
= hR,PQ .
R
--==; Figure 3-3
Multiplying the two formulas together and noticing that hA and hR have the same sign as SABC an dSRPQ, we have ~-~ IBCI"'" - IPQI"R·
I
Corollary 3.14 For a triangle ABC. we have
Proof Putting 6.RPQ to be 6.BCA and 6.CAB in Proposition 3.13 respectively, we obtain the result. I By Proposition 3.13, we have SABC = khAIBCI = khBIACI = khclABI where k is a constant which is independent of the triangle ABC. Setting k = 1/2, we obtain the usual formula for the areas of triangles.
Proposition 3.1S For a triangle ABC. SABC
111
= 2hAIBCI = 2h B IACI = 2hc1ABI .
Proposition 3.16 (The Herron-Qin Formula) For a triangle ABC. we have 16S~BC ...,..-;::2~
4AB AC -
p2
BAC ·
=
106
Chapter 3. Machine Proof In Plane Geometry
Proof Let F be the foot of the perpendicular line drawn from point A to lien BC. By Proposition 3.4, ~ = BC. Then PABF BF c
F
B
Figure 3-4
Then 16S~Bc
-2 -2 -2 = 4AF . BC = 4(AB -
-2 -2
BF )BC
~
= 4AB
--:--=2
2
. AC - PBAC .
Proposition 3.17 (The Herron-Qin Fonnula for Quadrilaterals) For any quadrilateral 2 ABCD, we have 16S~BCD = 4AC . BD2 - P1BcD .
Proof Take a point X such that CXDB is a parallelogram. Then CX Propositions 2.11. 3.7, and the Herron-Qin formula for triangles,
= BD .
By
S~ACX =slAc 1 - 2 --:--=2 2 16(4AX . AC - PXAC ) 1
-2
--:--=2
2
1
-2
--:--=2
2
16(4BD . AC - PXAAC ) 16(4BD . AC - PBADC )· You may compare the proofs for the two the Herron-Qin formulas based on trigonometric functions on page 37. The proof given here is independent of the concept of angles.
Exercises 3.18 1. Prove the following forms of the Herron-Qin formula.
• 16S~Bc •
= PACBPABC + PBCBPBAC. 16S~BC = PBACPACB + PACAPABC.
• 16S~Bc
= PCABPCBA + PABAPACB.
2. The absolute value of the area of a square is equal to the square of its side. Use this result to prove the Pythagorean theorem. (Hint. Use Figure 3-5.) Figure 3-5
107
3.l CoDStructive Geometry Statements
3.2
Constructive Geometry Statements
By constructive geometry statements. we mean statements which are assertions about configurations that can be drawn using only a ruler and a pair of compasses. More precisely. these configurations can be constructed by first taking arbitrary points. lines and circles and then taking the intersections of two lines and of lines and circles in a prescribed manner. From the constructed points, we can fonn new lines and circles. By forming the intersections of these new lines and circles, we can obtain new points, etc. Finally. we obtain a configuration consisting of points. lines and circles. The class of constructive statements is denoted by C. It is clear that the Hilbert intersection point statements belong to class C. In this section. we introduce a new subset of C, i.e., the linear constructive geometry statement C L • which is larger than C H and has the advantage that on one hand it contains most of the commonly used geometry theorems and on the other hand its readable proofs can be obtained efficiently by a mechanical method.
3.2.1
Linear Constructive Geometry Statements
Now we have three basic geometric quantities: • the area of a triangle or a quadrilateral. • the Pythagoras difference of a triangle or a quadrilateral. and • the ratio of parallel line segments. By Proposition 3.10, the ratio of parallel line segments can be represented as expressions in Pythagoras differences. Points are the basic geometry objects. From points, we can introduce two other basic geometric objects: lines and circles. A straight line can be given in one of the following four forms . (LINE U V) is the line passing through two points U and V . (PLINE W U V) is the line passing through point W and parallel to (LINE U V). (TLINE W U V) is the line passing through point W and perpendicular to (LINE U V). (BLINE U V) is the perpendicular-bisector of UV. To make sure that the four kinds of lines are well defined, we need to assume U is called the nondegenerate condition (ndg) of the corresponding line.
:F V which
108
Chapter 3. MachIne Proof In Plane c-.etry
A circle with point 0 as its center and passing through point U is denoted by (eIR 0 U) .
A construction is one of the following ways of introducing new points. For each construction, we also give its ndg condition and the degree of freedom for the constructed point. Cl (POINT[S] Yl,'" , Yi) . Take arbitrary points Y1 ,'" degrees of freedom.
,
Yi in the plane. Each Y; has two
C2 (ON Y In) . Take a point Y on a line In. The ndgcondition ofC2 is the ndg condition of the line In. Point Y has one degree of freedom. C3 (ON Y (eIR 0 P)) . Take a point Y on a circle (CIR 0 P). The ndg condition is o ¥ P . Point Y has one degree of freedom. C4 (INTER Y Inlln2). Point Y is the intersection of line Inl and line ln2. Point Y is a fixed point. The ndg condition is Inl IY In2. More precisely, we have 1. If Inl is (LINE U V) or (PLINE W U V) and In2 is (LINE P Q) or (PLINE R P Q), then the ndg condition is UV IY PQ. 2. If Inl is (LINE U V) or (PLINE W U V) and In2 is (BLINE P Q) or (TLINE R P Q), then the ndg condition is -,(UV l..PQ). 3. If lnl is (BLINE U V) or (TLINE W U V) and ln2 is (BLINE P Q) or (TLINE R P Q), then the ndg condition is UV IY PQ. C5 (INTER Y In (eIR 0 P)) . Point Y is the intersection ofline In and circle (eIR 0 P) other than point P . Line In could be (LINE P U), (PLINE P U V), and (TLINE P U V). The ndg conditions are 0 ¥ P, Y ¥ P, and line In is not degenerate. Point Y is a fixed point. C6 (INTER Y (eIR 0 1 P) (eIR O2 P)). Point Y is the intersection of the circle (eIR 0 1 P) and the circle (eIR O 2 P) other than point P The ndg condition is 0 1 , O2 , and P are not collinear. Point Y is a fixed point. C7 (PRATlO Y W U V T). Take a point Y on the line passing through W and parallel to line UV such that WY = TUV. where T can be a rational number. a rational expression in geometric quantities, or a variable. If T is a fixed quantity. then Y is a fixed point; if T is a variable then Y has one degree of freedom. The ndg condition is U i: V. If T is a rational expression in geometry quantities then we will further assume that the denominator of T could not be zero.
C8 (TRATlO Y U V T) . Take a point Y on line (TLINE U U V) such that T =
1;;';; (=
~~). where T can be a rational number, a rational expression in geometric quantities. or a variable.
3.1 COIIstrudive Geometry Statements
109
If T is a fixed quantity then Y is a fixed point; if T is a variable then Y has one degree of freedom. The ndg condition is the same as that of C7. The point Y in each of the above constructions is said to be introduced by that construction. Since there are four kinds of lines, constructions C2, C4, and C5 have 4, 10, and 3 possible fonns respectively. Thus, in total, we have 22 different fonns of constructions. Now class C L , the class of the linear constructive geometry statements, can be defined similarly as C a , i.e., a statement in class C L is a list
where Ci , i = 1, . . . , k, are constructions such that each Ci introduces a new point from the points introduced before; and G = (E J, E2) where EJ and E2 are polynomials in geometric quantities of the points introduced by the Ci and EJ = E2 is the conclusion of the statement. Let S = (Cit C2 , . . . , C (E J , E 2 » be a statement in C L • The ndg condition of S is the set of ndg conditions of the C i plus the condition that the denominators of the length ratios in EJ and E2 are not equal to zero.
'n
We call the statements in CL linear, because each of the constructions CI-C8 introduces a unique point. For the constructions involving circles, this fact may not be obvious. See the next subsection for more discussions. Example 3.19 The orthocenter theorem on page 31 can be described in the following constructive way. (POINTS A B C) (INTER E (LINE A C) (TLINE B A C)) (INTER F (LINE C B) (TLINE A C B» (INTER H (LINE A F) (LINE B E)) (PACH = PBCH »
ABl..CH.
The ndg condition: A::/: C , C::/: B, AF ~ BE.
3.2.2
A Minimal Set of Constructions
There are a total of22 constructions and three kinds of geometry quantities. So to provide an elimination method for each construction and each geometry quantity, we need to consider 22 * 3 = 66 cases. Instead of considering all these cases, we introduce a minimal set of constructions which are equivalent to all the 22 constructions but much fewer in number. A minimal set of constructions consists of Cl, C7, C8 and the following two constructions.
110
Chapter 3. Machine Proof in Plane Geometry
C41 (INTER Y (LINE U V) (LINE P Q)). C42 (FOOT Y P U V), or equivalently (INTER Y (LINE U V) (TUNE P U V) ). The ndg condition is U '" V.
We first show how to represent the four kinds of lines by one kind: (LINE U V). For In = (PLINE W U V), we first introduce a new point N by (PRATIO N W U V I). Then In = (LINE W N). For In =(TLINE W U V), we have two cases: ifW, U, V are collinear, In =(LINE N W) where N is introduced by (TRATIO N W U I); otherwise In =(LINE N W) where N is given by (FOOT N W U V). (BUNE U V) can be written as (LINE N M) where N and M are introduced as follows (MIDPOINT M U V) (i.e., (PRATIO M U U V 112», (TRATIO N M U I). Since now there is only one kind of line, to represent all the 22 constructions by the constructions in the minimal set we need only to consider the following cases. • (ON Y (LINE U V» is equivalent to (PRATIO Y U U V r) where r is an indeterminate. • (INTER Y (LINE U V) (CIR 0 U)) is equivalent to two constructions: (FOOT N o U V), (PRATIO Y N N U -I). • C6 can be reduced to (FOOT N P 0 1 0 2 ) and (PRATIO Y N N P -I). • For C3, i.e., to take an arbitrary point Y on a circle (CIR 0 P), we first take an arbitrary point Q. Then Y is introduced by (INTER Y (LINE P Q) (CIR 0 P». Proposition 3.20 The existence of the point introduced by each of the 22 constructions follows from Axiom A.2 on page 55.
Proof We can limit ourselves to the five minimal constructions. Constructions CI, C7, and C41 have been discussed on page 61. Let Y be introduced by (FOOT Y P U V). Then by Proposition 3.2 point Y is a point on UV with the position ratio ;:;;UY. Hence Y does exist by Axiom A.2. Let Y be introduced by (TRATIO Y U V r). Th~n Y can also be introduced as follows (check this). (POINT M); (FOOT N M U V); (PRATIO BUM N I); (PRATIO Y U U B illlJaL ). 4S UVB
Thus Y exists. Exercises 3.21 I . Show that constructions C I, C7, and C8 can also serve as a minimal set of constructions. The reason we use a larger minimal set is that constructions C41 and C42 are used frequently and special treatment of them will lead to short proofs.
III 2. We introduce a new construction (LRATIO Y U V r) which means taking a point Yon UV such that UY = rUV. Show that Cl, CS and the above construction also form a minimal set of constructions. (See Example 2.13). 3. Show that constructions Cl , C7, and C42 could form a minimal set of constructions.
3.2.3
The Predicate Form
The constructive description of geometry statements can be transformed into the commonly used predicate form. In addition to the three predicates POINT, COLL, and PARA introduced on page 63, we introduce two new predicates. I. Perpendicular (PERP P I , P2 ,P3 ,P4 ) : [(PI = P2 ) V (P3 dicular to P3 P4 )]. It is equivalent to Pp,p,p-:'P. = O.
= P4 ) V ( PI P2 is perpen-
2. Congruence (CONG P I ,P2 , P3 , P4 ): Segment P I P2 is congruent to P3 P4 • It is pp,P.P, . equivalent to Pp,p-:'P,
=
To transform constructions into predicate forms, we need only to consider the minimal set of constructions introduced in the preceding subsection. Also constructions CI, C41, and C7 have been discussed in Section 2.3.2. We thus need only to consider C42 and CS. C42 (FOOT Y P U V) is equivalent to (COLL Y U V), (PERP Y P U V), and U C8 (TRATIO Y U V
T)
is equivalent to (PERP Y U U V),
Now a constructive statement S lowing predicate form
= (CI ,· ·· ,Ck, (E , F»
\fP;[(P(Cd" "' " P(Ck ))
T
=I V .
= 4PuS IlYY , and U =I V. vu
can be transformed into the fol-
'* (E = F)]
where P( C i ) is the predicate form for C i and Pi is the point introduced by C i . We will now discuss what geometry properties can be the conclusion of a geometry statement in e L , i.e., what geometry properties can be represented by polynomial equations of geometry quantities. To illustrate that, let us give an algebraic interpretation for the area and Pythagoras difference. Let A , B, C , and D be four points in the Euclidean plane. Then SABe D and PABeD are propositional to the exterior and inner product of the diagonals AC and BD of the quadrilateral ABC: (see Chapter 5 for details) SABeD
1--
= 2[AC, BD] ,
PABeD
--:-::t= 2(Al.: , BD}.
So any geometry property that can be represented by an equation of the inner and exterior products can be the conclusion of a geometry statement. As examples, we show how to represent several often used geometry properties by the geometry quantities.
112
Chapter 3. Machine Proof 10 Plane Geometry
(COLLINEAR A B C). Points A, B, and C are collinear iff SABC variants, see the comments after Example 2.36 on page 74. (PARALLEL ABC D). AB is parallel to CD iff SACD
= O. For other
= SBCD.
(PERPENDICULAR ABC D). AB is perpendicular to CD iff P ACD (MIDPOINT 0 A B). 0 is the midpoint of AB iff ~~
= PBCD .
= 1.
= PCDC. A, B and C, D are harmonic points iff ~~ = g~.
(EQDISTANCE ABC D). AB has the same length as CD iff PABA (HARMONIC ABC D).
(EQ-PRODUCT ABC D P Q R S). The product of AB and CD is equal to the = ±~ if AB II PQ and product of PQ and RS, which is equivalent to RS II CD; P ACBD = ±PPRI.JS if AB II CD and PQ II RS; PABAPCDC = PPQPPRSR otherwise.
;g
(TANGENT 0 1 A O2 B). Circle (CIR 0 1 A) is tangent to circle (CIR O2 B) iff --2 ~ ~ Jl+T?+T~-2dTl-2dT2-2T1T2 = Owhered = 0 10 2 ,r1 = 01A ,T2 = 02B .
3.3 3.3.1
Machine Proof for Class CL The Algorithm
In Chapter 2, we have seen that the process of proving geometry theorems using the area method actually eliminates points from geometry quantities. To prove geometry theorems in class c L , we need to eliminate points introduced by constructions: Ct, C7, C8, C4t. C42 from three geometry quantities : the area, the Pythagoras difference, and the length ratio. Let G(Y) be one of the following geometry quantities: SABY, SABCY, P ABy , or PABCY for distinct points A, B, C, and Y. For three collinear points Y, U, and V, by Propositions 2.9 and 3.5 we have
(1)
G(Y)
= ~~ G(V) + ~~ G(U).
We call G(Y) a linear geometry quantity for variable Y . Elimination procedures for all linear geometry quantities are similar for constructions C7, C41. and C42. Lemma 3.22 Let G(Y) be a linear geometry quantity and point Y be introduced by construction (PRATIO Y W U V T). Then we have
G(Y)
= { (~~ + r)G(V) + (';j~ G(W)
+ T(G(V)
r)G(U) ifW is on line UV. - G(U)) otherwise.
113
3.3 A Proor Method for C L
Proof If W , V, and V are collinear, we have ~ = 1fJt + r ; ~~ = ~ - r . Substituting these into (0. we obtain the first fonnula. For the second one. take a point S such that WS = VV . By (0 G(Y)
WY
YS
= WS G(S) + WSG(W) = rG(S) + (1- r)G(W) .
By Propositions 2.11 and 3.7. G(S) = G(W) + G(V) - G(V) . Substituting this into the above equation. we obtain the result. Notice that in both cases. we need the ndg condition V:/; V. I Lemma 3.23 Let G(Y) be a linear geometry quantity and Y be introduced by (INTER Y (UNE V V) (UNE P Q» . Then
G(Y)
= SUPQG(V) -
SVPQG(V). SUPVQ
Proof By the co-side theorem. uuYv we prove the result.
= ssupq , YUVv = _ssvpq . UPVQ UPVQ
Substituting these into (0.
I
Lemma 3.24 Let G(Y) be a linear geometry quantity and Y be introduced by (FOOTY
P V V). Then G(Y)
= PpuvG(V) + PpvuG(V) . 2VV2
Proof By Proposition 3.2. ~ the result. Let G(Y)
(II)
= :::;~~, ~~ = ~y~ .
Substituting these into <0. we prove
I
= PAYB . By Proposition 3.5, for three collinear points Y. V. and V G(Y)
VY
YV
VY YV
= VVG(V) + VVG(V) - vv . VV Puvu .
We call PAYB a quadratic geometry quantity for variable Y. Since in the above three lemmas we have obtained the position ratios ~~. ~~ for Y when it is introduced by constructions C7. C41. C42. we can substitute them into (m to eliminate point Y from G(Y). Notice that in the case of construction C7, we need to use the second fonnula of Proposition 3.7. The result is as follows. Lemma 3.25 Let Y be introduced by (PRATJO Y W V V r). Then we have
P AYB
= PAWB + r(PAvB -
P AUB
Construction C8 needs special treatment.
+ Pwuv) - r(l- r)Puvu .
Chapter 3. Machine Proof in Plane Geometry
114
Lemma 3.26 Let S ABP -
Y
be introduced by (TRATIO
Y
P
Q
r).
Then we have
SABY
=
~PPAQB ' B
Proof Let Al be the orthogonal projection from A to PQ . Then by Propositions 2.10 and 3.2
p
Q
Figure 3-6
=
~~:~SPQY
Thus
SPAY
SABY
= SABP + SPBY -
Lemma 3.27 Let
=
~PAPQ ' Similarly, = SABP
SPAY
-
be introduced by (TRATIO
Y
~~=~SPQY
SPBY
~PPAQB ' Y
= ~PBPQ'
Now
I
r). Then we have
P Q
P ABy
=
P ABP - 4rSPAQB .
Proof Let the orthogonal projections from A and B to PY be Al and B I . Then P BPAY
PB,PA,y
Pypy
Pypy
SPA,QB,
AIBI
=
=
PY
SPQY
=
SPAQB
A
Q
P
SPQY .
Figure 3-7 2
1 -2
-=;-:2
By the Herron-Qin formula, SpQY = "4PQ . PY . Then Therefore P ABy = P ABP - P BPAY = P ABP - 4rSPAQB .
-=;-:2
Pypy
= 2PY = 4rSpQY. I
Lemma 3.28 Let Y be introduced by (TRATIO Y P Q r). Then we have PAYB
= P APB + r 2 PpQp -
4r(SAPQ
+ SBPQ) '
Proof By Lemma 3.27, P APY
= 4rSAPQ. P BPY =4rSBPQ'
By the Herron-Qin formula, pYPY
Then
PAYB
= P APB -
P APY -
= 2W = 4rSpQY = r 2 PpQP . 2 P BPY + pYPY = P APB + r PpQp -
4r(SAPQ
+ SBPQ).I
By Proposition 3.10 the ratios of parallel line segments can be represented by Pythagoras differences. Thus, we have given a complete method of eliminating points from geometry quantities. But usually, we consider the length ratios separately in order to obtain short proofs. The methods of eliminating point Y introduced by C41 and C7 from length ratios have been given by Lemmas 2.25 and 2.26. For other constructions, we have
3.3
115
A Proof' Method for CL
Lemma 3.29 Let Y be introduced by (FOOTY P U V). We assume D interchange U and V.
= DY = {PA"FDt S
G
EF
...iUl.ll:L
S EUFV
Proof If D E UV , let T be a point such that DT G = DY EF
i: U; otherwise
if D E UV. if D ¢ UV.
= EF.
By Propositions 3.2 and 3.8
= DY = PPDT = PPEDF . DT
PDTD
P EFE
The second equation is a direct consequence of the co-side theorem. Lemma 3.30 Let Y be introduced by (TRATJO Y P Q r).
G- DY - {:DPQ EPFQ - EF -
ifD ¢ PY. sDPQ-iPpQP lifD E PY SEPFQ
Proof The first case is a direct consequence of Proposition 3.3. If D E PY, then DY -- E'F DP - E'F' YP BYthe CO-SI'de theorem, E'F DP EF
SDPQ YP
SYPQ
rPpQP
= SEPFQ ; EF = SEPFQ = 4SEPFQ '
Now the second result follows immediately. For a geometry statement S = (CI , C2 , ... , CIt, (E, F)), after eliminating all the nonfree points introduced by C i from E and F using the above lemmas, we obtain two rational expressions E' and F' in indeterminates, areas and Pythagoras differences ofJree points. These geometric quantities are generally not independent, e.g., for any three points A, B, and C we have the Herron-Qinformula (Proposition 3.16):
16S~Be
= 4AIr~ -
(~ + AIr - BC
2
?
We thus need to reduce E' and F' to expressions in independent variables. To do that, we first introduce three new points 0 , U, and V such that UOl.OV. We will reduce E' and F' to expressions in the area coordinates of the free points with respect to OUV. Lemma 3.31 For three points A, B, and C. we have
I. SABe
= So~v
SOUA SOVA SOUB SOVB Soue Sove
1 1 1
116
2.
Chapter 3.
PABC
~
= AB + ==2 CB -
Madtioe Proof 10 Plane c-oetry
v
~
AC .
M
2
4. Souv
1 -2 -
= 40U
2
OV .
4-------u
o
Figure 3-8
Proof Case 1 is Lemma 2.31. Case 2 is the definition of the Pythagoras difference. For case 3, we introduce a new point M by construction (INTER M (PLINE A 0 U) (PLINE B 0 V)). ~
~~
Then by the Pythagorean theorem, AB = AM + BM . By the second case of Lemma 226 ~ = ~ = S.ov Ssoy.~ = s.O[(-sso". We have proved case 3. Case4isa . 'au Soouv Souv' ov Souv consequence of Proposition 3.13. I Using Lemma 3.31, E and F can be written as expressions of OU, OV, and the area coordinates of the free points.
Remark 3.32 In Lemma 3.31, we actually use the Cartesian coordinates of points to represent areas and Pythagoras differences. For a point P, let x p = yp = Then the formulas in Lemma 3.31 become
21tJtfr,
1'. 3'.
SABY
=~
XA
YA
XB
YB
Xy
YY
A1f = (XA -
xBf
2(gv{-
1 1 1
+ (YA -
YB)2.
Algorithm 3.33 (PLANE)
INPUT: S
= (Cl , C2 , ... , C/o (E, F)) is a statement in C L •
OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof for S. S1. For i
= k, · .. ,1 , do S2, S3, S4 and finally do S5.
S2. Check whether the ndg conditions of C i are satisfied. The ndg condition of a construction has three forms: A '" B, PQ II' UV, or PQ t UV. For the first case, we check ~ whether PABA = 2AB = O. For the second case, we check whether Spuv = SQuv, For the third case, we check whether P puv = PQuv . If a ndg condition of a geometry statement is not satisfied, the statement is trivially true. The algorithm terminates.
3.3
117
A Proof Method tor CL
S3. Let G\,· ·· , G. be the geometric quantities occurring in E and F . For j S4.
= 1"
" ,s do
S4. Let H j be the result obtained by eliminating the point introduced by construction C j from G j using the lemmas in this section and replace G j by H j in E and F to obtain the new E and F . S5. Now E and F are rational expressions in independent variables. Hence if E is true. Otherwise S is false.
= F, S
Proof of the correctness. Only the last step needs explanation. If E = F, the statement is obviously true. Otherwise, by Proposition 2.33 we can find specific values for the free parameters in E and F such that when substituting them into E and F, we obtain two different numbers, i.e., we have found a counterexample. I
For the complexity of the algorithm, let n be the number of the non-free points in a statement which is described using constructions CI-C8. By the analysis in Section 3.2, we will use at most 5n constructions in the minimal set to represent the hypotheses (we need five minimal constructions to represent construction (INTER A (BLINE U V) (BLINE P Q))). Then we will use at most 5n minimal constructions to describe the statement. Notice that each lemma will replace a geometric quantity by a rational expression with degree less than or equal to three. Then if the conclusion of the geometry statement is of degree d, the output of our algorithm is at most degree 3Sn d. In the last step, we need to represent the area and Pythagoras difference by area coordinates. In the worst case, a geometry quantity (Pythagoras difference) will be replaced by an expression of degree five. Thus the degree of the final polynomial is at most 5d3 5n • Remark 3.34 In Section 2.6, we showed that the area method is for affine geometry over any field. Algorithm 3.33 is actually for metric geometry over any field whose characteristic is not 2. We have to exclude the case of characteristic 2. Otherwise, for three collinear points A , B , and C we have PABC 2AB· CB 0, i.e., ABl.BC. For more details, see Chapter 5.
=
3.3.2
=
Refined Elimination Techniques
We have presented a complete method for proving geometry statements in class C L by considering a minimal set of constructions. But if we use only those five constructions, we must introduce many auxiliary points in the description of geometry statements. More points usually mean longer proofs. In this section we will introduce more constructions and more elimination techniques which can be used to obtain shorter proofs. The elimination lemmas in Section 3.3.1 can be refined in two ways: first we may consider more constructions instead of the minimal set; second for each elimination lemma
118
Chapter 3. Machine Proof In Plane Geometry
we may give the elimination results in some special cases of the configuration. As an example of the second way of refinement, prove the following result. Exercise 3.35 Let point Y be introduced by construction (FOOT Y P U V) . Prove the following results. if AB II UV; if AB.l.UV; if U, V, and A are collinear; if U, V, and B are collinear.
SABU
=
SABY
{ {
SABP Stlfll::::Pelldl::::
Sd[;~Se~lBl:::: Suvu
ifAB II UV; if AB.l.UV; if U, V, and B are collinear.
P ABP P ABU
P ABY
PdllJlPeBfl
Pu su
~
if A = B =
Pu vu p2
=
PAYB
P;
if A = B = U;
~
{
P'1 vu
~
if A = B = V;
::'~~~[lPe[ll:::: if A = U, B = V. Pu vu
To use the above elimination technique, similar to in Section 2.5.1, we need to find the collinear point sets, parallel lines, and perpendicular lines which are obvious from the constructive description of the geometry statement. See the following example. Example 3.36 The following ITUlchine proof of the orthocenter theorem on page 31 uses the above exercise. Constructive description
The machine proof
( (POINTS A B C ) ( FOOT E B A C)
EM= PSCH
( FOOT F A B C) ( INTER H ( LINE A F)
!!.. E.a.c.a -
( LINE BE))
(PERPENDICULAR AB C H) )
The eliminants PSCH!J.PACS PACHf!,PACS
PACS
sim!!!:.i!y 1
From the description of the statement, we can find the sets of collinear points:
{H,A , F}, {F,G,B}, {H,E,B}, {E,A,G}, and the sets of perpendicular lines:
H AF .l.BG F, HE B.l.E AG. Then by Exercise 3.35, P BC H
= PACB and PACH = PACB since AB.l.GH and GA.l.BH.
119
3A The Rado CoDStrucdoas
Exercise 3.37 Let G( Y) be a linear geometry quantity ofY , and Q(Y) a quadratic geometry quantity of Y . If Y is on line UV then
= ~G(V) + ~~G(U); Q(Y) = !!x'Q(V) + ~Q(U) uv uv m:: = {:ED~FYy if D ~ UV;
G(Y)
~+~
EF
~
2
2 . !!X. uv . ~ uv . UV .
ifDeUV.
uv
For several constructions, we need to compute the position ratio of Y with respect to UV and substitute them into the above formulas to eliminate Y. Precisely, we have I. IfY is introduced by (INTER Y (LINE U V) (PLINE R P Q» then ~ = SUPRq ~ = _ SVPRq UV
SUPVQ' UV
SUPVQ'
2. If Y is introduced by (INTER Y (LINE U V) (TLINE P P Q» then UY
-..!3!..£SL
iiV -
YV _
PuPVQ'
iiV - -
Py Pq PuPVQ '
.
3. If Y is introduced by (INTER Y (LINE U V) (BLINE P Q» then UY _
iiV -
PuPq-PC/' PuPVQ
YV _
'iiV - -
PyPq-PC/' PUPVQ .
4. If Y is introduced by (INTER Y (LINE U V) (eIR 0 U» then ~ - 2. fuJlJL ~ _ Poyo-PO!lO UV -
PuVU'
uv -
PuVU
•
Exercise 3.38 If Y is introduced by (INTER Y (PLINE W U V) (PLINE R P Q» then
G(Y) Q(Y) where r
=
= G(W) + r(G(V) = Q(W) + r(G(V) -
G(U»; G(U» - 2r(1- r)UV2
SWPRq. SUPvQ
Exercise 3.39 Of the 22 constructions, there are still eight which are not discussed (Take points on a BLINE, a PLINE, or a :rLINE; Take the intersections of two TLINEs, two BLINEs, a TLINE and a PLINE, a TLINE and a BLINE, and a PLINE and a BLINE; Take the intersection of a circle and a PLINE or a TLINE.) Try to eliminate a point introduced by one the eight constructions from a geometry quantity.
3.4
The Ratio Constructions
By the ratio constructions, we mean the constructions PRATIO and TRATIO and other constructions which can be reduced to them. These constructions are the most subtle
120
Chapter 3. Machine Proof In Plane Geometry
constructions; appropriate use of the ratio constructions may lead to elegant proofs for geometry statements.
3.4.1
More Ratio Constructions
We first introduce the following constructions for convenience. C9 (MIDPOINT Y U V). Y is the midpoint of UV. It is equivalent to (PRATIO Y U U V 112).
CIO (SYMMETRY Y U V). Y is the symmetry of point V with respect to point U. It is equivalent to (PRATIO Y U U V-I).
Cll (LRATIO Y U V r). Y is a point on UV such that ~~ (PRATIO Y U U V r).
= r.
It is equivalent to
C12 (MRATIO Y U V r). Y is a point on UV such that ~~ (PRATIO Y U U V 1~r).
= r.
It is equivalent to
Four collinear points A, B, C, and D are said to form a hamwnic sequence if
CA DA CB - - DB· Given two points A and B, we have the following ways of introducing points C and D such that A , B, C, and D form a harmonic sequence. £4.
(ON C (LINE A B», (LRATIO DAB tt-); 08+ 1
(MRATIO CAB r), (MRATIO DAB -r). For convenience, we can introduce a new construction
CI3 (HARMONIC DCA B) introduces a point D such that, for three collinear points A, B, and C, A, B, C, and D form a harmonic consequence. Another important geometry concept related to ratios is inversion. Suppose that we have given a circle with a center 0 and a radius r # o. Let P and Q be any two points collinear with 0 such that
OP·OQ=r 2 Then P is said to be the inversion of Q with regard to the circle and 0 is called the inversion center. We introduce a new construction as follows
121
3.4 The RIotio Constructions
CI4 (INVERSION P Q 0 A) means that P is the inversion of Q with regard to circle (CIR 0 A). This construction is equivalent to (LRATIO P 0 A g~) (LRATIO P 0 Q
ifQ E
~A.
otherwise.
fiJMJ.p, )
OQO
The ratio r in the ratio constructions could be rational numbers. variables. or expressions in geometry quantities. Now we allow r to be any algebraic numbers by adding a new special construction.
CIS (CONSTANT p(r) where p(r) is an irreducible polynomial in the variable r. This construction introduces an algebraic number r which is a root of p( r) = O. With the help of construction CONSTANT. we can deal with statements involving special angles such as 30 0 • 45 0 • and 60 0 , etc. In the rest of this section. we will use several examples to show how to solve geometry problems using the ratio constructions. The construction TRATIO can be used to express geometry statements involving squares easily.
Example 3.40 The following is the machine proof of Example 1.69 on page 32. Constructive description
The machine proof
The eliminants
«POINTS
EBi1E..
PBGCgPBAC+PACA-4SABC
A
8 C)
(TRA1l0 E
A 8 1)
(TRA1l0 G
A C -1)
(pERPENDICULAR E C G 8) )
PBGC
Q -
PACA±4SAG&-4SAAC PBAC+PACA-4SABC
~ PBAC±PACA-4SAB C
-
PBGEgPACA+4SACE-4SABC
SACE~~(PBAC)
PBAC+PACA 4SABC
.im~ifJl 1
Example 3.41 On the two sides AC and BC of triangle ABC. two squares ACDE and BCFG are drawn. M is the midpoint of AB. Show that CM is perpendicular to DF. (Figure 3-9) Constructive description
The machine proof
The eliminants
«POIIm A 8 C)
!:JJiat..
PFCM~~(PAGF)
PFCM
(TRATIO DCA 1) (TRATIO F C 8 -I)
M:
i
PBCD .PACF
(MIDPOINT M A 8)
E.~
(PERPENDICULAR D F C M) )
-
4SABC
Q~ - (4) ,SABC .im!!4ifJl
PDCM~HpBCD) PACF~4(SABC) PBCDg4(SABC)
122
Chapter 3.
Machine Proolln P1aDe Geomeby
o
~--------~----~A A
M
B
Figure 3-9
Figure 3-10
Example 3.42 12 Let M be a point on line AB. Two squares AMCD and BMEF are drawn on the same side of AB. Let U and V be the center of the squares AMCD and BM EF . Line BC and circle VB meet in N. Show that A , E, and N are collinear. (Figure 3-10)
Constructive description «POINTS A B) (ON M (LINE A B)) (TRAll0 C M A I) (TRAll0 E M B -I) (MIDPOINT V E B) (INTER N (LINE B C) (CIR V B)) (INTER T (LINE B C) (LINE A E)) (~ = ~) )
The machine proof (~Z)/(~) 'L~ BN - -SABE . eN
The eliminants u.!:~ cr- SACE u;.N PqRY CN (f) ·(2PCBV-PBCB)
-
SABE'(PCBv-tPBCB)
PCBV~!(PCBE) SABE~ - HPABM)
~
(tpCBE),SACE SABdtPCBE-tPBCB)
PCBE~PMBC+4SBMC
!:!. -
fd. -
f. -
PCBy,SACE
(PMBCHSBMcl,(}PMABC-SAMC) (-!PABM) ,(PMBC-PBCB+4SBMC) -(PBMB-PAMB),(PBMB+PAMA-PABM) PABM ,(-PAMB-PAMA)
M ( -PA BA · M1.+PA BA)·(2PABA·(4M)2_PA BA· 4M ) . ~ ~ AlL (-PABA·~+PABA)·(2PAB ...-( ~ )2-PABA ' ~)
=
SACE~HpMABC-4SAMC) PBCB',i,PBMB+PAMA SAMC',i, -
HPAMA)
PMABC',i,PBMB-PABM SBMC',i, -
~(PAMB)
PM BC',i,PBMB
PABM~
-
(~-l)'PABA)
PAMA~PABA ·(~)2
PAMB~(~-l)'PABA'~ AB AB PBM B~( ~-1)2 ' PABA
For more examples involving squares, see Section 9.4.
Example 3.43 Given four points A, B, C, and D which form a harmonic sequence and a point 0 outside the line AB, any transversal cuts the four lines OA, OB, OC, and OD in four harmonic points. 12This is a problem from the 1959 International Mathematical Olympiad.
123
3A The Ratio Constructions
Constructive description
The machine proof
The eliminants
( (POINfS
(-§t)/(~)
u,!&uu!
0 A B X Y)
(MRA1l0 CAB r) (MRA1l0 DAB -r) (INTER p (LINE 0 A) (LINE x (INTER Q (LINE 0 B) (LINE x (INTER R (LINE 0 C) (LINE x (INTER S (LINE 0 D) (LINE x (HARMONIC P Q S R) )
~
Y)) Y))
QR- SOCQ
~.SODQ
Y)) Y))
QS- SODQ
u!!fuI:..I!..
-SOUP
1!
S
-
S
S
SOCp ,( -SOXy ,SOBD),SOXBY
S
SOCP,SOBD (
SOXy ,SOAc),SOBD ,SOXAY
A
c
B
-
Figure 3-11
52
SOXAY SOXAY
C~ SOAC= r+1
SOAC ,(-SOAB)·(-r+1)
sim1!!i!y
D
SOXBY
SOADg~ r-1
SOAC ,SOBD
Q -( -SOAB·r) ,SOBC ·( -r+1) -
- SOXX ,S06C
SOBDg~ r-1
.im1!!i!y -SO ,IV ,SOBa -
f.
P -SOXy ,SOAO
ODP
E. -( -SOXy ,SOAD),SOBC ' SOX AY -
-Soxy ·SoaG
OCP-
sim1!!i!y -SODP ,SOBC -
SOXBY Q
OCQ
~ -SOpp ·( -SOXy ,SOBC) ,SOXBY -
Q -S01(y ,SQSQ
ODQ
-SODP ,SOCg SOCP ,SODQ
C~ SOBC= r+1
=.!:..5..a..aJ: SOAC
-r-( -SOABHr+1) .im1!!i!y
-
SOAB ·r ·(r+1)
-
For more examples involving harmonic sequences, see Section 6.3. Example 3.44 The inverse of a circle passing through the center of inversion is a line,
Constructive description «points 0 A X) (lratio P 0 A r,) (inversion Q P 0 A) (midpoint u P 0) (inter R (lox) (cir U 0)) (inversion G R 0 A) (perpendicular G Q 0 A) )
The machine proof EA= PAOQ _
P60R·POAQ AOG
-
PAOQ ,PORO
1!
(2PXOU ,P AOX) ,POAO ' POXO PAOQ·( 4Plou) ' Poxo
P
simf1:ify
!l. -
PORO
R(4)·(Pxou)' ORO POXO
PAQX ,POAQ (2) , PAOQ'Pxou
PAQx ,PflAQ (2) ,PAOQ '( ;PxoP)
= ~ ' POAO ' PXOP PAQX oPQAQ
.imgJi!y
PAQ/{
~ ' Pxop
Figure 3-12
P
PAQR , POAQ
Po
-
p
The eliminants
E.
~ .im1!!i!y
-
PAox·r,
-
AOR
R(2) ,PXOU ,P AOX POXO
Pxoug~(PXOp ) PAOQ=li!4 ,POAO OP PXOP!:,PAOX ' r,
124
Chapter J. Machine Proof in Plane Geometry
For more examples involving inversions, see Section 10.3. The ratio constructions are used extensively in the following example. Also notice that the construction CONSTANT is used to describe equilateral triangles.
Example 3.45 Three equilateral triangles AIBC, ABIC, ABCI are erected on the three sides o/triangle ABC. Show that CAICIB I is a parallelogram.
Constructive description ( (points A B C) (constant r' -3) (midpoint E A C) (tratio Bl EAr) (midpoint F B C) (tratio Al F C r) (midpoint G A B) (tratio C l G B -r) (parallel Al C l C B l ) )
A
SCB)A) SCB)C l
-
SCBlCl
'2~(PBlBCG ·r+4SCBlG)
SCBl Gg~(SBCBl +SACB l ) PBlBCGg -
SCB,A!
!PBIBco or+SCBlG SCBlAl
Q (4)·SCB)A) - -tPBCBl'r+tPACBl 'r+2SBCBl +2SACBl
~
-
~(PBCBl-PACBl)
~(PBlCF'r-4SCBlF)
SCBlF;}(SBCB l )
-
PBCBl ·r-PACBl·r-4SBCBl-4SACBl
PBlCF=2(PBCBl) I ( SACB l Bl = - 4 PCAE ·r )
f..
(2)·( iPBCB) 'r-2SBCB)) PBCBl .r-PACB l .r-4SBcB l -4S ACB l
SBCBl Bl
11
-
(-8)·( -
t PB) C F ·r+Sc B) F)
PACBl
~ PCARf{"rtPacg ·r 4S8CB±4SABE; r 2 - PCABE ·r+PCAE ·r+PBCE ·r-PACE ·r-4SBCE+4SABE·r' o
Ii
B
The eliminants
The machine proof
<2.!
G
Figure 3-13
i P BCB ·r+!PACB·r-tPA Bc ·r+2SABc ·r"-2SABC
- t P BcB·r+t PACB ·r-t PABc·r+2SABc·r'-2SABC
PBCBl
~PACE = - 4I ( PCABE·r-4SBCE ) ~PBCE+4SABE·r
PACE~}(PACA) PCAE~¥(PACA) SABE=¥(SABC)
SBCE~¥(SABC) PBCE~2(PACB) PCABE~~(PBCB-PABC)
Note that the condition r2 = 3 is not needed in the proof. i.e .• the result is true if triangles BIAC. AIBC, and CIAB are similar isosceles triangles. The above proofis used to illustrate the use of ratio constructions. For a much shorter proof of this example. see Example 5.61 on page 250.
304
125
The Ratio Constructions
3.4.2
Mechanization of Full-Angles
As an application of the construction TRATIO. we will present an automated theorem proving method for geometry theorems involving full-angles. The fonnal definition of full-angles is as follows. DermitioD 3.46 An ordered pair of lines AB and CD determines a full-angle. denoted by L[AB, CD]. which satisfies
1. L[AB, CD]
= L[PQ,UV] if and only if
Thus the tangent function for the full-angle. tan(L[AB, CD])
= 4SACBD PADBC
is a well defined geometry quantity.
= L[AB, PQ] is a constant. For all perpendicular lines ABl.PQ. L[1] = L[AB, PQ] is a constant.
2. For all parallel lines AB 3.
II
PQ. L[D]
4. There exists an operation •. + .. for full-angles which is associative and commutative. Furthermore, we have
• Ll1] + L[1] • lfPQ
= L[D].
II UV then
L[AB,PQ]
+ L[UV, CD] = L[AB,CD].
• The tangent function of the sum of two full-angles is defined as follows tan(L[AB, CD]
tan(L[AB, CD]) + tan(L[PQ,UV]) + L[PQ,UV]) = 1- tan(L[AB,CD])tan(L[PQ,UV])"
You can find the geometric background for the above definition in Section 1.10. For three points A, B, and C.let L[ABC] L[AB, BC].
=
Remark 3.47 According to the above definition. L[AB, CD] one of the following conditions holds.
= L[PQ, UV] if and only if
1. (PARA ABC D) and (PARA P Q U V); 2. (PERP ABC D) and (PERP P Q U V);
3. A "I B, C "I D, P "I Q. U i- V and the full-angle L[AB, CD] is equal to the full-angle LlPQ, UV].
126
Chapter 3.
Machine Proof in Plane Geometry
Proposition 3.48 (The Co-angle Theorem) In triangles ABC and XYZ. if L[ABC) L[XYZ) . L[ABC)/neL[I). and L[ABC)/neL[O). then SABC SXYZ
= PABC
where
=)..
P XYZ
Proof By Definition 3.46 if L[ABC) Herron-Qin formula
=
L[XYZ) then ~;~;
=
:;~; ~
2 2 + PXYZ = 4XY = 4AB · B, 16Sxyz Set SABC = )"SXYZ , PABC = )"PXYZ in the first equation we have
2
2
16SABC + PABC
....,.-=2
-==2C
4AF . CB
)..2
=
2
4AF· CB
= )...
By the
~
. ZY .
2
= 16SAyx 2 2 = 4XY· ~ 2· + PAyZ ZY
With the concept of full-angles. the constructive geometry statements can be extended as follows . First we have a new geometry quantity: the tangent function of full-angles. Since the tangent function can be represented by the area and Pythagoras difference, we do not need to introduce new elimination techniques to eliminate points from it. Its main contribution is that we may now prove assertions like L[AB,CD) = L[PQ,UV) and L[AB, CD) = L[PQ , UV)
+ L[XY, WZ).
The second extension is more interesting: we can introduce a new representation for lines. (ALINE P Q U W V) which is the line I passing through P such that L [PQ , I) = L[UW,WV).
With this new type of lines, we can introduce seven new constructions: 1. (ON Y (ALINE P Q L M N». Take an arbitrary point on an ALINE; the ndg conditions are P =/: Q, L =/: M, and N =/: M.
2. (INTER Y in (ALINE P Q L M N». Take the intersection of in and (ALINE P Q LM N).
• If in = (LINE U V) or In L[PQ , UV) =/: L[LM,MN) . • If in = (BLINE U V) or in L[UV, PQ) + L[LM, M N] =/:
= (PLINE W = (TLINE W
U V), then the ndg condition is U V), then the ndg condition is
L[l].
• If in = (ALINE U V X Y Z), then the ndg condition is L[UV, PQ] L[NM,ML] + L[XY,YZ] .
=/:
127
3.4 The Ratio Constructions
3. (INTER Y (CIR 0 P) (ALINE P Q L M N». The ndg conditions are Y ::j: p. P # o. P ::j: Q. L ::j: M. and N # M . To provide methods of eliminating points introduced by these new constructions from geometry quantities. we need only to reduce an ALINE to a LINE.
Proposition 3.49 If UW is not perpendicular to WV. line I =(ALINE P Q U W V) is the same as (UNE P R) where R is introduced by construction (TRA TJO R Q P ~) f'uwv
.
p
If UW -LWV.line I is (TLINE P P Q).
Q
x w
u
Figure 3-14
Proof. Let the line passing through point Q and perpendicular to PQ meet line I in R. Then R is introduced by construction (TRATIO R Q P r). where T
= 4SRQP = 4SQPR = tan{L[RPQJ) = tan{L[VWUJ) = 4Suwv . PQPQ
PQPR
Puwv
I
Remark 3.50 From the above proposition we see thilt the construction (TRATIO Y P Q T) is actually to take a point Y such thilt Y P -LPQ and tan( L [Y PQ)) = T. Now. we introduce the following predicates. (EQANGLE ABC DE F). L[ABC) = L[DEF) iff SABCPDEF = SDEFPABC. (COCIRCLE ABC D). Points A, B, C, and D are co-circle iff L[CAD] equivalently. SCADPCBD = PCADPCBD.
= L[CBD). or
Example 3.51 If N, M are points on the sides AC, AB of a triangle ABC and the lines BN, C M intersect at a point J which is on the altitude AD, show thilt AD is the bisector of the angle MDN. A
Constructive description ( (POINTS
A B C)
DAB C) (ON J (LINE A D» (INTER M (LINE A B) (LINE C (INTER N (LINE A C) (LINE B (FOOT
(EQANGLE M D A A D N) )
J» J»
Figure 3-15
128
Chapter 3. MadJine ProoIln Plane Geometry
The ndg conditions: B
=1=
C, A
=1=
D, AB IY C J, AC IY BJ.
The machine proof (-SADM),PADN SADN,PADM
!:!..
(-SADM ).( -PADJ"SABC) ,SABCJ (-SACD ,SABJ),PADM,(-SABCJ)
-
simgliJy -
1:1. -
N-SAgQ,SAB [
ADN
p
SABCJ NPAn,.SdRG
ADN
SAQM , PAP[,SABC
SACD ,SABJ , PADM
sifflJ!!.ify
S
SABCJ
p
(-SACj"SABD) , PADJ"SABg-{-S.WBJ) SACD ,SABJ,PADJ,SABC,SACBJ
-
The eliminants
M PAQ "SABa
ADM
-SACBJ
M,-SAO,.SA8D
S
ADM-
SACBJ
SABJ~SABD · tt;
SAC "SA8D
SACD ,SABJ
SACJ~SACD · tt;
J SACD · 4I. AO ,S ABD
_
-
SACD ,SABD · tt;
si"'!1ify
Example 3.52 (The Inscribed Angle Theorem) Let A, B, C, and D be four points on a circle with center O. Then L[ACB] = L[ADB] and L[AOB] = 2L[ACB].
Proof. We first use our program to compute tan(L[ACBJ). p
Constructive description ( (POINTS A B)
(ON 0 (BLINE A B) (TRATIO P BAr)
(INTER C (LINE A P) (elR 0 A)) «TANGENT A C B)) )
Figure 3-16
The machine proof
The eliminants
(-4) ,SABC
P
PACB
f. -
(
(-4)·( 2POAP,SABP),PA PA 2POPO ,POAP+2POAP ,PAPB+2POAP ,PAOA) , PAPA
simJ?!i fy
(4) ,SABP POPO-PAPB-PAOA
1:..
(4) ·( - t P ABA ·r) PBOB-PAOA+8SABo ·r
Q =..fA.B.A:.! -
8SABo ·r
simJ?!ify ~ (8),SABO
ACB
c(-2),(POPO-PAfB-PAOA),POAP PAPA
ABC
c(2) , POAp'SABP PAPA
S
PAPB~PABA '(r)2
POPO~PBOB+PABA·r2+8SABo·r
SABP~
-
i(PABA ·r)
129
3..4 The Ratio CoIlStnlctloBa
From the above computation, it is clear that tan( L[AC BJ) is independent of point P and = tan(L[ADBJ).
C, i.e., tan(L[ACBJ)
tan(U[ABCJ) =
2tan(L[ABCJ) 1 - tan(L[ABC])2
8AITS AOB AB4
= 16S~oB _
~
= = i.e., U[ABCJ
8AB SAOB 2 40)(2. OB - PloB - AB4 ~AOB 2 = 4SAOB = tan(L[AOBJ) 2AO -AB PAOB
= L[AOBJ.
Example 3.53 (Morley's Theorem) The three points of intersection of the adjacent trio sectors of the angles of any triangle form an equilateral triangle.
Constructive description BC L)
«P()UNnrS
(ON x (ALINE B L L B C» (ON Y (ALINE B x L B C» (ON z (ALINE C L L C B» (ON w (ALINE C Z L C B» (INfER A (LINE B Y) (LINE C W» (MIDP()UNnr 0 L C)
(CONSTANT r2 3) B (TRATIO T 0 C r) Figure 3-17 (INfER 0 (LINE B L) (LINE T C» (ON H (ALINE A BTL 0» (INfER M (LINE B X) (LINE A H» (ON G (ALINE A COL T» (INfER N (LINE C Z) (LINE A G» «TANGENT M L N) = r) )
Morley's theorem is among the most difficult problems proved by our method. The proof for it is still too long to be considered readable. Our method need further improvement to produce a readable proof for this theorem. Exercises 3.54
I . Note that the sine and cosine functions for full-angles are meaningless. But we can 2 · squares. For a full -angIe G, Iet sm . 2() 1&8 (0) 2() 1 de fi ne thelf G = 1+\&02(0)' cos G = 1+1&8'(0)· Then
S~BCD = ~E . BD2sin2(L[AC,BDJ), P~BCD =4E.Bdcos2 (L[AC,BDJ) . (Use the Herron-Qin formula.)
130
Chapter 3.
MachIne Proof in Plane Geometry
2. Try to eliminate Y from SABY if Y is introduced by (INTER Y (LINE U V) (ALINE P Q L M N» . Show that the ndg conditions ensure that all the geometric quantities occurring in the elimination have geometric meaning. 3. If Y is introduced by (INTER Y (TLINE U U 0) (TLINE V V 0» then
S
BU
S~BV
PorfyPOYll
{
p~~~fJ?gcjo ---- ----
p~~~%iQ 16Sovu
PABU {
if AB.l..OU; if AB.l..OV; if A = U, B = V; if A
= O,B = U;
if A
= O,B = V;
if AB
II OU;
PABV if AB II OV; Pouv if A = U, B = V; -PuoyPoyuPouy if A = U, B 16Sbuv
PovoPfwy 2
= 0 , B = U or if A = 0, B = Vor
POllO'W,':"OPUYll
if A
1
2 165OllV
{
= V;
of A
PouoPf,yu
=
o Figure 3-18
165
16Sbuv
A=B=U; A=B=V;
= B = o.
In the general case, this construction can be reduced to the following construction (TRATIO Y U 0 .is'O'VU)' Notice that this construction is actually to introduce the antipodal point of 0 with respect to the circumcircle of triangle OUV.
3.5 3.5.1
Area Coordinates Area Coordinate Systems
In Lemma 3.31, we use an orthogonal coordinate system. In order to do this, we have to introduce three auxiliary points 0, U, and V . In this section, we will develop a skew area coordinate system in which any three free points can be used as the reference points. As a consequence, we obtain a new version of Lemma 3.31 and consequently a new version of Algorithm 3.33. We will also prove some interesting features of the skew area coordinate system. Let 0, U, and V be three non-collinear points. For any point A, let SOUA - SOU V ,
XA---
SVOA YA=-S ' OUV
SUVA ZA=-SOUV
3.5
131
Area Coonllnates
be the area coordinates of A with respect to auv. It is clear that XA are some results proved before.
+ YA + zA = 1. Below
Proposition 3.55 The points in the plane are in a one to one correspondence with the triples (x, y , z) such that x + Y + z = 1. Proposition 3.56 For any points A, B, and C. we have SABC
= -Souv
XA YA 1 xB YB 1 Xc Yc 1
= -Souv
XA YA ZA XB YB ZB Xc Yc Zc
As a consequence of Proposition 3.56. we can give the line equation in the area coordinate system. Let P be a point on line AB. Then the area coordinates of P must satisfy
XA YA 1 XB YB 1 Xp yp 1
=0
which is the equation for line AB. Notice that this is the same as the line equation in the Cartesian coordinate system. Another interesting fact is the position ratio formula
= ;~ and T2 = ~. Then T1YQ + T2YP; ZR = T1ZQ + T2 Zp.
Proposition 3.57 Let R be a point on line PQ and Tl XR = T1XQ
+ T2Xp;
YR =
Proof This is a consequence of Proposition 2.9. We will now prove the distance formula between two points.
Proposition 3.58 For two points A and B. we have
Alf = aV2(XB -
XA)2
+ OU\YB
- YA)2
+ (XB
- XA)(YB - YA)PUOV .
Proof Let M and N be points such that MA II aU.MB II aV,NA II aV,and NB II au. Then by Proposition 3.6. (I)
Alf = AM2 + BM2 -
== au2 CAM)2 ou + aV
2(BM)2 Ov
By Lemma 2.28, ~~ =
PNAM = ~~ . ~~ prove the result.
A
PAMB
+ PNAM·
M
u
o Figure 3-19
s/UOoAvY = YB - YA, ~~ = S/O'!,~l = XA - XB. By Example 3.9,
. Puov = (YB -
YA)(XB - XA)PUOV . Substituting these into (I). we I
132
Chapter 3.
Macblne Proof in Plane Geometry
Corollary 3.59 Show toot
2AJf
= POVU(XB -
XA?
+ POUV(YB -
Proof We need only to observe that zA - ZB
YA)2
= (XB -
XA)
+ PUOV( ZB + (YB
ZA)2.
- YA).
In Algorithm 3.33, let E be an expression in areas and Pythagoras differences of free points. Instead of using Lemma 3.31, we can use the following procedure to transform E into an expression in independent variables. If there are fewer than three points involved in E then we need do nothing; if there are more than two points, choose three free points 0, U, and V from the points occurring in E , and apply Propositions 3.56 and 3.58 to E to transform the areas and Pythagoras differences into area coordinates with respect to 2 2 OUV. Now the new E is an expression in area coordinates offree points, OU , OV , W, and Souv. The only algebraic relation among these quantities is the Herron-Qin formula (Proposition 3.16):
Substituting S6uv into E, we obtain an expression in independent variables. If we assume that OU 1. OV and 10UI = becomes the Cartesian coordinate system.
10VI
= 1 then the area coordinate system
Exercises 3.60 1. The process of transforming the areas and Pythagoras differences of free points into expressions in independent variables presented in this section becomes particularly simple if there are only three free points in a geometry statement. Let the three free points be 0 , U, and V . Show that any polynomial 9 in the areas and Pythagoras differences involving 0, U, and V can be transformed into a polynomial of the following form 9' = f + Souv h where f and h are polynomials of OU 2 , OV 2 , and UV2 • We also have that 9 = 0 iff h = f = O. 2. If in a geometry statement there are only two free points U and V, and the third point 0 is introduced by the construction (ON 0 (BLINE U V» or (ON 0 (TLINE U V», design a simple method of transforming a polynomial in the areas and Pythagoras differences 2 involving points 0, U, and V into a polynomial of independent variables. (Chose OU and as independent variables.)
W
3.5.2
Area Coordinates and Special Points of Triangles
We introduce a new construction.
133
3.5 Area Coordinates
CI6 (ARATIO A 0 U V TO TU TV) ' Take a point A such that SAUV TO--- SOUV'
SOAV TU=--, SOUV
SOUA TV=-SOUV
are the area coordinates of A with respect to OUV. The TO, TU, and TV could be rational numbers. rational expressions in geometric quantities. indetenninates. or algebraic numbers. The ndg condition is that 0, U, and V are not collinear. The degree of freedom for A is dependent on the number of indetenninates in {TO, TU, TV} . Lemma 3.61 Let G(Y) be a linear geometry quantity and Y be introduced by (ARATlO YOU V TO TU TV). Then G(Y)
= ToG(O) + TuG(U) + TvG(V) .
Proof Without loss of generality. let OY intersect UV at T . If OY is parallel to UV. we may consider the intersection of UY and OV or the intersection of VY and OU since one of them must exist. By Proposition 2.10.
U
T Figure 3-20
By the co-side theorem • II !£I. = ~. ~ = ~. Substituting OT = T O.• m: OT = ~. Souv • uv SOUyv' uv SOUYV these into the above formula, we obtain the desired result. I Lemma 3.62 Let G(Y) be a quadratic geometry quantity andY be introduced by (ARATlO YOU V TO TU TV). Then G(Y)
= TOG(O) + TUG(U) + TVG(V) -
==2
2(TOTUOU
==2
~
+ TOTVOV + TUTVUV
).
Proof Continue from the proof of Lemma 3.61. By (IT) on page 113 G(Y)
= g~G(T) + ~~G(O) -
g~~~POTO
G(T)
= ~~G(V) + ~~G(U) -
~~~~Puvu.
Substituting G(T) into G(Y). we have G(Y) - T
where T = ToG(O) POTO
= -g~~~~~Puvu -
g~~~POTO
+ TuG(U) + TVG(V) . By (II). = ~~Povo + ~POUO - ~~~~Puvu.
= -TV~~Puvu -
TAg~POTO.
134
Chapter 3. Machine Proof in Plane GeooIeIl'y
Then
G(Y) - r TV OYUT OYTV -rV UVPuvu - rO OT UV Povo - rO OT UV Pouo SyUV -rorvPovo - roruPOUo - rurv(--S-OUYV -rOrVPOVO - roruPouo - rUrVPUVU · I
OYUTTV
+ rO OT UV UV Puvu
SOUV + -S--)Puvu OUYV
If Y is introduced by construction ARATIO, then we need rarely to eliminate Y from G = ~~. Because points D, E, and F are introduced before point Y, we generally do not know whether DY is parallel to EF or not. But in some special cases, we still need to eliminate point Y from G. This can be done as follows. One of 0, U, and V, say 0, S satisfies the condition that D, Y, and 0 are not collinear. Then G = ..2.C1.UY.. sOEDF • Now, we can use Lemma 3.61 to eliminate Y.
By using the construction ARATIO, we can treat some special points of triangles easily. Proposition 3.63 The area coordinate of the centroid of a triangle ABC is
(lll).
Proof Let G be the centroid of !:::.ABC and M be the midpoint of BC. Since Mis the midpoint of BC, we have SABM = SAMC and SaBM = SaMC. Then SaAB = SABM - PaBM = SAMC - SaMC = SacA. Similarly SaBc = SaAB = SaCA = lSABC .1 Proposition 3.64 The area coordinate of the orthocenter of triangle ABC is (PABCPACB
PCABPCBA)
16S~BC
16S~BC
Proof As in the figure, let H be the intersection of the altitudes CD and AE. Then
rC
AD
PCAB
CE
PBCA
SAHC SHBC
= DB = P ABC
SAHC SABH
= EB = PABC
.
c E
B
Figure 3-21
Then r A : rB : rC
= PABCPBCA : PCABPBCA : PABCPCAB .
By the Herron-Qin Theorem (Exercise 3.18), PABCPBCA + PCABPBCA 2A1f P BCA + PABCPCAB = 16Sl Bc . Now the result is clear.
+ PABCPCAB = I
135
3.5 Area Coordinates
Exercises 3.65 I . The area coordinate of the circumscribed center of triangle ABC is (PBCBPBAC
PACAPABC
32S1BC
32S~BC
PABAPACB) 32S1BC ·
(Use full-angles.)
2. Let I be the incenter or one of the excenters of 6.ABC. The area coordinate of C with respect to I BC is (_ 2PIABPIBA
PIABPIBI
PIBAPIAI ) .
PAIBPABA
PAlBPABA
PAlBPABA
(Use full-angles.) 3. Construction C6 on page 108 is equivalent to (ARATIO Y PO I O2
-
1
Now we can use the following new constructions:
CI7 (CENTROID GAB C). G is the centroid of triangle ABC. CI8 (ORTHOCENTER H A B C). H is the orthocenter of the triangle ABC. CI9 (CIRCUMCENTER 0 A B C). 0 is the circumcenter of triangle ABC.
C20 (INCENTER C I A B). I is the center of the inscribed circle of triangle ABC. This construction is to construct point C from points A, B , and I . Construction C20 needs some explanation. If three vertices of a triangle are given and we need to find the coordinates of the incenter, we generally have an equation of degree four in the coordinates of the incenter. The reason is that we can not distinguish the incenter and the three excenters without using inequalities. What we do here is to reverse the problem: when the incenter (or an excenter) and two vertices of a triangle are given, the third vertex is uniquely determined and can be introduced using the constructions given before. Remark 3.66 In this section. we actually use the centroid theorem (Example 1.12 on page 12). the orthocenter theorem (Example 1.67 on page 31). and the incenter theorem (Example 6.144 on page 327) in the proof of more complicated theorems. The four theorems themselves can be proved using the basic propositions. In general. we can use different constructions to describe the same geometry statement. and the more basic the constructions used the less prerequisite is needed in the proof of the theorem, and generally. the longer the proof is. On the contrary. the more complicated the constructions used. the more prerequisite is needed, and generally. the shorter the proof is. For constructions C 17, C 18, C 19, and C20, the eliminating results for some special cases
are very simple. As an example, we have
136
Chapter 3. Madtlne Proof ia Plaae Geometry
Exercise 3.67 For (CIRCUMCENTER 0 A B C), we have
2.
PABO
= Alf
4. SPQo = SPt<
+ SPf B , if PQl.AB.
Example 3.68 Let H be the orthocenter of triangle ABC. Then the circumcenters of the four triangles ABC, ABH, ACH, and BCH are such that each is the orthocenter of the triangle formed by the remaining three.
Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter 0 A B C) (circumcenter A, B C H) (circumcenter B, A C H) (circumcenter c, A B H) (parallel B, C, B C) )
Figure 3-22
The machine proof
The eliminants
SBCB, SBCC,
SBCC,
Qj
SBCB, =
C,l (
=2
SBCH+SABC
B'l( 2 SBCH+SABC )
SBCB, ·(2) SBCH+SABC
~ (2) ,(SBCH+SdBC)
(SBCH+SABC)·(2) simJ!lify
The above machine proof uses the fourth result of Exercise 3.67.
Figure 3-23
)
Figure 3-24
Figure 3-25
3.5 Area Coordlnales
137
Example 3.69 A line is drawn through the centroid of a triangle. Show that the sum of the distances of the line from the two vertices of the triangle situated on the same side of the line is equal to the distance of the line from the third vena. (Figure 3-23) Constructive description ( (POints ABC X) (centroid GAB C) (foot (foot (foot
D A G X)
The eliminants
-(%+%)
~!.~
_F
E B G X)
AD- SAXQ
-U An ,SAXC-Scxc
-
FC G X)
(M+~ =
The machine proof
( SAXC)
~ -(-SCXC ,SAXC-SBXC 'SdX C )
-1»
-
- (SC X C+SBXQ) SAXQ
-
AD- SA X Q
SA X Qg -
~(SACX+SABX)
SBXQg -
i(SBCX-SABX)
SAXC '( SAXC)
.im.!!!.i/~
Q
U~~
SCXCg!(SBCX+SACX)
-(3SACX+3SABXH3) (-SACX-SABX) ·«3))2
.i~i/Y 1
Example 3.70 Two tritangent centers divide the bisector on which they are located. harmonically (Figure 3-24). Constructive description
The machine proof
The eliminants
( (POINTS B C I)
(-~)/(~)
~~.fuu.
(INCENTER A I C B) (INTER D (LINE A J) (LINE B C»
~ .fuuJ. . _ ~
Q (INTER IA (LINE A I) (lLINE B B I» (HARMONIC A D I IA) )
PI BA ID -(-SBICd),PcBrSBIA PIBA ,SBICA ,SBCI
.im1!!.i/~ PCRrSRIA PIBA ,SBCI
DIA - PIBD
P IBD
OPe R,.S8l' SBICA
~£~
ID- SBCI p A PCRI ' PSls,Pacf IBA PBIC,PBCB S A -PSIR ,PAc,.Sscr BIA PBIC ,PBCB
~ PCBd
PBIWPRCl,SRCl) ,PBIC ,PRCR (-PCBrPBIB ,PBC/) ,SBcrPBIC ,PBCB
-
.im!1i/" 1
Example 3.71 (Euler's Theorem) The centroid ofa triangle is on the segment determined by the circumcenter 0 and the onhocenter H of the same triangle. and divides 0 H in the ratio of 1:2 (Figure 3-25). Constructive description ( (POINTS A B C)
PCBH
(CIRCUMCENTER 0 A B C) (CENTROID M A B C)
(LRATIO HMO
The machine proof .f.4.ac..
-2)
(PERPENDICULAR A H B C) )
!!..
PAAG
-
3PCBM-2PCBO
!!!.
PABd3) -6PCBO+3PBCB+3PABC
-
Q -PABd2) -
-2PABC
.im!1ih
The eliminants PCBH!f,3PCBM-2PCBO
PCBM~i(PBCB+PABC) PCBOg~(PBCB)
138
3.6
Chapter 3.
Machine Proof In Plane G_try
Trigonometric Functions and Co-Circle Points
The aim of this section is to provide an efficient method for dealing with co-circle points. We will prove the co-circle theorems given in Section 1.9 without using trigonometric functions; on the contrary, we will develop properties of trigonometric functions by using the area and Pythagoras difference only. The readers who are interested primaryly in machine proof may skip the following subsection and proceed directly to Subsection 3.6.2.
3.6.1
The Co-circle Theorems
Let J, A, and B be three points on a circle with center O. In what follows, we will fix J as a reference point. An oriented chord, AB, is a directed line segment such that J A (A '" J) is always positive and AB > 0 iff SJAB > O. Proposition 3.72 Let 6 be the diameter of the circumcircle of the triangle ABC. Then -EA • S ABC -- A"8-E8 2d c
Proof Let CE be a diameter of circle O . By Proposition 3.48, B
S~BC
S~c E = 2 Since S ACE
~2.~2 AE2. 62 -
l-C2 - 2 = 4A . AE ,we have -2==2~
2 _ AB . BC . AC . S ABC 46 2 ,I.e.,
Figure 3-26
IS
-
-
-
I _ IABI·ICBI·ICAI
ABC -
26
.
We still need to check whether the signs of both sides of the conclusion equation are the same. At first, it is easy to see that when we interchange two vertices of the triangle ABC, the signs of both sides of the equation will change. Therefore, we need only to check a particular position for A, B, and C, e.g., the case shown in Figure 3-26. In this case, we
haveSABC>O,AB>O,CB>O, andCA>O.
I
Let BC be an oriented chord on circle 0 and BB' be a diameter. We define the cochord of BC to be fiG whose absolute value is equal to IC B'I and has the same sign as PBJC . It is clear that
Proposition 3.73 For points A , B, and C on circle 0, we have PABC
= 2A"8-?&.
139
3.6 Co-Clrde Points
Proof By the Herron-Qin formula and Proposition 3.72, we have _ 4AIr CB 2
p2
ABC -
•
-
16S2
~-==2t'2~
_ 4AB . CB (0· - AC ) {j2
ABC -
Then IPABcl = 2 IABI'I?I'ICAI. As in Proposition 1.100, we can check that the signs of the two sides of the equation are equal. I
Proposition 3.74 Let A, B, C, and D be four cocircle points. E the intersection of the circle. and the line passing through D and parallel to AC. Then SABCD = AC'~'EB where 8 is the diameter of the circle.
A
B
Figure 3-27
Proof. If AC II BD. we have E = Band SABCD = O. The result is clearly true. Otherwise, since DE II AC, we have
Proposition 3.75 (ptolemv Theorem) Let A, B , C, and D befoureo-circle points. Then AB · CD+ BC· AD = AC · BD . Proof Let E be the intersection of the circle and the line passing through D and parallel to AC (Figure 3-27). Then by Propositions 3.72 and 3.74,
AC·BD·EB ---=-2-=-8--
BC·EC·EB+EA · BA · BE
= SABCD = SBCE + SEAB = - - - - - - 2 - 8 - - - - -
Let B be the reference point. Notice that EB we prove the result.
= -BE,AE = -CD, and CE = -AD,
Proposition 3.76 Let AB = d be the diameter of the circle. and P, Q be two points on the circle. Then
d· PQ
= AQ.AP-AP.AQ.
Proof. Apply Ptolemy's theorem to A, P, B, and Q: AB · PQ
= AP . BQ+AQ.PB.
A
Figure 3-28
Let J be the reference point. If SJAB < 0 (Figure 3-28), we have AB = -d,AQ and AP = BP. If SJAB > 0 or J = A, we have AB = d, AQ = QB, and AP The result is always true.
= BQ, = fiB.
140
Chapter 3. Machine Proof In Plane Geometry
Proposition 3.77 Let AB
= d be the diameter of the circle, and P, Q be two points on the
circle. Then d · PQ = AP· AQ+AP · AQ. Proof. Let S be the antipodal of Q (Figure 3-28). By Ptolemy's theorem, we have AB · PS
= AP· BS+AS· PB.
Let J be the reference point. If AB and QS have the same sign. we have
AB· PS
= dPQ,BS = AQ,AS · PB = AQ· AP
If AB and QS have different signs, we have
AB· PS
= -d· PQ,BS = -AQ,AS· PB = -AQ· AP.
The result is true in both cases. Propositions 3.76 and 3.77 are called decomposition formulas for chords. They can be used to reduce any chord or cochord to a polynomial in chords and cochords with a fixed point (A) as an end point. For points A and B on the circle, let us define
• sin(A)
= '1;
• sin(AB)
= At;
cos(A)
= "if.
cos(AB)
= At·
Then we have derived the following properties of trigonometric functions.
sin(A)2 + cos(A)2 = 1 sin(AB) = sin(B) cos(A) - sin(A) cos(B) cos(AB) = cos(B) cos(A) + sin(A) sin(B). The sin(AB) and cos(AB) are actually the sine and cosine of the inscribed angle in arc AB. But in the above definition of trigonometric functions, we did not mention the the concept of angles.
3.6.2
Eliminating Co-Circle Points
We introduce a new construction. C21 (CIRCLE Y1 •• • Y.) , (s ~ 3). Points Y 1 ••• Ys are on the same circle. There is no ndg condition for construction C21. The degree offreedom of this construction is
s+3.
141
3.6 c.arde Points
By Propositions 3.72-3.77, we have
Lemma 3.78 Let A, B , and C be points on a circle with center 0 and diameter O. Then SABC -- AiI·CiJ·EA 26 • AC
PABC -_ 2Ai1·CiJ·6A 6 •
= 0 sin(AC). At = Ocos(AC).
Using Lemma 3.78, an expression in the areas and Pythagoras differences of points on a circle can be reduced to an expression in the diameter 0 of the circle and trigonometric functions of independent angles. Two such expressions have the same value iff when substituting, for each angle a, (sin a)2 by 1 - (cos a)2 the resulting expression should be the same. We thus have a complete method for this construction. The reader may have noticed that this construction must always come first in the description of a statement. Otherwise, in the next step, we do not know how to eliminate these trigonometric functions . The proofs of many interesting geometry theorems use this construction. Example 3.79 (Simson's Theorem) Let D be a point on the circumscribed circle of triangle ABC. From D three perpendiculars are drawn to the three sides BC. AC. and AB of triangle ABC. Let E . F. and G be the three feet respectively. Show that E. F and G are collinear. Here is the input to our program. «CIRCLE
ABC D)
(FOOT ED B C) (FOOT
FDA
C)
(FOOTG D A B)
(INTER G, (LINE E F) (LINE A Bll Figure 3-29
(~=~» BG BG,
The ndg conditions: B
¥
C, A
¥
C, A
¥
B, EF IY AB, B
¥
G, B
¥
G1•
Here is the machine proof. The last step of the proof uses Lemma 3.78. The machine proof
(~)/(*) ~~.~ SAEY
Q -
.f.. -
BG
PAID ,SAFiE SAEY-(
PABO)
-P8AQ , PACQ , SABF: , P"CA
(-PCAO ·SACE) -PABO -PACA
6i~i/Jl
The eliminants ~~~ BG, -
SBEY
~Q..£.a..w...
BG--PABO F-PCAP ·S,CE AEY PACA
s
S BEY
S
F P,CP , SARli PACA _ E -PRCP ,SAsa
ACE
PBCB
S PSAQ , PACQ , SABfj
PCAO · SACS · PABO
EPCBP ,SABC ABE PBCB
PABO= -
2(BD-Ail-coo(AD»)
142
§..
Chapter 3. Machine Proof in Plane Geometry
The eliminants
PaAQ ' PA C O , PCRO , SARC , PBCR
PCAD '( - PB C D'S ABC ),PABD ,PBCB simJ!!.iJy
PBCD= -
PRAQ,PACO,PCRO
PCAD,PB CD,PABD co- cir
2(Cb.Bc.cos(BD»)
PCAD=2(AD.Ac.cos(CD»)
(2AD. AB ·cos( B D» ·( - 2CD ·Ac ·cos( AD)).(2iiD .Bc ·co.( CD» -(2AD.AC .cos(C D))·( -2ED.i;C .cos(BD» .( -2BD .AB.co.(AD»
s imJ!!.iJy
PCBD=2(iiD .Bc.cos(CD») PACD= -
2(Cb'Ac'cos(AD»)
PBAD=2( AD.AB.cos(BD»)
Example 3.80 (Pascal's Theorem on a Circle) Let A , B, e, D , E, and F be six points on a circle. Let P are collinear.
= AB n DF. Q = Be n EF. and S = eD n EA. Show that P,Q. andS
Here is the input to the program. ( (CIRCLE ABC D F E) (INTER P (LINE D F) (LINE A B» (INTER Q (LINE F E) (LINE B C» (INTER S (LINE E A) (LINE C D» (INTER
s,
(LINE P Q) (LINE C D»
(cs =~» DS
Figure 3-30
DS 1
The ndg conditions: DF II' AB, EF II' Be, AE The machine proof (~~)/(~)
II' eD, PQ
lYeD, D
i- s, D i- S1. The eliminants ~~SCPQ
DS, - SDPQ
~2.~ DS- SADI>
~ SDP9
-
2.. -
g, -
S CPQ •
-
Os
SACE,SDPQ SCPQ,SADE SACE,(-SDEP ,SBCF),SBFCE ( SCFE ,SBCP) ,SADE'( SBFCE)
simJ!!ify -
.f.
cs
-SACS ,Soge-SaGp SCFE,SBCP,SADE
-SACE,(-SDFE,SABD) ,SBCF ,SADBF SOFE '( -SBDF ,SABc) ,SADE '( -SADBF)
sim.!!!.iJy
SACE ,SOfe,SARQ ,SRCP
SCFE,SBDF ,SABO ,SADE
S
Q -Seffj ,Saep
CPQ S DPQ
S BOP S
SBFCE Q
SBFOE p -SAOp ,SARa SADBF !:SDFE,SA8Q
DEP-
SADE
SQEP,SRCF
SADBF DE.As.AD (-2).d
Bc ·Ac·AB (-2) .d fjp ·8F.iiD SBDF= (2) .d SABO
SOFE
JiE·CE·CF (-2) .d
SBOF
CF·8F·Bc (-2) .d
143
~e PoiDta
3.6
The eliminants
- -BD·AD·AB
The machine proof
(- 2 )·J
co=.cir (-EE.A'E.A"C).(- FE·DE·5FH -iiD ..w.A"8).(- EF.iiF.iiC)'«2d)j<
-oc .AC.Ail)'(- DE.A'E.AD).«2d)'
(- F'E.c"E·ch·(- D'F.iiF.BDH
-
.im!!f. i f~ 1
FE.DE·5F
SOF E SACE
( -2 )·J
EEoA'E.A"C ( -2)·J
Example 3.81 (The General Butterfly Theorem.) As in the figure. A , B . C. D , E . Fare six points on a circle. M = ABn CD ; N = AB n EF;G = AB n CF ; H = ABn DEo L_ M G BH AN Sho w t'lUt 'AG 'Nil MB - 1. Constructive description
A B C D E F) DC) (LINE A Bll N (LINE E F) (LINE A Bll
( (CIRCLE
(INTER M (LINE (INTER
(INTER G (LINE A B ) (LINE C Fll (INTER H (LINE
D E)
(LINE
A Bll
(~*=~~»
The machine proof
AN
S~E:
•
~ AG
-~FM"S8Q" ~.~ . SOEN"SAC F
!i:
AG - SACF !iSnEr,SABE OEN- -SAESF
S
-~ . ~ . SOEN
g
IJJJ.!!' §.aD..E. ~~~
-UL .4JL
!£
The eliminants NH- SDEN
~IJJJ.
~
AS
Figure 3-31
4JL!:!.&E..a£. AN- SAEF
UL ~ .h.c.IJ.... AB - SACSD
S
~ SCaE ,SASa
CFM-
:!£FWSSOE ' ( -SAUF)oSAfiF
~ . SAESF · SOEF oSASE o SACF .im~i/ll
SGEM ,S8Qfj ,SAEi f
SACF SASE=
~ o SOEF oSASE o SACF SDEF=
M.
(-SCOF ,SASC) oSSO" oSdu o(-SACSO) (-SSCO) oSOEFoSASE oSAcr(-SACSO )
-
. imI?J.i/Ji -
SCQf 'SABC,SBDF;-SAft£ SSCO,SOEF ,SABE oSdCF
co=.cir (- iiF·EF.CO)o(-iiCoA"CoA"8H -DEoiiEoBb)o(-iF.M oA'E)o«2d)·
-
(- CDoiiDoOC)o(-EF.D'F.DEJ.( -BEoA'E.Ail)o(-C'FoA'FoAC)o«2d)'
.i"'!1ifll
SSCD SAEF= SBDE SASC= SCDF=
SACSD
EFoM oA"C (-2) oJ
---
BEoAEoAB (-2) oJ
iFoiiFoDE (- 2 )oJ
COoiiDoiiC (-2) oJ
iFoMoA'E (-2) oJ
iiE.iiEoiiD (-2) oJ
iiCoA"CoA"8 (-2) oJ
iiF·fFoco (-2) oJ
Example 3.82 (Cantor' s Theorem) The perpendiculars from the midpoints of the sides of a cyclic quadrilateral to the respectively opposite sides are concurrent.
144
Chapter 3. Machine Proof In Plane Geometry
Constructive description
The machine proof
«CIRCLE ABC D)
~
(ClRCUMCENTER 0 A B C) (MIDPOINT
G A D)
PCBN !::!...
PaBa
-
PCBE+PCBF-PCBO
(MIDPOINT F A B)
~
(MIDPOINT E C D)
-
PaBa PCBF-PCBO+tPCBD+fpBCB
(PRAnO N EO F I) (PERPENDICULAR G NBC) )
E. -
Q -
The etiminants PCBN~PCBE+PCBF-PCBO
PCBE~l(pCBD+PBCB) PCBF~¥(PABC); PCBogk(PBCB) PCBCg2(PCBD+PABC) A
(2),PCBC -2PCBO+PCBD+PBCB+PABC (-2),(}PCBD+}PABC) 2PCBO-PCBD-PBCB-PABC
Q -(PCBD+PABC)·(2) -
-2PCBD-2PABC
simgli!y 1
Figure 3-32
The ndg conditions: A, B, C are not collinear; A '" D; A", B; C '" D; 0", F .
3.7
Machine Proof for Class C
The class C, or the class of constructive geometry statements, are the geometry statements which are assertions about the configurations that can be drawn by rulers and compasses only. To describe the statements in class C, we need two new constructions. We first introduce a new kind of circle: (CIR 0 r) is the the circle with center 0 and radius Jr. As before, r could be an algebraic number, a rational expression in geometry quantities, or a variable. Thus (CIR 0 P) is the same as (CIR 0 OP\ C22 (INTER Y (LINE U V) (CIR 0 r)). Point Y is one of the intersections of tine (LINE U V) and circle (CIR 0 r). The ndg conditions are r '" 0, U '" V. Point Y is a fixed point and has two possibilities. C23 (INTER Y (CIR 0 1 rIl (CIR O2 r2)). Point Y is one of the intersections of the circle (CIR 0 1 rl) and the circle (CIR O2 r2) . The ndg conditions are 0 1 '" O2, rl '" 0, and r2 '" O. Point Y is a fixed point and has two possibilities. Each of constructions C22 and C23 actually introduces two points, and we generally can not distinguish the two points. This is the main difficulty of dealing with these two constructions.
3.7.1
Eliminating Points from Geometry Quantities
Proposition 3.83 Let Y be one of the intersection points of line UV und circle (CIR 0 r). Then we have
Proof. Let X be another intersection point of the line UV and the circle (CIR 0 r), and M the midpoint of X Y . B y Proposition 3.2.
B y Proposition 3.5,
-
XUm = =OY XY
Figure 3-33
-
UY+=OX XY
--
-= o x + U-Y . U X = r + U Y . U X . (2)
XU U Y -=-=XY XY XY
From (1) and (2), it is easy to obtain the result.
I
Now we can give the elimination methods for construction C22 easily.
Lemma 3.84 Let Y be introduced by (INTER Y ( M E U V ) (CIR 0 r)). Then
where
-
satisfies (III).
-
+
-
7
+
Proof. B y Proposition 2.9. SABy= ~ ~ S A B(1V- ~ ) S A B = ( I~ S V A U B SABU.The second and third cases are consequences of Proposition 3.5. I
Lemma 3.85 Let Y be introduced by (INTER Y (WNE U V ) (CIR 0 r)). Then
--
--
ifDeUV. otherwise.
Proof: The first case is trivial. The second case is a consequence of the co-side theorem. I
146
Chapter 3. Machine Proof In Plane Geometry
Proposition 3.86 Construction C23 is equivalent to the following two constructions (LRATIO 00 1 0 2 r)
(TRATIO Y 0 0 1 s)
Proof Let 0 be the foot of the perpendicular dropped from point Y upon the line 0 10 2 • By Proposition 3.2,
Figure 3-34
~
Fors, we have S2
=~ = ~-1 = ~-l. 00 , 00 , r'O,O,
I
Lemma 3.87 Let Y be introduced by (INTER Y (CIR 0 1 rl) (CIR O2 r2». Then SABY
= SABO, + rSO,AO.B -
rs 4PO,AO.B
where r and s are the same as in Proposition 3.86. Proof Let 0 be the foot of the perpendicular dropped from point Y upon the line 0 10 2• By Lemma 3.26,
=
s
SABO -
rSABO,
4' POAOIB
+ (1 -
s
r)SABOJ - 4(rPo ,AB rs SABO, + rSO,AO.B - 4P02AOIB . I
+ (1 -
r)PO• AB - PO,AB)
Lemma 3.88 Let Y be introduced by (INTER Y (CIR 0 1 rl) (CIR O2 r2» . Then P ABy
= PABO, + rP02AO,B -
4rsSo,Ao JB
where r and s are the same as in Proposition 3.86. Proof Let 0 be the foot of the perpendicular dropped from point Y upon the line 0 10 2 • By Lemma 3.27, PABy
= =
PABO - 4SS0AO, B
+ (1 - r)PABO• - 4S(rSO,AB + (1 + rPO,AO,B - 4rsSO.AO.B. I
rPABo. PABOI
r)SOJAB - SO.AB)
The methods of eliminating Y from ~~ and PAYB can be found in Lemmas 2.26, 3.30, 3.25, and 3.28.
3.7
147
Madlioe Proof ror Class C
3.7.2
Pseudo Divisions and Triangular Forms
In this section, we introduce some algebraic tools which will be used in the machine proof for statements in class C . These tools are actually parts of Wu's method of automated geometry theorem proving, more details of which can be found in [36,12]. Let K be a computable field of characteristic zero (e.g., Q) and A = K[Xl , ... , xn] be the polynomial ring of the variables Xl,'" , Xn . For P E K[x]- K, we can write
P
= CdX= + ... + ClXp + Co
where c, E E[Xl ' ..., xp-d, p > 0, and Cd '" O. We call p the class, Cd the initial, xp the leading variable, and d the leading degree of P respectively, or dass(P) = p, init(P) = cd,lv(P) = x p' Id(P) = d. If P E K, we have dass(P) = O. Let p = dass(P) > O. A polynomial Q is said to be reduced with respect to P if deg(Q , xp) < Id(P) . Let 1= an v n + ... + ao and h = b"v" + ... + bo be two polynomials in A[v], where v is a new indetenninate. Suppose k, the leading degree of h in v, is greater than O. Then the pseudo division proceeds as follows: Definition 3.89 (pseudo Division) First let r = f. Then repeat the following process until m = deger , v) < k : r = bkr - Cmvm-"h. where Cm is the leading coefficient ofr.lt is easy to see that m strictly decreases after each iteration. Thus the process terminates. At the end. we have the pseudo remainder prem(f, h, v) = r = ro. Proposition 3.90 Continuing from the above definition. we have the following formula,
biJ
= qh + ro.
where s $ n - k + 1 and deg(ro, v) < deg(h, v).
(1)
Proof We fix polynomial h and use induction on n = deg(f, v). If n < k, then we have ro = I and 1= O· h + ro. Suppose n ~ k and formula (1) is true for those polynomials I, for which deg(f , v) < n. After the first iteration, we have r = bioi - anvn-kh, where an is the leading coefficient of I. Since deger, v) < n , we have btr = qlh + ro by the induction I hypothesis. Substituting r = bioi - anvn-kh in the last formula, we have (1).
=
Definition 3.91 A sequence of polynomials TS AI, ... ,Ap in K[X] is said to be a triangular form, if either p = 1 and Al '" 0 or 0 < dass(Ai) < dass(A j ) for 1 $ i < j .
=
For a triangular form TS AI, ... , A p, we make a renaming of the variables. If Ai is of class mi, we rename x m ; as Xi. other variables are renamed as Ul, .. ., tlq, where q = n - p. The variables til , . . . , tlq are called the parameter set of T S . Then T S is like
Al(Ul,· · · ,tlq,xd
148
Chapter 3.
Machine Proof In Plane Geometry
(IV)
For another polynomial G, we can define the successive pseudo division : Rp
= prem( G, Ap, xp), . . . , Rl = prem(Rt, AI , Xl) '
R = Rl is called thefinal remainder and is denoted by prem(G,At, .. . ,Ar). It is easy to prove the following important proposition:
Proposition 3.92 (The Remainder Formula) Let TS and R be the same as the above. There are some non-negative integers 51, .. . , 5 p and polynomials Ql , . . . , Qp such toot 1. I :'· ·· I; pG
= QlAl + .. . + QpAp + R where the Ii are the initials of the Ai.
2. deg( R, Xi) < deg(A i , xi),for i
= 1,' "
, po
Proof We use induction on p. The case p = 1 is actually Proposition 3.90. Suppose that > 1 and the proposition is true for p - 1. Thus we have:
p
with deg(R, xd < deg(Ai' xd, for i I' pG - Q A p, we have 1 and 2.
=
1, . . . , p - 1. Combining this with Rp-l I
=
Theorem 3.93 Let T S = AI, ... , Ap be a triangular set, G a polynomial. Ifprem(G, TS) = 0 then 'v'X;[(Al
= 0/\ ... /\ Ap = 0/\ It ::f: 0/\ .. . /\ Ip ::f: 0) => G = 0] .
Proof Since prem( G , TS)
= 0, by the remainder formula
Now it is clear that Ai = 0 and Ii
::f: 0 imply G = O.
Definition 3.94 A triangular set AI, ... , Ap of the form (W) is called irreducible ifforeach
1. the initial Ii of Ai does not vanish in the polynomial ring Ri = K(U)[Xl,· · · , x;]!(A l , ... ,Ai- l ), and 2. Ai is irreducible in R i.
3.7
149
Machine Proof tor ClaM C
Thus the sequence
is a tower offield extensions. Example 3.95 Let TS be the triangular set Al = x~ - Ul> A2 is irreducible over Ao Q[ud; but A2 is reducible over Al A2 = (X2 - xd 2 under UI = O. Thus TS is reducible.
= xi -
= x~ - 2XIX2 + UI . Al = Ao[xl]/(A I) because
Proposition 3.96 Let T S = A I , ... , Ap of the form (N) be irreducible, and G be a polynomial in K[u, x]. Then the following conditions are equivalent: (i) prem(G, TS) =
o.
(ii) Let E be an extension field of K. If p. = (771!' . . , 77q, (I! ... , (p) in Ed+r is a common zero of AI! ... , Ap with 77. transcendental over K, then p. is also a zero ofG, i.e., G(p.) = O.
Proof For any polynomial h,let it be the polynomial obtained from h by substituting u., x. for 77. , ( •. We use induction on k to prove the following assertions (for 0
~ p):
(U) For any polynomial P = a.xk + ... + ao (0 < k ~ p, 1 ~ S, a. E K[u, x\, ... , Xk-I], a. # 0) reduced with respect to AI, ... , A p , if p. is a zero of P, then P = O.
=
= a.xt + .. . + ao, with all aj
E K[u] . p. is a zero of P means Since P is reduced with respect to AI> S < deg(AI' XI)' By the uniqueness of representation in algebraic extension, we have all iij = O. Since 77. are transcendental over K, all aj = O. Hence P = O.
If k
I, then P
P = ii.(I + ... + iio = O. S
Now we want to prove (U) is true for k assuming it is true for k - 1. Since p. is a zero of P = ii.(k" + ... + iio = O. Since S < deg(Ak,xk), by the uniqueness of representation in algebraic extension again, all iij = O. Thus p. is also a zero of all aj . Since all aj are also reduced with respect to AI , ... , A p , aj = 0 by the induction hypothesis. Hence P = o.
P,
(ii) ~ (i). Suppose p. is a zero of G. Let R remainder formula
I:' ... I;-G
= prem( G, A\, ..., Ap).
We have the
= QIAI + ... + QpAp + R.
Hence p. is a zero of R. Since R is reduced with respect to A I , . .. , A p , R (i) ~ (ii). Suppose prem(G, AI, ... , Ap)
= O.
= O. Then by the remainder formula, we have
where the Ik the are initials of the Ak. Since prem(lk, TS) (ii) ~ (i» . Hence p. is a zero of G.
# 0, p. is not a zero of h I
(by
Chapter 3.
150
Madllne Proof in Plane Geometry
We call fl in (ii) a generic point of that irreducible triangular fonn in field E. The theorem is no longer true if AI, ..., Ap is reducible. We can find such an example by letting AI , A 2 be the same as in example 3.95 and G = X2 - XI· Theorem 3.97 Let TS
= AI , ..., Ap be an irreducible triangular form. G a polynomial. If
is true in any extension field of K then prem( G, T S)
= o.
Proof. Let 1}1 , • .• , 1}q be some elements which are transcendental over K. By the definition of irreducible triangular set, we can find a generic zero fl = (1}I , ••• , 1}q , (I, .. . , (p) of TS such that I i (fl) '" O, i = 1,···, p. Since Ai(fl) = O,Ii (fl) '" O, i = 1,···,p,wehave G(fl) = O. By Proposition 3.96,prem(G,TS) = O. Exercise 3.98 Let K be the field of the rational numbers, T S triangular fonn, and G a polynomial. If
is true in the field of complex numbers then prem( G , T S)
3.7.3
= A I , . . . , Ap an irreducible
= O.
Machine Proof for Class C
We now return to the theory of machine proof. First, let us restate Algorithm 3.33 using the language of triangular fonns and pseudo divisions. Let S = (CI , · ·· , C., (E , F)) be a statement in C L . We denote by UI ,··· , u q, XI ,·· · , xp the geometry quantities occurring in the proof of S such that the Ui are the free parameters and the Xi are those quantities from which a point will be eliminated. We arrange the subscripts such that
Let Ai
= IiXi -
Ui . Then TS
= AI, · ·· , Ap
is a triangular fonn with I i as the initials of A i . Furthennore T S is irreducible since I i '" 0 is true under the ndg conditions of S. Theorem 3.99 Use the same notations as above. The statement S is true F,TS) = O.
iff prem(E-
3.7
151
Machioe Proal f .... Class C
It.
Proof Notice that the proving process of S is as follows: first replace Xi by i = p ," . ,1. in E and F to obtain two polynomials E' and F' in the Ui only. Since the Ui are free parameters. the statement S is true iff E' F'. The above process is equivalent to taking the pseudo remainder of E-F withrespecttoTS. ThereforeS is true iffprem(E-F, TS) = o.
=
If S forms :
= (CI , "
.
,Cr , (E, F)) is a statement in C. then for each i. Ai has two possible
= IiXi + Ui. or (2) Ai = IiXr + UiXi + Vi (1) either Ai
where Ii. Ui , and Vi are polynomials in a triangular form. Let
R
UI,'"
,Uq, Xl,"
= prem(E -
' , Xi-I.
TS
= AI,'"
,Ap is still
F, TS).
Then we have Theorem 3.100
2. If R
1. If R
= 0, S is true.
# 0 and TS is irreducible then S is not a theorem in the complex plane.
Proof For the first case, by Theorem 3.93 we have
Under the ndg conditions of S, we have Ii # 0, i second case is a consequence of Theorem 3.97.
= 1""
,po Then E I
=F
is true. The
Remark 3.101 In practice, we do not have to take the pseudo remainder of E - F with respect to TS. The better way is to elimiTUlte Xi from E and F separately as usual, so that we can take the advantage of removing the common factors from E and F during the proof To elimiTUlte xifrom E, if Ai = IiXi - Ui , we need only to replace Xi in E by if Ai = IiXr + UiXi + Vi, we need to keep replacing in E by - Ui"'J;+V; until the degree of Xi in E is less that two.
xr
It;
If R # 0 and TS is reducible, we need to factorize TS into irreducible triangular forms. We will not discuss the factorization method in this book; those who are interested in this topic may refer to [36, 12]. Let us assume that for some Ai we have Ai = IiX~
+ UiXi + Vi
=
= (Ii ,IXi - Ui,I)(Ii,2Xi - Ui,2)
which is true under the condition Ak 0, k triangular forms TS I and TS2 as follows
= 1,, "
,i - 1. Then TS is factored into two
152
Chapter 3. Madtine ProoIln Plane Geometry
where A i I = Ii IXi - Ui h Ai 2 = Ii 2Xi - Ui 2. Geometrically, this means that the two poinlS introduced by a co~stru~tion of type C22 or C23 can be distinguished and the two triangular forms TS I and TS2 correspond to the two intersections. If R ;;j:; 0 and T S is reducible, let T S be factored into several irreducible triangular forms TS I ," . ,TSm . Then we have three possible cases:
• For each T Si' prem( E - F, T Si) ;;j:; 0, i.e., E = F is not valid for all triangular forms. In this case, we say that the statement S is generally false. • For some TS i, prem(E- F,TSi ) ;;j:; 0, whilefor other TSj , prem(E-F, TSj ) In this case, the statement S is true only for some configurations.
= O.
• prem(E - F , TSi ) = 0 for all i. In this case, S is still true in the Euclidean geometry. This will happen only when we want to introduce the intersection points of two circles or a line and a circle which are tangent to each other. With the help of algebraic tools, we have a complete method of machine proof for geometry statemenlS in class C. G
G
D 1E"---+_-1
F
A~----"B
Figure 3-35
Example 3.102 Let ABCD be a square. CG is parallel to the diagonal BD. Point E is on CG such that BE = BD. F is the intersection of BE and DC. Show that DF = DE.
Proof As shown in Figure 3-35, let G be a point on line AD such that AD = DG. Then point E has two possible positions. The following proof shows that P DFD = P DED is true for both positions of E. By Lemmas 3.22 and 3.27, we have P BCG = -
P DBC
PDBC=PADB=PDGD=PBCB=PADA PCGC=PBDB
Let r
= g~ . _ P DFD -
= 2PABA
= PDCD=PABA
By the co-side theorem, we have
PDCDS~DE S2 BDEC
PABAS~DC (SBDC - rSDCG)2
=
PABAS~DC (SBDC
+ rSBDc)2
PABA
= (1
+ r)2'
3.7
M8dlhle Proal for a.. c
153
By (II) on page 1l3,
PDED = (1 - r)PDCD Then PDFD
+ rPDGD -
(1 - r)rPCGC = ((1 - r)
+r -
2(1 - r)r)PABA .
= PDED is true iff (1
+ r)2((1 - r) + r -
2(1 - r)r)
=1
and this is the case since by Proposition 3.83, r
2
=
2rPBCG - PBCB PCGC
+ PBDB
-2r + 1 =--2
.
Example 3.103 Let ABC be a triangle such that AC = BC. D is a point on AC; E is a point on BC such that AD = BE. F is the intersection of DE and AB. Show that DF=EF.
If we describe the statement as follows, it becomes reducible. «POINTS A B) (ON C (BLINE A B)) (ON D (LINE A C)) (1NlER E (LINE B C) (elR B W)) (1NlER F (LINE A B) (LINE D E)) (MIDPOINT F D E)) E
By Proposition 3.83,
BE 2 (BC) . PBCB - PADA Since PADA=PACA· (~g)2
BE BC
= o.
(1)
= PBCB . (~g)2, {l)becomes AD AC
2
(=) . PBCB - PBCB · (=)
2
BE = (BC =-
AD AC
BE BC
=) . (=
AD AC
+ =)PBCB = O.
Then we have
BE BC
AD
= AC
or
BC
= - AC
which correspond to points El and E in Figure 3-36. In the first case, we have AB II DEl; the nondegenarate condition needed to construct point F is not satisfied. In the second case, the conclusion is true. Here is the proof of the example.
Chapter 3.
154
DFf..
DF
FE - -
f..~
S ABE
FE -
-
S ABE
f!.._~£~
:~SAB C
-
Q
SAB C i *
-
~SABC
si"'!1i!y
3.8
~gSAB C
Machine Proof In Plane Geometry
~ SABE
E BE
= Bc ' S ABC BE _ AD Bc --::w D S AD SABD = ABC ' ::W
1
Geometry Information Bases and Machine Proofs Based on Full-Angles
In many traditional proofs, the relationships among angles are always used directly if possible. This may be one of the main reasons that traditional geometric proofs are very short, skillful and interesting. But the angle is a concept involving the relation of orders, and is thus very difficult to fit into our machine proof system. In Section 1.10, we introduce the concept of full-angles as the basis of machine proof of geometry statements involving angles. In Subsection 3.4.2, we present a machine proof method for geometry statements involving full-angles based on the property of the tangent function of full-angles. But this approach loses some of the unique character of the traditional proofs based on angles. This section will be devoted to another approach to mechanical generation of proofs based on full-angles . The basic idea for this new approach is that we will build a geometry information base (GIB) based on the constructive description of the statement. The GIB for a statement contains some basic geometry relations about the configuration of the statement such as collinear points, parallel lines, perpendicular lines, and cyclic points, etc. In Subsections 2.5.1 (see the paragraph after Exercise 2.40) and 3.3.2, we have touched unon the idea of building some kind of geometry information bases. The purpose of building the GIB in those two cases is that the refined elimination techniques need these geometry relations. The GIB actually has much potential in the automated production of traditional proofs for geometry statements. In the following subsections, we will show that elegant proofs for many geometry theorems can be obtained by merely checking a good GIB. This section reports our initial study of this promising approach.
3.8.1
Building the Geometry Information Base
We will use Example U18 to illustrate our method. First, we will check every step in the proof of Example 1.118 to find out how to eliminate
155
3.8 Geometry lDIonnadoD Bases
points O,D and E from full-angles LIAD , AOI and LIAC,CEI. We first use rules Q7 and Q9 (on page 46) to eliminate point E from LIAC, CEI .
LIAC,CEI = LIAC, BCI + LIBC, CEI · Since BE = C E and that E is on line AB, by Q9 we have (2) LIBC,CEI = LlBE,BCI = LIBA,BCI . To eliminate the points 0 and D, we first use Q7 to divide LIAD, AOI into two parts: (3) LIAD , AOI = LIAD , ACI + LIAC,AOI. Since AD .1 BC, by Q7 we can eliminate D .
(I)
LIAD,ACI
(4)
Figure 3-37
= LIAD , BCI + LIBC,AGj = LIII + LIBC, ACI·
The next step is to eliminate 0 from LIAC, AOI. (Q9 and AO = CO) (5)LIAC, AOI = LICO,ACI (Q7) = LICO , MOI + LIMO,ACI (QI2, MB = MC and AO = BO = LIAC, ABI + L[MO, ACI LIAC,ABI + LIMO , BCI + LIBC,ACI (Q7) (Q7, MB = MC and BO = BC) = LIBC, ABI + LIII·
=
= CO)
Finally, replacing L[AD , AGj, LIAC, AOI in (3) by (4),(5) and LIBC, CEI in (1) by (2), we have the conclusion:
(6)LIAD,AOI + LIAC,CEI = LIAC, BCI + LIBA,BCI + LIII + LIBC,ACI = LIAC, ACI + LIBA , ABI + LIII + LIII
= LIOI
+ LIBC,ABI + LIII (Q7) (QI,Q3 and Q4)
Suppose that a program named QAP could prove the geometry theorem as above. Now we are going to check what kinds of information would be needed for designing the program. Obviously, the basic properties QI-Q12 (on page 46) about full-angles are necessary in each step of proof. But we will soon see that it is not enough to prove this geometry theorem by using these rules only. Much more geometric information is needed. Let us check the above proof step by step. Step (1) seems very easy. But from the point of view of mechanization, it is actua1Iy difficult to start this step. The question is why the full-angle LIAC, CEI should be divided into LIAC, BCI+LIBC, CEI but not anything else, for example, LIAC, ADI+LIAD, CEJ · Why could our program QAP foresee that the point E will bee1iminated from LIBC, CEI?
Chapter 3. MachIne ProoIln Plane Geometry
156
To make QAP work like this, we can imagine that a geometric information base (GIB) will be generated automatically before QAP proves the theorem. The program will know that L[BC, C E] = L[BA, BC] by checking Gffi. So it chooses step (1) so that point E will be eliminated in the next step. How do we generate the Gffi? Of course, the hypotheses of the proposition should be put into the Gffi first. There are four conditions in Example 1.118:
(Gl) The circumcenter of triangle ABC is O . (OA
= OB = ~C)
(G2) AD is an altitude of triangle ABC. (AD .L BC and DE BC) (G3) M is the midpoint of BC. (BM
= MC and ME BC)
(G4) E is the intersection of AB and MO. (E E AB and E E MO) But here (GI)-(G4) are only a small part of the Gffi. We must put more information into the Gffi. To see what is still needed in the Gffi, we check step (2). In this step, the rules on full-angles are not sufficient. We need the hypotheses of the proposition (Le., conditions (Gl)-(G4». Furthermore, we need some geometric facts which are derived by applying some geometry knowledge to the hypotheses of the statement, (such as deriving BE = CE from M B = MC,OB = OC and E is on the line OM). So we also need a geometry knowledge base (GKB) to build the Gffi. As an example, in order to obtain BE = C E , our GKB should include the following propositions:
(a) If PB = PC and QB = QC then PQ is the perpendicular bisector of BC. (ndg condition: B '" C and P '" Q). (b) If P is on the perpendicular bisector of BC, then BP
= CPo
= PB then L[PC, BC] = L[BC, PB]. If P is on line AB, then L[PB, XYI = L[AB, XYI.
(c) If PC
(d)
Applying (a) to Gl and G3, QAP obtains a new information and puts it into the Gffi:
(G5)
MO is the perpendicular bisector of BC. Applying (b) to G4 and G5, we have
(G6)
BE
= CEo
Applying (c) to G6, we have
(G7)
L[BC,CE]
= L[BE, BCI .
Finally, applying (d) to G4 and G7, we have
157
3.8 Geometry InfonnadoD Bases
(G8)
LIBE,BC]
= LIAB,BC].
GS is the deductive basis of step (2). For step (3), QAP has to foresee that points D and 0 can be eliminated as in steps (4) and (5) and as a consequence, it decides to split the full-angle LIAD, AO] into LIAD, AC] and LIAC, AO]. The following additional geometry knowledge should be included in GKB: (e) If PQ is perpendicular to UV, then LIPQ,XY]
= L11] + LIUV,XY] .
(f) (Another fonn of Q12) If 0 is the circumcenter of triangle ABC, then LIAC, AO] LIBC, AB] + LI1] .
=
Proof of (f). Let D be the intersection of AO and the circle. By the inscribed angle theorem, LIAB , BC] = LIAD,CD] . Since AC.lCD, we have LIAC,AO] = L11] + LIDC,AD] = L11] + LIAB,BC]. Applying (e) to G2 and (0 to GI, QAP will put the following infonnation into GIB:
(G9) (GIO)
LIAD , AC]
= L11] + LIBC,AC] .
LIAC,AO]
= LIBC,AB] + LI1] .
When QAP finds the infonnation GI-GlO, step (6) will be done according to rules of full-angle (QI-QI2) easily. Here we mentioned only the infonnation GI-GIO which are useful for proving the statement. In fact, much more infonnation about this statement will be put into the GIB, because QAP does not know what infonnation will be used when it generates the GIB. It will keep applying every rule in the GKB to all the infonnation in GIB to get new infonnation and to put the new infonnation into GIB until nothing new can be obtained. What geometry knowledge should be included in the GKB? Is it complete? At the present stage, the choices of the rules in the GKB are based on our experience of proving geometry theorems. Its completeness is still not considered. But if the method in this section fails to prove or disprove a statement, we can always use Algorithm 3.33 which is complete for constructive geometry statements. Thus, the GIB-GKB method is actually an expert system of proving geometry statements. Up to now, the following rules have been put into our GKB :
KI Two points A and B detennine one line. (ndg condition: A", B) K2 Three points A, Band C detennine one circle. (ndg condition: SABe '" 0) K3 L[PQ,XY]
= L[UV,XY] if and only if L[PQ,UV] = L[O]. (ndgcondition:
X'" Y)
Chapter J.
158
Machine Proof in PlaDe Geometry
K4 Four points A, B, C and D are cyclic if and only if L[AC, BC] condition: A, B, C and D are not collinear)
= L[AD, BD].
= AC if and only if L[AB,BC] = L[BC,AC]. (ndgcondition: L[AB,XY] + L[XY,UV] = L[AB,UV]. (ndgcondition: Xi- Y)
K5 AB K6
SABC
(ndg
i- 0)
K7 AB .l BC if and only if AC is the diameter of the circumcircle of triangle ABC. (ndg condition: S ABC i- 0) K8 AB is the perpendicular bisector of XY if and only if AX (ndg condition: A i- B and X i- Y)
= AY and BX = BY.
K9 If point 0 is the circumcenter of triangle ABC then L[OA, AB] (ndg condition: SABC i- 0)
= L[l] + L[AC, BC]
Suppose that for a geometry statement, the am has been generated by using OKB. The next step is how to generate a proof based on the am. The key idea is still to eliminate points introduced by constructions of the given statement. But here the eliminating rules are based mainly on the am rather than construction. Following are the rules for eliminating point X from the full-angle L[AB, PX]:
= L[AB, PQ]. If PX is parallel to UV, then L[AB,PX] = L[AB,UV].
QEl If X is on line PQ, then L[AB, P X] QE2
QE3 If P X is perpendicular to UV, then L[AB, P X]
= L[l] + L[AB, UV].
QE4 If X is on line UV and U, P, Q and X are cyclic, then L[AB, PX] L[UQ,PQ]. (because L[AB,PX] = L[AB,UV] L[UQ,PQ].)
+ L[UV,PX],L[UV,PX]
= L[AB, UV] +
= L[UX,PX] =
QE5 If X is on line UV and PX = PU then L[AB,PX] = L[AB,UV] (because L[PU,UV] = L[PU,UX] = L[UX,PX] = L[UV,PX].)
+ L[PU,UV].
QE6 If X is on line UV and PU is the perpendicular bisectorofQX, then L[AB,PX] = L[AB,UV] + L[PQ,QU]. (because L[PQ,QU] = L[UX,PX] = L[UV,PX].) QE7 If X is the circumcenter of triangle PQU, then L[AB, P X] L[UQ,UP] + L[l]. (by K9, L[PQ,PX] = L[l] + L[UQ,UP]) QE8 If L[UV,PX]
=
L[AB, PQI
= L[!] is known, then L[AB,PX] = L[AB,UV] + L[!].
Here we assume that points A, B, P, Q, U and V are introduced before point X. Exercise 3.104 Prove propositions QE1-QE8 based on Ql-QI2.
+
159
3.8 c-try lnformadOD Bases
3.S.2
Machine Proof Based on the Geometry Information Base
We will use the following new version of Example 1.118 to illustrate how the program works. Example 3.105 The circumcenter of triangle ABC is O. AD is the altitude on side BC. Show that L[AO, DA] = L[BA, Be] - L[BC, CAl . The constructive description is «POOO'S 8 C A)
(FOOT D A 8 C) (MIDPOINT M 8 C) (MIDPOINT N A 8)
(INreR E (LINE A 8) (PUNE MAD»
Figure 3-38
(INreR 0 (LINE M E) (lUNE N N 8»)
The conclusion is L[AO, DA] + L[BC, CAl alent to L[AO, DA] = L[BA, BC]- L[BC, CAl.
+ L[BC, BA]
= L[O] which is equiv-
The Gm for this example contains several groups of geometry relations which will be explained separately below. (11) p-list: includes all the points in the statement. listed in the introducing order. i.e.• p-list=(BC ADM N EO) . (12) free-points: includes all the free points in the statement. i.e .• free-points= (B C A) (13) all-lines: Gm can list all 13 the lines in the statement.
«0 N) (0 E M)(E N A B) (D A)(M D B C)) which means that points 0, E and M are collinear. etc. (14) p-lines: contains all the parallel lines: DA.
«0 E M) (D A)) which means that OEM 1\
(15) t-lines: contains all the perpendicular lines:
«0 N) (E NAB)) «(0 E M) (D A)) (M D B C)) which means that line ON is perpendicular to line EN AB; and lines OEM and DA are both perpendicular to line M DBC. \3 As we mentioned before. these lines are diose which can be obtained directly from the hypotheses of the geometry statement
Chapter 3. MachIne Proof 10 Plane Gametry
160
(16) circles: contains all the circles in the statement; «E 0) N EO) «A 0) N A 0) «B 0) N M B 0) «D 0) M DO) «C 0) M C 0) «D E) M DE) «B E) M BE) «C E) M C E) «A M) DAM) «B A) DBA (N)) «C A) DCA) «B C) B C (M)) (B C (E)) (C B A (0))
which means EO is a diameter of the circumcircle of triangle N EO; B A is a diameter and N is the center of circumcircle of triangle DBA; BC is a diameter and M is the center of the circle through points B and C; E is the center of the circle through points B and C; and o is the circumcenter of triangle C BA.
Using this information, QAP gives the following proof. L[BC,BA] + L[BC,CA] + L[AO,AD] = L[BC, BA] + L[BC, CAl + L[l] + L[BA, BC] + L[CA, AD] (By Q7, L[AO, ADJ = L[AO, CAJ + L[CA, ADJ . Since 0 is the circumcenter of ABC, L[AO, CAl L[lJ + L[BA, BC].)
=
= L[OJ + L[BC, CAJ + L[lJ + L[CA, BCJ + L[lJ (L[BC, BAJ + L[BA, BCJ = L[BC, BCJ = o.
Since AD.i BC, L[CA,ADJ = L[CA,BCJ + L[BC, ADJ = L[CA,BC] (L[lJ + L[lJ = L[OJ and L[BC, CAJ + L[CA, BCJ = L[OJ.)
+ L[lJ.)
= L[OJ.
In what follows, we will use more examples to illustrate the Gffi-GKB method. The examples in Section 1.10 were all produced according to the above method by our program.
Example 3.106 (Simson's Theorem) The same as Example 3.79 on page 141.
Constructive description «CIRCLE ABC D) (FOOT E D B C) (FOOT FDA C) (FOOTG D A B)
The conclusion: L[EF,FG]=L[O]. Figure 3-39
161
3.8 <>-try lnIonnadOD IIasa
Proof The machine proof L[EF,GFJ
= L[EF,DFJ + L[DF,GFJ = L[EC, DCJ + L[DA, GAJ
(Q7) (QlO (D, C, E, F; A, D, G, F cyclic.» L[BC,DCJ + L[DA,BAJ (Q8 (E E BC; G E AB». = L[BA, DAJ + L[DA, BAJ (QIO (A, B, C, D cyclic.»
=
= L[BA, BAJ = L[OJ.
There is a traditional proof which proves the theorem by showing that LEFC = LGF A. This proof is not strict: the fact that points E and G are in different sides of AC is used but not proved. Before presenting the next example, we introduce a new geometry object: (CIRC A B C) which stands for the circle passing through points A, B, and C .
Example 3.107 (Miquel Point) Four lines form four triangles. Show that the circumcircles of the four triangles passes through a comnwn point. Constructive description ( (POOO- A D E Q)
(INTER 8 (UNE D E) (CIRC A Q E» (INTER C (UNE A E) (CIRC D Q E» (INTER P (UNE A 8) (UNE C D»
L(Qc,c P)+L(AP,AQ)= L[o)
The machine proof
L[QC,CPJ + L[AP,AQJ = L[QC, DCJ + L[AB, AQJ (because C P II DC and AP II AB.) =L[EQ,EDJ + L[EB,EQJ = L[EB,EDJ
Figure J.4O
= L[OJ .
(because E, Q, D, C and A, B, E, Q are cyclic points respectively.)
Example 3.108 In a circle. the lines joining the midpoints of two arcs AB and AC meet line AB and AC at D and E . Show that AD = AE. Constructive description ( (POINI'S A M N)
(CIRCUMCem!R 0 A M N)
(POOT P A 0 N) (POOTQ A 0 M) (INTER D (UNE N M) (UNE A Q» (INTER E (UNE N M) (UNE A P» L[AD,DE)+L(AE,ED(=L (O(
FigUR: 3-41
162
Chapter 3.
MachIne Proof In Plane Geometry
The machine proof L[AE,DE]
+ L[AD, DE]
=L[AP,MN] + L[AQ,MN] (becauseAE II AP,DE II MN,andAD II AQ.) = L[AP,MN] + L[l] + L[MO,MN] (because AQl.MO.) = L[l] + L[NO,MN] + L[l] + L[l] + L[AM, AN] (APl.NO. L[MO,MN] = L[l] + L[AM, AN], because 0 is the circumcenter of 6.AMN) = L[l] + L[l] + L[AN, AM] + L[AM, AN] (since 0 is the circumcenter of 6.AM N, L[NO, M N] = L[l] + L[AN, AM].) =0
Example 3.109 From the midpointC ofarc AB ofacircle. two secants are drawn meeting line AB at F. G. and the circle at D and E. Show that F. D. E. and G are on the same circle. Constructive description ( (CIRCLE A
C D E)
(CIRCUMCENTER 0 A
C D)
(FOOT MAO C)
C»
(INTER F (LINE A M) (LINE D (INTER G (LINE A M) (LINE C E»
L(CE,FG]+L(CD,DE]=L(o]
The Machine proof L[CE,FG]
+ L[CD, DE] + L[AC,AE]
Figure 3-42
= L[CE,AM]
(FG II AM; since A,C,D, and E are cyclic, L[CD, DE] = L[AC,AE].) = L[l] + L[CE,CO] + L[AC,AE] (since AMl.CO, L[CE, AM] = L[l] + L[CE, CO].) = L[l] + L[l] + L[AE,AC] + L[AC,AE] (since A, C, D, and E are on the circle with center 0, L[CE, CO] L[AE, AC].) =0
=
L[l]
+
Example 3.110 Let Q. SandY be three collinear points and (0, P) be a circle. Circles SPQ andY PQ meet circle (0, P) again at points Rand X. respectively. Show that XY and RS meet on the circle (0, P).
3.11 Geometry InfonnadOD a..es
163
Constructive description «CIRCLE R P
Q S) (POINT X) Q S)(ClRC P Q X»
(IHTER Y (UNE
(IHTER I (UNE X Y) (UNE R P»
L[x I ,R/]+L[RP,X PI=L[OI
The machine proof
L[PR,PX]
+ L[IX, RI]
=L[PR, PX] + L[XY,RS] (because X I II XY and RI II
RS.)
= L[PR,PX] + L[XY,QS] + L[QS,RS] = L[PR,PX] + L[XP,QP] + L[QS,RS] (because X, P, Y, and Q are cyclic. and QS (because X, P, Y, and Q are cyclic, and I X
Figure 3-43
II QY.) II Y I .)
=L[PR,PQ] + L[QS, RS] = L[O]. (Points R, Q, P, and S are cyclic.) Example 3.111 Let ABC be a triangle. Show that the six feet obtained by drawing perpendiculars through the foot of each altitude upon the other two sides are co-circle.
c Constructive description ( (POINTS A B C) (FOOT F C A B) (FOOT DAB C) (FOOT E B A C) (FOOTG F B C) (FOOT I D A B) (FOOT H F A C) (FOOT K E A B)
L[GH,GI]+LIAK,HKI=L[OI
Figure 3-44
The machine proof
L[GH, GI] + L[AK, H K] = L[GH,GI] + L[AB,FI] + L[FE,EH] (because AK II AB. and K , H, F, and E are cyclic.)
= L[FE,EH] + L[GH,GI] (because AB II Fl .) = L[FE,AC] + L[CE,GI] + L[FG,CF] (because EH II AC. and H, G, C, and F are cyclic.)
=L[FE,AC] + L[CE,BF] + L[FD,DG] + L[FG,CF] (because I, G , F, and D are cyclic.)
Chapter 3. Machine Proof In Plane Geometry
164
=L[FG,CF) + L[FE,AC) (because A, D, C, and F are cyclic.) = L[AD,CF) + L[FE,AC) (because FG II AD.) L[AD,CF) + L[BF,BC) (because E, F, C, and B are cyclic.)
=
= L[l) + L[BC,CF) + L[BF,BC) (because AD.1.BC.)
= L[l) + L[BF, CF) = L[O) (because BF .1.CF.) Example 3.112 The nine-point circle cuts the sides of the triangle at angles IC - AI. and IA - BI· Constructive description «POINTS A B C) (FOOT F C A B) (MIDPOINT M B C) (MIDPOINT Q A C) (MIDPOINT P B A) (MIDPOINT L F P) (MIDPOINT S Q P) {INTER N (lUNE L L P) (lUNE SSP))
A
L(BC,AB)+L(AC,AB)+L(FN,LNj=L(o) Figure 3-45
The machine proof L[FN,LN) + L[AC,AB) + L[BC,AB) = L[l) + L[FN,PF) + L[AC,AB] + L[BC,AB] (because LN II CF and CF .1.PF.) = L[l] + L[l) + L[FQ,PQ] + L[AC,AB] + L[BC,AB] (Points F, Q, P are on the circle with center N.)
= L[FQ,CF] + L[CF,PQ) + L[AC,AB] + L[BC,AB] =L[l] + L[FP,QP] + L[FQ,CF] + L[AC,AB] + L[BC,AB] (because LN II CF and CF .1.PF.) =L[l] + L[MQ,BC] + L[FQ,CF] + L[AC,AB] + L[BC,AB] =L[l] + L[AC,AB] + L[FQ,CF] (because F P II MQ.QP II BC, and MQ II AB.) = L[l] + L[AC,AB) + L[l] + L[CF,AC] + L[AF,CF] (because F, A, C are on the circle with center Q.) = L[AC,AB] + L[AF,AC]
IB - q,
165
=L[AC. AB] + L[AB . ACJ = L[O]
Summary of Chapter 3 • We have the following fonnulas for the areas of triangles.
t.
SABC
= !IBClh A = !IACJh B = !IABlhc .
2.
SABC
= Souv
xA
YA
XB
YB
1 1
Xc Yc 1 where XA. YA. XB. YB. Xc. and Yc are the area coordinates of points A. B. and C with respect to OUV. 2 2 3. 16~Bc = 4AB2~ - (AC + AB2 - BC )2 = 4A'1f~ - P'iJAC'
• The following basic propositions and the ones on page 99 are the basis of the area method. 1. ABl..CD iff P ACD
= P BCD or P ACBD = O.
2. Let R be a point on line PQ with position ratios rl to PQ. Then P RAB P ARB
= =
r1PQAB r1PAQB
+ r2 P PAB + r2PAPB -
= ~~. r2 = ~ with respect rlr2 P pQP.
3. Let D be the foot of the perpendicular drawn from point P upon a line AB. Then we have
4. Let AB and PQ be two nonperpendicular lines and Y be the intersection of line PQ and the line passing through A and perpendicular to AB. Then PY
PPAB
QY
P QAB
==--,
PY
PPAB
QY
P QAB
= = - -, = = - -. PQ
PPAQB PQ
PPAQB
5. Let ABCD be a parallelogram. Then for any points P and Q. we have PAPQ PPAQ
+ P CPQ = + PPCQ =
P BPQ PPBQ
+ PDPQ or PAPBQ = PDPCQ + PPDQ + 2PBAD .
6. Let ABC D be a parallelogram and P be any point. Then PPAB
=
PPDC - PADC
= PPDAC
PAPB
=
PAPA - PPDAC·
166
Chapter 3.
Machine Proof in Plane C-try
• A constructive configuration is a figure which can be drawn using a ruler and a pair of compasses only. In other words, a constructive configuration can be obtained by first taking some arbitrary points, lines, and circles in the plane, and then taking the intersections of these lines and circles in a prescribed way. A constructive geometry statement is an assertion about a constructive configuration, and this assertion can be represented by a polynomial equation of three geometry quantities: the ratios of parallel line segments, the signed areas of triangles or quadrilaterals, and the Pythagoras differences. The set of all constructive statements is denoted by C. We also introduced a subclass of C, i.e., the class of the linear constructive geometry statements, which is denoted by C L • • A mechanical theorem proving method for class C was presented. The key idea of the method is to eliminate points from geometry quantities. The method can be used to produce short and readable proofs for geometry statements efficiently. • We report some initial results on how to use the geometry information base to generate readable proofs for geometry statements using full-angles.
Chapter
4
Machine Proof in Solid Geometry This chapter deals with the machine proof in solid geometry. Similar to plane geometry, we will consider those geometry statements that can be described constructively using lines, planes, circles, and spheres. In the first three sections, we deal with geometry statements involving collinear and parallel of lines and planes. More precisely, we deal with constructive statements in affine geometry of dimension three. Starting from Section 4, the Pythagoras difference is employed to deal with constructive statements involving perpendicular lines, circles, and spheres.
4.1
The Signed Volume
As before, we denote by AB the signed length of the oriented segment from A to B; we denote by S ABC the signed area of the oriented triangle ABC.
In solid geometry, we have a new basic fourfold relation among points, coplanar, which will be characterized by Axioms S.l-S.5 about signed volumes. We assume that Axioms A.I-A.6 (on page 55) are still valid, provided that all the points involved are coplanar. The signed areas of coplanar triangles can be compared, added, or subtracted. For instance, if A, 0, U, and V are four coplanar points, by Axiom A.5 (4.1)
SOUV
= SOUA + SOAV + SAUV·
A tetrahedron ABC D has two possible orientations. We use the order of its vertices to represent its orientation. If we interchange two neighbor vertices, the orientation of the tetrahedron will be changed. The signed volume V ABCD of an oriented tetrahedron ABCD is a real 14 number which satisfies the following properties. Axiom S.l When two neighbor vertices of an oriented tetrahedron are interchanged. the signed volume of the tetrahedron will change signs. e.g.• V ABCD = -VABDC . 14 Here.
we can use any Dumber field and the results in this chapter are still valid.
167
168
Chapter 4. Madtioe Proof 10 Solid Geometry
Axiom S.2lf A, B, C, and D are four non-coplanar points, we have VABC D Axiom S.3 There exist at least four points A, B, C, and D such that VABCD
#- O. #- O.
Axiom S.4 For five points A, B, C, D, and 0 (Figure 4-1), we have V ABCD
= V ABCO + VABO D + VAOCD + VOBCD ·
Axioms S.3 and S.4 are called dimension axioms. They ensure that we are dealing with a proper three dimensional space: Axiom S.3 says that there are at least four non-coplanar points; Axiom S.4 says that all points must be in the same three dimensional space.
c
oJC:..--------'4 Figure 4-1
Figure 4-2
Axiom S.5lf A, B, C, and D are four coplanar triangles and SABC pointTwe have V TABC = )..VTABD . (Figure 4-2)
= )"SABD thenforany
We extend the coplanar to be a geometry relation among any set of points: a set containing fewer than four points is always coplanar, and a set of points is coplanar if any four points in it are coplanar. We thus can introduce a new geometry object, the plane, which is a maximal set of coplanar points. Proposition 4.1 Four points A, B, C, and D are coplanar ijfVABCD
= O.
Proof If V ABCD = 0, by Axiom S.2 A, B, C, and D are coplanar. Let us assume A, B, C, and D to be coplanar points. If A, B , C, and D are collinear, we have SABC = O. Let X be a point not on line AB. Then by Axiom S.5, V ABCD = ~ss V = O. If A, B, and C ABX ABXD are not collinear, we have VABC D = ~s V AABC = O. I ABC Corollary 4.2 For three non·collinear points A, B, and C, the set of all the points D satisfying V ABC D = 0 is a plane and is denoted by plane ABC. Proof Let P, Q, R , and S be four points in plane ABC. We need to show that V PQRS = O. We first show that two of the points, say P and Q, are coplanar with A, B, C . By Axiom S.5, V ABPQ
=
SABP - S V ABCQ ABC
= 0,
169
4.1 The Siped Volume
i.e., A , B , P, and Q are coplanar. Similarly, we can show that three of P, Q, R, and S are coplanar with A , B , C, and finally P, Q, R, and S are coplanar. In what follows, when speaking about a plane ABC, we always assume that A , B , and C are not collinear. Similarly, when speaking about a line AB, we assume Ai-B.
4.1.1
Co-face Theorem
In this and the next subsections, we will derive some basic properties about volumes which will serve as the basis of the volume method. First, Axiom S4 can be written in the following convenient way.
Proposition 4.3 (The Co-vertex Theorem) Let ABC and DEF be two proper triangles in the same plane and T be a point not in the plane. Then we have ~ = Ua.a. s DEF . rTDEF Proof By Axiom S.S,
VTABC VTDEF
= VTABC VTABF VTAEF = SABC SABF SAEF = SABC VTABF VTAEF VTDEF
SABF SAEF SDEF
SDEF·
Before proving the co-face theorem, we need to define the signed volume of a special polyhedron with five vertices. p
B A
A
Q
c
c
Figure 4-3
The polyhedron fonned by five points in space is complicated. By PABCQ, we denote the one with faces P AB, P BC, PAC, QAB, QBC, and QAC. Figure 4-3 shows that several possible shapes of P ABCQ. The volume of P ABCQ is defined to be
(4.2) By Axiom S4, we have
(4.3)
170
Chapter 4. Machine Proof in Solid Geometry
Proposition 4.4 (The Co-face Theorem) A line PQ and a plane AEC meet at M. If Q i= M, we have
PM QM
PM -=PQ
V PABC VQABC
==--;
c
V
PABC = VPABCQ ;
QM PQ
VQABC
= VPABCQ '
Figure 4-4
Proof Figure 4-4 shows that several possible configurations of this proposition. Take points A' and E' such that MA' = CA, ME' = CE. Then SABC = SA'B'M. By Propositions 4.3 and 2.8, we have, J:::e.a..a.c. = ~ = ~ = !:.!:!.. Other equations are consequences of VQABC VQA'B'M SQB'M QM the first one. I v " ) is The above proof is a dimension reduction process. A quantity in the space (!..e..4!.B!..ILv.
QA'B'M
reduced to a quantity in the plane (~s ' ) which is further reduced to a quantity in a line QB'M
(~~). p
Proposition 4.5 Let R be a point on a line PQ and AEC be a triangle. Then we have
Proof By Proposition 4.4,
VpRBC V PQBC
Figure 4-5
-- =
PR VPARC PQ V PAQC
=, - -
PR VPABR == ,-- = PQ V PABQ
PR PQ
=.
By AxiomS4, VRABC
= =
VPABC - V pRBC - V PARC - V PABR PR VPABC - PQ(VPQBC + VPAQC + VPABQ )
=
PR V pABC - PQ VPABCQ
4.1
171
The Siped Volume
Proposition 4.6 Let R be a point in the plane PQS. Then for three points A , B , and C we have VRABC
=
SPQR - S VSABC PQS
SIUJS + -PQS S VPABC
SPRS + -PQS S VQABC .
Proof For any point X.let Vx = VXABC . Without loss of generality. let M be the intersection of PR and QS. By Proposition 4.5. PR RM V R = PM VM + PM VP
PR QM
= PM(QSVs +
L~l /A~c7 Figure 4-6
MS RM QSVQ)+ PMVP .
(1)
By the co-side theorem !!J::!. = SRqS ~ = SPqR ~ = hBS... J:.!l. = SpqRS Substi• PM SPQS' QS SPQRS' QS SPQRS' PM SPQs · tuting these into (I). we obtain the result. I
4.1.2
Volumes and Parallels
Two planes or a line and a plane. are said to be parallel if they have no point in common. Two lines are said to be parallel if they are in the same plane and do not have a common point. By the notation PQ II ABC. we mean that A, B, C, P, and Q satisfy one of the following conditions: (I) P Q. (2) A , B, and C are collinear. or (3) A. B. C. p . and Q are on the same plane. or (4) line PQ and plane ABC are parallel. According to the above definition. if PQ It' ABC then line PQ and plane ABC have a nonnal intersection. For six points A , B , C,P, Q, and R.ABC II PQRiff AB II PQR.BC II PQR.and AC II PQR .
=
Proposition 4.7 PQ
II ABC ijJVPABC =
VQABC or equivalently V PABCQ = O.
Proof If V PABC =I VQABC • then P =I Q and A , B , and C are not collinear. Let 0 be a VqABC . By Proposition 4.5• point on line PQ such that !:!l PQ = .Xu.a.a.... VPABCQ Thus ~ PQ = - VPABCQ VOABC = ~gVQABC + ~VPABC = o. By Axiom S2. point 0 is also in plane ABC. i.e .• line PQ is not parallel to ABC. Conversely. if PQ It' ABC then P =I Q and A , B, and C are not collinear. Let 0 be the intersection of PQ and ABC. By Proposition 4.4. !2f = QABC = 1. Thus P = Q. which is a contradiction. OQ fuBav.
172
Chapter 4. Machine Proof In Solid Geometry
Proposition 4.8
PQR
II ABC ijfVPABC = VQABC = VRABC.
Proof. By Proposition 4.7, V PABC = V QABC = VRABC iff lines PQ and PR are parallel to plane ABC. We must show that for any point D in plane PQ R, line P D is also parallel to ABC. By Proposition 4.6, VDABC
i.e., PD
II
=
SPQD - S VRABC PQR
=
SPQD VPABc( - S PQR
SPDR
SDQR
+ - S VQABC + - S V PABC PQR
+
SPDR -S PQR
PQR
+
SDQR - S) PQR
= V PABC
ABC.
A figure P 1P 2 . . . P n is said to be a translation of QIQ2 . .. Qn if P;PH1 = QiQi+!. Let triangle XY Z be a translation of triangle ABC. Then for any points P, Q, and R in plane XY Z , we define SPQR _
SPQR
SABC -
SXyz·
For convenience, we use the symbol SPQR
= >.
or
SABC
to denote the fact that plane PQR is the same as or parallel to plane ABC, and ratio of the signed areas SPQR and SABC.
>. is the
The following propositions about translations of line segments and triangles are often used in the machine proof method, in order to add auxiliary translations of line segments and triangles.
Proposition 4.9 Let PQT S be a parallelogram. Then for points A, B, and C, we have V PABC
+ V TABC = V QABC + V SABC or VPABCQ = VSABCT.
Proof. This is a consequence of Proposition 4.5, because both sides of the equation are equal to 2VOABC where 0 is the intersection of PT and SQ . I
Proposition 4.10 Let triangle ABC be a translation of triangle DEF. Thenfor any point P we have V PABC = VPDEFA. Proof. By Proposition 4.9 and (4.3), V PADC
=
V PAEF - V PAED - V PADF
V PABC
=
=
V PAEC -
VPDEFA.
V PADC
= V PAEF -
VPAED -
I
Corollary 4.11 1. For two parallel planes ABC and PQ R and a point T not in ABC we have~=~. SPQR
VTPQRA
4.1
173
The Signed Volume
2. For two different parallel planes ABC and PQ R we have &a.c. s PQR
= - ~y . APQR
Proof Let XYZ be a translation of RPQ to plane ABC. By the co-vertex theorem and the preceding proposition,
S ABC SPQR
S ABC
V T ABC
V T ABC
= SXYZ = V TXYZ = V TPQRA ·
Replacing T by P in the above equation, we prove the second result. Proposition 4.12 Let triangle ABC be a translation of triangle DEF. Thenfor two points P and Q we have V PABC
+ V QDEF
=
V QABC
+ V PDEF or VPABCQ = VPDEFQ.
In other words. when ABC moves by a translation the volume of P ABCQ remains the same. Proof By Proposition 4.10, V PABC = V PDEF which we obtain the result immediately.
-
V ADEF ; VQABC
= VQDEF -
V ADEF
from
I
From Propositions 4.9 and 4.12, we have the following interesting property for the volume VPABCQ . <;:orollary 4.13 Let PQ
4.1.3
= rST and SABC = SSEFG then VPABCQ = rsVSEFGT.
Volumes and Affine Geometry of Dimension Three
This subsection has two purposes. First, we will show how to prove geometry theorems using the basic propositions about volumes. Second, we will derive some basic properties of lines and planes in space using the volume method.
Example 4.14 If points P and Q are in plane ABC then line PQ is also in plane ABC. Proof For any point Ron line PQ,by Proposition 4.5, VRABC = ~~VQABC +~VPABC O. By Proposition 4.1, R is in ABC. I
=
Example 4.15 If two planes have a point in common then they have a line in common. Proof Let planes ABC and RPQ have a point X in common. Without loss of generality, we assume that A , B , C, R, P, and Q are not common to both planes. By Proposition 4.8, two of AB, BC, and AC, say AB and BC could not be parallel to RPQ. Let AB and AC meet RPQ in two distinct points Y and Z respectively. We must show that X, Y, and Z are collinear, i.e., SXYZ O. By Propositions 4.3 and 4.1 , &x..z.. 0, i.e., s ABC = Yax=y RABC
=
SXYZ
= O.
=
I
174
Chapter 4. Machine Proof In Solid Geometry
Remark 4.16 The above two examples and Corollary 4.2 are the incidence axioms in th£ usual axiom system for solid geometry [6]. Therefore. the geometry defined by Axioms A.I·A.6 and S.I-S.5 is an affine geometry of dimension three. The results related to affine plane geometry proved in Section 2.6 are also true for affine geometry of dimension three. So the volume method presented in Section 4.3 below is for constructive statements in affine geometry of dimension three associated with any field. We will now prove some basic properties for the parallel. Example 4.17 Through any point P not in plane ABC there is one and only one plane parallel to plane ABC. Proof. By the Euclidean parallel axiom (Example 2.13 on page 58), there exist points Q and R such that P ABQ and PAC R are parallelograms. By Propositions 4.7 and 4.8, PQR II ABC. To prove the uniqueness, let PTS be another plane parallel to ABC. By Proposition 4.10, V TPQR = V TABC - VPABc = 0, i.e., T E PAR. Similarly we have
S E PAR.
I
Example 4.18 If any two lines are cut by a number ofparallel planes. their intercepts are proportional. Proof. Let the parallel planes (t, /3, 1 cut two lines in the sets of points A, B, C and P, Q, R respectively. Let X, Y, and Z be three noncollinear points in plane /3. By the co-face theorem,
AB CB
VAXYZ
PQ
VPXYZ
= VCXYZ = VRXYZ = RQ '
Figure 4-7
Example 4.19 If a straight line is parallel to a straight line in a plane. it is in or parallel to the plane. Conversely, if a line in one plane is parallel to another plane. it is parallel to their line of intersection. Proof. Let AB be parallel to CD and E another point in plane CDE. By Propositions 4.3 and 2.10, F V EACD V EBCD
=
SACD
= 1
SBCD
'
i.e., AB II CDE. Conversely, let AB be parallel to plane CD E and CD be the intersection of the two .. 43 l 'I.e., 1 By Proposillon panes. . , §..w.u. S = Yu=. v =, AB II CD . BC~ EBCD
A
Figure 4-8
4.1
175
The Signed VoIUJne
Example 4.20 If two lines are each parallel to a third. they are parallel to one another. Proof Let ABC D and ABEF be two parallelograms. We first prove that C. D , E , and F are coplanar. By Proposition 4.9, VCDEF = VBDEF VADEF = VBDAF - VADBF = 0, i.e., C, D , E , and F .. 43 ~ are cop Ianar. By PropoSltlOn ., ~ SD E F = V.WEF = I , i.e., CD II EF. I
I!
o Figure 4-9
Example 4.21 If a plane cuts two parallel planes. the lines of intersection are parallel. Proof Let plane ABPQ cut planes ABC and RPQ at lines AB and PQ respectively. By Proposition 4.3, S "PQ = V" PQR = 1 i.e. AB II PQ. I S8PQ V8PQR ' ,
c
Figure 4-10
Example 4.22 (Co-trihedral Theorem) ~_ow ou OV VD"8 C VA . DB . oc'
If OW II DA. OU II DB. and OV II DC then
Proof Let R, P, Q be points such that DR = OW, DP = OU, DQ = OV. By the co-face theorem,
DA DA DA DB DC VDABC = DR VDRBC = OW VDRBC = OW . OU . OV VDRPQ . By Propositions 4.9 and 4.10,
VDRPQ Since RW
II
PU
= VORPQ -
II QV. VRWUV =
VWRPQ
= Vowuv -
VRWUV - VWRPQ .
VRWPV = VRWPQ which proves the result.
The above examples are about the basic properties of lines and planes. In what follows, we will prove some relatively non-trivial theorems. The proofs of these theorems are actually modifications of the proofs produced by our program.
c
c
c B
Figure 4-11
c. Figure 4-12
Figure 4-13
Chapter 4. Machlne Proof In SolId <>-nelry
176
Example 4.23 (Menelaus' Theorem for Skew Quadrilaterals) If the sides AB, BG, G D, and D A of any skew quadrilateral are cut by a plane XY Z in the points E, F, G, and H respectively, then ;~ . ~~ . g~ ~~ = 1. (Figure 4-11)
.
Proof By the co-face theorem DH
V
AH
V AXYZ ' DG
DXYZ ==---
Then it is clear that ~ . EB example. see Section 4.2
GG
V CXYZ
BF
V BXYZ AE
V AXYZ
VCXYZ
V BXYZ
==---,==---,==---. V DXYZ GF
BF . ~ . ~ FC GD HA
= 1.
BE
For the non-degenerate conditions of this I
Example 4.24 Let AIBl G 1 be the parallel projection of any triangle ABG in any plane. Show that the tetrahedra ABGA 1 and A 1 B 1G 1A are equal in volume. (Figure 4-12) Proof Since GG1 is parallel to plane Similarly. VAA,B,C = VAA,BC.
AAIBb
by Proposition 4.7. I
VAA,B,C,
= VAA,B,C.
Example 4.25 If a plane divides proportionally one pair of opposite sides of a skew quadrilateral. it also divides proportionally the other two sides. (Figure 4-13) Proof This statement is a direct consequence of Example 4.23. The following is a proof produced by our program. 15 Let Tl = ;~ = ~~ and T2 = ~~. We need to show that BG = T . By the co-face theorem !2Q = VRI>fH • By Proposition 4.5 BC 2 • BC VBEFH-VCEFH
= (T2 - I)VACEF = h -1)h -1)VACDE = (T2 -1)(Tl -1)TI V ABCD; V BEFH = T2 V BDEF = T2 T I V BCDE = T2 T l(Tl -1)VABcD . V CEFH
Then flSi. = BC
4.2
r2 r ,(r,-1) r2r,(r,-1)-r,(r,-1)(r2-1)
=
T
2·
Constructive Geometry Statements
In this section. we will introduce a class of constructive geometry statements which is a generalization of the constructive statements in plane geometry. 15This may serve as an example to show the difference between people and the computer in proving theorems. People may use all results available to make the proof short. But the computer does ooly prescribed steps according to the input.
4.1
177
Coastruc:tive Geometry Statemenls
4.2.1
Constructive Geometry Statements
In this chapter. by a geometric quantity we mean one of the following quantities: 1. the ratio of the signed lengths of two oriented segments on one line or on two parallel lines; 2. the ratio of the signed areas of two oriented triangles in the same plane or in two parallel planes; or
3. the signed volume of an oriented tetrahedron.
In Section 4.4. we will introduce more geometry quantities. We now introduce constructions in space. First, it is clear that most plane constructions can still be used if all related points are in the same plane. We will choose several of them as basic constructions in space. Definition 4.26 A construction is one of the following ways of introducing new points in space.
81 (POINTS Y1 ,' .. , Yj). Take arbitrary points YJ,' .. ,Yj in the space. Each Y; has three degrees offreedom. 82 (PRATIO Y W U V r). Take a point Y on the line passing through W and parallel to line UV such that WY = rUV. where r can be a rational number, a rational expression in geometric quantities. or a variable.
If r is a fixed quantity. Y is a fixed point; if r is a variable. Y has one degree of freedom. The non-degenerate (ndg) condition is U =/: V.lfr is a rational expression of geometry quantities then we will further assume that the denominator of r could not equal to zero. S S 83 (ARATIO Y L M N rl r2 r3). where rl = ~s ,r2 = :u.x.JL are SLMN , and r3 = SfZ.J.JCL LMN LMN the area coordinates of point Y with respect to LM N . The ri. r2 and r3 could be rational numbers. rational expressions in geometric quantities. or indeterminates satisfying rl + r2 + r3 = 1. The degree of freedom of Y is equal to the number of indeterminates in {rJ, r2, r3}' The ndg conditions are that L, M, and N are not collinear and the denominators of rl , r2 , and r3 are not equal to zero.
84 (INTER Y (UNE U V) (UNE P Q». Point Y is the intersection of line PQ and line UV which are in the same plane. The ndg condition is PQ It' UV. Point Y is afixed point.
85 (INTER Y (UNE U V) (PLANE L M N». Point Y is the intersection of a line UV and a plane LM N . The ndg condition is that UV
It' LM N. Point Y is afixed point.
178
Chapter 4. Machine Proof ill Solid Geometry
86 (FOOT2L1NE Y P U V) Point Y is the foot from point P to line UV. The ndg condition is U i- V . Point Y is a fixed point. Proposition 4.27 Let Y be introduced by one of the six constructions SJ-S6. Slww that the existence ofY follows from Axiom A.2.
Proof Constructions S I, S2, S4, and S6 have been discussed in Proposition 3.20 on page 110. The existence of the point introduced by S3 follows from Proposition 2.30 on page 68. Let Y be introduced by S5. By the co-face theorem 'u ~. Since UV It'LMN, yULMNV uYv we have VULMNV i- O. By Axiom A.2, Y does exist. I
=
Definition 4.28 A constructive statement is a list S
= (CI , C2 , •.. , Ck , G) where
1. Each C i , introduces a new point from the points introduced by the previous constructions; and
2. G = (E I , E 2) where EI and E2 are polynomials in geometric quantities about the points introduced by the C, and EI = E2 is the conclusion of S. The non-degenerate condition of S is the set of non-degenerate conditions of the constructions C; plus the condition that the geometry quantities in EI and E2 have geometry meanings, i.e., their denominators are not zero. If the constructions are limited to Sl-85, the corresponding statements are called
Hilbert's intersection point statements in space. The set of all Hilbert's intersection point statements is denoted by 8 H • The constructive description of geometry statements can be transformed into the commonly used predicate form. Following are several basic predicates. 1. Point (PO I NT P): P is a point in the space. 2. Collinear (CaLL PI P2 P3 ): points Pj, P2, and P3 are on the same line. 3. Coplanar (COPL PI P2 P3 P4 ): PI ,P2, P3 , and P4 are in the same plane. 4. Parallel between two lines. PI P2 1i P3 P4 •
(PRLL PI P2 P3 P4 ): (COPL PI P2 P3 P4 ) and
5. Parallel between a line and a plane. (P RLP PI P2 P3 P4 Ps): PI P2 II P3 P4 Ps. 6. Perpendicular (PERP PI P2P3 P4 ): [(PI to P3 P4 )]
= P2 )V(P3 = P4 )V(PI P2is perpendicular
179
4.2 eo.trudIve Geometry StalmleDls
We will now transfonn constructions into predicate fonns. S2 (PRATIO Y W U V r) is equivalent to (PRLL Y W U V), r S3 (ARATIO Y L M N
:2M~ ' r3
rl r2
= ~~, and U i
r3) is equivalent to (COPL Y L M N),
= ~2:~' and ..,(COLL L M
rl
V.
= ~~:z , r2 =
N).
S4 (INTER Y (LINE U V) (LINE P Q)) is equivalent to (COLL Y U V), (COLL Y P Q), and ..,(PRLL U V P Q). S5 (INTER Y (LINE U V) (PLANE L M N)) is equivalent to (COLL Y U V), (COPL Y L M N), and ..,(PRLP U V L M N). S6 (FOOT2LINE Y P U V) is equivalent to (COLL Y U V), (PERP Y P U V), and UiV. Now a constructive statement S = (C1 , . .. , C" (E, F)) can be transfonned into the following predicate fonn
where Pj is the point introduced by C j and P( C j ) is the predicate fonn of C j • Example 4.23 can be described in the following constructive way.
((POINTSABC DXY Z) (INTER E (LINE A B) (PLANE X Y Z)) (INTER F (LINE B C) (PLANE X Y Z)) (INTER G (LINE C D) (PLANE X Y Z)) (INTER H (LINE A D) (PLANE X Y Z)) = 1)) ( ~!ll:.~!llL BECFDGAH The ndg conditions:
AB IY XYZ, BC IY XYZ, CD IY XYZ, AD B i E, C i F, DiG, and A i H.
IY XYZ,
The predicate fonn of this example is:
'
CONC) where
HYP
=
((COLL E A B) 1\ (COPL E X Y Z) 1\ ..,(PRLP A B X Y Z) 1\ (COLL F C B) 1\ (COPL F X Y Z) 1\ ..,(PRLP C B X Y Z) 1\ (COLL G C D) 1\ (COPL G X Y Z) 1\ ..,(PRLP C D X Y Z) 1\
180
Chapter 4. MachIne ProoIln SoIJd c - t r y
(COLL H A D) II (COPL H X Y Z) II -,(PRLP D A X Y Z) II B '" Ell C '" F II D '" G II A '" H) ;
CONC = (AEBFCGDH=I) . BECFDGAH
4.2.2
Constructive Configurations
A geometric figure which can be described by constructions S 1-56 is called a constructive configuration. Constructions S 1-56, though simple, can be used to describe most of the commonly used configurations about lines, planes, circles, and spheres. To illustrate, we will introduce more geometry objects. • We will consider four kinds of lines: I. (LINE P Q). 2. (PLINE R P Q). 3. (OLINE S P Q R): the line passing through point S and perpendicular to plane PQR. The ndg condition is -, (COLL P Q R). • We will consider six kinds of planes:
1. (PLANE L M N). 2. (PPLANE W L M N) : the plane passing through a point W and parallel to plane LM N . The ndg condition is -,(COPL L M N). 3. (TPLANE W U V) : the plane passing trough a point W and perpendicular to line UV. The ndg condition is U '" V . 4. (BPLANE U V): the perpendicular-bisector of line UV. The ndg condition is U",V . 5. (CPLANE A B P Q R): the plane passing through line AB and perpendicular to plane PQR. The ndg condition is -,(ABl.PQR). 6. (DPLANE A B P Q): the plane passing through line AB and parallel to line PQ . The ndg condition is AB II' PQ . Now we can consider more constructions: • (ON Y In). Take an arbitrary point Y on a line In . Line In could be one of the three kinds of lines. • (ON Y pl). Take an arbitrary point Y in a plane pl. Plane pi could be one of the 6 kinds of planes.
181
4.1 Constructiye Geometry Sta_DIs
• (INTER Y Inlln2). Take the intersection of two lines Inl and In2 in the same plane. • (INTER Y In pI). Take the intersection of line In and plane pI. • (INTER Y pll pl2 pI3). Take the intersection Y of three planes pll. p12. and p13. Combining all the possible cases. there are totally (3 + 6 + 6 + 18 + 56 =) 89 constructions. Conveniently. all these constructions can be described by constructions Sl-S6. To show that. we need only to reduce all kinds of lines to the fonn (LINE P Q) and reduce all kinds of planes to the fonn (PLANE R P Q). We first introduce a construction frequently used. S7 (FOOT2PLANE Y P L M N) Point Y is the foot of the perpendicular from point P to plane LM N, i.e., Y is the intersection of line (OLINE P L M N) and plane (PLANE L M N) . The nondegenerate condition is that L, M, and N are not collinear. Example 4.29 Construction S7 can be represented by constructions S1-S6. Point Y can be introduced by the following sequence of constructions. (FOOT2LINE T (FOOT2LINE F (PRATIO S T F (FOOT2LINE Y
P M N) L M N) L 1) P T S)
p
M
Figure 4-14
Example 4.30 Find two points P and Q in plane (TPLANE W U V) such that W , P, and Q are not collinear. Then plane (TPLANE W U V) is the same as plane (PLANE W P Q). u
Proof If W ¢ UV. let P be introduced by (FOOT2LINE P W U V). Take an arbitrary point R in the space. Then Q can be introduced by constructions: (FOOT2PLANE TRW U V); (PRATIO Q P T R 1). It is clear that we need a nondegenerate condition R i- T, or R is not in plane WUV .
v
w Q
Figure 4-15
If W E UV, let R be an arbitrary point. We introduce point P as follows (FOOT2LINE T R U V); (PRATIO P W T R 1). Now point Q can be introduced as in the first case. I
Exercises 4.31 1. For an OLINE In. find two distinct points U and V such that In =(LINE U V).
182
Chapter 4. Machine ProoIln SolId Genmetry
2. Let pi be a plane of the form PPLANE, BPLANE, CPLANE, or DPLANE. Find three noncollinear points W, U, and V such that pi =(PLANE W U V). Example 4.32 The following results are clear. 1. Construction (ON Y (LINE U V» is equivalent to (PRATIO Y U U V r) where r is
an indeterminate. 2. Construction (ON Y (PLANE L M N» is equivalent to (ARATIO Y L M N 1 - rl - r2) where rl and r2 are indeterminates.
rl r2
3. Construction (INTER Y (PLANE L M N) (PLANE W U V) (PLANE R P Q» is equivalent to (INTER Y (LINE A B) (PLANE R P Q» where A and B are introduced as follows (INTER A (LINE L M) (PLANE W U V» (INTER B (LINE L N) (PLANE W U V» The ndg conditions are LM IYWUV,LN IYWUV,AB
IY RPQ.
From Exercise 4.31 and Example 4.32 it is clear that all 89 constructions can be described by constructions SI-S6. We may also consider circles and spheres. We define (CIR 0 P Q) to be the circle in the plane 0 PQ which has 0 as its center and passes through point P - We define (SPHERE o P) to be the sphere with center 0 and passing through point P. Then we can introduce the following new constructions. 88 (ON Y (CIR 0 U V». Take an arbitrary point on the circle. 89 (ON Y (SPHERE 0 U». Take an arbitrary point on the sphere.
810 (INTER Y in (CIR 0 W P». Take the intersection of line in and circle (CIR 0 W P) which is different from W . Line in could be (LINE W V), (PLINE W U V). and (OLINE W L M N). We assume that line in and the circle are in the same plane. 811 (INTER Y in (SPHERE 0 W» . Take the intersection ofline In and sphere (SPHERE OW) which is different from W. Line in could be (LINE W V), (PLINE W U V). and (OLINE W R P Q).
812 (INTER Y (CIR 0 1 W U) (CIR O 2 W V». Take the intersection of circle (CIR 0 1 W U) and circle (CIR O 2 W V) which is different from W. We assume that the two circles are in the same plane. 813 (INTER Y (CIR 0 1 U V) (SPHERE O 2 U». Take the intersection of circle (CIR 0 1 U V) and sphere (SPHERE O2 U) which is different from U.
4.3
183
Machine Prout for eta. SH
Here, we introduce another 10 new constructions. In total, we introduced 100 new constructions including 89 constructions about lines and planes, 10 constructions about circles and spheres, and construction S7. Example 4.33 All ten constructions involving circles and spheres can be represented by constructions SJ -S6. Proof For constructions S8, SlO, and S12, see Section 3.2.2. By what we have discussed above, we may assume that a line is always of the form (LINE W U). For construction SII, let Y be introduced by (INTER Y (LINE U V) (SPHERE 0 U» . Then point Y can also be introduced as follows
(FOOT2LINE N 0 U V); (PRATIO Y NUN 1). Let Y be introduced by construction S13. Then Y can be constructed as follows (FOOT2PLANE M O2 0 1 U V); (FOOT2LINE N U MO l ); (PRATIO Y NUN I). For S9, we need to take an arbitrary point Y on (SPHERE 0 U) . To do that, we first take an arbitrary point P , and then Y can be introduced by construction S 11 : (INTER Y (LINE P U) (SPHERE 0 U» . In summary, we have
Proposition 4.34 All J()() constructions introduced in this subsection can be reduced to constructions SJ -S6.
4.3
Machine Proof for Class SH
In this section, we will present a mechanical proving method for Hilbert's intersection statements in the affine space of dimension three. It is clear that the volume method is a natural generalization of the area method presented in Chapter 2. As with the area method, we must eliminate points from geometry quantities.
Eliminating Points from Volumes
4.3.1
The method of eliminating points from volumes is the basis of the volume method. Two other geometry quantities, the area ratio and the length ratio, will ultimately be reduced to volumes. In this subsection, we will discuss four constructions S2-S5. S 1 will be discussed in Subsection 4.3.4. Lemma 4.35 Let Y be introduced by (PRATIO Y W U V r). Then we Iulve V
_ { ABC Y -
(~~
+ r)VABc v + (~~ - r)VABcu + r(VABC V - V ABCU )
V ABCW
ifW is on line UV. in all cases.
184
Chapter 4. MadUne Proof in Solid Geometry
Proof If W, U, and V are collinear, by Proposition 4.5 we have V ABCY
UY YV UW WV = =V ABCV + =VABCU = ( = +r)VABcv + ( = UV UV UV UV
r)VABcu ·
Otherwise, take a point S such that W S = UV. Then we have VABCY
WY
YS
= WS VABCS + WS VABCW = rVABcs + (1 -
r)VABcw.
By Proposition 4.9, we have VABCS = VABCW + VABCV - V ABCU . Substituting this into the above equation, we obtain the result. Notice that in both cases, we need the ndg condition U =1= V. I
Lemma 4.36 Let Y be introduced by (ARATIO Y L M N rl r2 ra). Then we have
Proof This lemma is a direct consequence of Proposition 4.6.
Lemma 4.37 Let Y be introduced by (INTERY (liNE U V) (liNE I VABCY
SUlJ = -s--VABCV UIVJ
J». Then we have
SVlJ -S--VABCU. UIVJ
Proof By Propositions 4.5 and the co-side theorem, V ABCY Sur ,vAscv-Syr ...J. O. SUIVJ ,vASCI'. Since UV 1/'1 J • we have S UIVJ ..,..
= ~~VABCV + ~~VABCU = I
Lemma 4.38 Let Y be introduced by (INTER Y (liNE U V) (PLANE L M N». Then we have
Proof By Proposition 4.5 and the co-face theorem. V ABCY = ~~VABCV VCI! MNVARCY-VyC V. N V ..,.. ...J. O. VULMNV MNVASCU . Since UV I/'LM N ,we Uhave LM
+ ~~VABCU = I
Example 4.39 Let Y be the intersection point of three planes WUV. LM N and RPQ. Then Y can be constructed as follows (INTER X (liNE L M) (PLANE R P Q» (INTER Z (liNE L N) (PLANE R P Q» (INTER Y (liNE X Z)(PLANE W U V» By Proposition 4.38, we have
4.3 MachIne Proof tor a- Sa
vABCY --
185 ~V
...!::'.u:wL. V Vxwuvz ABCZ
V
ABCZ -
- VXWUVZ ABCX VL/l P9 V VNRP9 V VLRPQN ABCN - VLRPQN ABCL
V
ABCX -
VL/l P9 V VLRPQM ABCM -
V
XWUV -
V
-
ZWUV -
VLRP9 VLRPQU
v:MWUV -
VLRP9 V, VLRPQN NWUV -
VURP9 V VLRPQM ABCL VM RP9 V, VLRPQU LWUV VNRP9 V, VLRPQN LWUV
From the above formulas. we can express formed by the known points.
V ABCY
as a rational expression of volumes
Example 4.40 (Steiner's Theorem) If two opposite edges of a tetrahedron move on two fixed skew lines in any way whatever but remain fixed in length. the volume of the tetrahedron remains constant.
Constructive description «POINT'S ABC D) (ON X (UNE A C»
(PRAllO Z X A C 1) (ON Y (UNE B D»
(PRAllO W Y B D 1) (VXyZW = VABCD»
The ndg conditions: A
=/: C, B =/: D.
Figure 4-16
Proof. By Lemma 4.35, we can eliminate W
By Lemma 4.35 again, VBXZy=VBDXZ' :~; VDXZY=(:~ -1)· V BDXZ • Then V Xyzw = - V BDXZ V XBZD • Similarly we can prove V XBZD VABCD .
=
=
Example 4.41 Show that a plane which bisects two opposite edges ofa tetrahedron bisects its volume.
Constructive description «POINT'S ABC D) D
(MIDPOINT P A D) (MIDPOINT S B C)
(\.RAllO Q B D t)
(1NTl!R R (UNE A C) (PUN!! P S (VPCSR-VPDCS-VPDSQ
=
s
B
Q»
Figure 4-17
tVABCD»
The ndg conditions: A=/: D; B
=/: C; B =/: D; AC It' PSQ.
c
186
Chapter 4. MachIne Proof In Solid Geometry
Proof Using Lemma 4.38, we can eliminate point R V
_ RC _ AC VPCSA -
V ACPS ' VCPSQ VCPSQ - V APSQ
PCSR -
We can eliminate the remaining points by using Lemma 4.35:
== VPCSR
-VcPsQ' V ACPS _ - VCPSQ
+ V APSQ
-
VCDPS ' VCDPS'
r
r·
V ACPS
+ V ABPS ' r
-
VABPS
(-!VBcDP)·r,(!VABCP) - ! VBCDP '
r -
!VABCP '
r
1
(-!VABCD)·r,(!VABCD) (2). (-!VABCD )
Similarly, we can compute V PDCS == -~VABCD' Then V PCSR
- V PDCS - V PDSQ
== (~
+~ -
+ !VABCP 4(r,VABCD)
VPDSQ
== ~(r-1).VABCD'
1
r4 )VABCD
==
!VABCD .
This proof seems to be a little complicated, but the idea behind it is quite simple: one need only to eliminate the points from volumes using Lemmas 4.35-4.38.
4.3.2
Eliminating Points from Area Ratios
The following lemmas provide methods of eliminating Y from G == ~ss . The proofs of CDE the lemmas are similar: if all the points are in one plane, methods of eliminating Y have been given in Chapter 2. Otherwise, there is a point T which is not in the plane ABY. By Corollary 4.11,
G ==
(4.4)
SABY SCDE
==
V TABY VTCDEA'
Now the lemmas in Section 4.3.1 can be used to eliminate Y from V TABY .
Lemma 4.42 Let Y be introduced by (PRATIO Y W U V r). Then we have ABY
__
_
CDE -
{
~
VUCDEV
VUABw±rVUARV
SABYJ'.frfff:Bv-SABUl SCDE
ifW is not in plane ABY ifW is in plane ABY but line UV is not. ifW, U, V, A, B, and Yare coplanar.
4.3
187
Machine Proo( for CJass SH
rt
Proof If W ABY, let W S be a parallel translation of UV to line WY . By (4.4), VWABY CDE=VWCDEY
1 VWABY TVUCDEV '
By Propositions 4.5 and 4.9, VWABY = ~~VWSAB = TVUABWV . We prove the first case. The second case can be proved in a similar manner by replacing W by U . The third case is from Lemma 2.21 on page 65. 1 Figure4-IS
Remark 4.43 The first case of Lemma 4.42 is rarely used in practice, since in this case Y is actually the intersection of(PUNE W U V) and (PPLANE A C D E), i.e., Tis afixed quantity and is not easy to find.
Lemma 4.44 Let Y be introduced by (ARATIO Y L M N
ABY {,.,VlABMtr.VCABN VL C DE;A CDE "SARCt"~AB.,t"SARN
__ _
S C DE
Tl
T2 T3)' Then we have
lifoneo*L M andN , sayL, isnotinABY r.J " lifL M and N are inp/ane ABY. '
,
Proof If L is not in ABY, C ABDYE = ~VV . Now the result comes from Lemma 4.36. The LCDEA second case is Lemma 2.31. 1 Lemma 4.45 Let Y be introduced by (INTER Y (UNE U V) (UNE I J
ABY { CDE -
__ _
SCII cV,rARY
S U I V J..VUCDE..A.
SI!lY~ARc-~rrrySARC SCDES1UJ V
Proof. If U is not in ABY. ~ CDE case is Lemma 2.20.
». Then we have
lifone o*U V " I and J , say U, is not in ABY. r.J , lifU' V , I , JAB andY are cO'Pianar. , , ,
= VUCDEA ~ = UV !Ll::~ VUCDEA = ..§.u.u...~ SUIVJ VUCDEA .
The second
1
Lemma 4.46 Let Y be introduced by (INTERY (UNE U V) (PLANE L M N» . Then we have
ABY _ _ _ CDE -
Vi {.EJ..J.M.J:L ~ViVi UCD A Vi LOIN
if one of U and V , say U, is not in ABY.
W,CMN~ARY ~YCMNSAR(( ifUandVareinABY. SCDEVULMNV
=
Proof. If U is not in ABY , ~ ~ !Ll::~ ~...lli.uu:... CDE = VUCDEA UV VUCDEA = VULMNV VUCDEA . The second case is a consequence of Proposition 2.9 and the co-face theorem. 1
188
Chapter 4. Maddne Proof In SolId Geometry
Example 4.47 (Centroid Theorem for Tetrahedra) The four medians of a tetrahedron meet in a point.
Constructive description «POINTS ABC D) (MIDPOINT
S B C)
(LRATIO Y D S 2/3) (LRATIO Z A S 2/3)
(Y is the centroid of f:::.DBC.) (Z is the centroid of f:::.ABC.)
(INTER G (LINE D Z) (LINE A Y))
(~!:g
Figure 4-19
= 1/3))
The ndg conditions: B =I- C, D =I-
B
A
(INTER H (LINE C G) (PLANE A B D))
S,
A =I-
S,
DZ
I/' AY, CG I/' ABD, and SABD =I- O.
Proof Points will be eliminated in the order H, G, Z, Y, S, D, C, B, and A. By Lemma 4.46, SABH
VCABD
V CABG
VABCG
= VCABDG VCABDA = V ABDG + V ABCD . S By Lemma 4.37, V ABDG = ~ss V , V ABCG = ...2Z.A.L V ABCD • By Lemma 4.37 again, SZADY DAZY ABDZ SABD
V ABDZ
= 2/3VABDS = -1/3VABcD • Now S ABH SABD
h.u... SZADY
_
-
S·
1 - 1/3~s ZADY
NowallpointsareinplaneADS. ThenbyLemma4.42,SzAY = lSSAY = iSSAD;SDAY lSDAS = -lSSAD; SZADY = SZAY - SDAY = ~SSAD' Then ~!:~ = 1/3. I
=
Exercise 4.48 We can also use the following general method to eliminate points from G = ~s . If A, B, Y, C, D, and E are coplanar, we need to find a point T that is not in ODE plane ABY. Then G = SABY = V TABy . SCDE
VTCDE
Otherwise, by Corollary 4.11
G
= V CABY . V ACDE
The advantage of this method is that we do not need to use the volume of the polyhedron involving five points in all cases. Prove Lemmas 4.42-4.46 using the above method.
4.3.3
Eliminating Points from Length Ratios
The following lemmas present methods of eliminating point Y introduced by construction C from the length ratio G = ~~.
4.3
M8dt1ne Proal for a- SH
189
= ~~.
Lemma 4.49 Let G
C =(PRATIO Y W U V T).
Then
.uzr-
if DEWY.
-~
if D ¢ WY. U ¢ DWY. ifD ¢ WY. E ¢ DWY. if all points are coplanar.
~+r
G=
I
~i~VYF ~~~;V
Figure 4-20
Proof The first and the last cases have been proved in Lemma 2.26. If U ¢ DWY. take a point S such that DS = EF. By the co-face theorem G = DDYS = ~vv = ~vv . DWUVS EWUVF If E ¢ DWY. take a point T such that WT = UV. By the co-side and co-face theorems = ~ = .Y.cui:l:£:.. . By Propositions 4.9 and 4.12 G = !2l: DS SDWST VDWTES V DWTE
= V DWVE -
VDWTES
=
VEWTEF
V DWUE
=
= VUDEWV,
-VFWTE
=
-VFWVE
+ V FWUE
=
VUEFWV.
ThenG=-~. VUEFWV
Lemma 4.50 Let G
= ~~. C =(ARATIO Y L M
N
Tl T2 T3).
Then we have
ifD ¢ LMN. if D E LM N . E ¢ LM N ,and DY It' N M. if all points are coplanar and DY It' N M. Proof If D is not in plane LM N, the result is a direct consequence of Propositions 4.4 and 4.12. For the second case, by Corollary 4.13, G
= VDMNEY = V DMNE VDMNES
Tl VLMNE.
VEMNEF
The third case can be proved similarly. Lemma 4.51 Let G
= ~~. C =(INTERY (UNE U V) (UNE I J)). Then we have ..YmLn. D ¢ UV IJ and ...,(COLL U V I).
G-
~UVIF
~
- { 'I:;r:
SEUFV
DE UVIJ. EF ¢ UVIJ. and D ¢ UV. D,E,F are in UVIJ, and D ¢ UV.
190
Chapter 4. Machine Proof in Solid Geometry
Proof The first case is a consequence of the co-face theorem. For the second case, by Corollary 4.13,
G = DY =
VDUVEY
EF VEUVEF If all points are coplanar, see Lemma 2.25.
Lemma 4.52 Let G
= ~, C =(INTER Y
VOCMri VELMN-VFLMN
G{
= _ VDUVE. V FUVE
(liNE U V) (PLANE L M N)). Then we have
IfD is not in plane LMN. If D E LM N and one of L,M, N, say L ¢ DUV.
vEUVLVOUyT FUVL
Proof If D is not in plane LM N, the result is a direct consequence of the co-face theorem. For the second case, take a point S such that DS = EF. Then we have G=m::=~=~ DS VDUVLS VEUVLF'
I
Example 4.53 (Ceva's Theorem for Skew Quadrilaterals) The planes passing through a point 0 and the sides AB, BC, CD, and DA of any skew quadrilateral meet the opposite sides of the quadrilateral at G, H, E, and F respectively. Show that ~~ . ~~ . g~ ~~ = 1.
.
A E
Constructive description «POINTS A B G D 0 ) (INTER E (LINE A B) (PLANE 0 G D»
G
(INTER F (LINE B G) (PLANE 0 A D» (INTER G (LINE G D) (PLANE 0 A B» B
(INTER H (LINE D A) (PLANE 0 B G» (AEtM:"CGllli EB Fe OD HA
= 1»
D
Figure 4-21
The ndg conditions: AB
C "# F; D "# G, A "# H.
II' OCD; BC II' OAD; CD II' OAB; AD II' OBC;
B
"# E;
Proof By Lemma 4.52 or just by Proposition 4.4, we have BF CF = Therefore ~ {g Qfi. !JJ;! EB FGGD HA
=
- VABDO -V ; AGDO
CG V ABGO ==---; DG V ABDO
1.
Example 4.54 (Centroid of Tetrahedra) The four medians of a tetrahedron meet in a point which divides each median in the ratio 3:1, the longer segment being on the side of the vertex of the tetrahedron.
4.3 M8chinc ProoI lor a.. Su
191
Constructive description «POoos ABC D) (MIDPOII'n' S B C)
c
(!.RAllO Z A S 2/3) (LRA110
Y D S 2/3)
(INTER G (LINE D Z) (LINE A Y » A'"~:"':":"'~-~"--~B
(~
= 3»
Figure 4-22
c. A =I s. D =I s. DZ IY AY. G =I Y. ~ Proof By Lemma4.S1. AyGG = ~s • By Lemma 4.42. ~s = ~2/3; ~s = AZZS§..uJ.s.S = f. Then ¢~ = 3. I The ndg conditions: B =I
-
DZY
DZY
-
DSZ
ADS
Exercise 4.55 Let point Y be introduced by construction (INTER Y (UNE U V) (PUNE R P Q)). Show that
• V
=
ABCY
{
SUPRqV SVRP9V SUPV9 ABC V - SUPvQ ABCU VPURqVABcv-VPVRqVABCU VURQPV
if P, Q, R, U, V are coplanar. otherwise.
• IfD is on UV SDPRq SEPFQ VDPqR VEPQRF VPEDRq VQEFRP
If all points are coplanar. IfD
f/. PAR.
If D E PQR and E, F
f/. RUV.
• SCDE
where r'
+ (1 -
r')~
AB E RUV and PQ E RUV. AB E RUV and PQ f/. RUV. = $CDE SCDE r' ~ 1 r'...r.u.au.. AB f/. RUV and PQ E RUV. { VXCDEA + ( - )VXCDEA r"~ + (1- r',)...r.u.au.. AB f/. RUV and PQ f/. RUV. VXCDEA VXCDEA
r' §..un:. SCDE
S ABY
=
SC.DE
r"~+(I-r")~
VUPRq VUPVQ'
r"
=
VUPRq VVPRQU
and X is U
•
V
or R depending on which is not in
planeABY.
Exercise 4.56 Try to eliminate Y from the three geometry quantities ifY is introduced by the following constructions. (INTERY (PUNEW U V) (PUNE R P Q)) (INTERY (PUNEW U V) (PLANE L M N))
192
4.3.4
Chapter 4. MachIne Proof In SolId c - I r y
Free Points and Volume Coordinates
After applying the above lemmas to any rational expression E in geometric quantities, we can eliminate the non-free points introduced by all constructions from E. Now the new E is a rational expression in indeterminates and volumes of free points in space. For more than five free points in the space, the volumes of the tetrahedra formed by them are not independent, e.g. the following equation is always true: VABCD
= VABCO + VABO D + V AOCD + VOBCD .
To express E and F as expressions in free parameters, we introduce the volume coordinates. Definition 4.57 Let X be a point in the space. For four noncoplanar points 0, W, U, and V, the volume coordinates of X with respect to OWUV are rl
V OWUX Vowxv Vo xuv V xwuv =- - , r2 = - - - , r a = ---,r4 = - - -. V V V V owuv
It is clear that rl
owuv
owuv
owuv
+ r2 + ra + r4 = 1.
Since the sum of the volume coordinates of a point is one, we sometimes omit the last one to obtain an independent set of coordinates. Exercise 4.58 Show that the points in the space are in a one to one correspondence with thefour-tuples (x, y, z, w) satisfying x + y + w + z = 1. Lemma 4.59 Let G
= V ABCY , and 0, W, U, V
G = V ABCO
+
VOABCV VOWUY
be four noncoplanar points. Then we have
+ VOABCU V OVWY + V OABCW V OUVY TT
"owuv
Proof We have V ABCY
= V ABCO + V ABOY + V AOCY + V OBCY .
(1)
Without loss of generality, we assume that YO meets plane WUV at X. (Otherwise, let YW meet plane OUV at X, and so on.) By Proposition 4.4, we have V, OABY
OY v,
= OX
OABX
=
VOWUVY V OABX V owuv
(2)
By Proposition 4.6, V OABX
SWUX Swxv Sxuv = -S--V OABV + -S--VOABU + -S--VOABW wuv WUV wuv
(3)
43
Machine Rmlfor Clsas SH
By Lemma 4.46. we have
swux ---Vowuv . s w x v - -.Vovwv -Sxuv --vouvv -swuv Vowuvy ' Swuv Vowuvy Swuv Vowuvy ' I
Substituting them into (3) and (2), we have
VOABY=
+
VOWUYVOABV + VOVWYVOABUVOUVYVOABW vowuv
(4)
Similarly, we have
vowuv VOWUYVOCAV + VOVWY VOCAU VOUVYVOCAW Vocav = vowuv Substituting them into (1) and with (4.3)
+
we can obtain the result.
I
Corollary 4.60 Use the same notations as in Lemma 4.59. For any point P let x p = v v Y p = v:WWUP :t zp = v:c; . Then the formula in Lemma 4.59 can be written as
E,
which is quite similar to the formula of the volumes in terms of the Cartesian coordinates of its vertices. Now we can describe the volume method as follows: for a geometry statement in Sg: S = ( C 1 , . . ,Crr( E l ,E2)), we can use the above lemmas to eliminate all the non-free points and express the volumes of free points as their volume coordinates with respect to four fixed points. Finally, we obtain two rational expressions R1 and R2 in independent parameters respectively. S is a correct geometry statement if R1 is identical to R2.For the precise description of the algorithm, see Algorithm 4.86 on page 205. We end this section by a methodological comment. One of the most important ideas in the traditional method of solving problems in solid geometry is reducing a problem of
194
Chapter 4. MachIne Proof In SolId c-try
higher dimension to a problem of lower dimension and then solving it using knowledge from plane geometry. This is the so-called "dimension reduction method." But our volume method always does the converse, i.e., it reduces length ratios and area ratios to volumes, because in space the volume is easy to deal with. This can be seen from the three groups of lemmas in Sections 4.3.1, 4.3.2, and 4.3.3. This is also the reason why the method is called the volume method.
Working Examples
4.3.5
Example 4.61 For a tetrahedron ABCD and a point 0 , let P = AO n BCD, Q = BOnACD, R = COnABD, andS = DOnABC. Showthat~~ +~+g:+~~ = 1.
Constructive description «POINTS A B G D 0)
(INTER P (LINE A 0) (PLANE B G D» (INTER Q (LINE B 0) (PLANE A G D» (INTER R (LINE G 0) (PLANE A B D)) (INTER S (LINE D 0) (PLANE A B G))
(~+gg+~+~ AP
BQ
CR
DS
= 1» Figure 4-23
The ndg conditions: AO IY BCD, BO IY ACD, CO IY ABD, DO IY ABC, A"# P, B"# Q, C"# R, D"# S .
Proof By the co-face theorem
OS DS
OP AP
V ABGO
= = - - -; V ABGD
V OBGD
= VABGD ·
By Lemma 4.59,
OP AP
OQ
OR
OS
+ BQ + C R + DS
V OBGD
=
+ V AOGD + V ABOD + V ABGO V
ABGD
= 1.
Example 4.62 Let ABC D be a tetrahedron and 0 a point. Let line DO and plane ABC meet in S; line AD and plane OBC meet in P; line BD and plane OAC meet in Q; and line CD and plane 0 AB meet in R. Show that !2!l +~ + !2.l! os = !J.!:. PA QB Re · Constructive description «POINTS A B G D 0)
(INTER S (LINE D 0) (PLANE A B G» (INTER P (LINE D A) (PLANE 0 B G)) (INTER Q (LINE D B) (PLANE 0 A G)) (INTER R (LINE D G) (PLANE 0 A B»
(~ =*+~+~))
The eliminants IlJ!. !! !::".!Juw. CR-VABCO
£i2~ BQ- -VABCO
lZ&. !. Y.a.aJ2ll.
AP-VABCO ~ S VAROO-VARGO OS VABCO
VBCDO=VACDO-VABDO+VABCO-VABCD
195
Machine Proof for Class SH
4.3
The machine proof ~
~
-(~+~+*)
!J
VAR~
-(VABDO+VABCO ' ~+VABCO ' *)
• ~ os
c • ~
VABCO ,(-VABCO)
-(VACDO . VABCO-VABDO.VABCO-~BCO ' *)
.im!!1Y~
. ~ os
VARGa VACDO-VABDO-VABCO ' *
t;,
-
Figure 4-24 • ~
(VABCO)' VBCDO.vABCO+VACDO ,VABCO
VABDO ,VABCO
os
. ~ (VBCDO VACDO+VABDO) os
.im!!f.ify
1.
A
OS
VARGa
(VABCO-V,1BCD),VABCO -(VBCDO-VACDO+VABDO)-(-VABCO)
.i~iIJl
volu~e-co VARGO-VARGa tlimJ!!.i/S/
VARCo-VARcn
VBCDO
VACDO+VABDO
-
VABCO-VABCD
-
In the above proof, a volu~.-co b means that b is the result obtained by replacing each volume using the volume coordinates with respect to four fixed points.
Example 4.63 Let a line I meet four coaxial planes in A, B, C, and D respectively. The cross-ratio of I for the four planes is defined to be (ABC D) = ~; . Show that for any line I the cross ratio is fixed.
:g.
The eliminants
Constructive description «POINTS
x
~l!! VXYBD,
Y A 8 C, D.)
(INTER C (LINE A 8) (PLANE
c,
c, B, X Y»
(INTER D (LINE A 8) (PLANE D, X Y» (INTER A, (LINE C, D,) (PLANE A X Y»
(INTER
B,
(~u AD BC
=
(LINE
c,
Dtl
(PLANE 8
A,C, B,D,» A,D, B,C,
The machine proof u.~ "G AQ
~.~ CIS l
~ -
D.A I
-VXYBC, CA
(-VXYBD,) · ~
1lR
,4£
' Bc ' AD
~ VXYBC, ,(-VXYAD,) • u . ~ VXYBD, ' (
VXYAC,)
BC AD
x Y»
-
VXY BC,
C,A, ~ VXYAC, D,A, - VXYAD, ~ D VXY D, c+VXY AD, AD VXYAD, uD VXYBD, BC VXYD,C+VXYBD, V C VXYBD, ,VXYAC, -VXYBC, ,VXYAD, XYD,C VXYBC, - VXYAC,
196
Chapter 4. Madlioe Proof ID SoUd Geometry
Q
VXY BD) ·(VXY DJ C+VXYAD J ) ,VXYBC , ,VXYAD 1
-
VXY BD, ,VXY AC, ,VXY AD, ,(VXYD, C+VXY BD,)
simJ!!.ify (VXY D, C+VXY AD, ) ,VXYBC, -
VXY AC, ,(VXY D, C+VXY BD,)
£
(-VXY BD, YXY AC, +VXY AD, ,VXY AC, ) ,VXY BC, ,( -VXY BC, +VXYAC,)
-
VXYAC"(-VXYBD, ,VXYBC, +VXYBC, ,VXYAD,)-(-VXYBC, +VXYAC,)
sim~ifY
c
Example 4.64 16 Let ABC D be a tetrahedron and G the centroid of triangle ABC. The lines passing through points A. B , and C and parallel to line DG meet their opposite face in P, Q, and R respectively. Show that VGPQR = 3VABCD , Figure 4-26
Constructive description
The eliminants
«POINTS ABC D)
VCDGP=-VACDG
(CENTROID GAB C)
(INTER P (PUNE A D G) (PLANE BCD))
VBDGP=-VABDG
(INTER Q (PUNE B D G) (PLANE A C D))
VBCGP=VABCD
(INTER R (PUNE C D G) (PLANE A B D))
VDGPQ=-VBDGP
(3VABCD
=
VGPQR))
V.
Vanap ,VA RCn-VRCGp ,VACPG
CGPQ VGPQR
VACDG VDGPq,VABCD-VCGPq,VABDG VABDG
The machine proof (3),VABCD VGPQR
!!:. -
(3) ,V A BCD ,VABDG VDGPQ,VABCD-VCGPQ ,VABDG
2. -
(3) ,VABCD ,VABDG'(VACDG)2 -vCDGP,vACDG ,vABDG,vABCD-vBDGp ·vlcDG,vABcD+vBCGP'v lCDG,VABDG
si"'J!!ify
(-3) ,VABCD,VABDG,VA9DG VCDGP ,VABDG ,VABCD+VBDGP ,VACDG ,VABCD-VBCGP'VACDG ,VABDG
.f.. (
3)YABCD ,V6BDG,V,
-
-3VJCDG ,V A CDG,VA BDG ,VABCD
sim~ifY 1
In the above proof the fact that G is the centroid of triangle ABC is not used. We thus have the following extension of Example 4.64. Example 4.65 The result of Example 4.64 is still true ABC.
if point
G is any point in plane
We further ask whether the result of Example 4.64 is true or not if point G is an arbitrary point. 16This is a problem from the 1964 International Mathematical Olympiad.
4.4
197
Pythagoras Differences III Space
The eliminants
Constructive description «POINTS
ABC D G)
VCDOP=-VACDO
(INTER P (PUNE A D G ) (PU.NI! BCD» VBDOP=-VABDO
(INTER Q (PUNE B D G) (PU.NI! A C D» (INTER R (PUNE CD G) (PU.NI! A B D» (3VABCD
VBCOP=-(VABCO-VABCD)
= VOPQR»
VDOPQ=-VBDOP VCQOP -VARco-Vacap ,VACPQ VCOPQ VACDO VDOPQ oVABCD-VCOPo oVABDO VOPQR VABDO
The machine proof ( 3) oV A B C D VOPQR
!i -
g
(3) oV A BCD-VABDO VDOPQ oVABCD VCOPQ oVABDO (3) oVABCD oV A BDo o(VACDO ).
VCDOp o VACDo o VABDo o VABCD-VBDOp o V~CDO o VABCD+VBCOp o V~CDO o VABDO
.imgliJII
E. -
(-3) oV A BCD oVABpo oV A CDO VCDOp oVABDo oVABCD+VBDOpoVACDo oVABCD VBCOp oVACDo oVABDO
(-3WABOP OVABPo ov1cpooCVBCPO)3 VJCDo OVAODo oVABDo oVABCO-3VBCDO oVACDo oVABDooVABCD
.im~ iJy
(-3)oVA BCp 3VA BCD
VABCO
We thus obtain the following extension of Example 4065: VGPQR planeABCo
4.4 4.4.1
= 3VABC iff G is in
Pythagoras Differences in Space Pythagoras Difference and Perpendicularity
From now on, the concept of the square-distance between two points will be usedo The definition for the Pythagoras difference is the same as in the plane geometry, ioeo, for b.ABC
For a skew qUadrilateral ABCD, we define PABCD
= P ABD -
PCBD
=='.! = --:-=2 AB + CD -
~
~
BC - DA .
Properties of the Pythagoras difference in space are quite similar to those in a plane. Propositions 3.4, 3.2, and 303 are true in space, since all the points involved are in the same plane. Surprisingly, Propositions 3.1, 3.5, and 3.7 are also true in space even if the involving points are ~ot coplanar.
198
Chapter 4. Machine Rod in Sdid Gcomehy
-
Proposition 4.66 Let R be a point on line PQ with position ratio rl = respect to PQ. Thenfor any points A and B in space, we have
-
g,r2 = % with
Proof: Since points P, Q, R , A and P, Q, R, B are two groups of coplanar points, we may use Proposition 3.5 on them separately:
== 2
-4
-
---2+*
--2
--2
---2
ThenPRAB + A B - R B ~ = T ~ ( Q +AA B - Q B )+r2(PA + A B - P B ) = rlPQas + r2PPAB.The second equation can be proved similarly. I Proposition 4.67 Let R be a point in the p h e PQS, and rl = r3 = Thenfor points A and B we have
e.
e,e, r2 =
Proof: The proof for proposition is similar to that for Propositions 3.61 and 3.62.
and
PQS
1
Proposition 4.68 Let ABCD be aparallelogram. Thenfor any points P and Q, we have
Proof: The proof is the same as for Proposition 3.7.
I
As in plane geometry, we use the notation A B l C D to denote the fact that four points A, B , C , and D satisfy one of the following conditions: A = B , or C = D, or line AB is perpendicular to line CD. Proposition 4.69 A B l C D iff PACD= PBCDor PACED= 0.
- -
Proof: Let E be a point such that AE = CD. By Proposition 4.68, PACED= PAABE= PBAE.By the Pythagorean theorem, A B l C D iff PACED= PBAE= 0. I
As consequences, Proposition 3.10 and Example 3.9 are still true in space.
4.4
199
1'ythacoI'M Differences In Spare
Just as the volume makes the machine proof for geometry theorems involving collinear and parallel possible, the Pythagoras difference will make the machine proof for geometry theorems involving perpendicular possible. Before presenting the method of machine proof, let us first get acquainted with the Pythagoras differences by proving some basic properties of the perpendicular.
Example 4.70 If a given line is perpendicular to a pair of non-parallel lines in a plane. it is perpendicular to every lines in the given plane, i.e.• the line is perpendicular to the plane. Proof As in Figure 4-27, line PQ is perpendicular to line OU and OV. Let W be a point in plane OUV. We need to show that PQ.J..OW. By Propositions 4.67 and 4.69
Souw Sowv Swuv -S--PPQv + -S--PPQU + -S--PPQo ouv ouv ouv + Sowv + Swuv)P Souw = ( - - - - - - PQO = PPQO· Souv Souv Souv
=
Figure 4-27
By Proposition 4.69 again, PQ.J..OW.
Example 4.71 (The Theorem of Three Perpendiculars) A line PQ is perpendicular to a line AB iff PQ is perpendicular to the orthogonal projection of AB to a plane containing PQ. Proof Let 0 be the orthogonal projection of point A to plane PQB. Then AO.J..PQ. If PQ.J..BO, we have PPQA PPQO = PPQB , i.e., PQ.J..AB. Conversely if PQ.J..AB, we have PPQB = PPQA = PPQo, i.e., PQ.J..BO. I
= Figure 4-28
Example 4.72 (The Orthocenter Theorem for Tetrahedra) Iftwo pairs ofopposite edges of a tetrahedron are at right angles to one another. the third pair are at right angles; and the altitudes are concurrent, and pass through the orthocenters of the opposite faces. Proof Let AB.J..CD and AC.J..BD. Then PABC = PABD = PCBD = PDBC , i.e., AD.J..BC. Let the two altitudes AQ and DR of triangle AC D meet in F. Then PBAQ - PF AQ = PBAD - PF AC = PCAD - PDAC = 0, i.e., BF .J..AQ. Similarly, BF .J..DR. Therefore BF is the altitude from B to plane ACD.
Figure 4-29
200
Chapter 4. Machine Proof In SolId Geometry
To prove that the four altitudes are concurrent, we first show that BR.lAG which follows from PACR = PACD = PACB . Let altitudes AP and BR of triangle ABG meet in E and DE and BF meet in H. We need to show that AH.lBGD. By Proposition 4.69, PH DC = PBDC = PADC, i.e., AH .lDG. Similarly AH .lBG. Thus AH .lBGD. Example 4.73 Show tlult the construction (FOOT2PLANE A P L M NJ is equivalent to the construction (ARATIO A L M N rl r2 raJ where
M
Figure 4-30
Proof We construct point A as in Example 4.29. Then SYMN
YT
PPTS
PPTS
= ST = P TST = PLFL'
SLMN
By Propositions 4.66 and 4.69, P PTL
= PPMNPPNL + PPNMPPML P MNM
--2
PPMNPPNL
+ (2MN
-
PPMNPPNM
PPMN)PPML -
PPMNPPNM
P MNM -PPMNPNML
+ 2MJf PPML
P MNM 16SEMN _ 4LFf . P MNM
-
~P
P'itLN
MNM
We have proved the first case. Other cases are similar.
4.4.2
Pythagoras Difference and Volume
With the concept of perpendicularity, we can obtain the exact measurement for the volume of a tetrahedron. Definition 4.74 Let F be the foot of the perpendicular dropped from the point R upon the plane LM N . The distance from R to LM N , denoted by hR,LMN, is a real number which has the same sign as VRLMN and IhR,LMNI = IRFI .
4.4
201
Pythagons Dilre",Dca 111 Space
Proposition 4.75 For any two tetrahedra ABCD and RLMN.let hA h
R,LMN ·
Sho
W
thot
VASCp
-
ISBCDlh A -
= hA,BCD.
hR
=
VRCMN
ISLMNlhR .
Proof Without loss of generality. we assume R
that points B , C, D, L, M, and N are in the same plane. As in Figure 4-31, let RF be the altitude of the tetrahedron RLM N and S be a point on RF such that AS II BCD . Then VABCD = VSBCD . By Propositions 4.3 and 4.4.
VABCD
VALMN
SBCD
= SLMN ;
Figure 4-31
Multiplying the two formulas together and noting that hA and hR have the same sign as VABC D and VRLM N. we prove the result. I
Coronary 4.76 For a tetrahedron ABC D. we have
Proof Replacing the tetrahedron RLMN in the Proposition 4.75 by BCDA. we obtain the first equation. etc. I By Proposition 4.75. we have
= khA ,BCDISBCDI = khB,CDAISCDAI = khc,DABISDABI = khD,ABCISABCI where k is a constant which is independent of the tetrahedron ABC D. Setting k = 1/3. VABCD
we obtain the usual formula for the volumes of tetrahedra.
Proposition 4.77 We have
Proposition 4.78 (The Herron-Qin Formula for Tetrahedra) Prove the following formula 2
144VABCD
...,-=2 ...,-=2 - 2 ...,-=2 2 ...,-=2 2 = 4AB ·AC ·AD -AB PDAC-AC PBAD -
...,-=2AD p2 P P P. BAC+ BAC BAD CAD·
202
Chapter 4. Madlioe Proof 10 SolId Geometry
Proof. Similar to Example 4.29, we construct the altitude BO as follows.
(FOOT2LINE P D A C) (FOOT2LINE H B A C) (PRATIO G H P D 1) (FOOT2LINE
°
F
A
B H G)
H
Figure 4-32
Then
Bd = BH2 _ H0 2 = 4~c _ H02. AC
(1 )
By Propositions 3.1, 3.2, and 4.66
OH)2 HG2 ( HG
= (PBHG )2 Hd = P~HG = P~ PHGH
41Rf
4DP
+ PBCAPBAD - PBACPBCA)2/(4AC4 • 4DP2) 2 (PBACPBCD + (2AC - PBAC)PBAD - PBACPBCA?/(64ESbAC)
(PBACPBCD
(-PBACPCAD
+ 2E PBAD )2 /(64ESbAC)
Substitute this into (1). By the Herron-Qin formula for triangles (on page 105), we have
Corollary 4.79 (The Cayley-Menger Fonnula) We have the following commonly used version of the Herron-Qin formula.
288Vp,P2P3P4
where rij
= Ip;Pjl.
=
0 rr2 rra rr4 1 r~l 0 r~a r~4 1 r~l r~2 0 r234 1 r~l r~2 r~a 0 1 1 1 1 0 1
4.5
203
'The Volume Method
Proof In the above detenninant, subtracting the first row from the second, the third, and the fourth rows and subtracting the first column from the second, the third, and the fourth columns, the detenninant becomes 0
r~l r~l r~l
r?2 -2r?2
-P213
r?3
- P312
-2r~3
- P412
- P413
1
0
0
1 0 0 -2r?4 0 0 0
r?4
-P214 -P314
=
2r?2
P 213
P 214
P 312
2r?3
P 314
P 412
P 413
2r~4
Expanding the last detenninant and comparing the fonnula in Proposition 4.78, we prove the result. I
The Volume Method
4.5
Since we have a new geometry quantity, the constructive statements can be enlarged in the following way: the conclusion of a geometry statement could be the equation of two polynomials of length ratios, area ratios, volumes and Pythagoras differences.
4.5.1
The Algorithm
Now we have six constructions SI-S6 and four geometry quantities. We need to give a method to eliminate the point introduced by each of the constructions S 1-S6 from each of the four quantities. This section deals with the cases which are not discussed in Section 4.3. Lemma 4.80 Let G P ABW
= PABY . Then + r(PABV -
P ABU )
ifY is introduced by (PRATIO Y W U V r) rl P ABL
+ r2 P ABM + r3 P ABN
ifY is introduced by (ARATIO Y L M N rl r2 r3)
G=
2.w..Lp SU/V) ABV
-
..i.Jt..J..Lp Su/V) ABU
ifY is introduced by (INTERY (UNE U V) (UNE I J» _,,_l_(VULMNPABV YULMNV
VVLMNPABU)
ifY is introduced by (INTERY (UNE U V) (PLANE L M N» PPllyPAB~y({PA8(J
wv
ifY is introduced by (FOOT2UNEY P U V) Proof We only need to find the position ratio of Y with respect to UV and substitute it into the first equation of Proposition 4.66. For the second case, see Lemma 4.36. For other cases, see Subsection 4.3.1 for details. I
Chapter 4. MachIne Proof In SolId Geometry
204
From the above lemma and Proposition 4.66, it is easy to eliminate Y from P AYB . We leave this as an exercise.
Exercise 4.81 Try to eliminate Y from P AYB ifY is introduced by each ofthe constructions S2-S6.
Lemma 4.82 IfY is introduced by (FOOnUNEY P U V) then VABCY
Ppuv
Ppvu
'UVU
'UVU
= -;;-- VABCV + z:;-- VABCU'
Proof This is a consequence of Propositions 4.5 and 3.2.
Lemma 4.83 Let Y be introduced by (FOOnUNE Y P U V). Then
?,~~V DY _ -
EF -
.LD..EIL.JL
{
~PUVF
~v
'l,UVF ..2.LllLlL
s EUFV
if D E UV. D ¢ PUV ifD E PUV andE ¢ PUV if all points are coplanar
In all cases, we assume P is not on line UV; otherwise P Proof of the DY _ EF -
=Y
and ~~
= ~;.
The first and last cases are from Lemma 3.29. The second case is a consequence co-face theorem. For the third case, let T be a point such that DT = EF. Then DY _ ~ _ ~ _ ~ __ ~ DT - SDUTV - VDUVET - VEUVEF VFUVE' I
Lemma 4.84 Let Y be introduced by (FOOnUNE Y P U V). Then PPllyV~±PPYllVpARrr
S
ABY SCDE
=
2UV VPCDEA ~ VUCD v { pprry~~±ppyrrSARrr 2UV SCDE
if P is not in ABY. ifUV It' ABY. if P, U, and V are in ABY. .
Proof If P is not in ABY, by Proposition 4.3, ~sS CDE from Lemma 4.82. For the second case S ABY _ Vu ABYV SCDE - VUCDEV
= ~vv . PCDEA
Now the result comes
Vu ABV VUCDEV'
The last case is Lemma 3.24. By now, we have given methods of eliminating points introduced by constructions S2S6. We still need to deal with free points. By Lemma 4.59, volumes of tetrahedra can be reduced to volume coordinates with respect to four non-coplanar points. The following lemma will reduce the Pythagoras difference of free points to volume coordinates.
205
4.5 1be VoIlIIM Method
Lemma 4.85 Let O. W. U. and V be four points such that OW.lOUV. OU .lOWV. and OV.lOWU. Then
= OW2(~)2 + mr(~)2 + OV2(~)2 . Vowuv Vowuv Vowuv 2 2 2 (2) V6wuv = iliOW 0U 0V . (1) AB2
v
Proof (2) is from Propositions 4.77 and 3.15. For (1), let D, E, and F be points such that AD II OW, BE II OV, BF II OU, DE II OU, and DF II OV. Then
D
E
u Figure 4-33
By the co-face theorem, .:iQ. ow
= -~ =~ !!.§. = Yaawu and!}.!. = ~ I Vwouv Vowuv ' ov Vowuv ' ou Vowuv .
Now we have the main algorithm.
Algorithm 4.86 (Solid)
INPUT: S
= (Cl , C2 , . • • , Ck , (E, F)) is a constructive geometric statement.
OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof forS . S1. For i
= k, ··· , 1. do S2, S3, S4 and finally do S5.
S2. Check whether the nondegenerate conditions of C i are satisfied. The nondegenerate condition of a construction has three foons : A =/: B, PQ IY UV, PQ IY wuv . ....,.-=2 For the first case, we check whether P A BA = 2AB = O. For the second case, we check whether VPQuv = 0 and Spuv = SQuv. For the third case, we check whether Vpwuv = VQwuv . If a nondegenerate condition of a geometry statement is not satisfied, the statement is trivially true. The algorithm tenninates. S3. Let G l , ' S4
.,
,G. be the geometric quantities occurring in E and F . For j
= 1"
.. , 5 do
S4. Let H j be the result obtained by eliminating the point introduced by construction C i from G j using the lemmas in this chapter and replace G j by H j in E and F to obtain the new E and F .
206
Chapter 4. Machine Proof in Solid Geometry
85. Now E and F are rational expressions of independent variables. Hence if E = F, S is true under the nondegenerate conditions. Otherwise S is false in Euclidean solid geometry. Proof. The algorithm is correct, because the volume coordinates of free points are indeI pendent parameters.
For the complexity of the algorithm, let n be the number of the non-free points in a statement. If the conclusion of the geometry statement is of degree d, the output of our algorithm is at most of degree 5d3 n
Working Examples
4.5.2
Example 4.87 If a line divides two opposite sides of a skew quadrilateral in direct proportion, and a second line divides the other two opposite sides in direct proportion, the two lines are coplanar.
Constructive description «POINTS ABC D)
c
(LRAll0 E A B r,) (LRAll0 F D C
r, )
(LRAll0 HAD r,) (LRAll0 G B C r,) (VEFHG =
0»
The machine proof
Figure 4-34
The elirninants VEFHGg -
=
-VCEFWr,+VBEFWr,-VBEFH
~ -(-VBDEF ·r~+VBDEF·r,+VACEF ·r~-VACEF ·r,) simplify
=
(r2-1)-(VBDEF-VACEF) ' r,
n
=
(VCEFWr,-VBEFWr,+VBEFH)
VBEFH~VBDEF'r,
n
(r,-I),(VBCDE·r,-VACDE ·r, +VAcDE)·r,
=n (r,-I)·(O)·r,
VCEFH~(r,-l).VACEF
VACEF~(r,-l).VACDE VBDEF~VBCDE·r, VACDE~VABCD'I') VBcDE~(rl-l)·VABCD
sim!!J.ify 0
Example 4.88 Continue from Example 4.87. Let the two lines EF and GH intersect at I. Then !iJ.. =~ EF AD' Constructive description «POINTS ABC D) (LRAll0 E ABrIl (LRAll0 F D C (LRAll0 G B C r,) (INTER J (LINE E F) (LINE H G»
rIl
(LRAll0 HAD r,)
(ft = r,»
207
The Volume Method
..05
The eliminants
The machine proof l1.
-I
~
EF
...
.EL
I
-
!!. -
~=VSCE:H VSCFHg( ... -I),V A SCF
1
VSCEHg -
-( -VSCEH) r, ,(-VSCFH+VSCEH)
(VSCDE: ·... )
VASCF~ - (rl-l) ,VASCD) VSCDE~(rl-l).VASCD
-(-VSCPE ·,.,) ... ·(VSCDE · ... +V,u c r ... VASCF)
.im!!1i/v
VRCn& VSCDE · ... +VASCF ·... -VASCF
E. -
- ( ~-1)
~GYaa£.a.
= ....(-!eHa+l) ~ Q
1 -5....-''---
VaCQfi
VSCDE· ... -VASCD ·... ·rl +VASCD ·... +VASCD·rl-VASCD
~ VARcp-rJ -VASCO
-
VASCD·rl
.im!!1i/y
VASCD
1
Example 4.89 The sides AB and DC of a skew quadrilateral are cut into 2n P 2n and Ql Q2n respectively. Show that segments by points PI I •• • I
(2)
If sides BC and AD are cut into and SI I PnQn.
+ 1 equal
I ••• I
2m
+ 1 equal segments
by points
Rl,' "
I
R2m
S2m respectively. then the area of the quadrilateral formed by the lines Pn+lQn+l. RmSm. and Rm+lSm+l is (2n+l)1~2m+l)1 V ABCD• •• • I
Figure 4-35 shows the case n = m = 2. Note that in the following machine proof for (I). we use some different names for points P nl Pn+l l Qn+ll Qn. Constructive description
The eliminants
«POINTS ABC D)
v -(VpxYV-n+Vcxyv-n+Vcxyu) XYUV 2n+l V. u(n+l) ,VcDXY CXYU 2n+l
(UATIO X A B (UATIO Y A B (UATIO U D C (UATIO V D C
(VXYVU
2."ti)
f4'r)
2."ti)
f4'r)
= VASCD»
The machine proof =!::L=
v
v - VeDrY'n
VDXYU 2n+l V. Y -(Vsqpx ·n+VSCPX+VAcpx ·n) CDXY 2n+l x~
VACDX= 2n+l II: x -(n+l),VAsqp SCDX 2n+l p, p,
VASCD
~ -(-VDXyu ·n-Vqxyu ·n-Vqxyu)
-
VASCD·(2n+l)
Q
4VcDKv ·n2t4VcQKy ·ntVcpKY
-
VASCD·(2n+l)'
s, s. 1---.r--f--t--t-'1
Figwe 4-35
R.
208
Chapter 4. Madtine Proof In Solid Geometry
aimJ!.!.ify
!:. -
!i
Vcnxy VABCDo(2n+I)
-VacDx ·n-Vacnx-VAcox -n VABcDo(2n+l)2 ° -(-4VABco on 2 -4VABco on-VABco)
-
VA Bco o(2n+l)·
simElify
1
(2n+l)2
By Example 4.88, PnQn and Pn+IQn+I are cut into 2m + 1 equal segments by R;S;, i 1, 000, 2m respectively. Now (2) comes from (1) directly. I
=
A line joining the mid-points of two opposite edges of a tetrahedron will be called a bimedian of the tetrahedron relative to the pair of edges considered. The common perpendicular to the two opposite edges of a tetrahedron is called the bialtitude of the tetrahedron relative to these edges.
Example 4.90 The bialtitude relative to one pair of opposite edges of a tetrahedron is perpendicular to the two bimedians relative to the two other pairs of opposite edges.
/Y
• :x
Constructive description «POINTS x Y A e) (FOOT2LINE S A X Y) (ON B (LINE SA)) (FOOTILINE T
ex
B
Y)
(ON D (LINE T e)) (MIDPOINT N B
e)
(MIDPOINT Q A D) (PERPENDICULAR N Q
D
x
V))
The machine proof
T
Figure 4-36
The eliminants
fr..x..t:J..
PvXS=PyXA
PvXQ
PyXB= -
9.
Pyl(N tPvXD+tPvXA
PyXT=PyXC
(2)oPvXN PvXO+PvXA
PyXN=t(PyXB+PyXC)
-
!:.. -
!:!.. -
M. -
!2.
(2) o(tPvXB+iPvxc) PvXO+PvXA PyXBtPy"C PvXO+PvXA PYKatPy)(Q
-PvXT o ~+PyXT+PvXco~+PvXA _ -
-(PvXB+PyXC) -PyXC-PyXA
Pvxo= -
(Pyxs oU-PyXS-PyXAoU) SA SA oU ) (PvXT·U-PyXT-PvXC TC TC
PvXQ=2(PyXO+PyXA)
209
• ..5 The Volume Method
_B
-PyXS · u/'+PyXS+PyXC+PyXA · u/' SA SA
-
PyXC+PyXA
_ -
(
PyXC PyXA) PyXC+PyXA
.im~if~ I
Example 4.91 17 A plane parallel to AB and CD meets the edges AD. AC. BD. and BC in p. Q. and R respectively. The plane divides the tetrahedron into two parts. Let r be the ratio of the distances between AB. CD and the plane PQS. Find the ratio of the volumes of the two parts,
s.
First by the co-face theorem.
VAPQS AP r=---== VDPQS PD ' Since the volume VI of AB - PQRS is equal to V ABSR + V APRQ + V APSR • we will compute ~ VAPRq and ~ separately VABCO' VABCO'
VABCO
'
The eliminants
Constructive description
~g;~=r+l
ABC D) (LltA1l0 PAD if;) (INTER Q (UNI! A C) (PUNE P C D))
«POINTS
VBCOP=
~ r+1
(INTER S (UNI! B D) (PUNE P A B))
~=!:±!
(INTER R (UNI! B C) (PUN!! P Q S»
VABCP=
SABP
(~1;~~ ))
r
~ r+1
(~-I) .VBCOP VBDPQ
Scn~ SCOP
VBCPQ=
-r:gg
e
COP V:OPq VBPQS= .:!4B..Il SABP
B
VABCS=
A
Q
Vf:B~D ABP
R
(~-I)' VBCPQ S,AP
c
Figure 4-37
The machine proof ...!A.u.&.... -VABCO
.!.! -
§.. -
-VBPqS,VABCS -VABCO,(-VCPQS+VBPQS)
-(-VBOPQ)·(-VABCO),«_~))2 5
~l
~
VABCO ' ( -VBOPQ · ~-VBCPQ ' SABP +VBCPQ ' SABP
H( -
~»2 SABP
11Tbis is a problem from the 1965 International Mathematical Olympiad.
210
Chapter 4. Machine Proof 10 SaUd Geometry VBPPQ (VBPPQ+VBCPQ · !.ull.-VBCPQ)·!.ull.
simg1ify
SABP
SABP
(V BCDP ' ~ sene - V BCDP )(~)' . sene
Q .~2_V
-(V
BCDP SCDP
. ~-V
BCDP SCDP
. ~ . ~+V
ASCP SCDP SABP
. ~) . ~ . ~
ASCP SCDP
SABP
SCDP
(~-I).yBCPP
siml!!.ify
(VBCPP ' ~S -VBCPP-VABCP ' SSASD +VABCP)' !ABP CDP ABP ASP
1:.
(r)3 ,(-VABcp) ·(r+1)' (-VABcp ·r 3 -2VABcp ·r'-VA Bcp ·r) .(r+l )2
-
sim:e!.i!y ~ (r+1)2
Similarly. we can compute
VAPRQ VABC D
(r)2
= (r + 1)3
!
VAPSR VABC D
(r)2
= (r + 1)3 .
Thus
Let
TT
-
V2 -
VABCD
-
V; - .k±.LV 1 -
(r+1)3
ABCD·
p.InallY we have YL _ V. -
r'(r+3) 3r+1 .
Example 4.92 (Monge's Theorem) The six planes through the midpoints of the edges of a tetrahedron and perpendicular to the edges respectively opposite have a point in common. This point is called the Monge point of the tetrahedron.
Constructive description «POINTS ABC D) (MIDPOINT L A B) (FOOTILINE L, L C D)
D
(FOOTILINE A, A C D) (PRATIO L2 L , A, A 1)
(MIDPOINT R A C) (FOOT2LINE R, R B D)
Figure 4-38
(FOOTILINE A2 A B D) (PRATIO R2 R, A. AI)
(INTER P (LINE L L,) (PLANE R R, R2» (INTER Q (LINE L L2) (PLANE R R, R, »
s, S, B, B 1) S2» (MIDPOINT N C D) (PERPENDICULAR N M A B»
(MIDPOINT S B C) (POOT2LINE 5, S A D) (FOOT2UNE B, B A D) (PRATIO (INTER M (LINE P Q) (PLANE S
s,
The proof for this theorem is too long to print here. We need more elimination techniques to produce short and readable proofs for problems like this one.
4.6
211
Volume CoordIDllIe System
4.6
Volume Coordinate System
In Lemma 4.85, we use an orthogonal coordinate system, which is essentially the same as the usual Cartesian coordinate system. In order to do this, we have to introduce four auxiliary points 0, W, U, and V . In this section, we will develop some properties for a skew volume coordinate system in which any four free points can be selected as the reference points. As a consequence, we obtain a new proof of Lemma 4.85 and hence a new version of Algorithm 4.86. Let 0, W, U, and V be four non-coplanar points. Then for any point A, we will denote its volume coordinates with respect to OWUV as V OWUA XA=---, V owuv
It is clear that
XA
VOWAv YA=---, V owuv
V OAUV ZA= - - - , V owuv
V AWUV WA=--- · V owuv
+ YA + ZA + WA = 1. Following are some known results.
Proposition 4.93 The points in the space are in a one to one correspondence with the four-tuples (x, y, z, w) such that x + Y + z + w = 1. Proposition 4.94 For any points A, B, C, and D. we have YA
ZA
XB
YB
ZB
Xc
Yc
Zc
XD
YD
ZD
XA VABCD
= V owuv
1 1 1 1
As a consequence of Proposition 4.94, we can give the equation for planes in the volume coordinate system. Let P be a point in plane ABC. Then the volume coordinates of P must satisfy XA YA ZA 1 XB YB ZB 1 =0 Xc Yc Zc 1 Xp yp Zp 1 which is the equation for the plane ABC. The position ratio fonnula is still true Proposition 4.95 Let R be a point on line PQ and Tl ration of R with respect to PQ. Then
= ~ and T2 = ~ be the position
212
Chapter 4. Machlne ProoIIn SolId <>-try
Proof This is a consequence of Proposition 4.5. We now develop the fonnula for the square distance between two points.
Proposition 4.96 Let OXZIY - ZYIOIXI be a parallelepiped. We have 00 1
2
= OX2 + Oy2 + 7'JZl + PXOy + Pxoz + Pyo z . . ..-_...: . ..
Proof By Proposition 3.6,
'
,
2
OX I = 27'JZl + 2D? - zYl. 2 OY/ = 20X2 + 20Z - XZ2. 2 OZI = 20X2 + 20y2 _ Xy2. (1) XX 12 + YY/ = 27'JZl + 2Xy2. 00 1 2 + Z Z1 2 = 27'JZl + 20 ZI 2 = 40X2 + 40y2 + 27'JZl- 2Xy2.
,.:.--.,- __ . y¥--:.:.=---------.;.: --",·zI Figure 4-39
Let C be the center of the parallelepiped. By Proposition 4.66, 001 2 !Oy1 _ ~YYI2). Then
2
= 40C2 = 4( !OY2 +
2
+ yyI 2 = 40X2 + 20y2 + 47'JZl _
2Xy2.
(2)
2
+ XXI 2 = 20X2 + 4D? + 47'JZl- 2zYl.
(3)
00 1 Similarly, we have 00 1
Solving the linear system (1), (2), (3), we have
00 12
30x2 + 3D? + 30Z2 - zYl- ZX2 _ Xy2
=
-2
OX
~
==2
+ Oy + OZ + PXOY + Pxo z + Pyoz .
Proposition 4.97 Let 0, W, U, and V be four free points. Then -2
AB
-2
2~
2==-:2
2
OU (XB - XA) + OV (YB - YA) + OW (ZB - ZA) (YB - YA)(XB - XA)PUOV + (ZB - ZA)(YB - YA)PWOV + (ZB - ZA)(XB - XA)PWOU.
=
+
Proof We fonn a parallelepiped AM LN RPBQ such that AR II OW,AM II OU,and AN II OV. By Proposition 4.96,
Alf = AM2 + AN2 + AR2 +PRAN + PRAM + PNAM .
M
u
By Lemma 4.49, Figure 4-40
4.6
213
Volume CoordInate System
Alf = OW2( AR)2 = OW 2( VBOUVA? = OW2(ZB _ OW
VWOUV
ZA)2.
AN2 and AM2 can be computed similarly. By Proposition 3.9,
We can compute
P RAN
and
PRAM
similarly.
Once again with Algorithm 4.86, let E be an expression in volumes and Pythagoras differences of free points. Instead of using Lemma 4.85, we can use the following procedure to transfonn E into an expression in independent variables: if there are fewer than four points occurring in E then we need do nothing. Otherwise choose four free points 0, W, U, and V from the points occurring in E and apply Propositions 4.94 and 4.97 to E to transfonn the volumes and Pythagoras differences into volume coordinates with respect to OWUV . Now the new E is an expression in volume coordinates of free points, 2 2 2 2 OW , OU , OV , UV , and Vowuv . The only algebraic relation among these quantities is the Herron-Qin fonnula (Proposition 4.78). Substituting V6wuv into E, we obtain an expression in independent variables. Example 4.98 The above process of transforming an expression in volumes and Pythagoras differences of free points into an expression in free parameters becomes very simple when there exist exactly four free points 0, W, U, and V in a geometry statement. In this case. we first transform the square of the volume Vowuv into Pythagoras differences using the Herron-Qin formula (Proposition 4.78). and then express the Pythagoras differences 2 2 2 2 2 2 in terms of the six square-distances. Ow , OU , Ov , Uv , UW , and WV • which form a set offree parameters for this geometry statement. Exercises 4.99
2
2
2
l. If OV l.OU, OV l.OW, OW l.OU, and OU = OV = OW = I , the volume coordinate system developed in this section becomes the standard Cartesian coordinate system. Prove the following fonnulas in the Cartesian coordinate system. l. AB2 (XB - XA)2 + (YB - YA)2 + (ZB - ZA?
=
2. PABC
= 2«XB -
XA)(XB - xc) + (YB - YA)(YB - Yc)
3.
XA YA ZA 1 XB YB ZB VABCD = 6 Xc Yc Zc XD YD ZD
1 1 1 1
+ (ZB
- ZA)(ZB - zc)) .
214
Chapter 4. Macblne ProoIln Solid Geometry
2. Prove the Herron-Qin formula for tetrahedra (see Propositions 4.78 and 4.79) using the three formulas in the preceding exercise. (First notice that V ABC D
1
XB - XA
YB - YA
ZB -
= -'6
Xc - XA
Yc - YA
Zc - ZA
XD - XA
YD - Y A
ZD -
Let M be the matrix in the above formula. Then the transpose of M .)
V1BCD
ZA ZA
= islM x
M Ol
where
MO
is
Summary of Chapter 4 • Signed volumes and Pythagoras differences are used to describe some basic geometry relations in solid geometry: collinear, coplanar, parallel, perpendicular, and congruence of line segments. 1. Four points A , B, C , and D are coplanar iff V ABC D
2. PQ
II
V QABC
or equivalently iff V PABCQ = O.
= V QABC = V RABC • PQ.l..AB iff PPAQB = PPAB - P QAB = O.
3. PQR 4.
ABC iff V PABC =
= O.
II
ABC iffVPABC
• We have the following formulas for the volumes of tetrahedra. 1. V ABCD = ihA,BCDISBCDI where hA,BCD is the signed altitude from point A to plane BCD. 2. VABC D
= V owuv
XA
YA
zA
1
XB Xc
YB Yc
ZB Zc
11
xD
YD
ZD
1
where X A, YA, Z A are the volume coordinates of point A with respect to points 0 , W, U, and V . 3. (The Herron-Qin Fonnula)
= 4rf2rf3rf4 - rf2 P 314 where rij = IPiPjl , Pijl, = PP, PjP. , 144VAp2P,p.
rf3 P 214 - ri4 P 312
4. (The Cayley-Menger Fonnula)
288VAp,p.p.
=
0 r~l r~l 1
rf2 rfa rf4 0 r~ r~4 r~2 0 rL 1
1
1
1 1 1 0
+ P314P214P312
4.6
215
Volume CoordIoale System
• The fonowing basic propositions are the basis of the volume method.
1. (The Co-vertex Theorem) Let ABC and DEF be two proper triangles in the same plane and T be a point not in the plane. Then we have ~ = ~s . rTDE,.. DE,..
2. (The Co-face Theorem) A line
PQ
and a plane ABC meet at M. If Q
i- M.
we have PM
V PABC
PM
V PABC
QM
V QABC
PQ
V PABCQ
= = - -; -=-=
3. Let R be a point on line
PQ .
4. Let R be a point in the plane V RABC
QM
V QABC
PQ
V PABCQ
; -=-=
.
Then for any points A . B , and C , we have
PQS.
Then for any points A, B, and C we have
SPQR SRQS SPRS =S V SABC + - S V PABC + - S V QABC . PQS PQS PQS
5. Let PQTSbe a parallelogram. Then for points A,B, and C. we have VPABc + V TABC
= VQABC + V SABC • or VPABCQ = VSABCT .
6. Let triangle ABC be a parallel translation of triangle DEF. Then for points P and Q we have V PABC = VPDEFA and VPABCQ = VPDEFQ ' Also notice that an the propositions on page 165 about Pythagoras differences are also valid if the points involved are in the space. • We present a mechanical proving method which can produce short and readable proofs for many constructive geometry statements in the space.
Chapter
5
Vectors and Machine Proofs In Section 2.6, we mentioned that there are two approaches to defining geometries: the geometric approach and the algebraic approach. In this chapter, we will show how to prove geometry theorems automatically in a geometry that is defined using the algebraic approach. The modern source for affine and metric geometries based on the algebraic approach is linear algebra or the theory of vector spaces (see [5, 16,33]). In this approach, the metric is introduced by the inner product of vectors, while the areas and volumes are represented by the exterior product of vectors. It is interesting to note that two of the most important concepts in this linear algebra approach to geometry, the inner and exterior products of vectors, are essentially the same as the two basic geometry quantities used by us: the Pythagoras difference and the area (or volume). This strongly suggests that the method based on areas, volumes, and Pythagoras differences can be stated in the language of vectors. Moreover, the vector approach based on inner and exterior products has the advantage that it is easy to develop; it uses more geometry quantities such as the vector itself; and it covers more geometries such as the Minkowskian geometry. But on the other hand, the vector approach needs more algebraic prerequisites. Also, the proofs produced by the vector approach generally do not have such clear geometric meaning as do those produced by the volume-Pythagoras difference approach.
5.1
Metric Vector Spaces of Dimension Three
Let £ be afield with characteristic different from two. A vector space over £ is a set V with two structures:
• V x V
• £ xV
--+ ---+
V, denoted by (x , y) V, denoted by (a , x)
---+ ---+
x
ax
which satisfy the following properties:
VI x+y =y+x
+ y and
(commutative law)
217
218
Chapter S. vectors and MachIne Proofs
V2 (x
+ y) + z = X + (y + z)
V3 There exists a zero-element origin ofV.
(associate law) 0
such that x
+0 =x
for every x E E.
V 4 To every element x there exists an inverse element: -x such that x V5 (Cif3)X V6 (Ci
is called the
+ (-x) = o.
= Ci(f3X)
+ (3)x = CiX + f3x; Ci( x + y) = CiX + Ciy
V7 1 · x
0
(distributive laws)
= x.
The elements of V are called vectors and are denoted by x, y, z,·· ·; the elements of £ are called scalars and are denoted by Ci, f3, a, b, . . .. Scalars are always written on the left of the vectors. The vector space V is called n-dimensional if there exist n elements e" . .. ,en in V such that • for any x E V there exist scalars • if Cilel
Cil, •. • ,
Ci n such that x
= Cil e l + ... + Ci nen and
+ ... + Cine n = 0 then Ci; = 0, i = 1, ·· · ,n.
The n elements e I, • •• ,en satisfying the above property form a basis for the vector space V. If x = Cil e l + .. .+Cine n, we say that x is a linear combination of the vectors e" . .. , en and (Cil, ... , Ci n ) are the coordinates of x with respect to the basis e" ... , en. Given an ordered basis for V, we can associate with each vector x the unique n-tuple of coordinates (Cil , ' . . , Ci n ) such that x = Cilel + ... + Cine n. This establishes a one-to-one correspondence between the vectors in V and the elements in £n which is the Cartesian product of £ with itself n times. It is easy to show that this correspondence preserves the structure of the vector spaces, Le., it is an isomorphism between vector spaces. So we may safely assume that V is actually £n . A nonempty subset W of a vector space V is called a subspace of V if the following conditions hold:
1. If x E Wand YEW, then x
+ YEW .
2. If x E W. then CiX E W for Ci E £. Let /1' ... , / m be vectors in V. Then the set of all the vectors like m
L Cid;,
Cii
E
£
;=1
is a subspace of V and is called the subspace generated by vectors /" . . . ,/m. This chapter uses some basic knowledge from linear algebra, e.g., the first five chapters of [22].
5.1.
219
MdJ'k Vector S..,.. 01 DimcDslOD Thnc
Inner Products and Metric Vector Space
5.1.1
In what follows, we assume that n = 3 and an ordered basis for V has been given. For x = (Xl, X2 , X3), Y = (Yl , Y2 , Y3), and 0 E E, we thus have
Dermition 5.1 An inner product on V is a map
V xV
--+
E, denoted by (x, y)
--+
(x, y)
which satisfies the following properties
11 (x , y) 12 (ox
= (y , x) .
+ fJy, z) = o(x, z) + fJ(y, z) where 0
and fJ are scalars.
Proposition 5.2 Let (ell e 2, e 3) be a basis of V, x + Y2Y2 + Y3Y3' Show that
= Xl e. + X2e2 + x3e3'
and y
=
Yl y.
(z,o)
~ (x"x"x,)M ( ~ )
where M = ((ei ' ej)) is a symmetric matrix. Proof
where M
= (( ei, ej) ) is a symmetric mattix.
Dermition 5.3 A vector space with an inner product is called a metric vector space. Definition 5.4 Two vectors x and yare perpendicular if (x, y)
= O.
A metric vector space is called nonsingular if its origin is the onJy vector which is orthogonal to all vectors. Proposition 5.5 Metric vector space V is nonsingular ifand only iflMI
= det(M) =f. O.
220
Chapter 5.
Proof If x
VcctDn ad Macblae Proofs
= (x }, X2, X3) and Y = (Yl, Y2, Y3), (z,O)
~ (x"x"x,)M (= ).
Then x is perpendicular to all vectors in V iff
(x"x"x,)M (
E) ~
0
for all possible Yi, i.e., iff
= (0,0, 0). The above linear system has nonzero solution iff 1M I = O. (Xl,X2 , X3)M
A vector x is called isotropic if x is perpendicular to itself, or equivalently if (x , x)
= O.
The origin 0 is always isotropic. Even if V is nonsingular, there may be many nonzero isotropic vectors. This is seen by observing that a vector is isotropic iff its coordinates satisfy the equation n
L
miJxiXj
=0
i.,j=1
where M = (mi,j) ' The solutions for the above equation (if exist) consist of a cone which is referred as the light cone. The word light cone is originated from physics. More physics background can be found in [38]. For x
= (Xl , X2 , X3), let the square of x be x2
= (x , x).
Suppose that V is a metric vector space, but that we only know the value of x 2 for each vector x E V. Can we compute the inner product (x, y) for all x and y? The answer is affirmative.
Proposition 5.6 In a metric vector space, the square function x 2 determines the inner product completely. Proof For x , y E V, by Il and 12, (x
+ y)2 = x 2 + 2(x, y) + y2 .
Since & is not of characteristic 2, we have
221
Metric: Vector S~oIDimeDs1oD 'Ibree
5.1.
CoroUary 5.7 (Pythagorean Theorem) z.Ly
iJf z 2 + y2
- (z
+ y )2 = O.
Definition 5.8 A coordinate system e" e,. , e 3 of V is called a rectangular coordinate system if e j.L e j for j :I i. For a rectangular basis e" e,. , e 3 • the matrix defining the inner product is diagonal. i.e .•
(e"e,) ) .
Proposition 5.9 A metric vector space V always has a rectangular coordinate system.
Proof We prove the proposition using induction on the dimension of V. H V is of dimension one. any basis of it is a rectangular basis. Suppose that the result is true for all vector spaces with dimension less than n. Let V be a vector space of dimension n. Hall the vectors in V are isotropic. by the Pythagorean theorem any basis is a rectangular basis. Otherwise. let e, •... , en be a basis of V such that e, is a non-isotropic vector. Let
e~ = •
j
ej - (e ,e , ) e"i = 2" " ,no e, 2
Then it is clear that ell e~, . .. , e~ are also a basis of V and e,.Le~, i = 2"" , n. By the induction hypothesis. the vector space generated by e~, i 2, . . . , n. has a rectangular basis Ij , i = 2"" , n . Then it is easy to check that ell 1,."" , In fonn a rectangular basis for
=
V.
I
Exercises 5.10 1. H £ is the field of real numbers and the inner product of z (Yt, Y2 , Ya) is defined to be
= (Xt,X2,X3) and y =
the resulting space is the Euclidean space of dimension three. Show that the Euclidean space as defined above is a nonsingular metric vector space satisfying 13 (z, z) ~
°
and (z, z)
= 0 iff z = (0,0,0).
222
Chapter 5. vectors and Madtlne Proofs
2. The n-dimensional (n ~ 3) Minkowskian space is a metric vector space whose inner product for x = (Xl, .. . , Xn) and y = (Yl>" " Yn) is
(:e, y)
= XlYI + .. . + Xn-lYn-1 -
XnYn '
Show that the Minkowskian space is a nonsingular metric vector space in which there exist nonzero isotropic vectors. 3. Show that if E is the field of real numbers and n = 3, every nonsingular metric vector space has a coordinate system such that its matrix is one of the following fonn.
00) 1o 01 , M2 = (100) 00 10 -10 o o
-1
0)
0 -1
,M 4
=
(-1 0 ~ ). 0 0
-1 0 -1
The matrices Ml and M2 determine, respectively, the Euclidean and the Minkowskian spaces. We call the geometries determined by M4 and M3 the negative Euclidean space and the negative Minkowskian space respectively. 4. Let (e" ... , en) and (/" ... , In) be two different bases for V. Then there is anonsingular matrix 'P such that
Let M and M' be the matrices of the inner products corresponding to the bases (e" ... , en) and (/" .. . , In). Show that
M'
= 'P' M'P
where 'P' is the transpose of 'P. The two matrices in the above proposition are called congruent. Thus two matrices are congruent iff they represent the same metric of V relative to different coordinates systems. Therefore, the study of the metric vector space is equivalent to the study of the symmetric matrices under the equivalent relation of congruence. 5. In the algebraic language, Proposition 5.9 is equivalent to the following fact. Let g be an n x n symmetric matrix. Then there exists an n x n nonsingular matrix 'P such that 'P' G'P is a diagonal matrix. Prove the above fact directly.
5.1.
223
Metric Vector S.-es 01 Dlmeasloa Three
Exterior Products in Metric Vector Space
5.1.2
In what follows, we always assume that V is a nonsingular metric vector space with a rectangular basis (e" e" e 3). Thus the matrix that defines the inner product is
Definition 5.11 An exterior product on V is a map V xV
--+
V, denoted by (z, y)
--+
[z, y]
which satisfies lhe following properties
= -[y,z]. [oz + fly, z] = o[z, z] + fl[y, z] where 0
El [z,y] E2
and fl are scalars.
E3 z.l[z, y]. Note that property E3 is not in the definition for the exterior product in the general case. We add it to make the relation between the inner and exterior products simple. From E I and the fact that £ is not of characteristic two, we have
[z,z] =0. Proposition 5.12 Let (e" e" e 3) be a rectangular basis of a nonsingular metric vector space V . Then
Taking the exterior products of e. - e, and the vectors on both sides of the above equation, we have s3(e" e.) = s2(e" e,), i.e., (e" [e 3, e,l) = (e" [e., e 3 l) . Similarly, we can prove that (e 3 l [e" eol) (e., [e 3, e,l). I
=
224
Chapter S. Vecton aDd MachlDe Proofs
Remark 5.13 The constant 0 = (e" [e" e 31) is a basic quantity related to the exterior product. We alway~ assume that 0 :I O. Proposition 5.14 Let (e" e 2, e 3) be a rectangular basis of a nonsingular metric vector space V. x = Xl e 1 + X2e2 + X3e3. and y = Yl e 1 + Y2e2 + Y3e3' Then [x, y] = 0(_1 ml
I
X31 , 2..1 X3 Xl 1,2..1 Xl Y3 m2 Y3 Yl m3 Yl
X2 Y2
X21) Y2
Proof. By EI and E2, [x , y]
=
t
x ;y; [e; , ei]
;,;=1
= I X2 Y2
I
X31 [e 2,e3] + X3 Xl Y3 Y3 Yl
I[e3, e + I XlYl 1]
X21 [e"e 2]. Y2
Now the result follows immediately from Proposition 5.12.
Proposition 5.15 If the metric vector space is not singular. then x = oy
iff [x, y] = O.
Proof. By Proposition 5.9, we can chose a rectangular basis for V. If x = oy then [x, y] = o[y , y] = O. Conversely, let us assume [x , y] = O. By Proposition 5. 14,we have
Xl Yl
for a scalar
>.. Thus x
=
= X2 = X3 = >. Y2
Y3
>.y.
Definition 5.16 The triple scalar product for three vectors x , y. and z in V is defined as follows.
(x, y, z)
(x , y , z)
= ([x , yJ, z) .
= (e" [e 2, e 31)
Xl X2 X3 Yl Y2 Y3 Zl
Z2
Z3
Proof. Since V has a rectangular basis, this is a direct consequence of Proposition 5.14. I
We thus have
TI (x,y , z)
= (y , z,x) = (z , x , y) = -(x , z , y) = -(z,y , x) = -(y,x , z).
5.1.
225
Metric c-tne. 01 DImmsIoD Three
=
T2 In a nonsingular mettic space, (z, y, z) 0 iff vectors z, y , and z are coplanar, i.e., iff there exist scalars 01,02 , 03 not all zero such that 01 X + 02Y + 03Z O.
=
Proposition 5.18 • (The Lagrange Identity) ([x ,
• [[x , Yl.
zl = o«x , z)y -
YI, [u, v]) = o( (x, u}(y, v) -
(x , v}(y , u» .
(y , z)x)
Proof The two fonnulas can be obtained by direct computation. We leave them as exercises. I
Exercise 5.19 Show that [[r., r21, [r3' r,,11 is the same as in Proposition 5.18.
5.2
= o( (r" , [r., r
2
])r3 - (r3 ' [r., r 2 ])r,,) where 0
The Solid Metric Geometry
Let & be a field with characteristic different from two. The vector space &3 is also called the affine space associated with field &. Dermition 5.20 A non-singular metric vector space &3 is called a solid metric geometry. As usual, elements in &3 are called points. Let A and B be two points. Then line passing through A and B is the set
{oA + .BBI
0
+.B = I} .
The plane passing through three points A , B, and C is the set
{oA+.BB+,,),CI
0+.B+")'
= I}.
Two points A and B in &3 determine a new vector, --+
AB=B-A A being the origin and B being the endpoint. Thus two vectors
and only if A + Q = P Thus we also have
+ B.
AB and PQ are equal if
--+ Let 0 be the origin of V. For any point A, let -A = OA. --+ -B - -A . AB=
226
Chapter S.
Vecton aDd Machine Proofs
It is easy to show that line AB can also be written as follows
--
{A + ,6ABI _2
Line AB is called isotropic if AB
,6 E t'} .
= O. Similarly, the plane ABC can be written as
{A+,6AB+,),ACI
,6,,),Et'} .
- -
>. such that AB = >'PQ. AB = >.PQ, we say that the ratio of the parallel line segments AB and PQ is >., i.e.,
Line AB is called parallel to line PQ if there is a scalar
AB PQ
If
= >..
Q
p
" ~c ,,
,,
A
Figure 5-1
Figure 5-2
To see the geometric meaning of the addition of two vectors point such that
AB and PQ.
Let C be a
BC = PQ. Then (Figure 5-1) ----;:::-;:t-;-::t
AB + PQ
= AB + Bt; = At;.
-
We will now give a geometric interpretation of the coordinates of vectors with respect to a basis. Let 0, W, U, and V be four points not in the same plane. For any vector AB, we form a parallepiped AMLN - RPBQ (Figure 5-2) such that AR II OW, AM II OU, and AN II OV. Then --+ --+ AR --+ AM --+ AN --+ AB = AR + AM + AN = =OW + =OU + =OV, OW . OU OV ---+~
~
~
i.e., OW , OU, and OV form a basis for t'3 and the coordinates of AB with respect to this · (AR AM · A N) b aslS are OW, au' OV .
Exercise 5.21 Show that point Y is on line AB iff there is a scalar a such that AY Point Y is on plane LMN iff there are two scalars
= aAB;
aand,6 such that IX = aIM + ,6LN.
5.1.
227
Metric Geomebies 01 DimeDSloa Thne
5.2.1
Inner Products and Cross Products
--+ and CD The inner product of vectors AB satisfies --+ 1. (AB , CD) ~~
2. (AB , CD) ~
3. (0' A
= 0 if and only if AB.lCD. ---+~
= (CD,AB),
- t ---+
----+
-t
- t .-..+
+ (3 B , CD) = 0'( A, CD) + (3( B, CD) where 0' and (3 are scalars.
The square distance between two points A and B, or the square length of the vector is defined to be
AB2
--+2
AB,
--+--+
= AB = (AB,AB).
By Proposition 5.6,
Then it is easy to check that P ABCD
--+ BD) = -2(AC, .
Proposition 5.22 (Pythagorean Theorem) For any points A , B, C , and D
• AB.lBC iff AB2 + BC 2 - AC2 = O• • AB.lCD iff PACBD
= AC2 -
CB 2 + BD2 - AD2
If four points A , B , C, and D are collinear or AB segments is --
AB· CD
= O.
II CD, then the product of the oriented
--+= (AB,CD)
and the ratio of the oriented segments is
~
AB
--+ (AB,CD)
CD
= (cD,cD)
---+
~
-+
The exterior product [AB , CD] of Al1 and CD satisfies the following properties -+-
1. [AB, CD]
= 0 iff AB II CD.
228
Chapter 5. Vectors and MadJlne Proofs ---+
-+0
2. [AB,CD] ~
3. [a A
- + ---+
= -[CD , AB] .
--+ ---+
--+ ::::-::t
--+ - +
+ f3 B , CD] = a[ A , CD] + f3[ B , C 1)] where a
and f3 are scalars.
By Lagrange's identity, --> -:-::t 2 --> -:-::t a [AB,Al;] = a(AB2 . AC2 - (AB,AGY) = '4(4AB 2 • AC2 - P~AC)
where a is the constant defined in Proposition 5.18. Comparing with the HelTon-Qin formula on page 105, we see that the length of [AB, AC] is proportional to the area of triangle ABC. Remark 5.23 We can determine the exact relation between the area and the exterior product as follows. In Euclidean geometry. a = 1. By the Herron-Qin formula, --> --> 2
[AB,AC] ---+ ---+
---+
= 4S~BC' ---+
Thus [AB, AC] is a vector AD such that AD is perpendicular to the plane ABC. -->-:-::t--+
and pointed in such direction as to make (AB , AG' , AD) a right handed triple and IADI = 2ISABcl· We thus define the signed area of triangle ABC to be a quantity with the same sign of --> -:-::t
[AB, A<..:] and S~BC
--> -:-::t 2 = HAB, At.:] . Then Heron-Qin's formula in any metric geometry is
S~BC
= ~ (4AB2 . AC2 - P~Ad.
Two planes ABC and PQR are parallel if [AB, AC]II -->-:-::t
[PQ, PR].
Let A be the scalar
-->-->
ratio of these two parallel vectors, i.e., lAB, AG'j = AIPQ, P R] . Then A = &.aa. sPQR . We thus have A = SABC
= (lAB, AC], [PQ, PRJ) ([PQ, PRJ, [PQ, PRJ)'
SPQR
The volume of the tetrahedron ABC D is defined to be one sixth of the triple scalar product: VABCD
1
--+
--> -->
= 6'(AD, [AB,ACJ).
As a consequence
1
-+
-+-+
VABC D = 6'( (D , [A , B J)
-+
-+-+
-+
-+-+
+ (D , [B , C J) + (D , [ C , A J) -
-+-+-+
(A, [B , C J»).
229
5.l. Metric Geometries 01 DlmensioD Three
Starting from several points in the space, we can form vectors and inner and exterior products of these vectors. Since exterior products of vectors are still vectors, we can further form the inner and exterior products of these new vectors. Expressions thus obtained are called recursive expressions in inner and exterior products of vectors. It is clear that a recursive expression in inner and exterior products of vectors can be a scalar or a vector. A --+
vector like AB for points A and B is called a simple vector.
Proposition 5.24 Any recursive expression in inner and exterior products of vectors can be represented as a polynomial in inner products of simple vectors, exterior products of simple vectors, and triple scalar products of simple vectors. Proof By Proposition 5.18, for any vector riO r 2, r3 and r 4
1. [[riO r 2], r 3] = a( (r" r3)r2 - (r2' r 3)r,). 2. (The Lagrange Identity) ([r" r 2], [r 3, r 4])
= a( (r" r 3)(r2, r 4) -
(r" r 4)(r2, r 3)).
By repeated use of the above two identities, any recursive expression in inner and exterior products of vectors can be represented as a polynomial of inner products of simple vectors, exterior products of simple vectors, and triple scalar products of simple vectors. I Exercises 5.25 I. Show that with the above definition for the ratio of lengths, signed areas, signed volumes, and the Pythagoras differences, Axioms A.I-A.6, S.1 -S.5, and the properties of the Pythagoras difference are true. 2. Prove the following formula of the distance from a point A to a line PQ
3. Prove the following formula of the distance from a point A to a plane LM N. --+
d2
_ A,LM N -
---+ --+ 2
(LA, [LM,LN]) ---+ --+ 2
•
[LM,LN] 4. Prove the following formula of the distance between two skew lines UV and PQ
cP.
_ 9(UV,[UP,UQ]))2
UV,PQ -
--+ --+ 2
[PQ,UV]
•
230
Chapter S.
5.2.2
vectors and Maddne Proofs
Constructive Geometry Statements
The constructive statement defined in Section 4.2 can be generalized by considering more constructions and more geometry quantities.
Definition 5.26 By geometric quantities we mean vectors, the inner or exterior products of vectors, or the quantities which can be represented by the inner and exterior products of vectors. With the geometry concepts introduced in the preceding subsection, constructions S 1-S7 on page 177 are still meaningful in our metric geometry, except for constructions S6 and S7 whose ndg conditions need modification. We will also introduce a new construction S8. S6 (FOOT2LINE Y P U V) Point Y is the foot from point P to line UV. Point Y is a fixed ----t
2
#
point. The ndg condition is UV ----t 2
UV
O. Notice that in the general metric geometry
# 0 is not equivalent to U # V.
S 7 (FOOT2PLANE Y P L M N) Point Y is the foot of the perpendicular from point P to plane LM N. The nondegenerate condition is [LM, _
LNj2 # O. _
----t
S8 (SRATIO A L M N r) Take a point A such that LA = r[LM, LN], where r can be a rational number, a rational expression in geometric quantities, or variables. If r is a fixed quantity, A is a fixed point; otherwise, A has one degree of freedom. The ndg condition is [LM,
LNj2 # O.
Two basic geometric relations, parallel and perpendicular, can be easily described by the exterior and inner products. For instance, to represent AB II CD we need two equations VABCD = 0 and SACD = SBCD. But using exterior product, we need only one equation ----t
---+
[AB,CD]
= O.
Proposition 5.27 We have 1. AB.l. CD 2. AB
II CD
----t
{=:::}
----t
{=:::}
---+
(AB, CD) ---+
[AB, CD]
3. A, B, and C are collinear
4. AB.l. PQR
{=:::}
= o.
= o. ----t
{=:::}
----t
[AB, AC]
[AB, [PQ, PHil
= O.
= O.
5.3.
231
Mxhiae ProoI by Vector c.Icu1atiOD
5. AB
II
--+
--+ --+
PQR <=> (AB , [PQ , PRj)
= O.
6. ABC.L PQR <=> ([AB , AG],[PQ,PRj)
7. ABC
II PQR
<=> [[AB , AG] , [PQ , PHil --+
= o.
= o.
--+
~
8. A. B. C. and D are coplanar <=> (DA , [AB , AGj)
= O.
Proof. The first two cases are from the definition. The other cases are consequences of the first two cases. I
Example 5.28 (The Centroid Theorem for Tetrahedra) Let G be the centroid of tetrahedra ABC D. Show that
G=
--+
--+
--+ --+
At B t C ± D .
This example can be described constructively as follows.
Constructive description « POINT'S AB C D ) ( MIDPOINT 5 B C)
c
( \.RAllO Z A 5 2/3 )
(!.RAllO Y D 5 2/3)
(00CIt G (LINE D Z) (LINE A Y» --+ --+ --+ --+ --+ ( G A±BtC±D »
A
=
The ndg conditions: B:f: C, A :f: S, D :f: S, and
B
Figure 5-3
DZ IY AY.
5.3
Machine Proof by Vector Calculation
As before, we will give methods of eliminating points from the geometry quantities.
5.3.1
Eliminating Points From Vectors
We first consider how to eliminate points from vectors.
Proposition 5.29 Let R be a point on line PQ (P :f: Q). Then -+
R
PR-+
RQ-+
= PQ Q + PQP '
Chapter s.
232 -
--to
--+
-
Proof We have ;~Q +~P
-
--+
= ;~(Q
--+
--+
- P)+ P
--+
--+
-
Vectors 8Dd MachIne Proofs
= ;~PQ+ P
---+
--+
=PR+ P
--+
= R·I
Lemma 5.30 Let Y be introduced by (PRATIO Y W U V r). Then --+
Y
----t
Proof Since WY
--+
=W
--+
- Y
--+
=W
--+--+
+r(V - U) -+
--+-
-+
= rUV. we have Y = W
Lemma 5.31 Let Y be introduced by (ARATIO Y L M N --+
Y
--+
= rl L
---+
- rUV. rl r2
r3). Then
--+
--+
+r2M +r3N.
Proof Since Y is in the plane LM N. we have
where
Cl
and C2 are some scalars. Then ---+---+
[LN,LY]
---+---+ --+---+ ---+---+ = cdLN, LM] +c2[LN,LN] = cdLN,LM].
Lemma 5.32 Let Y be introduced by (INTER Y (UNE U V) (UNE P
Q».
Then
-Y =SUPQ - - -SVPQ--V -U. SUPvQ SUPvQ
Proof By Proposition 5.29,
UY- VYY==V-=U. UV UV Let r
---+---+ = -~~. Then UY = rUV and --+ --+
-+ --+
[UY , PQ] = r[UV, PQ]. --+ ---+
Since [UY , PQ] Vy_~
iiii -
SUpvq·
-+ ---+ --+ --+ -+ = [UQ,PQ] + [QY,PQj = [U,-!,PQ]. we have r = sSUpq . upvq ~
I
Similarly
Lemma 533 Let Y be introduced by (INTER Y (LINE U V )(PLANE L M N)). Then
Proof: By Proposition 5.29.
-
. UY--, Y =v+ Let r =
UV
-
-
=YUV TV j
K.Then U Y = rUV and 4
d
d
-
4
4 - +
[ U Y ,[ L M ,LN]]= r[UV,[ L M ,LN]]. Since 4
d
4
+ - 4
[ U Y ,[ L M ,LNII = [UL,[ L M ,LNI] + [ L Y ,[ L M ,LNII = VULMN, = we have r = Similarly I VULMNV '
m.
Lemma 534 Let Y be introduced by (FOORLJNE Y P U V ) . Then
y =-PPUV7 Puvu
+
+ Proof: By Reposition 5.29, Y =
+
ppvu puvu
Tj
-. -+ V U . Let r =
g,or W = rUV. W e have 4
++
4
r ( U V ,U V ) = (W,U V ) = (UP,U V ) + ( P Y ,U V ) = ( U P ,U V ) . d
Thenr =
j
e.
4 4
Similarly
-
=
4 4
e.
4 4
I
Lemma 5 3 5 Let Y be introduced by (SRATIO Y L M N r). Then +
Y =
t+ r [ L M ,L N ] . - 4
Proof: This is the definition o f the construction SRATIO.
I
Lemma 5 3 6 Let Y be introduced by (FOORPL4NE Y P L M N). Then + ~VPLMN -' 4 Y=P+,, [ L M ,L N ] . [ L M ,L W 2
+
--
8 = r [ L M ,LN].
*-
+ 4
4 - 4
Then ( L P ,P Y ) = r ( L P , [ L M ,L N ] ) = 6rVLMNP. ( ~ p , p y=)-py2 = - P-. Thus r = [LM,LNp [LM,LNp
Proof: Let + 4
36V2
234
Chapter 5. Vectors ad MaddDe Proofs
Example 5.37 LetY be introduced by (INTERY (PUNEW U V)(PUNE RPQ». Show
tluJt -Y
=W U). - + SWPR9(V -- SUPV9
Proof Take points X and S such that ~~ ~
--+
~
= 1 and ~ = 1. By Proposition 5.29,
--+
--+
--+
--+
--+
--+
Y =rX +(l-r)W=r(X -W)+W=r(V - U)+W
where r
= wx WY = SWPR9. SUPV9
Example 5.38 Continue from Example 5.2S. We can actually derive the result of this example without knowing it previously. Constructive description «POINTS ABC D) (MIDPOINT S B (LRAll0 Z A S (LRAll0 Y D S (INTER G (LINE
C)
2/3) 2/3) D Z) (LINE A Y))
(0' ))
The eliminants
7!
7!G z
Q - 7 ,SAOY± 7l ,SAZY -
SAZY~
Y 17!' ,SADS+ 1 D .S ADZ iSDSZ+'SADZ
=' ~
("5 ,SADS+fD,SADS+fA ,SADS fSADS
simJ!lify 1 ( 2S 4
1.. -
-
+D+
.SA~:~y~ 'Suy
SADYZ~H2SDSZ+3SADZ)
-SADYZ
-
5.3.2
- -
The machine proof.
A)
-i(SADZ)
SADY~HsADS) SDsz~i(SADS) SADZ~~(SADS) 7!'~i(27 +A)
D+c'tIl+A 4
7G;Kc+ Il)
Eliminating Points from Inner and Exterior Products
Let Y be introduced by one of the constructions SI-S8. By Lemmas 5.30-5.36,
(1) for vectors r , and r, and scalars a and /3. To eliminate Y from the inner product, let us note that ----t
----t
(AB, CY)
--+ --+
--+ --+
= ( B, Y) + (A , C) -
--+ --+
--+ --+
(B , C) - ( A, Y). --+ --+
--+ --+
Then we need only to consider how to eliminate Y from ( A, Y) and (Y , Y). If A
i: Y,
235
MadlIae Proal by Vector Cakulatloa
5.3.
-+ -+
For (Y , Y), we have
(y , Y) = < or, + fJr~ , or, + fJr~ >= o2(r" r,) + fJ2(r~ , r~) + 2ofJ(r" r~) .
lAB,cYJ,let us note that
To eliminate point Y from the exterior product ---+ ---+
lAB , CY] If A
--+ -+ -+ = I-+ A , C J + IB , Y J -
--+---+
--+ --+
I A , Y J - I B , C J.
-+ = Y we have I-+ A , Y J = 0; otherwise we have ---+ --+
lA , YJ
---+
--+
= alA , r ,J+ fJlA , r~J .
Since other geometry quantities can always be represented as a rational expression in inner and exterior products, we can eliminate points introduced by constructions SI-S8 them. Here are some examples. Example 5.39 Let Y be introduced by (SRATIO Y L M N r). By Lemma 5.35, we have
= ----+ ----+ (YB ,CD) = = (YB,YC) = = ----+ --+ IYB ,CDJ = = VYBCD
---+
---+
~
(BY, IBC, BDJ) ---+
~
= VLBCD -
~
---+
BC
~
r(ILM , LNJ, IB ,BDJ)
---+ ---+
( B , CD) - (Y ,CD) ~
---+
----+ --+
---+
(LB , CD) - r(CD , ILM,LNJ) . ---+--+
--+--+
---+---+
--+---+
(B , C)+(Y , Y)-(Y,B)-(Y,C) ~ ~
(LB ,LC) --+ ---+
~
--+
----+ --+
----+ ---+ 2
+ r(BL+ CL,ILM ,LNJ) + r2ILM , LNJ --+---+
IB,CDJ -IY , CDJ --+ ---+
[LB ,CDJ
---+
---+--+
+ rICD,ILM,LNII.
Example 5.40 Let Y be introduced by (FOOT2PlANEY P L M N). By Lemma 5.36,
VABCD ---+ ---+
(AB ,CD) ---+ --+
(AB , AC) where r
=
= = =
~
~
---+--+
VPBcD+r(lPM, PNJ,IBC,BDJ) . --+ ---+
---+
~
~
(PB,CD)-r(CD , IPM , PNj) . ---+ --+ ---+ ---+ ~ ---+ ----+ --+ 2 (PB ,PC) +r(BP+CP, IPM,PNj) + r 2[PM , PNJ
~l~ • [LM.LN]2
To eliminate points introduced by constructions S4, S5, and S6, we do not need to break the inner and exterior products into the sum of several components. In these three
Chapter S. Vecton and MadUae ProoIs
236
constructions, Y is always on a line UV. Remember that a geometry quantity G(Y) is called a linear quantity of point Y if UY G(Y) = UV G(V)
YV
+ UV G(U).
A geometry quantity G(Y) is called a quadratic geometry quantity of point Y if
UY
G(Y)
YV
UY YV ==2
= UVG(V) + UVG(U) - uv UV UV --+
--+ -----+
.
-----+ -----+
Example 5.41 Show that YB. [YB,CDj. and (YB,CD) are all linear in Y; and --+ --+
(Y B, YC) is quadratic in Y . ->
Proof By Proposition 5.29, Y
Uy-> yv-> = uv V + uv U .
--+
YB
---+-----+
---+---+
Now it is clear that [Y B, CDj and (Y B, CD) are also linear in Y. ForG(Y) Uy YV Th 1et Tl = uv and T2 -- uv' en G(Y)
= (Y---+:-:-::t, B, Yc..:),
+ Tiffi, Tl IT + T2UC) T~G(V) + T~G(U) + TIT2((VB,UC) + (UB, IT)) --+ --+ = Tlh + T2)G(V) + T2(Tl + T2)G(U) - TIT2(UV, UV} --+ --+ T1G(V) + T2 G(U) - TIT2(UV,UV}. I
=
(Tl VB
Therefore to eliminate point Y from the geometry quantities, we need only to find the position ratios of Y with UV. which has been done in Lemmas 5.32. 5.33, and 5.34.
5.3.3
The Algorithm
Algorithm 5.42 (VECTOR) INPUT: S
= (C}, C2 , • •• , Ck, (E, F)) is a constructive geometry statement.
OUTPUT: The algorithm tells whether S is true or not, and if it is true, produces a proof forS.
5.3.
237
M8ch1ae Proof by Vector Cakuladoa
S1. For i
= k, .. . , I, do S2, S3, S4 and finally do S5.
S2. Check whether the nondegenerate conditions of Ci are satisfied. The nondegenerate conditions of a statement have five fonns: A
i-
B , AB2
i-
0, PQ
It' UV,
PQ
It'
[IM, LN]2 i- O. For the first case, we check whether Ii = B. For the second case, we check whether (AB , AB) = O. For the third case, we check whether --+ --+ [PQ , UV] = O. For the fourth case, we check whether Vpwuv = VQwuv. For the fifth case, we check whether [IM, LN]2 = O. If one of the nondegenerate conditions
WUV, and
of a geometry statement is not satisfied, the statement is trivially true. The algorithm tenninates.
S3. Let GlI ··· , G. be the geometric quantities occurring in E and F . For j
= 1"", s do
S4
S4. Let H j be the result obtained by eliminating the point introduced by construction C j from Gj using the lemmas in this section, and replace Gj by H j in E and F to obtain the new E and F.
S5. Now there are only free points left. E and F are rational expressions in indetenninates, inner and exterior products of free points. Replacing the inner and exterior products by their coordinate expressions, we obtain E' and F' . Then if E' = F', S is true under the nondegenerate conditions. Otherwise S is false.
In the above algorithm, we represent the ratio of lengths and the ratio of areas as expressions of inner and exterior products, and then eliminate points from the inner and exterior products. In order to obtain short proofs, we may also use the lemmas in Section 4.3 to eliminate points directly from the ratios of lengths or areas. We actually use a hybrid method: inner products, exterior products, length ratios, area ratios, volumes, and the Pythagoras differences are all used in the proof. Remark 5.43 Since the inner product. the exterior product, and the triple scalar product are proportional with the Pythagoras difference, the area, and the volume. the volume (area) and the Pythagoras difference method developed in Chapters 3 and 4 are valid for constructive geometry statements in metric geometries associated with any field with characteristic different from 2. Thus our method works not only for Euclidean geometry but also for non-Euclidean geometries such as the Minkowskian geometry.
The vector approach has the advantages that it is easy to develop, and more geometry quantities such as the vector itself can be used. But the proofs produced by the vector approach generally do not have the clear geometric meaning of those produced by the volume-Pythagoras difference approach.
238
Chapter S. Vectors and Machiae Proofs
Notice that the elimination results (Lemmas 5.30-5.34) for points introduced by Sl-S6 are exactly the same for all metric geometries. From this, one might wonder is there any connection among these metric geometries? But at the last step of the algorithm, we need to replace the inner and exterior products of free points by their coordinate expressions. This step depends on the specific geometries. Thus in order to obtain some meta theorems for different geometries, we need to limit the class of geometry statements. Definition 5.44 A constructive geometry statement is called pure constructive if it can be described by constructions SJ-S7 and its conclusion can be one of the geometry relations in Proposition 5.27. We further assume that the ratio r in the ratio constructions PRATIO and ARATIO can only be scalars or variables. In the predicate fonn for a pure constructive geometry statement, there are only affine invariants like the ratio of parallel line segments and geometry predicates like COLL, PRLL, PERP, etc.
Proposition 5.45 Let G 1 and G 2 be two geometries over the same base field pure constructive geometry statement is true in G 1 iff it is true in geometry G 2 •
c.
Then a
Proof By Proposition 5.9, we can assume that the matrices for the inner products of G1 and G 2 are
Let x, y, z and x', y', z' be vectors in Gland G 2 with the same coordinates respectively. Any geometry predicate can be represented by the following three quantities. (x , y) = a1x1Y1 + a2x2Y2 + a3x3Y3 in G1. [x, y] = 01 (a2a3(X2Y2 - X3Y2) , a1a3(X3Y1 - X1X3), a1a2(X1Y2 - X2X1)) in Gl . where 01 is the constant in Proposition 5.18 for G1. Xl X2 X3 (x, y, z) = 01 Y1 Y2 Y3 Zl Z2 Z3 (x' , y') = b1X1Y1 + b2X2Y2 + b3X3Y3 in G2. [x', y'] = 02(b2b3(X2Y2 - X3Y2), b1b3(X3Y1 - X1X3), b1b.z(X1Y2 - X2Xl)) in G2. where 02 is the constant in Proposition 5.18 for G2. (x', y', z')
= 02
Y1 Zl
Y2 Z2
Y3 Z3
5.4.
239
Machine Proor In Metric: Plane c-tric:s
Let F be the algebraic closure of field £ . Consider the following automorphism of the polynomial ring F[ X l ,X2 ,X3 , ... , Z l , Z2, Z3]
We have
TR«(x' , y')) = (x , y) . TR(!x' , y']) =0 ¢:::=* [x , y] =0. TR(x',y',z') =0 ¢:::=* (x . y , z)=O. Also note that an affine invariant is not changed under T R . Since each pure constructive statement can be described by affine invariants and the above geometry quantities. it is clear that a pure constructive geometric statement is true in G1 iff it is true in G2 • I
Machine Proof in Metric Plane Geometries
5.4
Since a metric plane can be seen as a subset of a metric space. the method presented in the preceding section is also valid for plane metric geometries. For an independent study of this topic. see (72). But for metric plane geometry. the method can be greatly simplified. Without loss of generality. we assume that the metric plane consists of aU the points x = (x\, X2, X3) such that X3 = O. Let
be the matrix defining the inner product. 1ben for vectors x = (x\, X2, 0), y = (Yb 112, 0), and z (zl, Z2. 0) in the plane. we have
=
(x. y)
= alxlYl + ~X2Y2 .
= (0,0, oal a2(xIY2 (x,tI,z) = O. [x, tI]
a is a constant.
X2Yl».
Since aU the exterior products of vectors in the same plane are parallel. we can just define [x,
y]
= oal~(xlY2 -
X2Yl) .
It is easy to check that some of the basic properties of the exterior products such as E I and E2 are still true. For points A, B, and C. the signed area of triangle ABC is defined to be
--+] AB,AC. 21[--+
240
Chapter 5. Vectors and MadUoe Proof.
5.4.1
Vector Approach for Euclidean Plane Geometry
For the Euclidean geometry, the matrix M defining the inner product is the unit matrix. Thus if the conclusion of a geometry statement is a polynomial of inner and exterior products, the proofs produced according to the vector approach and the area-Pythagoras difference approach are actually the same. But in the vector approach, we can use a new geometry quantity: the vector itself.
Example 5.46 (The Centroid Theorem) Show that the three medians of a triangle are concurrent and the medians are divided by their common point in the ratio 2:1.
Constructive description
The machine proof.
«POINTS A B C) (MIDPOINT F A C)
AG (t)·A1
(MIDPOINT E B C) (INTER G (LINE A E) (LINE B F»
(AG = ~A1»
The eliminants
g
A1.SARE
m· A1·S ABEF
8irrp!.'!y
--+
-
AGg~ SABEF
SABEF~~(SBCF+2SA BF ) SBCF~~(SA BC ) SABF~~(SA BC )
§.. -
1:.. -
(3).SABF (2 )·SABEF
(3) ·SABF (2) ·(fSBCF+SABF) (3 )·(tSABC) iSABC
Figure 5-4
8im!!1i!y
Example 5.47 Let G be the centroid ofa triangle ABC. Show that a
= leA + It + e).
Proof Using our method, we can actually compute this result without knowing it previously. The centroid G of the triangle ABC can be introduced as follows. (POINTS A B C) (MIDPOINT F A C) (LRATIO G B F 2/3) By Lemma 5.30,
a = 2/3"7 + 1/3It = He + A + It) .
You might think that the above introduction of G is too tricky. The usual way of introducing G is as follows.
241
SA. MKblae Proof I. Metric P1ue Geometries
Constructive description
«POOn'S A B C)
(MIDPOINT F A C) (MIDPOINT E B C) (~
(0
»
0 (UNE A E) (UNE B F»
- -
The eliminants
The machine proof. (/ Q -
-
(/0 F .S'~!~E~ ·S'fl>
7 .s AHd It .SUI> SABEF
~ } -;t .SABC+} It .SABF fSBCF+SABF L t C .SABC+} It .SABC +} 7 .SABC
-
ISABC
oimJ!!i/ll 1 ( -
-
3
-
SABEF~~(SBCF+2SABF ) SAFE~ - ~(SABF) SABE~~(SABC) SBCF~~(SABC) SABF~~(SABC)
-Fl(-) F =2 C + A
-)
C+B+A
Example 5.48 The triangle having for its vertices the midpoints of the sides of a given triangle has the same centroid as the given triangle.
The machine proof.
The eliminants
(MIDPOINT D B C)
:sl
It~H7+7+It)
(MIDPOINT E A C)
It
(MIDPOINT F A B)
~ ---.,.-..:::0--,.(",,3),--
Constructive description «POImS ABC X Y)
(CEImIOID 0 A B C)
--
(CEImIOID K D E F)
(O=K»
---
(/gHc +It +7) 7q(It+ 7)
F+E+D
Q (3) .(C +It +7) -
~
--+
--+
(F
+ E + D )-(3)
7
+ It +t It +t 7
L
7~Kc+7)
ItgKc+It)
c+It+7
-- - -
~ (2)·(C + It +7)
Figure 5-5
-
2D + C +
Q
(2)·(C + It +7)
-
--+
2 C +2
--+
B
B
+2
+2
A
--+ A
.im~i/ll 1
Example 5.49 The triangle formed by the three lines passing through the three vertices and parallel to the opposite sides of a triangle is called the anticomplementary triangle of the given triangle. Show that a triangle and its anticomplementary triangle have the same centroid.
242
Chapter S. vec:tors and Machioe Proofs
Constructive description « POINTS A B C) ( PRATIO ( PRATIO
PA C B 1) Q AB C 1)
The machine proof.
The eliminants
(J
lt~i{Jt + Q + 7)
It
(Jgi(c + If +7)
(PRATIO R B A C 1) (CENTROID
-+ ( G
C)
Jt!JC+B-A
Q R)
Q~C-B+A
GAB
(CENTROID K P
= -+ K ) )
7~
M ~
P
B
Figure 5-6
R
~
--+
- (c - If -7)
c±I1±A --+ --+
--+ --+
Q+P+ C +B-A
9.. c+If+7
-
-+
-+
P +2 C
.f.. c+If+7 - --+ --+ --+ C+B+A Bimg1i/y 1
With the help of the vector method. we can prove the following theorem about npolygons. The centroid of n points PI , ... , Pn is defined to be
;0>; + ... + p.:).
Example 5.50 (Cantor's First Theorem) For n points on the circle O. perpendiculars from the centroids of any n - 1 points taken from the n points to the tangent lines of circle o at the remaining n-th points are concurrent. -+
----+
Proof. Let the n points be Plo· · · ,Pn and P; = Op; . Let G 1 be the centroid of P2 ,'" ,Pn • G 2 the centroid of PI, P3 , • •• ,Pn • and Y the intersection of (PLINE G 1 0 PI) and (PLINE G 2 0 P2 ) . By Example 5.37,
which is a fixed point.
Example 5.51 (Cantor's Second Theorem) For n points on the circle O. perpendiculars from the centroids of any n - 2 points taken from the n points to the lines joining the remaining two points are occurrent.
5.4.
243
MadUoe hoof I. Metric PIaDe ~ --+
--+
Proof Let the n points be Pit ·· · ,Pn and Pi = Op;. Let G I be the centroid of Pa,'" , Pn, G2 the centroid of Pit P2, Ps '" ,Pn• M the midpoint of P I P2• N the midpoint of PaP4 • and Y the intersection of (PLINE GlOM) and (PLINE G2 N P2 ). By Example 5.37.
where r
= SO,ON-SOaON = (n-2)SMON 2SWON = _2_ . Therefore SOONN n-2 --+
Y
1
--+
= n_2(PI +"
--+
, +Pn )
which is a fixed point.
5.4.2
Machine Proof in Minkowskian Plane Geometry
In Minkowskian plane geometry. M and Y
= (~ ~1) ' Then the inner product of z = (Xit X2)
= (YI. Y2) is (z, y)
= XIYI -
X2Y2 ·
Thus zl.y iff
XIYI - X2Y2
= O.
We define the exterior product of z and y to be
[z. y]
= -XIY2 + X2YI.
For points A, B, C, and D in the Minkowskian plane the area for the quadrilateral ABC D. i.e .•
SABCD
[AC , ED] is also interpreted as twice
1 -:-:t --+ = 2[At.;, BD] .
We thus have the Herron-Qinformula in the Minkowskian geometry
16S!BCD
= P1BCD -
2 4AC . BD2.
In the Minkowskian plane. there exist isotropic lines (vectors). Vector z isotropic iff X~ - x~ = (Xl - X2)(XI + X2) = 0,
= (XI. X2) is
i.e .• the isotropic lines are those which are parallel to one of the following lines Xl -
X2 = 0
or
Xl
As a consequence of Proposition 5.45. we have
+ X2
= O.
Chapter 5. vectors and Madtine Proofs
244
Proposition 5.52 A pure constructive geometry statement is true in Euclidean geometry if and only if it is true in Minkowsldan geometry. As a consequence, most of the geometry theorems proved in this book are also valid in Minkowskian geometry. But there are do exist geometry statements in Euclidean geometry which are not true or do not have geometric meaning in Minkowskian geometry. For instance, in Minkowskian geometry there exist no equilateral triangles. If a geometry statement is in the affine geometry, i.e., the only geometry relations in the statements are incidence and parallel, then it is obvious that this statement is true in Euclidean geometry iff it is true in Minkowskian geometry. The reason is that both geometries are developed from the same affine geometry by adding different metric structures. In other words, the affine part of the two geometries are the same. But for a geometry statement involving geometry relations like perpendicular or measurement, it is not obvious that its validities for both geometries are the same.
Example 5.53 (Orthocenter Theorem in Minkowskian Geometry) The same as Example 3.19 on page 109. The following proof is essentially the same as the proof of Example 3.19. But it is for a different geometry theorem. Compare Figure 1-42 (on page 31) and Figure 5-7.
~
Constructive description «POINTS A B C) (FOOT E B A C) (FOOT DCA B) (INTER H (UNE C D) (UNE BE» (PERPENDICULAR B C A H) )
A
B D
Figure 5-7
The eliminants
The machine proof.
--+ ~ --+--+ (AC,AH)~(AB,AC) --+ ~ --+---+ (AB,AH)~(AB,AC)
--+--+
(AB,AH)
(AC,AH) --+--+
!!.. (AB,AC) -
--+ --+
(AB,AC)
simJ!!.ifll
-
Before giving examples involving circles, let us note that the method presented in Subsection 3.6.2 to eliminate co-circle points is still valid in Minkowskian geometry except that we need to use the hyperbolic trigonometric functions instead of the trigonometric functions. The "circle" in Minkowskian geometry is actually a hyperbola:
The diameter of the above circle is 0
= 2r.
245
Madtlae ProoIla Metric PIaae o-trla
SA.
Lemma 5.54 Let A , B, and C be points on a circle with diameter 8 in the Minkowsldan plane. Then
=
SABC
AB · CB · CA
28
'
PABC
_ -
2AB·
CJj. cosh(CA)
8
AC
'
= 8 sinh(AC)o
The proof for the above lemma can be carried out as in Section 3060
Example 5.55 (Simson's Theorem in Minkowskian Geometry) The geometry statement of Simson' s theorem is exactly the same as in Example 3079 on page 1410 E
Constructive description «CIRCU!
ABC D)
(POOT ED B C) (FOOT
FDA C)
(FOOTG D A B)
(~~~=-1»
Figure 5-8
The machine proof.
The eliminants
~ . u, . ~ AF C E BG
~G (AB,AD)
g
--4 - - 4
---+ --+ (AB ,AD) . ~ o u,
-<11.A ,BD)
AF CE
BG
---+ ---+
-(BA,BD) --4 - - +
~t; (CA,eD) AF
-;-::t - - 4
-(AG.AD)
u,E
-----+ --+ -(CB,CD)
- - 4 --+
(CB .CD)= - 2CDoBC oroeb(BD) --4 --+
(AC,AD)=2AD oAC ocooh(C D)
co=..ar -( -2COojiCocooh(AD))0(2BD.OC.cosb(C D)).(2ADo:;rnoroeh(BD)) -
--4--+
(-2BDoAB .cosh(AD)).(2AD .AC.cosh(CD))0(-2CDoBCoroeh(BD)) (BA,BD)= - 2BD.:;rn.coob(AD)
.im~ifll
--4 --+
(AB,AD)=2ADo:;rnoroeh(BD) (BC .BD)=2BDoBC oroeb(CD) --4 --+
(CA,CD)= - 2CDoAC ocoob(AD)
By Menelaus' theorem. E. F. and G are collinearo
Example 5.56 (Cantor's Theorem) The same as Example 3082 on page 143.
246
Chapter 5. vectors aDd MachIne ProorJ
Constructive description «CIRCLE ABC D) (CIRCUMCENTER 0 A B C) (MIDPOINT G A D) (MIDPOINT F A B) (MIDPOINT E C D)
(PRATIO N EO F 1) (PERPENDICULAR G NBC) )
The machine proof.
(BC,BG) ---+ --+
(BC,BN) ---+ ---+
t£ ___-.l.:(B=C.J.!,B~G'=!L)_ __ -+
-t
-+ ---+
---+-+
(BC,BE)+(BC,BF)-(BC,BO)
Figure 5-9
~
(BC,BG) (BC,BF)-(BC,BO)+t(BC]iD)+t(cB,cB) ~ (2HBC,BG) -2(BC,BO)+(BC,BD)+(cB,cB)+(BA,BC) g (-2)o(i(BC,BD)+}(BA,BC» 2(BC,BO)-(BC,BD)-(cB,cB)-(BA,BC) Q -«BC,BD)+(BA,BC»o(2) - -2(BC,BD)-2(BA,BC)
.i~ifll 1
In the last step,let T be the midpoint of BC. Then we have BCl.TO, and hence
--
1--
~- = "2«CB,CB}) . (BC,BO) = (BG,BT)
5.5
Machine Proof Using Complex Numbers
In addition to vectors, complex numbers may be used to generate readable proofs for geometry theorems. Complex numbers are often used to solve difficult geometry theorems and to interpret various different geometries. See [11, 38]. They have also been used to mechanical geometry theorem proving in [149] based on the Grabner basis computation. In this section, we will show that it is possible to obtain readable proofs for many geometry
5.5.
247
MM:hlae Prout UsiaC Complex Numbers
theorems using complex numbers. Complex numbers may be looked as vectors. But we can also multiply two complex numbers. We will start our discussion with this special property of complex numbers. Let
be two complex numbers, where i = A . The conjugate of fI = YI + Y2i is fI Then z · fI = XIYI + X21/2 - (XIY2 - x2ydi. If the complex number z is clear that
(z, y)
1/2i.
= Xl + X2i is seen as the same as the vector z = (Xl , X2), then it z . fI
Thus
= YI -
= (z, fI) -
= ~(z . fI + x . V);
[z, fI]i .
[z, y]
= ;i(X . y -
z . fI)·
That is, the inner product and the exterior product can be expressed by the multiplications
of complex numbers. -+
In geometric language, for each point P,let P be the corresponding complex number, -
-+
and P the conjugate of P . Then the above two equations become ~~-
-
-
-~~
PABC=(B - A)(C-A)+(B-A)(C - A) SABC
1 = :4;«B -
---+
--+
--+
--+
-
-
A)( C - A) - (B - A)(C - A)).
Therefore, the vector approach for Euclidean geometry can be translated into the language of complex numbers. But the proofs thus produced are generally longer than those produced by the vector approach. The reason is that in the vector approach the area and Pythagoras difference are treated like one-term variables, while in the complex number approach they are expressions of several terms. This complex number approach is essentially the same as the Wu's method in the case of constructive geometry statements [55,62]. But in many cases, the complex number approach does give short proofs. Example 5.57 As shown in Figure 5-10, on two sides AC and BC of triangle ABC, two similar triangles PAC and QC B are drawn. RPCQ is a parallelogram. Show thot triangle RAB is similar with triangle PAC. --+
For two points A and B, let AB and have the same orientation iff (5.1)
-+
=B
-+
- A. Two triangles P AB and QC B are similar
Chapter 5. Vectors and Machlne Proofs
248 We thus can use a new construction
(SIM-TRIANGLE Q C B P A C) which introduces a point Q such that (5.1) is true. Now Example 5.57 can be described constructively as follows. R
Constructive description
Q
«POINTS ABC P)
(SIM-1lUANGLE Q B C P C A)
(PRATIO R Q C P 1) --+ --+ --+ --+ (PA-AB RA ·AC) )
A
=
Figure 5-10
The machine proof.
The elirninants
--+ --+ Ap·AB --+ --+ AR·AC
A1~AQ+AP-AC
-+QcP.BC- If-AC+AC.A -AC AC=f!-A BC=f!-7!
--+ --+ R _---'-A,..,P-'.A.....,B<-,---:-
AQ
--+ --+ --+ --+ (AQ+AP-AC) ·AC
~ -
AP .~.(-AC)
--+ --+ --+ --+ --+ --+ --+ 2 --+ --+ ~ (C P·BC- B ·AC-Ap·AC+AC +AC· A )·AU
-+ -+ CP= P -+ -+ AB= B -+ -+ AP= P -
-+ -+
AP·AB --+ --+ --+ --+ --+ --+ --+ 2 --+--+ CP·BC- B ·AC-Ap·AC+AC +AC· A --+ --+ --+ --+
-( P - A
H
B - A
-+ C -+ A -+ A
1
--+ --+ --+ --+ --+ --+ --+:2 -P · B+P · A+B · A-A .im~ifY 1
Let w = e ~. Then the three points corresponding to the three complex numbers 1, W, and w 2 form an equilateral triangle with positive orientation (Figure 5-11).
C 1--------4 A
Figure 5-11
Thus ABC is an equilateral triangle with positive orientation (Figure 5-11) iff ~
w-l
AC=W2 -1 '
s.s.
249
M8cbIDe Proal Usia, Camplex Numbers
From the above equation, we have
(w 3
-
w2 }E/ + (W -l}e + (w 2
-
W}A
= O.
Dividing W - 1 from both sides of the above equation, we have that ABC is an equilateral triangle with positive orientation iff
e +w E/ +wA = O. 2
Similarly ABC is an equilateral triangle with negative orientation
e + wE/ +w A 2
iff
= o.
We thus introduce two new constructions. (PE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle -+
with positive orientation, i.e., C
-+
-+
+ w 2 B + W A = O.
(NE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle -+
with negative orientation, i.e, C
-+
-+
+ W B + w 2 A = O.
Example 5.58 (Echols' First Theorem 18 ) If ABC and PQR are equilateral triangles with the same orientation, then the triangle formed by the midpoints of AP. BQ. and CR is also an equilateral triangle. Constructive description «POJNI'S A
B P 0)
(PI!-1'IUANGUl C B A) (PI!-11UANGUl R
0
P)
(MlOPOOO' LAP) (MlOPOOO' M B 0) (MlOPOOO' NCR)
Figure 5-12
(PE·11UANGUl N M L) )
The eliminants
The machine proof.
N+M .w+L .w' ~ M .w+ L .w·+t It +t ~ (t)·(2L .w'+ It + ~
e
Ng~(1t +e}
e
e +Q ''''+ I! .w)
(tHIt + + Q .w+ -;.w'+ I!.w+ A.w') ~ (tHe + I! .w+ A.w') ~
Mathematical Monthly. 39. 1932. p.46
M~~(Q + I!} L£;~(-; + A} It~ - ({Q + -;.w).w) ({I! + A .w),w)
eg -
Chapter S. vectors and MachIne Proofs
250
Example 5.59 (Echols' Second Theorem) If ABC. PQR. and XYZ are equilateral triangles with the same orientation. then the triangle formed by the centroids of triangle APX. BQY. and CRZ is an equilateral triangle.
Constructive description «POINTS A B P
Q x Y)
( PE-l1IIANGLE C B A) (PE-l1IIANGLE R
Q P)
The eliminants
N';;K7 + It + c) M~KY' + Q + It)
(PE-l1IIANGLE Z Y X)
"TJ;l("x +f/ + 7)
(CENTROID LAP X)
7~
- ((Y' + x .w).w)
It~
- (Q + f/.w) .w)
(CENTROID M B
QY)
(CENTROID NCR Z) (PE-l1IIANGLE N M L) )
cg - (It +7 .w).w)
The machine proof.
N' +M .w+ "T .w' ~ 3M.w+3"T .w'+ 7 + It + C ~ 9"T .w2 +37 +3It +3C +3Y.w+3Q.w+3It.w ~
(3) .( 37 +3It +3C +3Y,w+3X .w'+3Q,w+3f/.w'+3It.w+37.w') (9).(It + C + Q .w+ f/ .w'+ It .w+ 7.w') ~ (9).(C + E!.w+ 7 .w2 )
~
Bim!!li/y 0
Example 5.60 (Echols' Theorem in General Form) Let P;,l ... P;,n, i = 1, .' " m be m regular n-polygons with the same orientation. Then the centroids of the m-polygons P1 ,j ... Pm ,j , j = 1, .. " n.form a regular n-polygon,
Proof From the proofs of Examples 5.58 and 5.59, we need only to show that Al ... An is a regular n-polygon iff points Al, .. " An satisfy some linear relations Rk(Al, .. " An) = 0, k = 1, ... 5. We leave the details to the reader. I
Example 5.61 The same as Example 3.45 on page 124, The following proof is much shorter. (Figure 5-13)
251
5.50 MadlIoe Proof UsiaC Compl" Numbers
ConsbUctive description «POIIITS A B C) (CONSTANT w 2 +w+l)
The eliminants
The machine proof.
~~
--+ --+ AICI-CBI
- A"": -cs: -It
(I'6-TlUANGU! BI A C)
~
(1'6-TlUANGU! A I C B)
--+ = -(CBI- --+ C ow+ A o w)
(~~CIBA)
(AICI-CBI = 0»
n
ow 2 -
A
A"":~ oW
- (A"": +It ow 2 + A o",)
- (C + It o",)o",) (-+ -+ '" ) C o w2 + -+ C + A 0
--+8 CBI ~ -
~
~ -(- C ow'-C ow-C) .imJ!iifll
-
(w'+w+1)o
-+ C
n
=0 .im!!f.ifll 0
x
B
t---:-----I--4__
y
D
Figure 5-14
Figure 5-13
Example 5.62 The converse of Example 50610 (Figure 5-14)
The eliminants
ConsbUctive description The machine proof. «POIIITS A B C)
(CONSTANT
w' +w+1)
(PRAno DAB C I) (PE-11UANGU! X B C) (PI!-TlUANGU! Y C D)
Y ow+ X ow2 +A ~ X ow'- [jow'- Cow'+ A ~ - [jow'- Cow'- C ow.- Itow'+ A
x~
- (C ow+ It)ow)
I1gC-~+A
~ -(Cow'+ C ow'+ C ow'+ A ow'- A) .imJ!iifli
(1'6-TlUANGU! A Y X) )
Y~ - ([j ow+ C)ow)
-
-+
-(w'+w+1H C oW'+
-+ -+ A °W- A )
Using complex numbers, we can also deal with some theorems involving squares easilyo For two points A and B, C is a point introduced by (TRATIO CAB I), or equivalently C satisfies CA = AB, CA.l AB, and SCAB> 0, iff
AC=i.:4B.
252
-
Cbapler S. Veeton and MachIne Proofs
-
Geometrically, the above equation means that AC is obtained by rotating AB by 90° counterclockwise, or CAB is an isosceles right triangle with positive orientation. Similarly, C is a point introduced by (TRATIO CAB -1) iff
- AC
= -i ·AB.
We thus can introduce two new constructions (PE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with positive orientation, i.e., -+ C
= -+ A + i . AB.
(NE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with negative orientation, i.e, -+ C
= -+ A -
i . AB.
Example 5.63 On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show tlult FC l..BH and FC = BH. Constructive description: «POINTS
AB
C) (CONSTANT i'-I) (PE-SQUARE
F A B)
(NE-SQUARE
The machine proof -+ --+
-(CF+BH.i)
~ -(cP-AC.i·-A8.i)
H A C) (FC-i .B#
= 0) )
The eliminants B#~ - (AC'i+A8) cPt; - (AC-A8.i)
~ -(-AC.i·-AC) .im!!J.ifll (i'+l)'AC n
=0
Example 5.64 On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. M is the midpoint of BC. Show tlult F H 2AM and F H l..AM. (Figure 5-15)
=
Constructive description «POINTS A B C) (PE-SQUARE F A B) (HE-SQUARE H A C)
2! B C)
(~POINT
(HF-2i·AM
= 0) )
The machine proof --+
--+
-(FH+2AM·i) --+ --+ --+ = -(FH+AC·i+AB.i) n
~ -(-AP+A8'i) n
=0
The eliminants AM~HAC+A8) F8~
- (AP+AC .i) AP~AB.i
5.5.
253
Madtlne Proof Uslne Complex Numbers
Example 5.65 19 Starting with any trian· gle ABC, construct the exterior (or interior) squares BCDE, ACFG, and BAHK; then construct parallelograms FC DQ and EBK P _ Show that P AQ is an isosceles right triangle.
G p
K;t--~H
Figure 5-16
Constructive description The machine proof
The eliminants
«POINTS A B C)
;:tP~AK+A£-AS AQ~Ab+At-AC AK~ - (i-I)-AS)
(CONSTANT
i>
-1)
F C A)
(NJ!.SQUARE
(PE-SQUARE D C B)
~ -(-AQ+AK-i+A£-i-AS-i)
~
(PRAll0 P K BEl)
-AK-i-A£-i+Ab+At-AC+AS-i ~ -(A£-i-Ab-At+AC-AS-i» ~ -(Ab-i-Ab-At-AC-i+AC-AS-i>+AS-i)
(AQ-i-;:tP =
~ -(-BC-i>+BC-i-At-AS-i>+AS-i)
(PRAll0 E DC B 1) (PE-SQUARE
K B A)
(PRAll0 Q F CD 1)
0) )
-+F -+ AF=(i+1)-AC -+ -+ -+ AB= B - A
AC=C-A BC=C-It
~ BC-i>-BC-i+AC-i+AC+AS-i>-AS-i -+ --+ -+ -+ --+ = -BC-i-BC+AC-i+AC-AB-i-AB - t -+ = -(i+1)-(BC-AC+AB) n
A£~Ab-AC+AS
Abg - (BC-i-AC)
~
6impli/u
~
~ -(i+1)-(O) 6im~i/u 0
Summary of Chapter 5 • The metric vector space is a vector space with inner and exterior products. The metric geometry of dimension three associated with field £ is the nonsingular metric vector space £3 . • Some basic geometry quantities can be described by the inner and exterior products:
I.
P ABC
= 2(AB, CB};
PABCD
= 2(AC, DB} _
2. If four points A, B, C, and D are collinear or AB --+ AB--+ AB= CDCD,
II CD,
(AR,cD) CD = (cD,cD) AB
Ifnis example is from Amer. Math_ MOD. 75(1968)_ p_899_
254
Chapter 5.
3. VABCD
Vedors and Macbfne Proo6
1 --+ ---+ ---+ = 6(AD, [AB , AC)) .
4. If six points A , B, C , p. Q. and R are coplanar or ABC
II PQR.
SABC
([AB,AGI, [PQ,PR))
SPQR
= ([PQ,PRI,[PQ,PRJ)
• We have the following criteria for parallel and perpendicularity. ---+ ---+
I. AB.l. CD
2. AB
II CD
¢::::::>
---+ ---+
¢::::::>
3. AB.l. PQR
4. AB
II PQ R
¢::::::> ¢::::::>
5. ABC.l.PQR 6. ABC
= O. [AB,CDI = o. (AB,CD)
II PQR
-+
---+ ---+
--+
---+ ---+
= o. (AB, [PQ, PRJ) = O. [AB,[PQ , PRII
---+ ---+
¢::::::>
--+ ---+
¢::::::>
--+--+
([AB,ACI,[PQ,PRJ) =0. --+---+
[[AB,AC],[PQ,PRII
= o.
• A mechanical proving method is presented for constructive geometry statements in metric geometry of dimensions two and three. This method works similarly to the area-volume-Pythagoras difference method developed in Chapters 3 and 4. The basis of the method is to eliminate points from vectors and inner and exterior products of vectors. • We have proved the following meta theorem. A pure constructive geometry statement is true in one metric geometry iff it is true in alI the metric geometries associated with the same field.
Chapter 6 Topics From Geometry A Collection of 400 Mechanically Proved Theorems
This chapter is a collection of 400 geometry theorems in plane geometry proved automatically by a computer program20 based on the method developed in the first part of this book, including 205 machine proofs (in LaTeX form) produced entirely automtically by the program. In addition, there are another 78 problems solved by o w computer program in Chapters 2-5. We include a collection like this for two reasons. First, it would show the power of the method/program: most theorems involving equalities only in geometry textbooks are in the collection. Second, many proofs produced by the program are very beautiful. Following the spirit of E. W. Dijkstra (p. 174 [19]),we believe that such beautiful proofs deserve special attention. Among the work consulted to collect these examples, special mention must be made of the following [I,3,4,13.23,391. The figures and inputs for many of the examples in this collection are from [12]directly. The examples are classified according to the types of constructions needed to describe them.
6.1
Notation Convention
As an example of how to read the examples in this collection, let us consider Ceva's theorem. As shown, a typical entry in this collection includes a description of the theorem in English, a diagram, a constructive description of the theorem as the input to the program, and a machine proof produced by the program.
2oTheprogram is available via ftp at emcity.cs.twsu.edu publgeometry.
255
Chapter 6. Topics from Geometry
256
Example 6.0 (Ceva's Theorem) (0.01,1,3) The lines joining the vertices of a triangle to a given point determine on the sides of the triangle six segments such that the product of three nonconsecutive segments is equal to the product of the remaining three segments.
A
F
Figure 6-0
Constructive description
The machine proof
The eliminants
«points AB e 0) (inter D (I B e) (loA» (inter E (I A e) (loB» (inter F (I A B) (10 e»
-% . ~~ . *
4.f:.f.§..s=
( ;~ ~g~
= 1»
BP- SBCO
~ -(-S..\Oo) . ~ . llJl, -SBCO
~ -
8i m1!li!y -
!2.. -
AE CD
-SaGo ·SAQO • I.!l SBCO ,(-SABO) CD
~~~ AE--SABO
u.£~ CD-SACO
§..s= . IlJJ. S A BO
CD
SABO ·S.WO 8im1!li!y 1 S A BO ,SACO -
Notation Convention. 1. We use a triple (time,rru.u:t,lems) to measure how difficult a machine proof is: time is the time needed to complete the machine proof in a NexT Turbo workstation (25 MIPS); rru.u:t is the number of terms of the maximal polynomial occurring in the machine proof; and lems is the number of elimination lemmas used to eliminate points from geometry quantities, i.e., the number of the eliminants. For Ceva's theorem, time =0.01 second, rru.u:term = 1, and lems =3.
2. We use some abbreviations in the constructive description of the geometry statements in order to save printing space. For instance, LINE is represented by I; PLINE is represented by p; BLINE is represented by b; and ALINE is represented by a. 3. The ndg conditions and the predicate forms of the geometry statements are not given, since they can be generated directly from the constructive description. See page lOS. 4. For explanation of the machine proof and the eliminants, see page 72. In Part I of the book, we have introduced 28 constructions for plane geometry, which are listed here for your convenience. 1. (ARATIO A 0 U V TO TU TV) ' Take a point A such that TU, TV, and TO are the area coordinates of A with respect to OUV. For the exact definition of this construction, see page 132.
2. (CENTROID GAB C). G is the centroid of triangle ABC. See page 135.
6.1.
257
NOladoo Coovendoo
3. (CIRCLE Y1 . . .
Yo) , (8
~
3). Points Y1 . . .
Yo are on the same circle.
See page 140.
4. (CIRCUMCENTER 0 A B C). 0 is the circumcenter of triangle ABC. See page 135. 5. (CONSTANT per)~ where per) is an irreducible polynomial in r . This construction introduces an algebraic number r which is a root of p( r) = O. See page 121 . 6. (HARMONIC DCA B) means that D is a point such that A, B , C , and D form a harmonic consequence. See page 120. 7. (INCENTER C I A B) I is the center of the inscribed circle of triangle ABC. This construction is to construct point C from points A , B , and I. See page 135. 8. (INTER Y Inl/n2) . Point Y is the intersection of lines Inl and In2. See page 108.
9. (INTER Yin (CIR 0 P)). Point Y is the intersection of line In and circle (CIR 0 P) other than point P . Line In could be (LINE P U), (PLINE P U V), and (TLINE P U V ). See page 108. 10. (INTER Y (LINE U V) (CIR 0 r )). Point Y is one of the intersections of line (LINE U V) and circle (CIR 0 r). See page 144.
II. (INTER Y (CIR 0 1 P) (CIR O 2 P)). Point Y is the intersection of the circle (CIR 0 1 P) and the circle (CIR O 2 P) other than point P - See page 108. 12. (INTER Y (CIR 0 1 rl) (CIR O 2 r2)). Point Y is one of the intersections of the circle (CIR 0 1 r l) and the circle (CIR O 2 r2). See page 144. 13. (INVERSION P Q 0 A) means that P is the inversion of Q with regard to circle (CIR 0 A). See page 120. 14. (LRATIO Y U V r). Y is a point on UV such that ~~
= r. See page 120.
15. (ON Y In) . Take a point Y on a line In. Line In could be one of
me forms below.
(LINE A B) is the line passing through two points A and B . (PLINE C A B) is the line passing through C and parallel to (LINE A B). (TLINE C A B) is the line passing through C and perpendicular to (LINE A B). (BLINE A B) is the perpendicular-bisector of AB. (ALINE P Q U W V) is the line I passing through P such that L(PQ, II L[UW,WVI · See pages 108 and 126. 16. (ON Y (CIR 0 P)). Take a point Y on a circle (CIR 0 P) . See page 108.
258
Chapter 6. Topics from c-netry
17. (MIDPOINT Y U V). Y is the midpoint of UV. See page 120. 18. (MRATIO Y U V r). Y is a point on UV such that ~~
= r. See page 120.
- - --
19. (NE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with negative orientation, i.e, C
=
A - i . AB. See page 252.
20. (NE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle with negative orientation. See page 249. 21. (ORTHOCENTER H A B C). H is the orthocenter of the triangle ABC. See page 135.
--
--
22. (PE-SQUARE C B A) introduces a point C such that CAB is an isosceles right triangle with positive orientation, i.e., C
=
A
+ i . AB.
See page 252.
23. (PE-TRIANGLE C B A) introduces a point C such that ABC is an equilateral triangle with positive orientation. See page 249. 24. (POINT[S)}Ii,"', }Ii). Take arbitrary points Y1 ,' " ,}Ii in the plane. See page 108. 25. (PRATIO Y W U V r). Take a point Y on the line passing through W and parallel to line UV such that WY = rUV . See page 108. 26. (SIM-TRIANGLE Q C B P A C) introduces a point Q such that triangle QCB and triangle PAC are similar and have the same orientation. See page 248. 27. (SYMMETRY Y U V). Y is the symmetry of point V with respect to point U. See page 120. Y U V r). Take apointYon line (TLINEU U V) such that r 28. (TRATIO _
~~). See page 108.
= 4PuS vu (= rrvy
The following are the predicates accepted by the program as conclusions. 1. (COCIRCLEABC D). PointsA,B,C,andDareco-circleiff L[CAD) or equivalently, SCADPCBD = PCADPCBD. See page 127. 2. (COLLINEAR A B C). Points A, B, and C are collinear iff SABC comments after Example 2.36 on page 74.
= L[CBD),
= O. Also see the
3. (EQANGLE ABC D E F). L[ABC) = L[DEF) iff SABCPDEF = See page 111. 4. (EQDISTANCE ABC D). AB has the same length as CD iff PABA page 111.
SDEFPABC.
= PCDC. See
6.2.
259
Geometry
5. (EQ-PRODUcr ABC D P Q R S). The product of AB and CD is equal to the product of PQ and RS. See page 111. 6. (HARMONIC ABC D). A, B and C, D are harmonic points iff ~~ page 120.
= g~ .
See
7. (INVERSION P Q 0 A). P is the inversion of Q with regard to circle (CIR 0 A). See page 120. 8. (MIDPOINT 0 A B). 0 is the midpoint of AB iff ~~
= 1. See page 111.
9. (NE-TRIANGLE C B A) ABC is an equilateral triangle with negative orientation,
C + w B + w 2 A = O. See page 249.
i.e,
10. (ON-RADICAL P 0 1 A O 2 B). P is on the axis of circles OlA and 02B iff PPOIP - PAOlA
= Ppo,p -
PBO,B .
11. (PARALLEL ABC D). AB is parallel to CD iff SAC D
= S BC D. See page 111.
12. (PE-TRIANGLE C B A). ABC is an equilateral triangle with positive orientation, .
l.e.,
-+C
-+ + w 2-+ B + w A = O. See page 249.
13. (PERPENDICULAR ABC D). AB is perpendicular to CD iff P ACD See page 111.
= P BCD .
14. (PERP-BISECT 0 P Q). 0 is on the perpendicular bisector of PQ iff Popo
= P OQo .
15. (TANGENT 0 1 A O 2 B). Circle (CIR 0 1 A) is tangent to circle (CIR O 2 B). See page 111.
6.2
Geometry of Incidence
In this section, we include those geometric statements that can be formulated and proved without the measurement or comparison of distances or of angles, i.e., geometric facts involving incidence only. Such problems include the transversal, properties of cross-ratios, projective configurations, etc. These problems belong essentially to affine geometry. To prove statements involving incidence only, we need only to use Algorithm 2.32. Actually, the co-side theorem would be enough for the proofs of most of the examples in this section.
6.2.1
Menelaus'Theorem
For the machine proof of Menelaus' theorem, see Example 2.35 on page 73.
260
Chapter 6.
Topics from C-try
Example 6.1 (Converse of Menelaus' Theorem) (0.033,2,6) lfthree points are taken, one on each side of a triangle, so that these points divide the sides into six segments such that the products of the segments in each of the two sets of nonconsecutive segments are equal in magnitude and opposite in sign, the three points are collinear.
Constructive description «points A B C) (mratio D B C ril (mratio E C A r2) (inter F (I D E) (I A B»
The machine proof
The eliminants
!!.!.
u:.!: fuuli;. AF- SADE
-AL
E~
r2'rl
L
(~: = -r,"r2 » C
-
r2'rl·SADE
!d.
SABD"r2"! r 2+1) r2"rd SACD)"(.,+1)
-
simJ!!ify -
Q -
F
SADE= .,+1
SROe
~
SBDE~~ r,+l D~
SACD= r,+1
S
r],SACD
!'.&aa:!1. r,+1
ABD-
-SABc"r,.(r, +l) r,"( -SABc)"( r, +1)
simJ!!ify
Figure 6-1
Example 6.2 (Menelaus' Theorem for a Quadrilateral) (0.033,1,4) A line XY meets the sides AB, BG, G D, and DA of a quadrilateral at Ai> Bi> G1, and D1 respectively. . !l!JJ. .~ . !J!1J. = 1. Show that ~ BA, CB, DC, AD,
Constructive description ( (points ABC D X Y) (inter A, (I A B) (I x Y» (inter B, (I B C) (I X Y» (inter c, (I C D) (I x Y» (inter D, (I A D) (I x v»~ (~".2I. " ££I. "~ BA] C B DC AD ] I
I
= 1) )
The machine proof
The eliminants
~ " ££I. " .2I. "~ AD1 DC CB BA1
~~§..u.u..
I
I
AD, -
SAXY
~
§..u.u.. . ££I. ".21. ".:!!L
a~§.cJa. DC, - SDXY
~
Soxy-SQ/(Y • .21."~ SAXy"SDXY CB, BA,
.2I.l!.!~
SAXY
DC, CB, BA,
simgJify §.cJa. . .21. . ~ SAXY
CB 1 BA1
CB, - SCXY
.:!!L~~ BA, - SBXY
~ Sery -Sary • ~ SAXY"SCXY
8imgJify ~ • SAXY
BA,
.:!!L BA]
~ SAry,SAliX SAXy·SBXY
Figure 6-2
simgJify
Example 6.3 (0.033, 2, 7) The converse of Example 6.2. The figure of this example is the same as Example 6.2.
6.l.
261
Geometry oIlnddence
The machine proof
Constructive description
~
«POints ABC D) (mratio B, B C r,) CDr.)
(inter D, (I
D A)
(I
B, Cd)
(inter
A B)
(I
B, Cd)
(l
~~SBB,C, AA, - SAB,C,
AA,
~ 'r"2'rl AD,
(mratio C, A,
The eliminants
~
~l!JSOB)C, AD] -
SBB)C)
C,SCOB,
~ .,., .r,.SAB ] C , AD] ~
SOB,C,
SBB,C, ,SAB,C,
SBB,c,
SOB,c, ·... ·r' .SAB,C, .img1i/~
SABle]
=
,.,+1
C l -SBDB, -r,
=
,.,+1
SBB,C, SOB]C] -"2'r,
<2!
(-SBOB, .,.,).(,.,+1) SCOB, ·... ·r,.(,.,+I)
-
.im1!!i/~ -
-SBOB, SCOBI'r]
~ -( -SBco ·r').(r,
+1) .im1!!i/y SBco·r,·(r, +1) -
-
Definition. Two points on a side of a triangle are said to be isotomic to each other. if they are equidistant from the midpoint of this side. Example 6.4
21
(0.083, 2, 9) The isotomic points of three collinear points are collinear.
Constructive description The machine proof ( (POints A B C) (lratio D B C rd (lratio E A C ,.,) (inter F (I A B) (I
(pratio (pratio
~ . ~.~ AE.
_~_I
BF
D, BCD -I)
.~
(if+1).(ii- l )
CO,
~ ~~--~~~~I)~----
F.B DIC EIA
E~' ~ B
F
(if+1) · (ii-I) .(-~-I)
~
SR~
(SAOBE+SBOEH ii-I).( ~+1)
~ ____~-~(~-~S~A~B~p~·,.,~+~S4A~B~PL)·?~~-- (-SABO.",+SABC.,.,H -,.,+1).( ~+1)
.imgi/~
Figure 64
SARa (SABO-SABcH ~+1)
Q -
~~-lBF, if+1 ~~~ AE, ii-I ~~-l~+I
co,
~!.~ BF- SBOE
~~.!. AE
...
SAOBE~ - (SABO-SABC ·,.,) SBOE~ - «,.,-I).SABO)
~gr'~1 SABogSABc·r,
SABC ·r,.(r,-l) (SABC·r,-SABC) ·rt
.img1ih 21ln this chapter. the
AE, CD,
-I
E, CAE -I)
(~O FLAB F -I) (~~~=-I»
10.
BF.
~ .....;;.!-. . ~ . ~
D E»
F, A
CD.
The eliminants
1
figure indices are always the same with the indices of the examples thaI they belong
262
Chapter 6. Topics from
~try
Ceva's Theorem
6.2.2
For the machine proof of Ceva's theorem, see page 72. For the generalizations of Ceva's theorem to arbitrary polygons, see Subsection 2.7.3.
Example 6.5 (The Converse of Ceva's Theorem) (0.066, 2, 8) If three points tak£n on the sides of a triangle determine on these sides six segments such that the products of the segments in the two nonconsecutive sets are equal. both in magnitude and in sign. the lines joining these points to the respectively opposite vertices are concurrent.
Constructive description ( (points A B C ) (meatio D B C r.) (meatio E C A r2) (meatio F B A "' 'r2) (inter 0 (I B E) (I A D» (inter z (I B A) (I 0 Cll
(*=%»
The machine proof
The eliminants
(*)/(~;)
u.~§..aaJl. AZ- SACO S 0 SAGo , SARE
!
Saa o
ACO
%,(-SACO)
g .J..-SBCE,SABD),SABDE % ,SACD,SABE'( -SABDE) sim!!J.ijy
SBGE ,SARD %,SACD,SABE
1:.
SBCE ·SABD ·(-I)
-
r2 -r l "SACD OSABE
!l.
-SABC· ... ·SABD·( ... +l) r2 ·r "SACD ,SABC ·( ... +I)
-
.imp!ij~ -
~
S
SABDE
OSaCfj -SARD
BCO
*~
SABDE
- (r2 l) or
E~
SABE= ... +1
E~
SBCE=
r2+1
SACDg~ r,+1 D~
SABD=
r,+1
r) -SACD
.Q -SABc·r"!rJ+l) - r,.( -SABc)·(r, +1) A
F
B
simg1ijy
Figure 6-5
Definition. If a line is drawn through a vertex of a triangle. the segment included between the vertex and the opposite side is called a cevian. The triangle DEF in Figure 6-5 is called the cevian triangle of the point 0 for the triangle ABC. Example 6.6 (0.050, 3, 12) If LM N is the cevian triangle of the point S for the triangle ABC. we have SL/AL + SM/ BM + SN/CN = 1.
6.2.
263
G_try oIlncideoce
Constructive description ( (POints ABC S) (inter L (1 B C) (1 A S» (inter N (1 B A) (1 C S» (inter M (1 A C) (1 B S» (*+i* +~
= I) )
The machine proof
The eliminants
UL+~+~
UL~~
BM
CN
AL
BM - -SA B C
-~.SABC-li .SABC+SACS eN It'
M_
~!!.§..u..s.. CN -SABC
-SABC
~!..§..as:a
!:!. -*'~BC+SACS·SABC-SABS·SABC -
AL -SABC
SABC,(-SABC)
.im.!!!.i/ll
¥,: ,SABC-SACS+SABS
-
!:
SBCS=SACS-SABS+SABC
SABC
S acs ,S ABa SACS ,S'8C+S'BS ,SABC (SABC)'
8im~i fJl Sacs S.GS+S.BS SABC
ore.!.,-co -
SABC
.im.!!!.i/y
Figure 6-{)
~
-
I
Example 6.7 If LM N is the cevian triangle of the point S for the triangle ABC (Figure 6-6) we have •
SAMCS8NMScrN SANLSBLMSCNU
= 1.
Constructive description: «POints (inter M (1 A C) (1 S B» (inter N (1
ABC S) A B)
(1
(inter
L
(1
(1
A»
S
S
N
!:!.
S
(-SCSL ' S,VC H -SBCS ,SABU) ·S ,"'M'SACBS·( -SACIS) (-Scsu ,SABC) ,SBLM'( -SACS,SABL) ,SACBS '( -SACBS)
-
oi"'.!!!.i/" -
-
8M
,S"M
CLN
.im~i/"
S
SACIlS -SA8CS
M Sacs ,SAcS
CSAI
S
SA8CS
M-S'C,.S'AS
S
SABCS MSARs ,SAsa
ABU-
SBCS ,SACS ,(SA8S)' ·( -SACS ,SABcH< -SABSC»' (SACS)' ,SBCS ,SABS ' ( -SABS ,SABC H( -SABSC
»2
SAC8S
AI SBn ' SABe
ALU
SCSdSABS)' ,SACL (SACS)' ,SBSL ,SABL
SaCS ,S68M
/II Sas' ,SASG
S
SCSU ,S8LU ,SACS ,SABL SCSL ,S BCS ,SASS ,SA8d SACL ,SA8SH SABCS),SABCS SBCS ,SACS ,S8SL ,SA8C,SACS ,SABL ,(SA8CS)'
-
f.
N
BUN
S
BLU
!1..
SACS ,SARC
SAC8S NSCSN,SAAG CMN SACBS
S
.im~iJy Scs,.Sse S , S'
SANLSBLUSCNM) )
ALN
M
SCMN ,SBLU ,SALN
-
=
The eliminants
The machine proof SCI H oSSAIN ' S"
B C)
S C» (SAULSBNMSCLN
S
SAIlCS
L $'8s·S,aG A8L
SABSC S L SsCS ·S.aS 8SL -SA8SC 5 L SACS ,SABG ACL SABSC S L SaCS ,SACS CSL -SABSC
Example 6.8 (0.033,3,4) The lines AP. BQ. C R through the vertices of a triangle ABC parallel. respectively. to the lines OA I• OBI. OCI joining any point 0 to the points AI. B l • Cl marked in any manner whatever. on the sides of BC. CA. AB meet these sides in the points P. Q. R Show that OAI/AP + OBI/BQ+ OCI/CR = 1.
264
Chapter 6. Topics (rom Geometry
The eliminants The machine proof Constructive description ~!!~ CR - SASC ( (points ABC 0) ~+~+~ CR SQ AP ~9.~ ~!.§..a= (on A, (I B C» SQ - -SASC' AP - SASC .SASC+~·SASC+SASO !! ~ S9 AP SSCO=SACO-SASO+SASC (on B, (l C A» SASC (on c, (l B A» SC+SACO,SASC-SASO ,SASC _Q -~.SA' de (inter p (I B C) (P A A, 0» SASc-(-SASC) (inter Q (l A (inter R (I A
C)
B)
(P B B, (P C c,
0» .imJ!lify
~.SASC-SACO+SASO
-
0»
E..
SASC Saco-SABC-SACO-SABctS6BO"SABQ
-
(SASC)'
simJ!!.iJy SaCO-SAGQ+SARQ SASC are!.,-co ~ SASC
C,
A
B
R
Figure 6-8
Example 6.9 (0.083, 2, 9) If LM N is the cevian triangle of the point S for the triangle ABC, we hnve AS/SL = AM/MC + AN/NB.
Constructive description ( (points ABC S) (inter L (I B C) (I A S» (inter N (I B A) (I C S» (inter M (I A C) (I B S»
The machine proof
The eliminants
-( ~*)
.:1..l!.A:!~ CM- -SSCS
.u SL
.41:!.!!.~
M -(-.4I:!.,SSCS+SASS)
=
(~+*=*»
u
it·(-sscs)
tf:.
-(Sscs:.§.6cs-SSCS,SASS)
SN- SSCS
~~~ SL -
i t ·(sscs)· simJ!lify -(S,wS-SASS) -
.f.. -
B
c Figure 6-9
L
i t ·sscs -(SACS-SASS)'SSCS SASSC 'SSCS
simJ!lify -
-(SACS-SASS) SASSC
are~-co
-(SACS-SASS) -SACS+SASS
-
SSCS
SASSC= -
8im~ifY
Example 6.10 (0.083, 1, 5) The Ceva' s theorem for a pentagon.
(SACS-SASS)
6.2.
265
G«metry oIlDddence
Constructive description ( (POints A 8 C D E 0) (inter A, (lOA) (I C D» (inter 8, (I 0 8) (I D E» (inter c, (I 0 C) (1 E A» (inter D, (I 0 D) (I A 8» (inte~i!.QR~ C» (~~~~~=I» AID BIEC1A 0 1 8 E 1 C
The machine proof
The elirninants
-~ - ~ - ~ - ~ - ~
~ ~2.a.F:o.
AC I
EB1
DA1
eEl
SOl
CE, - SCEO
~ -(-Suo) _ ~ _ ~ _ ~ _ ~ -SCEO
AC,
EB,
DA, BD,
~ -SUO -(-SApO) • ~ _ ~_~ SCEO -(-SBDO)
~
AC,
EB,
DA,
-Seeo -S81m- S ADO • ~ - ~ SCEO -SBDO -( SACO) EB, DA,
SBDO -SACO
EB,
BD, -
SBDO
~<:1~ AC, -
-SACO
!?!.I.l!.t&.ao. EB, -
SBEO
~~~
.im~'/!I SaBo -SAOO • ~ _ ~
D
~~~
DA, -
DA,
SADO
~ Sapo·SHgO ,SAPO • ~ SBDO -SACO -SBEO
"mg1'/Y
DA,
~.~ SACO
DA,
~ SACO ,SAPO 0,
A
B
Figure 6-10
-
SACO -SADO
.im~'/II I
Example 6.11 (0.067,3, 16) If the three lines joining three points marked on the sides of a triangle to the respectively opposite vertices are concurrent, the same is true of the isotomics of the given points. Constructive description ( (POints A 8 C 0) (inter D (I A 0) (I C 8» (inter E (I 8 0) (I A C» (inter F (1 C 0) (I A 8» (pratio D, 8 CD-I) (pratio E, CAE -I) (~OF~8F-I) (~~~=I»
Figure 6-11
FIB DIe E1A
The machine proof -~ . ~ - ~ A£.
CD]
BF
~
AE, CD,
-(-~
.~
(*+l) .(~-l)
CD,
*+1
~~~ BF,
8F]
~-±!L.~ . ~ -.ut-l
The elirninants ~~~ AE,
~-l AE
~~JCD,
~+l
.ut!..~ BF- SBCO
~
K -
1
(*+l) -(~-l)-(-~-I) Sa~
(SACBO+SBcoH~-l).(~+l)
~!.~ AE- SABO
~Q~ CD- SACO
266
Chapter 6. Topics from Geometry
The eliminants
The machine proof ~
SABOC= -
SaGo -SABQ
(SACBO+SBcoHSABCO-SABO)'( ~+l)
!2. -
(SACO-SABO)
SABCO=SACO+SABC
SBCO,SABO ,(-SACO) (SACBO+SBCO) ,(SABCO-SABO)'( SABOC
SACBO=SABO-SABC SACO)
SBCO=SACO-SABO+SABC
are!!...- co (SACO-SABO+SABcl ,SABO ,SACO -
SACO,(SACO-SABO+SABcl ,SABO
simJ!!ify
6.2.3
The Cross-ratio and Harmonic Points
Definition Let A, B, C, and D be four collinear points. The cross ratio of them, denoted by (ABCD). is defined to be
(ABCD)
= (CA)j(DA) . CB
DB
Example 6.12 (0.066, 2, 10) The cross ratio offour points on a line is unchanged under a projection.
Constructive description «points 0 ABC, D,J (inter C (i A B) (I 0 C,)) (inter D (I A B) (I 0 D,)) (inter A, (I c, D,J (lOA)) (inte~B, (!..fL~1 0 B» (..ll::l!2=~~» Be AD
A
AID) BICI
Figure 6-12
The eliminants
The machine proof M.A£
D,B,~SOBD,
AD lW:-a ~.~
CtB)
~
C,B, -
D}AI
SOBSi..... •
.I!.II. . ~
~ SOBD)'D1A 1
D1AI -
SOBD, ,SOAC,
I},JJ,E.SOBD, -IoD- SOAD,
!J£. AC
~QSOAC,
AD BC
(-SOBD, ) ,SOBC, ,SOAD, •
.4.l<.
SOBD, ,SOAC,'( -SOAD,)
BC
simglify
SOBC, SOAC,
•
.4£ BC
52
(-SOAC, ) ,SOBC,
-
SOAC, ' ( -SOBC,)
sim~ifY
SOAD)
AD BC
~ SOBC, ,SOAD, •
g
SOBC,
C,A, ~ SOAC,
BC- SOBC,
267
6.l. Geometry oIlDddcocc
Definition. For any five different points 0 , A, B , C, and D. the cross-ratio of any four collinear points on lines OA, OB, ~C, and OD isa constant and is denote by O(ABCD).
Example 6.13 (0.050,2, 10) Two lines OABC and OA1B1Cl intersect at point O. If (OABC) = (OA 1B 1 C l ) then AA I • BBl. and CCI are concurrent.
Constructive description ( (points 0 A Ad (mratio
(mratio
B A 0 rd
C 0 A ... )
(mratio B, 0 A, r3) (mratio C, 0 A, r' ·....r,) (inter x (I A Ad (I B Bd) (inter Y (I A Ad (I C Cd)
(ft, = fl;»
The machine proof
The eliminants
(~)/(-P,:;)
~!: SACC, ~~ SABS,
!.
s
-
SASS"SA,CC, SACC"SA,SS,
<} SASB, ,SOA,c'(r"""r,+l) - (-SAA,c 'r"""rd ,SA,Bs,'(r"""r, +1) .im!!4ify
!!J -
£. -
SASS, ,SOA, C SAA,C'r"""r"SA,sS,
(-SAA, B'r3) ,SOA, c ·(r3+1) SAA,c ·r3 ·,.,·r"SOA,B·(r,+1)
.im!!4ify
Figure 6-13
A,y-SA,CC,
~ SA,CC, • ~ SACCI A,X
SAA, S,SOA, C SAA,C'''' 'r"SOA,S
SAA,S,(-SOAA, ·,.,H,.,+1) SOAA, ·... ·r' ·SoA,B·( ... +l)
.imJ!!.ify -SAA,S r"SOA,B
!L -
A,X- SA,SS, ~ -SAA 1 c ·r3'r']·rt Acel rS ·r2 ·rl 1
+
~
S
AIGC) -
SOA,C r3 ' r 2" r l+1
~SOA'S SA,BS, - r.+l B, -SAA,B'r,
SABS, =
r.+l
CSOAA, SAA,C= r.+l C-SOA.A) ' ''2
SOA,C
... +1
S-SOAA, SOA,S= r,+1 SSOAA, ·r, SAA,S= r,+1
-SOAA, ·r..(r, +1) r..(-SOAA,)·(r,+1)
.im~ifY 1
Defmition. If(ABCD) = -1. we call A,B,C,D a hamwnic sequence. or C,D divide the segment AB harmonically. Four lines passing through a point 0 is said to form a harmonic pencil if another line meets them in a hamwnic sequence A, B, C, D. Each of the line is called a ray of the the pencil and 0 is called the center of the pencil. 0 A and o B. OC and 0 D are said to be a pair ofconjugate rays. or OC and 0 D are harmonically separated by OA and OB.
Example 6.14 (0.016,2,4) Let A, B, C, D be four hamwnic points and 0 be the midpoint ~ of AB. Then OC · OD = OA .
---
Chapter 6. Topics from Geometry
268 Constructive description ( (points A B) (meatio CAB r) (meatio DAB -r) (midpoint 0 A B) (~=~»
The machine proof (~)/(~)
(-*+tH-¥a+t)
Q
-
«¥»'
Q (2*-I).(-r-l) -
-r+l
52
(r-l)·(r+1) (r-l) ·(r+1)
-
simg1ify 1
If 0 is the midpoint of AB. and ijC, D
Example 6.15 (0.050,2,3) The converse theorem. are points of the line AB such that OC . 0 D =
M. then A, B, C, D are harmonic points.
Constructive description ( (points A B) (midpoint 0 A B) (lratio CO A r) (iratio DO A ~) (harmonic ABC D) )
The eliminants ..uLg....::.=L
The machine proof (-~)/(*)
BO
D -~ .r+1
-
=~ . -~
AO
~g -
C -(r-l)·(~ ·r-l) ~
=
~ . r-l
.4£ C r-l BC=~
(1)
(r-l) .(-~+r)
simplify
=
~'r-l ~ lIJl.- r AO
Q (_l r _1).
- (-Ir-h·(t)
simg1ify 1
Example 6.16 (0.266, 8, 6) The sum of the squares of two harmonic segments is equal to four times the square of the distance between the midpoints of these segments. Constructive description ( (points A B) (meatio CAB r)
(meatio DAB -r) (midpoint 0 A B) (midpoint M C D) «.4.1!.).+(~2) MO MO
= 4»
The eliminants ~
-
g~
OM
CQ.""
(-t) ' (2~-I)
~~ 1 OM (-t).(2~-I) -0(_12 )·(24!O.-1) ~= :...sA
co
-0 ~
AB ~O
co
_
l2£ AlL
~·r-4!O.-r AS A8 -(r-l)
-(r-:!l ~ 'r-*-r
~£_r_
AB-r+l
269 The machine proof
4(* +~) 1 -.
-.
- 41' ~'-~ - 41'-~ 141' 1 M ~' cn cn + cn CQ.."MQ + aD + ..
-
.
GO
4
(4) - (-~+t>'
~ +1
• imJ!!.i/y
(2~-1)'
41'+1
g
_
--"0
(-~-2ii+1)'
D (4£ =
_(~)' AB
_2~2 'r+4£ -24£ 'r 2 +24£ 'r+2r2 -2r+l)·( ~ .r_4£_r)2 .( -r+l)2 fA AJL. 4JL AB ..dtR fA 2
l.r'
AB
(- i i -r + i i - )' -( -r+ I)' -( i i -r - ii-r)' ' -2 .imJ!!.i/y i t -r'-2it -r+ii, -2~ - r'+2it -r+2r'-2r+1 -2
-2
-
£ -
(ii -r -ii+1)' (r<+2r'+1)-(r+I)' (r'+1)' -(r+1)'
. im~i/y 1
Example 6.17 (0.083,1,10) Let A, B, R be three points on a plane. Q and P be points on line AR and BR respectively_ S is the intersection ofQB and AP_C is the intersection of RS and AB. F is the intersection ofQP and AB. Show that (ABCF) = -1.
Constructive description «points A B R) (on p (I B R» (on Q (I A R» (inter S (I A P) (1 8 Q» (inter c (I R S) (I A B» (inter F (I P Q) (1 A 8» (4£ Be
The machine proof 4£
t;,~-4£ -SAPQ
C
B
Figure 6-17
F
Be
SARS -SBP9 -SAPQ -SBIIS
-
~ A
4££§..u..s.
-~ BF
= _~» BF .
§.. -SARP -SABq -SBPq -(-SABPq) -
SAPQ -( SBRQ -SABP) -SABPQ
BimJ!!.i/y
-
~!.SAPq
BF- SBPQ
~
52
The eliminants
-SARP-SABq -SBPq SAPQ-SBIIQ -SABP
2/i!!e. - 7W-::rn-P-Bii19fi- RQ-B!:i).«2»' (-RP -AQ-PH -RQ-8R-P)-BP-AR-P-«2»'
Be-SBIIS S S SBllq -SABP BRS SABPQ S S SARP'SABq ARS SABPQ
SABP=~(BP-AR-p) SBIIQ= -
l(RQ ·8R-P)
SAPQ= -l(RP-AQ-P) SBPQ= -
2(RQ-BP-P)
SABQ=t(8R-AQ.P) SAIIP= 2(7W-AR-p)
.im!!li/y I
Example 6.18 (0.033,7,6) Given (ABCD) = -1 and a point 0 outside the line AB. ifa parallel through B to OA meets ~C. OD in p. Q. we then have PB = BQ.
Chapter 6. Topics from c-try
270
Constructive description ( (POints ABO) (mratio CAB r) (mratio DAB -r) (inter p (I 0 C) (p BOA)) (inter Q (I 0 D) (p BOA)) (midpoint B P Q) )
Figure 6-18
The eliminants
The machine proof
IJ.J:.~~ BQ- SBOD S P -(SOCD-S6BO-SBOD -S60C)
-u, BQ 2~
-
BOPD
SBOD
!:..
SBOD= -(r-1)
-(-SOCD -SABO+SBOD -S •.aC) SBOD -SAOC
-
S
!!..SSQC-r-SAQC
OCD-
Q (SBOC-S6Bo-r'-SBoc-S6BO -r) -( -r+1) -
£. -
r
1
C~
S60C=
S6BO -S60c-(-r+1)2
sim1!!.i/~
SAOC
D~
r+1
C~
~
SBOC= r+1
S60C -S6Bo-r-(r+1) sim!!1i/y (-S6BO -r)-(r+1)
Example 6.19 (0.033,2,8) The converse theorem of Example 6.18. Constructive description ( (POints ABO) (on C (I A B)) (inter P (I 0 C) (P BOA)) (lratio Q B P -1) (inter D (I 0 Q) (1 A B» (harmonic ABC D) )
The machine proof
The eliminants
(-~)/(*)
AJil,£SAOg BD-SBOQ
g
g
SBOg • S60Q
S60Q~ SBOQ~
_01£ BC
-(-SBOP) • 01£ -S60P-2S6BO BC
~ -(-SBOC)-S6BO •
S6BO-S60C
01£ BC
.i"'!1:.i/~ §..a.aJ;;.. 01£ S60C BC
C 01£-(-S6BO-01£+S6BO) A8 68
_
-
(-S6BO-~)-(~-1)
-
(S60p+2SABO)
-
(SBOP)
SAOP!:, - (SABO) S P-5SQC·S'80 BOP SAOC SAOC£ SBOC£ -
(SABO-~) (~-1)-S6BO)
01££.JL BC
01£-1 6B
.im~i/J/ 1
6.2.4
Pappus' Theorem and Desargues' Theorem
For the background about na configurations. see Subsection 2.7.2.
271
Example 6.20 (pappus' Theorem) (0.166, 1, 14) Let ABC and AIBIC I be two lines. = ABI n Al B. Q = ACI n Al C. S = BCI n BI C. Then P. Q and S are collinear.
and P
For a machine proof of this example, see Example 2.42. The following proof is shorter.
Constructive description «POints A A, B B,) (on G (I A B» (on G, (I A, B,» (inter p (I A, B) (I A Bd) (inter 0 (I A G,) (I A, G» (inter S (I B, G) (I B G,» (inter T JLB, G) (I p Oll (W=W» c, B,
The eliminants
The machine proof
!I,Z:! -SB,QP
(W)/( W)
CT -
J::~.~ -SB,QP
E..
CS - SBCC, S P -SA, BB, ·SAB, Q B,QP SAA,B,B
CS
(-SBB,C, )·SqcP -SB,QP ·(-SBCC,)
-
SQCP
~~SBB,C,
S
pSBQC ·SAA,B, SAA,B,B S ~ -SBQC ·SAA, B, BCC, SAA,QB, QCP
J:..
-SBB,C, ·(-SBQC ·SAA,B,) ·S,U,B,B
-
(-SA,BB, ·SAB,Q) ·SBCC,·(-SAA,B,B)
6im~ifll
~
SBB,C, ·SBQC ·SAA, B, SA,BB, ·SAB,Q ·SBCC,
S
~SA,BB, · SAB,Q BB,C, -SAA,QB,
SA, BB, ·SAB, Q ·SBQC ·SAA, B, ·SAA,QB, SA,BB, ·SAB,Q ·(-SBQC·SAA,B, H-SAA,QB,)
-
6imJ!!. i fll B
Figure 6-20
Line PQ is called the Pappus' line and is denoted by [123] indicating the order of the points AIBICI . Note that there are six Pappus' lines for one Pappus configuration: [123], [231], [213], [321], [312], and [132].
Example 6.21 (0.116, 1,6) The dual of Pappus' theorem.
Constructive description ( (POints A B GOOd (inter A, (I 0, B) (lOG)) (inter B, (lOA) (I 0, G)) (inter G, (lOB) (1 0, A)) (inter 1 (1 B Bd (I A Ad) (inter zu!.G, G) (1 A Ad)
(..4l. = .!!.L.) ) A,I A,ZI
The eliminants
The machine proof
.!!.L. ~ -SACC,
(/!;)/(~)
A,ZI -
~ SCA,C, •
..4l.
-
-SACCI
A,I
L
SABB, ·SCA,C, -SACC, ·(-SBA,B,)
~
SCA,C, SABB, A,I -SBA,B, S ~ SACO, ·SABO ACC, SABO,O
..4l.4,
S
SABB, ·SBCA, ·SAOO, ·SABO,O
-
SACO, ·SABO ·SBA,B, ·(
6imJ!!.ifll -
~ SBCA, · SAOO, CAlC, BA,B, -
SABO,O)
SABB, ·SBCA, ·SAOO, -SACO, ·SABO·SBA,B,
S
-SABO,O
l!J SBCA, ·SAOO,
S
-SACOO,
'!J SACO, ·SABO ABB, -
SACOO,
~ SACO, ·SABO·SBCA, .SAOO, ·( -SACOO,) -SACO, ·SABO ·SBCA, ·SAOO, ·SACOO,
Figure 6-21
.imJ!!.ifll I
Example 6.22 (pappus Point Theorem) (0.550,2,26) The three Pappus lines [123]. [312] and [231] are concurrent. So are the Pappus lines [213]. [321] and [132].
Chapter 6. Topics from "-try
272
Constructive description ( (points A A, B Cd (on B, (I A, Cd) (on C (I A B» (inter E (I A, B) (I A Bd) (inter F (I A, C) (I A C,» (inter G (I B C,) (I A Ad) (inter H (I B Bd (I CAd) (inter I (I B Bd (I A Cd) (inter J (I B Ad (I C Cd) (inter K (I G H) (I E F» (collinear
B,
Figure 6-22 I J K) )
Example 6.23 (Leiseniog's Theorem) (1.233, 4, 32) Continuing from Example 6.20, let o be the intersection of line AB and line AIBl> and L I, L 2 , and L3 be the intersections of lines OP and CCI , lines OQ and BBI , and lines OS and AAI respectively. Show tlwt L I , L 2 , L3 are collinear. Constructive description ( (points A B A, Bd (on C (I A B», (on c, (I A, Bd) (inter p (I A B,) (I B A,» (inter Q (l A, C) (I A Cd) (inter s (I B C,) (I B, C» (inter 0 (I A B) (I A, B,» (inter L, (loP) (I C Cd) (inter L2 (I 0 Q) (I B B,» (inter z, (I L, L2) (I A Ad) (inter Z2.l!...f, L2) (los»
o
(~=~» L2Z1 L Z 2 2
Figure 6-23
s
Example 6.24 (Desargues' Theorem) (0.250,2,18) Given two triangle ABC, AlBIC.. if the three lines AA I , BBI , CCI meet in a point, S, the three points P = BC n BICI, Q = CA n CIA .. R = AB n AIBI lie on a line. Constructive description ( (points ABC S) (on A, (I SA» (on B, (I S 8» (on c, (I S C» (inter P (I 8 C) (I 8, Cd) (inter Q (I A C) (I A, Cd) (inter R (I A 8) (I A, B,» (inter Z2 (I A, B,) (I P (inter z, (I A B) (l P (~ . ~
Q» Q»
QZ, PZ2
~s==----4-+-='1
=
I» Figure 6-24
273 The machine proof
The eliminants
!:!.I. . ~
!:!.I.~~
QZ.
QZ. - SABQ
PZ2
~~ , ~
~~SA.B.Q
~ SABP"SA. B.Q SABQ ,SA . B.P
S
-
SABQ
~
pz. - SA.B,P
pz.
ABQ
SABP ,SA) 8,C, ·SACA, "SAA,CC,
-
SAA.C. ,SABC ,SA.B.P ·(
.imJ!l.i/u -
f.
A,B.Q S
-SABP ,SA. B. c. ,SACA.
-
QSA.B.C. ,SACA. SAA,CC, pSA)B,C"SSCB)
A.B.P
SAA.C"SABC ,SA,B.P
.imJ!!.i/Ji
S
-SBB.CC.
pSBB.C. ,SABC ABP SBB,CC,
c. = - ( SACA. · SO) ~
SB8,O, -SACA)
SAA,C.
SAA.C. ,SBCB.
SBB.C. ~ - -
( -SBCB • . ..u;. ~).SACA • .im~ih
S
S
SAA.CC.)
-SBB.C. ,SABC ,SA.B.C. ,SACA. ,(-SBB.CC.) SAA.C. ,SABC ,SA.B.C. ,SBCB. ,SBB.CC.
-
QSAA.C. ,SABC SAA.CC.
(SBCB. ' ~ SO)
(-SACA. ' ~ ),SBCB.
Example 6.25 (0.216,2,12) The converse of Desargues' theorem. Constructive description ( (points A B G Ad (on p (I B G» (on 0 (I A G» (inter R (I A B) (I PO» (on B, (I R Ad) (inter G. (I A. 0) (I B. P» (inter S (I A A,) (I B Bd) (inter z (I A Ad (I G G,» (ti; = tt» The machine proof (~)/('f!;) ~ -SCA.C, ,~ SACCI
.§...
A.S
-
- SABB. ,SCA.C, SACC. ,(-SBA,B.)
<2l
SABB. 'SA. PB. ,SCA.Q,SA. PQB.
-
SPQB"SACA. ,SBA.B. ,SA.PQB •
• imJ!!.i/u -
SABB. ,SA. PB. ' SCAt Q SPQB, ,SACA. ,SBA.B,
'iiif"
'iiif"
~ SABA· ' E'(-SA,PR ' ~+SA,PR) 'SCA.Q RB RB SA. PQ ' ~ · SACA.·(-SBA.R ' ~+SBA.R) .im1!! i /y ~
SA. PQ ,SACA. ,SBA.R SABA. ,SA. PQ,SABP"SCA. Q ,SAPBQ SA, PQ ,SACA, '( -SBPQ ,SABA. H -SAPBQ)
.imJ!!.i/ll -
9.
SABA. ,SA, PR ,SCA,Q
SABP"SCA. Q SACA. ,SBPQ
SABP"( -SACA. ' i!.+SACA,)
SACA. ·(-SABP, ~+SABP)
6imgf, i /ll
Chapter 6.
274
6.2.5
Topics from Geometry
Miscellaneous
Example 6.26 (0.133,1,8) In a hexagon AClBAlCBlo BB l , ClA, AlC are concurrent and CCl> AlB, BlA are concurrent. Prove that AA l , BlC, ClB are also concurrent.
Constructive description ( (points A A, B C C,) (inter 0 (I A C ,) (I A, C» (inter H (I A, B) (I C C,) (inter B, (I A H) (I B 0» (inter I (I B, C) (I A A,) (inter zLi!. c, B) (I A A,)
(.li A11
=
~»
The eliminants .:!!.L. ~ SABC,
The machine proof (fh)/(~)
A1Zr -
~ -SA,BC, •
oiL
-
All
.!... -
-SABC I
A1J- SA 1 C8 1 AICSI -
(-SACB,),SA,BC, SABC, ,(-SA,CB,)
-
SABC,-(-SA,BC,SAOH)-SABHO
.imJ!!ify -
!i -
l!.! SA,BC -SAOH
S
~ SACH-SABO,SA, BC, '( -SABHO)
A1Z/
SACH-SABO,SA, BC, SABC,-SA,BC,SAOH
l!J SACU , SABa
S
ACB, SABHO H SA, BC, -SACO AOH SA,CBC,
S
!!.SA,BC -SACC, ACH- -SA,CBC, OSACC, -SAA,C ACO
S
Q -
Figure 6-26
SAAlelC
0 SABO, ·SAA,C ABO
""A----"-C'-,-----:::,.,.O
SABHO
S
S
SA, BC -SACC, -SABO -SA, BC, -( -SA, CBC, ) SABC,-SA,BC -SA,BC, -SACO-( SA,CBC,)
.imE!.i fy -
SAIBO]
~.!. SACB,
SAA1C)C
SACC, -SABO SABC, -SACO
SACC,-SABC,-SAA,C ,(-SAA,C,C) SABC, -( SACC, -SAA,C) -SAA ,C,C
Bim!!l:ify 1
Example 6.27 (Nehring's Theorem (1942» (0.533,2,14) Let AA l , BB l , CCl be three concurrent cevian lines for triangle ABC_ Let Xl be a point on BC, X 2 = XlB l n BA, X3 = X 2 A l n AC, X 4 = X 3 C l nCB, X5 = X 4 B l n BA, X6 = X5Al n AC, X 7 = X 6 C l nCB_ Show X 7 = XlConstructive description «points B C A 0) (inter A, (I A 0) (I B C» (inter B, (I B 0) (l C A» (inter c, (l C 0) (I B A» (on x, (I B C» (inter X, (I B, X,) (I B A» (inter X, (I A, X ,) (I C A» (inter x . (l c, X,) (l B C» (inter x. (I B , X.) (I B A» (inter x. (I A, X.) (I C A» (inter Z, (I C A) (I c, X,» (inter hl A, X.) (I c, X,) (~ . ~=l» XIZI
CtZ'l
Figure6-Zl
275
6.l. Geometry oIlDddence
The machine proof ~- ~
x.z.
C 1 Z2
~ -SA,C,X, _ ~
c)
- SAl X 1 Xs
Zl
~ SA,CtX$,SCAX,
SA,X,X. -SCAC, ~ (-SB,C, X, -SBAA, )-SCAX, -SBB, AX,
SAA,X, -SBB,X, -SCAC, -SBB,AX. 'im~i fJl
-ss,o, x. ·SSAA1 ,SOAX,
~
SAA,X, -SBB,X. -SCAC.
x,
,Sscc, ,SSAA) ,SOAX, ,Ssc, ex;! SAA.X. -(-SBC,X, -SBCB. )-SCAC. -(-SBC,CX,) -58, C,
.im~i f'JI
-
-Ss,c,x" ,Sscc, , SSA"'] ,SOAX, S AA,X, -SBC,X, -SBCB.-SCAC.
~ -SAt B t X"SOAC, 'Sscc, ,SSAA, ,SOAX, ,SCA) AXa
-
SAA,X, -SAA,X, -SBCC, -SB CB.-SCAC. -S CA,AX,
.. mI!lif~ -SA, B, X, -SBAA, -SCAX, SAA.X. -SAA,X, -SBCB,
~
-SA, 8, X , ,SSAS, ,SSAA] ,SCAX"SSB,AX,
SAA,X, -( -SAB.X, -SBAA,)-SBCB, -(-SBB,AX,)
,im1!!.i/lI
-SA,B,X, ,SSAB"SCAX,
-
SAA,X, -SAB,X. -SBCB,
-(~ -SBCB 1 -SBCBL..40 -!!I.)-SBAB -( -SBCA -!!I.+SBCA) ~.JU;. I BC
X
(-~-SBCA+SBCA - ~H-SBAB, - ~+SBAB,)-SBCB, .i"'!1 if~ I
Example 6.28 (0.733,2,12) Let three triangles ABC. AIBICh A 2B 2C2 be given such that lines AB. AIBI . A2B2 intersect in a point P. lines AC. AICI • A 2C 2 intersect in a point Q. lines BC. BI C h B 2C 2 intersect in a point R. and P. Q. R are collinear. In view of Desargues' theorem, the lines in each of the triads AAh BBJ, CCI ; AA2• BB2• CC2; A I A 2• B I B 2• C I C 2; intersect in a point. Prove that these three points are collinear. Constructive description
( (POints A A, (inter p (I A,
B B, 8, C) Bd (1 A
B»
(on Q (1 A C» (inter R (l P Q) (1 B C» (inter C. (1 B. R) (l A, (on A, (1 B, P» (inter c. (1 B, R) (l A, (inter / (1 B B,) (1 A Ad) (inter J (1 B 8,) (1 A A , (inter K (1 B. 8,) (l A, A.» _ __ (inter Z K (1 J /) (1 B, 8,» (~= :!~:)
Q» Q» »
Figure 6-28
)
Chapter 6. Topics from Geometry
276 The machine proof (~)/(B'ZK) B2K
B 2 ZK
B 2 K- SAISlA]
J
BlK
K (-SAJBJAz) ·SBz/J SB, /J.( -SA,B,A,)
!::.
SA,B,A,·SBB,/·SAB,Az·(-SABA,B,) SB,B2rSABA2·SA,B2A2·( SABA 2B2)
simJ!lify -
1..
SA) 8, A2"SSB2/ -SABaA2 SSlB2/ oSABA2"SAtB2Al
SA,B,A,·(-SBB,B, ·SAA,B) ·SAB,Az ·(-SABA,B,) (-SBB,B2 ·SAA,B,)·SABA,·SA,B.A. ·( SABA,B,)
-
sim1!!.i!y -
~
SAl 8
J
/ta,SAA) S ,SABaA,
SAAtBt"SABA2"SAIB2A2
(-SA,,,, B•. ~+SA'B'B' ).SAA,B .SAB.P·V
SAA'B,·(-SABB, · ~+SABB,)·SA'B.P·~ simJ!lify
-
1:. -
B, ZK Z.lf SB,/J B.ZK - SB./J
~~SA,B'A'
~ -SBzIJ . ~ -SSIIJ
The eliminants
SA, B, B 2 ·SAA] S·SAH2P
SAA,B,·SABB2 ·SA,B,P
SB,IJ
58, BarSABA? -SABA.B,
) SSBal"SAB2Aa
SB./J
-SABA,B2 /SBB,B2"SAA,B,
SB,B,I
SABA,B,
1 SSB)Ba"SAA)B
SBB.I
SABA,B,
A· ~ SA,B,A. = S A,B,P" B.P SABA '2 ~ -
(~-I).SABB ) HlP 2
A2 SA SAB,A. =SAB.P"1if!
=-
SA,B,A. A. S
(~ B.P - I)S . A,B,B. )
pSA)B,Bz"SAA,B A1 8 2
P
S
-SAAIBBI
P-SABB, ·SAA,B, AB2P
SAA SB I 1
SA, B,B,·SAA,B·(-SABB, ·SAA,B, H-SAA, BB,) SAA,B,·SABB, ·SA,B,B,·SAA,B·SAA,BB,
sim!!4i!y
Example 6.29 (0.250, 1, 18) If (Q) is the cevian triangle of a point M for the triangle (P). show that the triangle formed by the parallels through the vertices of (P) to the corresponding sides of( Q) is perspective to (P).
Constructive description ( (points ABC M) (inter A, (I B C) (I A M)) (inter B, (I A C) (I B M» (inter c, (I A B) (I C M)) (inter c, (P B A, Cd (P A B, Cd) (inter A, (I c, B) (P C A, Bd) (inter B, (I c. A) (I A, C)) (inter p (I B B,) (1 A A.» (inter zillc, C) (I A A.)) (.oU:. =~» A2P
A'JZp
Figure 6-29
277
6.l. C-try allDddence
The eliminants
The machine proof ~ -Scc''''' , ~
-
-S... cc.
.f.
....p
S ... BB, ' SCC, ... , S ...CC,,(-SB ... ,B,)
-
.im2!i!1I -
S"'C"' t ,SBCM,S"'BBt S ...CM ,S ... B ... "SBCB,
~ S"'C"' t ,SBCM ,S"'BM ,S"'BC ,S"'BCM
-
S ... CM,S ... B ... "SBCM,S ... BC ,S ... BCM
.im2!t!1I -
S ... C ... "S ... BM S ... CM ·S ... B ... ,
~ (-S"'CWS ... BC) ,S... BM ,( -S ... BMC)
-
S"'CM'(-S ... BM ,S ... BCH-S... BMcl
.i"'J!!.i!1I -
1
Example 6.30 (0.066, 2, 7) If a hexagon ABC D E F has two opposite sides BC and E F parallel to the diagonal AD and two opposite sides CD and F A parallel to the diagonal BE. while the remaining sides DE and AB also are parallel. then the third diagonal CF is parallel to AB. Constructive description
( (POints
B C E)
(on D (P C B Ell (inter A (P B D E) (P D B Cll (inter F (P A C D) (P E B Cll (midpoint N, A C) (midpoint N, C E) (inter G (I A N,) (I E Nil) (inter H (I D F) (I G Bll (parallel C F A B ) )
D
B
Figure 6-30
278
Chapter 6.
The eliminants
The machine proof
S
~
BAF
-SSAF
L -
!!. -
SCAD
SSCA'(-SSCO) -SSECA,SSCAO
A (SSECO-SSEO) ,SSCD SSECD
SSECA4SsCD-SSCE
(SSCO)2 ,(-SSECO) (Ssco-SscEH-sSECO ,SSCO+SSEO'SSCO)
SSCA4SsCD SSEDfl, -
SaGP .SageD
(SSCE)
SSECDfl,( ~-l).SSCE
(Ssco-SsCEHSSECO-SSEO)
Sscofl,SSCE ' ~
_D SSCE'~'(SSCE'~-SSCE) Be ~
-
F S8lijQA·SaCAD -SBCD
S
simg1iJy
Topics from Geometry
(SSCE ' ~~ -SSCE),SSCE'n
sim~ifY 1
Example 6.31 (0.683, 4, 37) Prove that the lines joining the midpoints of three concurrent cevians to the midpoints of the corresponding sides of the given triangle are concurrent,
Constructive description ( (points ABC 0) (inter D (I B C) (I A 0)) (inter E (I A C) (I B 0)) (inter F (I A B) (I C 0)) (midpoint D, A D) (midpoint E, B E) (midpoint F, C F) (midpoint A, B C) (midpoint B, C A) (midpoint C, A B) (inter 1 (I
Figure 6-31 B, E,)
(I
A, D,)
(collinear 1 c,
F,) )
Example 6.32 (1.583, 8, 28) Let 0 and U be two points in the plane of triangle ABC. Let AO, BO, CO intersect the opposite sides BC, CA, AB in P, Q, R Let PU, QU, RU intersect QR, RP, PQ respectively in X, Y, Z . Show that AX, BY, CZ are concurrent.
Constructive description ( (points ABC 0 U) (inter p (I B C) (I A 0» (inter Q (I C A) (I B 0» (inter R (I A B) (I CO» (inter x (I Q R) (I p U» (inter Y (I R P) (I Q U)) (inter z (I P Q) (I R U» (inter I (I B Y) (I A Xl) (collinear C Z J) )
Figure 6-32
6.l.
279
Geometry fIllnddeDce
Example 6.33 (0.667, 4, 27) Let 0 be a point in the plane of a triangle ABC. and let AI. B I • CI be the points of intersection of the lines AO. BO. CO with the sides of the triangle opposite A. B. C. Prove that if the points A 2 • B 2 • C2 on sides BI C I• CIA I• AIBI of ~AIBICI are collinear. then the points of intersection of the liens AA2 • BB2 • CC2 with the opposite sides of ~ABC are collinear.
Constructive description ( (POints ABC 0) (inter A, (I B C) (I A 0» (inter B, (I A C) (I B 0» (inter c, (I A B) (I C 0»
(on
A.
(I
B, Cd)
(on
B.
(I
A, Cd)
(inter c. (I A. B.) (I A, B,» (inter A. (I B C) (I A A.» (inter B. (I A C) (I B B,» (inter c. (I
Figure 6-33 A B)
(I C
C.»
(collinear
A. B. C.) )
Example 6.34 (0.533, 5, 20) The sides BC. CA. AB of a triangle ABC are met by two transversal PQR. PIQIR I in the pairs ofpoints P. PI; Q. QI; R. R I . Show that the points X = BC n QR I • Y = CA n RPI • Z = AB n PQI are collinear.
Constructive description ( (POints
A B C)
(on P (I B (on Q (I C
C» A» (inter R (I P Q) (on P, (I B C» (on Q, (I C A»
(I
A B»
(inter R, (I P, Qd (I A B» (inter x (I Q R,) (I B C» (inter y (l R Ptl (I C A» (inter z (I
p,
Figure 6-34 P
Qd (l
A B))
(collinear x
Y Z) )
Example 6.35 (0.516, 4, 28) Through the vertices of a triangle ABC lines are drawn intersecting in 0 and meeting the opposite sides in D. E. F . Prove that the lines joining A. B . C to the midpoints of EF. F D. DE are concurrent.
Constructive description ( (POints ABC 0) (inter D (I B C) (I A 0)) (inter E (I A C) (I B 0» (inter F (I A B) (I C 0» (midpoint A, E F) (midpoint B, F D) (midpoint c, D E) (inter I (I B Btl (I A Atl) (collinear
A
Figure 6-35 I C Ctl )
Chapter 6. Topics rrom Geometry
280
Example 6.36 (0.500,4,25) A transversal cuts the sides BC. CA. AB of triangle ABC in D. E. F. P. Q. R are the midpoints of EF. F D. DE. and AP. BQ. CR intersect BC. CA. AB in X. Y. Z. Show that X. Y . Z are collinear. Constructive description «points A B C) (on D (I B C)) (on E (I A C)) (inter F (I E D) (I A B)) (midpoint P E F) (midpoint Q F D) (midpoint R D E) (inter x (I A P) (I B C)) (inter Y (I B Q) (I A C)) (inter z (I C
Z
Figure 6-36 R)
(I
A B))
(collinear x
Y Z) )
Example 6.37 (0.200,2, 19) The lines AL. B L. C L joining the vertices ofa triangle ABC to a point L meet the respectively opposite sides in Ai. Blo Cl . The parallels through Ai to BBl CCl meet AC. AB in p. Q. and the parallels through Ai to AC. AB meet BBl> CCl in R. S. Show that the four points p. Q. R. S are collinear. Constructive description ( (points ABC L) (inter A, (I B C) (I A L)) (inter B, (I A C) (I B L)) (inter c, (I A B) (I C L)) (inter P (I A C) (P A, B B,) (inter Q (I A B) (P A, C c,) (inter R (I B B,) (P A, A C))_ (inter ZR (I Q P) (I B, B)) (~
__ =~»
BR
B
c
AI
Figure 6-37
BZR
Example 6.38 (0.533, 3, 18) Starting from five points A. B . C. D and E with A. B. C collinear. new lines and points of intersection are formed as in Figure 6-38. (1) AB. GJ and HI are collinear; (2) AB. GK and IL are collinear; (3) AB. HL and JK are collinear. (The proofs for (2) and (3) are omitted.) Constructive description for (I) ( (POints A B D E) (on C (I A B)) (inter F (I E D) (I A B)) (inter I (I D B) (I A E)) (inter J (I C D) (I A E)) (inter L (I C D) (I B E)) (inter G (I B E) (I A D)) (inter K (I C E) (I D B)) (inter H (I E C) (I A D» (inter o.J!.g J) (I (inter ZO (I H J) (I F C)) (.t:2 = ~) ) co
cZo
B
A B»
Figure 6-38
281
6.3. Triangles
6.3 6.3.1
Triangles Medians and Centroids
Example 6.39 (0.001, 1, 2) The line joining the midpoints of two sides of a triangle is parallel to the third side and is equal to one-half its length.
c
B
Figure 6-39
We have to prove two results. Constructive description ( (POints A B C) (midpoint M A B) (midpoint N A C) (paraUel M NBC) )
The machine proof
The eliminants
SBCNt;;~(SABC)
§..a.c.M.. SBCN
SBCM~2(SABC)
N S
= f:.!'sMc M. -
Constructive description ( (POints A B C) (midpoint M A B) (midpoint N A C)
(2Hts.uc) .im2!.ifll SABC -
The machine proof 2(~)
!:!.
(~=1/2»
-
M. -
SACM (fH-SABC)
The eliminants 4Ut!:!~ BC--SABf SACM~ - 2(SABC)
(-2)·(-tSABc) .i"'J!ifli SABC -
Definition. The line joining a vertex of a triangle and the midpoint of the opposite side is called a median of the triangle. Example 6.40 (Theorem of Centroid) (0.016, 1, 4) The three medians of a triangle meet in a point, and each median is trisected by this point.
Constructive description ( (points A B C) (midpoint F A C) (midpoint E B C) (inter 0 (I A E) (I B F» (42 Oll
The machine proof
The eliminants
!(-~)
42 Q §.u.E.
Q ..=..§..u.L. -
~
=2»
-sr
D
Figure 6-40
F
-
(2) ,(-;SBCF)
t;;
i
SABC ,SABC
-
!(SBCF)
SBcF~t(SABC)
(2) ,SBFE
.imgpfll A
lIO-SBFlI
SBFE~
SABF~2(SABC)
282
Chapter 6. Topics from c-netry
For other fonns of the centroid theorem, see Examples 5.46 and 5.47 on page 240. A
Example 6.41 (0.083,2,9) With the medians of a triangle a new triangle is constructed. The medians of the second triangle are equal to the three-fourth of the respective sides of the given triangle. Constructive description ( (POints A 8 C) (midpoint M A 8) (midpoint N A C) (midpoint K 8 C) (inter P (P A C M) (PK 8 N» L A K)
(midpoint
(~~ = 3/4) )
p
K
B
Figure 6-41
The machine proof
The eliminants
l(-4). ~) .f.. (-4) ,SAKP
iC=-SABC S p SAQKu SA8KN AKP -SBCNM
-
(3)·( SABC)
E.
(4) ,SACKM-SABKN (3) -SABC-( SBCNM)
-
!i. -
(-4) :(-;SBCM+SACMH;SBCN+SABN) (3)-SABC -SBCNM
~ (SBCM-2SACM)-(iSABC) (3)-SABC-(SB CM-, SACM) .imJ?!i lll SSCM-2SAGM 2SBCM SACM
M iSABC
PLL~ o
SABKN~!(SBCN+2SABN ) SACKM~ - !(SBCM-2SACM) SBCNM~!(2SBCM-SACM ) SABN~HSABC) SBCN~!(SABC) SACM~ - !(SABC) SBCM~HSABC)
,SABC 8im~ifll 1
Example 6.42 (0.050, 2, 8) The area of the triangle having for sides the medians of a given triangle is equal to three-fourth of the given triangle (Figure 6-41). Constructive description ( (POints A 8 C) (midpoint M A 8) (midpoint N A C) (midpoint K B C) (inter P (P A C M) (PK BN» L A K)
(midpoint (4SAKP
= 3SABC) )
The machine proof
The eliminants
~ (3)-SABC
S
E.
SABKN~!(SBCN+2SABN ) SACKM~ - !(SBCM-2SACM) SBCNM~!(2SBCM-SACM) SABN~HsABC) SBCN~!(SABC) SACM~ - !(SABC) SBCM~!(SABC)
-
!i. -
!:!.. -
(4) -S,.cKWSABKN (3)-SABC-(-SBCNM) (-4)-(-tSBCM+SACMHtSBCN+SABN) (3)-SABC-SBCNM (SBCM-2SACM H tSABC) (3) -SABC -( SBCM-t S ACM)
.imJ!i/ll SSGM-2SAGM 2SBCM-SACM
M. -
tSABC ~SABC
8imJ?! i /1l
~SACKM · SARKN
AKP-
-SBCNM
283
6.3. Triaagles
Example 6.43 (0.016, 1,4) Show that the line joining the midpoint of a median to a vertex of the triangle trisects the side opposite the vertex considered.
Constructive description ( (POints A 8 C) (midpoint F A 8) (midpoint N C F) (inter K (I 8 C) (I A N» (il£ KC
= 2) )
The machine proof
The eliminants
~( - ~)
il£!S.~ CK- SACN
!i
SACNg~(SACF ) SABNg~(SABC) SACF~ - ~(SABC)
-
-(-SABN) (2)·( SACN)
N
-(lSABC)
= (2).{fSACF) 1:. B
K
-StRG (2H-;SABC)
.img1i/"
1
Figure 6-43
Example 6.44 (0.033,2,6) Show that a parallel to a side ofa triangle through the centroid divides the area of the triangle into two parts. in the ratio 4:5. Constructive description ( (POints A 8 C) (centroid G A 8 C) (inter p (I A 8) (p G 8 C» (inter Q (I A C) (P G 8 C)) (4SPQCB
p~--:o---~
B
= 5SAQP) )
Figure 6-44
The machine proof
The eliminants
(4) ,SBCQP (5) ,SAPQ
s
9.
(4),(SBCP,SABC-SBCG ,SACP ),SABC (5) ·( SABGC ,SACP ),SABC
-
.im1!!.if" -
(-4HSBcp,SABC-SBCG ,SACP) (5) ,SABGC ,SACP
(-4) ·(SABGC · SBCG ·SA BC+SBCG 'S~ RG) ,SABC (5) ,SABGC,(-SABGC ,SABC) ,SABC
1: -
.im1!!.i/" -
9. -SABGO,SAce
APQ-
5
SABC
Q SSCP, SA8 0-SacQ,SAcP
BCQP
SABC
SACP~-SABGC SBCP~SBCO SBcogl(SABC) SABOCgi(SABC)
(4) ,(SABGC+SABC),SBCG (5) ,(SABGC)'
Q (4)·(5SABC) ,SABd(3))· -
(5)-(2SABC))· ·«3))·
.img1i/"
1
Example 6.45 (0.001, 2, 5) If L is the harmonic conjugate of the centroid G of a triangle ABC with respect to the ends A. D of the median AD. show that LD = AD.
CIuIpter 6.
284
Constructive description ( (points A B C) (midpoint D B C ) (centroid GAB C) (harmonic L G D A) (midpoint D L A) )
The machine proof 1M;,
J=~.~ -1JSi.-1 AO
Q
A
AO
IJSi.
RJ.=~ AO -(~+l)
1JSi.£~ AO- SABC
-SAGo ,SABe
(SACO+SABc) ·SABC
~im!!!.if1l
tc;;:'
The eliminants -L
AO
-
Topics from G.mIeIl'y
-SAGa SACO+SABC
SABOg~(SABC) SACOg -
~(SABC)
Q -(-tSABC) -
C
ISABC
Bim!!l:iflJ 1
Figure 6-45
Example 6.46 (0.033, 1, 6) Show that the distances of a point on a median of triangle from the sides including the median are inversely proportional to these sides.
Constructive description ( (points A B C) (midpoint F A B) (on N (I C F» (foot K N A C) (foot J NB C) (eq-product N K A C N J
The machine proof
The eliminants
PNKN,PAGA
P
PNJN ·PBCB
.L B C) )
PNKN " PAGA · PacB
(16S~CN)·PBCB
sim!!!.ify PNKN oPAGA (16) ·( SBCN)'
Ii -
(16S!CN)·PACA (l6) ·(SBCN)' · PACA
Bim£!.iflJ ~ (SBCN)'
~(16) · (SBCN)'
NJNP
PBCB
~(16) ·(SACN)' NKNPACA
SBCN~SBCF"fc!f SACN~SACF"fc!f
SBCF~~(SABC) SACF~
-
!(SABC)
N (SACF · ~)' = lO,i;. (SBCF · fc!f ).
Bim£!.ify ~ (SBCF)'
!:.. Figure 6-46
-
«-tSABC»' «tSABC»'
Bim!!:.iflJ 1
Example 6.47 (0.066, 5, 7) Show that, if a line through the centroid G of the triangle ABC meets AB in M and AC in N, we have, both in magnitude and in sign, AN . M B + AM · NC=AM · AN.
6.3.
285
Triangles
The eliminaots
Constructive description «POints A B C) (centroid GAB C) (on M (I A B» (inter N (I G M) (l A C» (WI.+~ AM
AN
=
~!!.~ "N-S"GM
SCGM~SSCG ' *-S"CG ' *+S"CG S"OM~ - (S.. so·*)
1»
-M~-1
U4.=~ .. M
~ S
t S.. sog3(S"SC) S .. CGg -l(s.. sc) sscog1(s.. sc)
The machine proof -(~+~)
!:!. -
- ( * .S,IGM+SCGM) S"GM
M_
-(SSCG ' ~' -s"CG ' ~' +s"CG,~-s"SG' ~' +s"SG ' ~) AB AB 68 AB AA
B
D
Figure 6-47
S .. SG·* G (9S .. sc · ~)·(3) simplify 8
=
::......4 SABc·* ·«3»·
=
1
Example 6.48 (0.033, 1, 8) Two equal segments AE. AF are taken on the sides AB. AC o/the triangle ABC. Show that median issued from A divides EF in the ratio o/the sides AC.AB.
Constructive description «POints F E) (on A (b F E» (on C (l A F» (on B (I A E» (midpoint A, B C) (inter D (I E F) (l A A,» (~~P.. s ..
= p .. c .. ) )
The machine proof ~'(~)' p.. c.. FD
!2
(SE .... , )' ,p .. s .. p ..C .. ·(SF .... ,)'
-
~ «!SE .. C»' ,P.. BA - P.. C.. ·«fSFAB»'
-B (SE"C)"PE"E'a~' -
P"C" ,«-SFE .. ·*»'
simJ!!.ify (SE4C)' ,PE .. E
-
P"C" ,(SFE .. )'
C (SFE,, · ~)' · PE"E A~ PF"F"~ ,(SFE.. )'
= Figure 6-48
.im!1ify ~ ~!?uL oimJ!!.ifll PF .. F - PF .. F -
The eliminaots ~£SE"",
FD-SF .... ,
SF .... , ~~(SF .. S) SE .... ,~~(SE .. C)
SF"S~
- (SFE .. ·* )
P"S,,~PE"E'(*)' P"C .. gPF"F"(~)' .. F
SE"CgSFE" ' ~ .. F PE .. E4PF .. F
286
Chapter 6. Topics trom c-netry
Example 6.49 (0.250,4,8) Show that the (algebraic) sum of the distances of the vertices of a triangle from any line in the plane is equal to the sum of the distances of the midpoints of the sides of the triangle from this line.
Constructive description ( (POints ABC X Y) (midpoint D B C) (midpoint E A C) (midpoint F A B) (foot A, A X Y) (foot B, B x Y) (foot c, C x Y) (foot D, D x Y) (foot E, E x Y) (foot F, F x Y)
The machine proof ~+~+l AA, AA,
!!:I. +!!I. +~ ...tAl AAI AA.1 !!I. -SAXy+~-SAXy+SXYF AA.l
AA.I
A.A. J
A.A.)
AAI
(~+~+l) - (S AXY )' AA., AA.,
~
~-S~Xy+SXYF -SAXy+SXYE -SAXY
(l+~+~ = ~+£.~.I+~,!:i.) AA.I
( ~+~+l)-S AA.1 A...t.J..... AXY
~
AAI
) DD
~-SAXy+SXYF+SXYE
The eliminants a.~fua.L AAI -
(~+~+l) - (SAXY l' AAI AA.,
SAXY
~1E!ll=.
SXYF-SAXy+SXYE -SAXy+SXYD -SAXY
AA, - SAXY ~~§.Ja.u. AA, - SAXY ~S&:u. A.A. -
8imJ!!.ifll ( ~+~+l)S AA] AA] - AXY SXYF+SXYE+SXYD
SA,XY
Qj ( * -SAXy+SCXy+SAXy)-SAXY
~lb~
-
AA, - SAXY
SXYF~t(SBXy+SAXY)
(SXYF+SXYE+SXYD)-SAXY
simJ!!.ify ~S AA, - AXY+ SCXy+SAXY
SXYE~~(SCXy+SAXY )
-
SXYF+SXYE+SXYD
~ SCXy-SAXy+SBXy-SAXY+S~ "Y (SXYF+SXYE+SXYD)-SAXY
SXYDg2(SCXy+SBXY)
8imJ?!.ify SCXy+SAry+SHY SXYF+SXYE+SXYD
E..
x
Figure 649
SCXy±SSlCytSAXf
-
SXYE+SXYD+tSBXy+;SAXY
~
(2) -(SCXy+SBXy+SAXY) 2SXYD+SCXy+SBXy+2SAXY
g
(2)-(SCXy+SBXy+SAXY) 2Scxy+2SBXy+2SAXY
simgj/fy 1
Example 6.50 (0.016,3,1) Compute the square of the lengths of the medians_ Constructive description The machine proof ( (points A B C) ~(PAMA) (midpoint M B C) (MK) ) M -tPBCB+iPACA+iPABA 2
The eliminants
PAMA~
-
i(PBCB-2PACA-2PABA)
287
6.3. TriaDcJes
Example 6.51 (0.066, 2, 14) If K , KI are two isotomic points on tire side BC of tire triangle ABC. and tire line AK meets tire line N M of in K 2• wlrere N and M are tire midpoints of AB and AC. Show that line KIK2 passes through tire centroid G of ABC.
Constructive description ( (POints A B C) (midpoint N A B ) (midpoint M A C) (inter G (1 C N) (1 B M» (lratio K B C r) (!ratio K, C B r) (inter K, (1 M N) (! A K» (inter .uLB C) (1 K, G»
The eliminants
The machine proof
U!.SBGK, CZ-SCGK,
(~)/(~) ~
5
-SBGK,
S
~ -(-SCGK1 ) OK.
OK.
SBNQ -SAMK
5
~ -SCMG -SANK
SGNMMl(SACN)
SSNG , S68M
SABM M 2 (SABC)
SCMG -SACN
SBNMM -
Q ( SBNWSBCN) -SABW( SBCMN) B
K
Figure 6-51
K,
SCNM-SBCM -SACN -(-SBCMN)
!1.
~(SBCN) ~(SABC) SBCNg~(SABC)
SACNg -
SBNWSBCWSABM SCNM -SBCM -SACN
-(-tSBCN) -SBCN-(}SABC) (tSACN)-(tSABC)-SACN
-
.im1!!i/y -
N
5BCMN
SBCMMt(SABC)
SBNG -(SABW r SABM) -r (r - I) -SCMG -( -SACWr)
.imgi/y
-SBCMN OSBNU,SACN
BNG
SBNQ,SAMK ' r
C
OSCNU , SSCM
S
(r-I)-SCMG -SANK
.im~iJJI -
r
CMG
CK,
!f:
SANKM
SANK~ - (SACN-r) SAM K~(r-I)-SABM
!!.!!J..-SCMG -SANK -(-SANKM) CK,
.imgli/y
-SANKM
~SBNO · SAMK
BGK, -
~~!=l
~ l::§BNG -SAMK)-(-SANKM)
( u=~ » zc K.C
~SaMQ · SANK
CGK, -
~
(SACN)'
«tSABC»'
= «-lSABc»' .imgli/y
Example 6.52 (0.083, 5, 3) Tire sum of tire squares of tire medians of a triangle is equal to threefourth tire sum of tire squares of tire sides.
Constructive description ( (POints A B C) (midpoint E A C) (midpoint F A B) (midpoint D B C) (4A'if+4F(f+4W = 3AC' +3AB' +3BC') )
Figure 6-52
Cbapter 6. Topics from ~try
288
The eliminants
The machine proof
PADA g -
(4).(PCFC+PBEB+ PADA) (3)·(PBCB+PACA+PABA)
!2. -
~(PBCB-2PACA-2PABA)
PCFC~~(2PBCB+2PACA-PABA)
(4).(PCFC+PBEB-}PBCB+t PACA+t P ABA) (3)·(PBCB+PACA+PABA)
PBEB!iH2PBCB-PACA+2PABA)
K 4PRE8±PBca±4PAcAtPARA -
.K -
(3) ·(PBCB+PACA+PABA) 3P8QR±3PACA±3PAAA (3)·(PBCB+PACA+PABA)
sim!!1i!y
J
Example 6.53 (0.050, 4, 6) If two points are equidistant from the centroid of a triangle, show that the sums of the squares of their distances from the vertices of a triangle are equal.
Constructive description «points A B N M) (on G (b N M)) (midpoint F A B) (lratio C F G 3) (lJif' +NEf +Nc' = AM' +ME' +MC") ) The machine proof. PNONtPRNBtPANA
c
B
Figure 6-53
The eliminants PMCM'i,6PGFG-2PMFM+3PMGM
PMCM+PBMB+PAMA
PNCN'i,6PGFG-2PNFN+3PNGN
Q
6PQfO- 2PN f N±3PNGN+PRNStPAN6
-
6PGFG 2PMFM+3PMGM+PBMB+PAMA
f.
3PNGN+3PRQRt3PAQr P ABA 3PMGM+3PBGB+3PAGA-PABA
PMFM~~(2PBMB+2PAMA-PABA) PNFN~~(2PBNB+2PANA-PABA) PGFG~H 2PBGB+2PAGA-PABA)
Q
3PNGN±3P8GB±3PAQA-P6B6
PMGMgPNGN
-
3PNGN+3PBGB+3PAGA PABA
-
simgp!y
J
Definition. The triangle havingfor its vertices the midpoints of the sides ofagiven triangle is called the medial triangle of the given triangle. The triangle formed by the parallels to the sides of a given triangle through the opposite vertices is called the anticomplementary triangle of the given triangle. Example 6.54 (0.050,2,6) The median AAJ of the triangle ABG meets the side BIGI of the medial triangle AIBI GI in P , and G P meets AB in Q. Show that AB = 3AQ.
289
TrIaqIa
6.3.
Consttuctive description ( (POints A B C) (midpoint A, B C) (midpoint C, A B) (inter P (I A A,) (p c, B C» (inter Q (I A B) (I P C» (~ = AS
The eliminants
The machine proof
~~~
(3) - ~
AB- SACBP P -(SACCI -SABA, +S1scl ACBP SABC S P -SACCi ·SACA, ACP SABC
AS
5l -
E. -
-1/3»
S
(3)-(-SACP) SACSP (3)-( -SACCI -SACA, )-SASC (-SACC, -SABA,-S~BC) - SABC
.im~i/~
~
(3)-(-tSABC) -SACA,
-
-fSABA,-SABC+~BC
.im~i/~
4! A,
SACC, ~ -
~(SABC) SABA,~~(SABC) SACA, ~ - ~(SABC)
(3) -SACC, -SACA, SACC, -SABA, +51BC
(3) -SACA, SABA,-2SASC
(3)-(-tSABC) .im1!l.i/~ -iSABc 1
Figure 6-54
Example 6.55 (0.066, 4, 5) The sum of the squares of the distances of the centroid of a triangle from the vertices is equal to one-third the sum of the squares of the sides.
Consttuctive description «POints A B C) (midpoint E A C) (lratio G B E 2/3) (3AG' +3X +3BG' =
AC' +AF +BC')
B
)
D
Figure 6-55
The machine proof
The eliminants
(3)-(PCGC+PBGB+PAGA) PSCB+PACA+PABA Q (3) -(iPCEC+i PBCB+iPAEA+iPABA) PBCB+PACA+PABA
PAGAg -
!:
t(2PBEB-6PAEA-3PABA)
PBGBgt(PBEB) PCGcgf( 6PCEC-2PBEB+3PBCB)
PAEA~4(PACA).PCEC~~(PACA)
PBCB+PACA+PARA .im~i/~ 1 PBCB+PACA+PABA
Example 6.56 (0.050,7,8) If M is any point in the flane of the triangle ABC, and G is _ --2~ ~ . ==2==2 ~ the centroid of ABC, we have MA + MB + MC = GA + GB + GC + 3MG .
Consttuctive description «points ABC M) (midpoint E A C) (lratio G B E 2/3) (3W +AG' +X +BG' =
w
+Mlf +"f.i??) ) p
Figure 6-56
Chapter 6.
290
The eliminants
The machine proof
PAGA g -
3PMGMtPCGC+PBGBtPAQA
PCMC+PBMB+PAMA Q 2PMEM+i PCEC-i PBEB+PBMB+i PBCB+!PAEA+i PABA PCMC+PBMB+PAMA
§. 3PCMC+3PRMR+3PAMA -
Topics from Geometry
~(2PBEB-6PAEA-3PABA)
PBGBgt(PBEB) PCGC g g( 6PCEC-2PBEB+3PBCB)
PMGMg~(6PMEM-2PBEB+3PBMB)
PAEA~I(pAcA)
(3),(PCMC+PBMB+PAMA)
PBEB~t(2PBCB-PACA+2PABA) PCEC~4(PACA) PMEM~~(2PCMC+2PAMA-PACA)
simg1i!y
Example 6.57 (0.067,7,10) lftM pairs of points D. D 1 ; E. E 1 ; F. Fl are isotomic on tM sides BC. CA. AB oftM triangle ABC. tM areas of two triangles DEF. DIEIFI are tM same.
Constructive description «points A B C) (lratio D B C rIl (lratio E CAr,) (iratio FA B r,) (lratio D, C B r,) (lratio E, ACT,) (!ratio F, BAr,) (SD,E,F, = SDEF) )
The eliminants ~ - (SSDIE1 ' f'3-SSDIEl-SADlEl ' T3) ~ - (SACD] ora) SBD,E, ~ - (,,-I) ,SABD,) SABD, ~ - (h- 1),SABC)
SD 1 E 1 F 1 SADIE1
S ACD1'2 -
(SABC 'Tl)
SDEF~SBDE ·r,-SADE·T3+SADE SADE~(,,-I).SACD SBDE~SABD '" SABDgSABc ·r, SAcDg(r,-I) ,SABC
The machine proof SDJEJF, SDEF
F
~ -SSDJEJ 'r3±SBD, E, tSADtE, ·r3
SDEF ~ -(SACD, ' TJ'T2-SABD, ' T3 ' f'1±SABD, 'T3±SABD, '1'2-8.4.8D,)
Figure 6-57
SDEF
Qj -( -SABC·r,·" -s ABC·r, 'r, +SABC 'r, -SABO·,, ·r, +SABO'r, +SABC'r, SABC) SDEF
8im1!!.i!y (r"r,+r"r,-r,+r"r,-r,-" +1),SABO -
SDEF
L
(TJ ·ra tr3 ' r] -r3 tra ' r] -T2 - T ] +1 )·s "" Be
-
SBDE ·r, -SADE·r,+SADE
~
(r3 ' T2±r3'TI-T3±r2'Tt-r2-rl tl) ,SABe
-S"CD·r,·r2+S"OD ·r,+S"CD ·r.-S"OD+SABD·r,·r2
!2
-(r"r2+r"r,-r,+r"r,-,,-r, +1),S"BC -SABC ·r3 ·r2 -SABC ' T3 ·r1 tSABC'T3 -SABC ' T2'Tl tS.ABC ' f"2±SABC·Tl-SABC
Bim1!!.i!lI
6.3. Triangles
291
Example 6.58 (0.200, 3, 11) Show that the parallels through the vertices A. B. C of the triangle ABC to the medians of this triangle issuedfrom the vertices B. C. A. respectively. form a triangle whose area is three times the area of the given triangle. Constructive description «points A B C) (midpoint A, C B) (midpoint B, A C) (midpoint C, B A) (inter c, (P B C Cd (P A B (inter B, (P A B Bd (P C A (inter A, (P C A Ad (P B C (SA,B,C,
Bd) Ad)
C,»
= 3SABC) )
The machine proof - 5C 2 8
2-"2
(3),SA BC ~
Figure 6-58
-SSCBaCt,SACA,Ca
-
(3) ,SABC'( -SACA, C,)
~
(-SACA, C, ,SA BC+SABA, B, ,SBCC, ),SACA, C, (3 ),SA BC ,SACA , C, ,SA BA,B,
~
-
-(SACA, C, ,SABC-SABA, B, ,SBCC, ),(SBCB, C, ,SACA, -SABA, B, ,SA BC) (3) ,SABC ,SACA, C, ,SABA, B, ,SBCB, C,
~I -(-!SABA, B, ·$ABC+SACA, ·$ABC-iSABAt,$ABC) ,(-SABA,B, ·$ABC+SSCB J ,SACA , +!SACA, ,SA BS,)
-
(') ,SABC ,(SACA,-!SABA, ),SABA,B,,(SBCB, +fSABB,)
.im.1!!.i/lJ -(SABA, B, -2SACA, +SABA, H2SABA, B, ,SABC-2SBCB, ,SACA, -SACA, ,SABB,) (3)·(2SACA, SABA, ),SABA,B, ·(2SBCB, +SABB,)
!!l -( -tSACA, +2SABA, H-!SACA, ,SABC+2SABA, ,SA BC) -
(3) ·(2SACA, -SABA, H -JSACA, +SABA, ).(JSABC)
s, m1!li/ll
4!
(SSACA, -4SABA, )' (9)·(2SACA, -SABA, HSACA, -2SABA,)
«-ISABC»' (9).«-JSABC»2
.im~i/J/ 1
Dermition. If the corresponding sides of two similar polygons are parallel. the two polygons are said to be homothetic. The lines joining the corresponding vertices of two homothetic polygons are concurrent at the homothetic center.
Example 6.59 (0.500, 5, 29) Let L. L1 and M. M1 be two pairs of isotomic points on the two sides AC. AB of the triangle ABC. and L 2• M2 the traces of the lines BL. CM on
292
Chapter 6.
Topics from Geometry
the sides Al Cl , AIBI of the medial triangle AIBI Cl of ABC. Show tlult the triangles ALIMl , AIL2M2 are homothetic. Constructive description ( (points A B C) (midpoint B, A C) (inter A, (I B C) (P B, A B)) (midpoint c, A B) (on M (I A B)) (on L (I A C» (lratio L, B, L -1) (lratio M, c, M -1) (inter L. (l A, Cd (I B L)) (inter M. (I A, B,) (I C M» (inter 0 (I L, L.) (I A Ad) (collinear
A
Figure 6-59 0 M, M.) )
Example 6.60 (0.283, 6, 16) A parallel to the median AAI of the triangle ABC meets BC, CA, AB in the points H, N, D. Prove tlult the symmetries of H with respect to the midpoints of NC, BD are symmetrical with respect to the vertex A. Constructive description ( (points A B C) (on H (I B C» (midpoint A, B C) (inter N (I C A) (P H A A,)) (inter D (I H N) (I A B)) (midpoint K N C) (midpoint L B D) (lratio H, K H -1) (lratio H. L H -1) (midpoint
A H, H.) )
Figure 6-60
Example 6.61 (0.366, 4, 16) The parallels to the sides of a triangle ABC through the same point, M, meet the respective medians in the points P, Q, R. Prove tlult we have, both in magnitude and in sign, (GPIGA) + (GQIGB) + (GRIGC) = O. Constructive description ( (points ABC M) (midpoint D B C) (midpoint E C A) (midpoint F A B) (inter p (I A D) (P M B C)) (inter Q (l B E) (P MAC)) (inter R (I C F) (P M A B» (inter G (I A D) (I C F» (collinear G B E) (~+~ = ~) )
Figure 6-61
293
Triangles
6.3.
Altitudes and Orthocenters
6.3.2
For machine proofs of the orthocenter theorem, see Example 3.36 on page 118 and Example 5.53 on page 244.
Example 6.62 (0.033, I, 6) The dual of the onhocenter theorem.
Constructive description ( (POints A 8 C 0) (inter D (I 8 C) (t 0 0 A» (inter E (1 C A) (t 0 0 8» (inter F (I A 8 ) (t 0 0 C» (inter ~I E D) (\ A 8 » .!£t;. ) )
( .u; = SF
BZF
TIle eliminants
The machine proof
.!£t;.Z.f~
(*)/(m)
BZF -
SBOE
~!. !?ADJ:;..
~::..§.s..D.E.. . ~
E..
PA,QC , SBQE
BF- PBOC 5 E PAQA ,SAC a AOE -PCSAO
-
SAOE , PBOC
S
£
P A OC , PBOC,S ASO ,( -PCB60) PAOB ,SACO , PSOC ,( -PC BAO)
S
-SAOE
-
6imJ!l.iJy -
SF
! .PSQQ , SA80
BOE-
PCABO
ACO
S
PAQG ,SAAO PAOB ,SACO
D PAns ,SABo
ABO
Q
PAOC , PAQB , SABC · PC68Q
-
PAOS ,PAO C ,SABC , PCABO
-PCBAO
D PAQG,SABG
PCABO
.im~i/y 1
Example 6.63 (0.050, I, 8) In a given triangle the three products of the segments into which the orthocenter divides the altitudes are equal. Constructive description ( (POints A 8 C) (foot D CA 8) (foot E 8 A C) (inter H (I C D) (I 8 E» (eq-product C H H D 8 H
H E) )
A
D Figure~63
The machine proof =..fr.JuJ. -PBH E
!:!.
(- PCDC ,SBDE ,SBCE) ,S¥GEo (- PBEB ,SCOE,SBCD) ,SBCED
-
sim~ iJy -
£ -
PBEB ,SCDE ,SSCD PCDC'(
PBAC ,SBCD),PACB ,SABC,(PACA)'
(l6S~BC) · PACB · SA C D · SBCD ·( PACA )'
.i"'!1ifv
PGOG , PRAQ (16),SABC ·SACD
(16S~RG)·PBAC · PABA
Q -
Pe p e,S ROE ,SRGt:;
( 16) ,SABC ,( -PBAC ,SABC) ,P A BA
.im~i/v
The eliminants
B
294
Chapter 6. Topics from a-1I'y
Example 6.64 (0.066, 2, 8) The product of the segments into which a side of a triangle is divided by the foot of the altitude is equal to this altitude multiplied by the distance of the side from the orthocenter.
Constructive description ( (POints A B C ) (foot F C A B) (foot E B A C) (inter H (I C F) (I B E)) (eq-product A F F B C F H
A
F
The elirninants
F) )
B
Figure 6-64
The machine proof ~ PCFH
!!.
(-PAFB) -SBCEF PC FE -SBCF
-
12
-PA FB"(-PA C B-SACF+PAC6 -SBCrJ- P ACA PCFC -PBAC -SBCP-PACA
-
si"'J!!ify PAFB -(P,WB -SACF-PACA -SBCrl -
t;
PCFC-PBAC -SBCF (-PB ,W-PABC)-(-PB ,W -PACB -PABA-SABC-PACA -PABC -PABA -SABC) -(PABA)'
(16S~BC) - PBAC - PABC-SABC-(PABA)'
sim1!!i/y -
PBAC ' PAcatPAcAoPABC (16) -(SABCl'
he!:!:on (PBAC -PACB+PAC6 -P6BC) -(I6) 16PBAC-PA C B+16PACA-PA BC
(This step uses Herron-Qin's fonnula)
s imgi!y
Example 6.65 (0.050, 3, 4) If p. q. r are the distances of a point inside a triangle ABC from the sides of the triangle. show that (p/ha) + (q/hb) + (r/hc) = 1. Constructive description ( (POints ABC 0) (foot DAB C) (foot E B A C) (foot F C A B) (foot D, 0 B C) (foot E, 0 A C) (foot F, 0 A B) (~+~+~=1» AD BE CF A
Figure 6-65
295
Triangles
6.3.
The machine proof
The eliminants
~+~+~ CF BE liD
~~~
OD [! ~ sF; ,SABC+W ,SIIBC+SIIBO
~~ha:J....
-
CF - SABC BE -
SABC
fl -~ .S~BC+SACO ·SIIBC-SIIBO . SABC -
-SABC
~12~ AD -
SIIBC ,(-SABC)
SABC
SBCO=SACO-SABO+SABC
. . ~S .,m1!!'fu AD ' ABC-SACO+SABO -
e.
SABC SBCQ , SABC-SACO,SABctSARQ , SARC
-
(SABC)'
.'m1!! i IJl
SROo
-
SACO±SABQ
SABC
area-co &.as;;. .impJify
=
SABC
=
1
Example 6.66 (0.050, 2, 3) Show that the sum of distances of a point on the base of an isosceles triangle to its two sides is equal to the altitudes on the sides. ' ants·• Theelimm
n".~h&..D.... n .. !:.~ S S ~- -SBAC' ~- SBAC' ACD- BCD-SBAC.
Constructive description «POints B A) (on C (b A B» (on D
(I
A B»
(foot G B A C) (foot FDA C) (foot H A B C) (foot K D B C) (~+RL AH
BG
=
1»
The machine proof
*+~ Ii. -~.SBAC+SBCD -
-SBAC
t:..
S,cn ,SS.:fa-S8QP ·$BAC (SBAC)'
-
.imJ!!ifll SAOD Seon -
SBAC
c
Q §.Ju.a. .imJ!!ify -
SBAC
-
1
Figure 6-66
Defmition. The triangle having for its vertices the feet of the altitudes of a given triangle
is called the orthic triangle of the given triangle. Example 6.67 (0.016, 1, 2) The three triangles cut offfrom a given triangle by the sides of its orthic triangles are similar to the given triangle.
c
B
Figure 6-fJ7
Chapter 6. Topics from <>-try
296 Constructive description ( (points A B C) (foot FC A B) (foot E B A C) (eq-product A F A B A E A C)
The eliminants
The machine proof E.B..4E.. §. E.B..4E.. PCAE
-
L l:.a.M;;. )
-
PBAC
PCAE~PBAC PBAF~PBAC
PBAC 8imJ!!. i fll -
Example 6.68 (0.083, 1, 12) The sides of the orthic triangle meet the sides of the given triangle in three collinear points.
Constructive description ( (points A B C) (foot DAB C) (foot E B A C) (foot F C A B) (inter A, (I B C) (I E F» (inter B, (I A C) (I D F» (inter c, (I A B) (I D E» (inter z~ B, Ad (l D E» (~ = Eel
DZC I
The machine proof (~)/(DZCI Eel
Z,g'
-SEAl BI • ~ -SDA,B, EC,
~ SABD ·SEAI BI SDA, B, ·SABE ~ SABD ·SDEF·SACAI ·(-SADCF) (-SDFA, ·SACD) ·SABE ·SADCF
»
EZCl
.imJ!!.ify -
~ B
0
C
Figure 6·68
)
EZC 1
AI
4!
SABD ·SDEF·SACAI SDFA, ·SACD ·SABE
SARpoSnEf,SCBr,SARc ,S8fjGf (-SDEF ·SBCF) ·SACD·SABE·SBECF
simgj,ify
SARP ,SOEE,SARe -SBCF ·SACD · SABE
f..
SABO , PBAC ,SBCB ,SABC ,PABA -PABC·SABC·SACD ·SABE · PABA
-
simJ!!.ify -
SARD ,PRAO ,SaGe -PABC·SACD·SABE
~ SABO,PSAC , PAQR , SABC , PACA -PABC ·SACD · PBAC·SABC ·PACA
simJ!!ify
SARD,PAGS
-PABC·SACD
g
PARC ,SARC ' PACR"PaQR
8im~ifll
-PA B C· (-PACB·SABC) · PBCB
Example 6.69 (0.067, 4, 11) The altitudes of a triangle bisect the internal angles of its orthie triangle. Constructive description «points A B C) (foot F C A B) (foot E B A C) (foot DAB C) (eqangle E D C B D F) )
A
B
D
Figure 6-69
c
o.
297
TriaDcJes
The eliminants
The machine proof. SeEP ·PaDe (-SSPD),PCDE
!2.
PACB ,SBCE'!Pscp , PASC-PACS , PASC) ,(PSCB)' PABC ,SBCP ,(PCBE ' PACS PACS , PABcl' (Pscs)'
-
.im!4ifll
!i.
SBCE,(PBCP-PACB) SSCF'(PCSE-PABC)
P ,<\OB ,S ABc-( PSCP - P ACB)' P ACA SSCP ,(PBCB,PSAC+PACB , PABC PACA,PASC) , PACA
-
,im!fjYII
E.
PACB ,SASC ,(PSCP-PACB) SSCP ,(PBCB,PBAC+PACB , PABC-PACA , PABC)
PACS ,SABC ,(PBCS , PBAC+PACS , PASC-PACS , PASA),PASA PASC ,SABC,(PBCB , PBAC+PACS , PABC-PACA , PABC), PABA
-
8imJ!!. i fll
,imJ!!.ifll -
PACB,(PBCB,PSAC+PACS , PABC-PACS,PASA) PABC,(PBCB , PBAC+PACS , PABC-PACA , PASC)
1
Dermition, Let H be the orthocenter of triangle ABC. Then the four points A, B I C, H are such that each is the orthocenter of the triangle formed Uy the remaining three. Four such points will be revered to as an orthocentric group of points, or an orthocentric quadrilateral. Example 6.70 (0.067 4 10) Let H be the orthocenter of triangle ABC. Then the circumcenters of the four triangles ABC, ABH. AC H, and H BC form a triangle congruent to ABC; the sides of the two triangles are parallel.
Constructive description «POints A B C) (orthocenter H A B C) (circumcenter o. B C H) (circumcenter O. A C H) (circumcenter Oc A B H) (circumcenter 0 A B C) (~=1» cs
The eliminants ~~PBAO.-iPABA BC
P
-
PASC
'2 PBAH,PAHG , PAGAtPBAC , PAHA , PAGH
BAO. -
S
(32),(SACH)'
!!.PAAC,PACB
ACH- (-16),SABC
PACH~PACS
!!. PBCB,(PBAC)' AHA- (16) ,(SABC)' p !!.. PBAC,PAG8"PARG AHC- (-16) ,(SAscl' P
PBAH!!.PSAC PASC=t( PSCB -PACA +PABA) P A CB=2(PSCB+PACA-PASA) PBAC= -
~(PSCS-PACA-PASA)
Figure 6-70
298
Chapter 6.
Topics from c-netry
The machine proof _~ ~ -(PBAO.-}PABA) BC PABC ~ -(PBAH ' PAHC · PACA+PBAC·PAHA · PAC H-16PABA , S~aH) PABC · (32S~CH)
H = (- (-4096PBCB ' PAAC·PACB·S~BC+4096PAAC·P~CB·PABA · S~BC+4096PAAC·PACB·PACA · PABC · S~BC> •
«16S~BC> simplify
)2) / «32) . PABC • ( (-PBAC ,PACB ,SABC) )2. (
(16S~BC>
)3.
(-16S~BC»
(PBCB ,PBAC PACB'PABA-PACA ,PABC) (2) ,PABC,PACB
f1i -(-4P~GR+4P~GA-8PACA · PABA+4P~RA)·«2»' 8imJ!!ify 1 -
«2»4 ,(PBCB-PA CA+PA BA),(PBCB+PACA-PABA)
-
Example 6.71 (0.067,1,6) Continuing from Example 6.70. show that the point H is the circumcenter of the triangle OaObOc' Constructive description The machine proof «points A B C) PHOgH (orthocenter H A B C) PHO.H (circumcenter O. B C H) PHo.H'(64S~GH) (circumcenter o. A C H) ~ - PCHC,PAHA ,PACA (circumcenter 0, A B H) <2. (64) ,PCHC ,PBHB ,PBCB,(SACH)' (circumcenter 0 A B C) PCHC ,PAHA ,PACA ·(64S1(CH) (eqdistance H O. H 0.) ) 8imJ!!.ify H P ACA
The eliminants
PBHB,PBCB ·(S,ICH)' PAHA ,PACA,(SBCH)'
.p! ! c ,PSCB,«-PBAC ' P ACB"SABC»2'«16S~ Be»3
PBCB · PBAC · PACA · (PACB · PABC · SABC)' · «,·s~BC»3 sim!J.ify
Example 6.72 (0.350,10,17) Show that the three perpendiculars to the sides ofa triangle at the points isotomic to the feet of the respective altitudes are concurrent. c Constructive description ( (points A B C) (foot DAB C) (foot E B A C) (foot F C A B) (pratio D, BCD -I) (pratio E, ACE -I) (pratio F, A B F -I) (inter I (t E, A C) (t D, B C» (perpendicular I F, A B) )
F
B
Figure 6-72
Example 6.73 (0.100,4, 15) Show that the symmetries of the foot of the altitude to the base of a triangle with respect to the other two sides lie on the side of the orthic triangle relative to the base.
299
6.3. TriaDgies
Constructive description ( (POints A B C) (foot DAB C) (foot E B A C) (foot FC A B) (foot P DBA) (inter J (I E F) (I D (midpoint P D J) )
P» D
Figure 6-73
Example 6.74 (0.183, 3, 11) Shaw that the product of the segments into which a side of a triangle is divided by the corresponding vertex of the orthic triangle is equal to the product of the sides of the orthic triangle passing through the vertex considered. A
Constructive description ( (points A B C) (foot F C A B) (foot E B A C) (foot DAB C)
(eq-product
B D D C E D F D) )
D
Figure 6-74
Example 6.75 (0.100, 3, 14) If p. Q are the feet of the perpendiculars from the vertices B. C of the triangle ABC upon the sides DF. DE. respectively. of the orthic triangle DEF. shaw that EQ = F P . Constructive description ( (points A B C) (foot FC A B) (foot E B A C) (foot DAB C) (foot Q C D E) (foot P (eqdistance E Q F P) )
B D F) D
c
Figure 6-75
Example 6.76 (0.167, 6, 22) The four projections of the foot of the altitude on a side of a triangle upon the other two sides and the other two altitudes are collinear. Constructive description ( (POints
A B C)
(orthocenter H A B (foot F C A B) (foot P F A C) (foot T F B C) (foot Q F A H) (collinear P Q T) )
C)
A
F
Figure 6-76
B
Chapter 6. Topics from Geometry
300
Example 6.77 (0.083, 4, 9) D P, DQ are the perpendiculars from the foot D of the altitude AD of the triangle ABC upon the sides AC, AB. Prove that the points B, C, P, Q are cyclic. A Constructive description ( (POints A B C) (foot DAB C) (foot Q D A B) (foot P D A C) (cocircle B C P Q) )
Figure 6-77
The eliminants
The machine proof
P -(PCAD-PBAC) ,PACD PACA P P'CP,SARG BCP PACA S 0 PARO ,SARG BCQ PABA o -(PBAD-PBAC),PABD PBOC PABA
(-SBCP ),PBqC (-SBCO) , PBPC
1:.
PBPC
S
(-PACD,SABC) , PBqC ,PACA (-SBCO)-{ -PCAD-PACD+PBAC-PACD)-PACA
-
-SABC , PBQC
P. -
PABD -SABC-(PCAD
simJ!!.i-Jy -
DP8AC ' P'Qa±PACA , PASC
CAD
~ -SABC-( -PBAD -PABD+PBAC -PABD)-PABA
P
PBAC)-PABA
PRAD-PRAG
PBCB DP8AG ,PAsg+PAGa'PABA BAD PBCB
PACB=I(PBCB+PACA-PABA)
PCAD-PBAC PABC=2(PBCB-PACA+PABA)
Q
(-PBCB,PBAC+PBAC ' P ABC+PACB -PAB6),PBCB ( PBCB , PBAC+PBAC-PACB+PACA -PABC)-PBCB
-
simJ!!.iJy -
PRGB · Pa
PBAC= -
~(PBCB-PACA-PABA)
AC-PRdG , PA Be-P'cR,PABA
PBCB -PBAC-PBAC -PACB
PACA -PABC
~ (-2PiCR+2P~GA-4PACA - PABA+2P~RA)-«2»3 .im1!l i fJl 1 -
(-2PBCB+2PACA-4PACA -PABA+2PABA)-«2j)3
-
Example 6.78 (0.450, 6, 29) The perpendiculars D P, DQ dropped from the foot D of the altitude AD of the triangle ABC upon the sides AB, AC meet the perpendiculars BP, CQ erected to BC at B, C in the points P, Q respectively_ Prove that the line PQ passes through the orthocenter H of ABC. A
Constructive description ( (points A B C) (foot DAB C) (inter H (I A D) (t B A C» (inter P (t D A B) (t B B C» (inter Q (t D A C) (t C B C» (collinear H Q P) )
D
Figure 6-78
c
Example 6.79 (2.067, 68, 61) The algebraic sum of the distances of the points of an orthocentric group from any line passing through the nine-point center of the group is equal to zero.
6.3. Triangles
301 A
Constructive description ( (POints A B C) (orthocenter H A B C) (midpoint MI B C) (midpoint M, C A) (midpoint M. A B) (midpoint NI MI M,) (midpoint N, MI M.) (inter N (t NI NI MI) (t N, N, M I » (on x (I B C» (foot HI H x N) (inter AI (I x N) (P A H Hd) (inter BI (I x N) (P B H HJ» (inter C I (I x N) (P C
x Figure 6-79 H HJ»
(~+~ HHI
HH)
= ~-l) HH J
)
Example 6.80 (0.350, 6, 10) If through the midpoints of the sides of a triangle having its vertices on the altitudes of a given triangle, perpendiculars are dropped to the respective sides of the given triangle, show that the three perpendiculars are concurrent. A
Constructive description «POints A B C) (orthocenter H A B (on R (I A H» (on P (l B H» (on Q (I C H» (midpoint RI P Q) (midpoint Q. R P) (midpoint PI Q R) (inter I (P RI A H) (P Q. C H» (perpendicular I PI A C) )
C)
D
Figure 6-80
Example 6.81 (2.833,7,43) Show that the perpendiculars dropped from the orthocenter of a triangle upon the lines joining the vertices to a given points meet the respectively opposite sides of the triangle in three collinear points. Constructive description ( (POints ABC 0) (orthocenter H A B C) (inter AI (l B C) (t H 0 A» (inter BI (I A C) (t HOB» (inter CI (I A B) (t HOC» (inter cillA B) (I AI Bd) (~=~» BC I
BC,
Dermition. Two lines passing through the vertex of a given angle and marking equal angles with the bisector of the given angle are said to be isogoal or isogonal conjugates.
Example 6.82 (0.133, 5, 22) Show that the line joining a given point to the vertex of a given angle has for its isogonal line the mediator of the segment determined by the symmetries of the given point with respect to the sides of the angle.
302
Chapter 6. Topics from Geometry
c Constructive description ( (points A 8 C M) (foot P M A C) (foot Q M A 8) (!ratio P, PM-I) (!ratio Q, Q M -I) (midpoint N P, Q,) (perpendicular A N P, Q,)
B
)
Figure 6-82
Example 6.83 (1.550, 5, 25) If circles are constructed on two cevians as diameters, their radical axis passes through the orthocenter H of the triangle.
Constructive description ( (points A 8 C P) (orthocenter H A 8 C) (inter D (I 8 C) (I A P)) (inter E (I A C) (I 8 P)) (midpoint M A D) (midpoint N 8 E) (on-radical H MAN 8) ) Figure 6-83
Example 6.84 (0.183,3,12) In triangle ABC, let F the midpoint of the side BC, D and E the feet of the altitudes on AB and AC, respectively. FC is perpendicular to DE at G. Show that G is the midpoint of D E.
Constructive description «POints A 8 C) (foot DCA 8) (foot E 8 A C) (midpoint F 8 C) (midpoint G D E) (perpendicular G F D E) )
The eliminants PEDCgHpDED)
p \
PEDF~~(PCDE+PBDE)
E PAOB ,PACS Po
BDE P
PACA
E POnc ,PRIG
CDE
PACA
E PcnC·PSAQ-PSA.Q ,PAcstPAQA,PACR
, DED
PACA
D-PSAC· PARC P !!~ ADB PABA ADA- PABA Po D(l6HSABC)' CDC PABA P
16S1BC=PBAC·PACB+PACA · PABC PABC=t(PBCB-PACA+PABA) PACB=2( PBCB+PACA -PABA) PBAC= -
The machine proof EE..D.t:L Q tPDED .f.. PEDF
12 -
-
PEDF
-
~(PBCB-PACA-PABA)
PaRn (2)·(tPcDE+1PBDE)
(PCDC·PBAC-PBdC·PACB+P,WA · PACBHPACA)· (PCDC·PBAC·PACA+PADB·PACB·PACA)·PACA
B
F
Figure 6-84
303
6.3. Triangles ~im~.JJI
-
g
PCDC , PAAC-PBAC , PACS±PA Od , PACR
PCDC ,PSAC+PADS ,PACS (P~AC ' PACS ' PASA-PSAC ' PACS ' P!BA+16PSAC ' PASA ' S'!Bal ' (PASA)'
(-PSAC ,PACS ,PASC ' PASA +16PSAC,PASA , s~scHPASA)'
.im~ifll
-
PSAC ,PACS-PACS ,PA SA+l6S'!RC -(PAcs ,PA sc-16S1sc)
"e!!:oo (32PSA C, PACS-16PACS ,PASA+16PACA ,PASC) ,(l6) (16Ps Ac ,PACS 16PAcs ,PA sc+16PAcA ,PASC)·(16) e!! «2))4 ,(-2PAcB+ 2PSCS ,PACA+2PsCS ,PASA) .i"'l!f.ifll - (-4pAcs+ 4PsCS ,PACA+4PSCS ,PAS A ),«2»' I
Example 6.85 (0.083, 3, 10) Let BI and C I be the midpoints of AC and AB, D the foot from A to BC. Show that the triangle DBICI is congruent to the Euler triangle of triangle ABC. A
Constructive description ( (POints A B C) (foot DAB C) (foot FH B A C) (inter H (I A D) (I B FH)) (midpoint P A H) (midpoint Q B H) (midpoint C, B A) (eqdistance c, D P Q) )
D
c
Figure 6-85
The machine proof
The eliminants
PDC,D PpQP
PDC,D C'I( = 4 2PSDS+2PADA-PASA )
~ }PSDS+}PADA-tPABA
PpQP
t
~ 2PBOB+2PAOA- P BA - (4) ,(tPHPH+tpsPS-: PSHS)
f.. -
2PSQB±2PADA PABA PASA
12
-PACB ' PASA+2PSCs ' P!AC+32Pscs ' S~8C
.im~ifll
PASA '(PSCS)' -(PSCS · PASA-2P!AC-32S~AC)
PASA' PSCS "e!!:oo -16PSCS,PASA+32PSAC,PACS+32PACA ,PASC+32P!AC PSCS,PASA ·(16) ell -(-16PsCS ,P A SA) BimJ!!.ifll PSCS ,PAS A ,«2))4 -
6.3.3
PPQP~ ~(2PH PH+2PsPs-PSH s) Psps~~(2PSHS-PAHA+2PASA)
PHPH~HpAHA) P ADA
D(l6) ,(SASC)' P Q~ PSCS • SDS- PSCS
16S~sc=PSAC ' PACS+PACA 'PASC
PASC= ~(PSCS-PACA +PASA) PACS=~(PSCS+PACA-PASA) PsAc= - ~(PSCS-PACA-PASA)
The Circumcircle
Dermition The circle passing through the three vertices of a triangle is called the circumcircle of the triangle. The center of the circumcircle is called the circumcenter of
Chapter 6.
304
Topics rrom Geometry
the given triangle.
Example 6.86 (0.033, 6, 8) The angle between the circumdiameter and the altitude issued from the same vertex of a triangle is bisected by the bisector of the angle of the triangle at the vertex considered. A
Constructive description ( (points A B C) (circumcenter 0 A B C) (foot DAB C) (eqangle BAD 0 A C) ) The machine proof
P DPBAC ,PARC±PACB , PABA SAD PSCS
(-SASD),PCAO SACO,PSAD
!2..
S
(-PASC ,SASC),PCAO'PSCS SA CO ,(PSAC ,PASC+PACS,PASA)'PSCS
-
simEJ.ify
Q
sim!!1ify he~on
-
D PABo ,SAsa
ASD
5
PSCS 0 PAQd"PABG
ACO
(-32).SASC
PCAOg!(PACA)
-PASC ,SASC,PACA·(-32SASC) PACA ,PASC ,( PSAC,PASC+PACS ,PASA)·(2)
-
Figure 6-86
The eliminants
(l6),(SASC)' PSAC ,PASC+PACS,PASA
16S~sc=PSAC·PACS+PACA · PASC
PASc=l(PSCS-PACA+PASA) PACS=2(PSCS+PACA-PASA) PSAC= -
!(PSCS-PACA-PASA)
16PBAC ,PAc8±16PAGA oPARG
(PSAC ,PASC+PACS,PASA)·(16)
qca
~ (-2P +4PSCS,PA CA +4Pscs'PASA -2P~c A +4PACA ,PASA -2P; a .)·«2))' ( 2Pscs +4PSCS,PACA+4PsCS,PASA 2PACA+4PACA ,PASA-2PASA)·«2»3
Example 6.87 (0.001, 1, 2) The product of two sides of a triangle is equal to the altitude to the third side multiplied by the circumdiameter. Constructive description ( (points A B C) (circumcenter 0 A B C) (foot FC A B)
(AC"BC' = 40A'.CF')
)
The machine proof (t) ,PSCS,PACA PCFC ,PAOA
E. -
(}),PSCS ,PACA,PASA
(16S~sc)·PAOA
Figure 6-87
6.3.
305
Triangles
Q -
PBC B ·P,WA · p,
.i m~ifY
Example 6.88 (0.050, 5, 14) Prove that the circumcenter of a triangle is the orthocenter of its medial triangle.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (midpoint A , B C) (midpoint B, A C) (midpoint C, A B ) (perpendicular 0 A, B, C , )
)
The machine proof
Figure 6-88
The eliminants
POB, C, PA,B,C,
~
iPBB,O+}PAB,O tPBB,A, +!PAB,A,
~
iPBCO+}PBAO+tPACO-tPACA tPBCA, +tPBAA, +tPACA, -tPACA
~
Q -
PRfQtPRA.f+PACQ-PACA
t PBCB+;PBAC+;PAC B -PACA +t PABA (2H4PB CB-4PACAHPABA) ( PBCB+PBAC+PACB-2PACA+PA BA)·«2))'
~ (PBCa-PACA+PABA) ·«2»'
-
4PBCB-4 PACA+4PABA
. i~ifY 1
Example 6.89 (0.150, 4, 11) The area of a triangle is equal to the product of its three sides divided by the double circumdiameter of the triangle. Constructive description «POints A B C) (midpoint N A C) (midpoint L A B) (midpoint M B C) (inter 0 (t L L B) (t M M B» (AB"AC'cB' = 16S~BC '0A')
)
Example 6.90 (0.016, 1, 4) The radii of the circumcircle passing through the vertices of a triangle are perpendicular to the corresponding sides of the orthic triangle. Constructive description: ( (POints A B C ) (circumcenter 0 A B C) (foot Fe A B) (foot E B A C ) (perpendicular
E F A 0) )
306
Chapter 6.
Topics from Geometry
The eliminants
The machine proof fr:J..4E.. POAF
K
PCAO "PBAC
-
POAF ,PACA
E.
PCAO -PRAC ·PARA
-
PBAO,PBAC,PACA
simJ!!.i/y
-
PCAO · PABA
PBAO,PACA
Q PACA ,PAB,d2) .im1!! i fy 1
Figure 6-90
-
PABA,PACA ·(2)
-
Example 6.91 (0.016, 1, 3) Let ABC be a triangle with AC = AB. D is a point on BC. Line AD meets the circumcircle of ABC at E. Show that AB2 = AD . AE.
Constructive description ( (POints B M) (on A (t M M B» (inter 0 (I A M) (b A Bll (\ratio C M B-1) (on D (1 B Mll (inter E (I A D) (cir 0 All (eq-product A B A BAD A
The machine proof Eu.B. PDAE
§. -
PBAa-PAQA
2POAD ,PADA
simJ!!ify ~ (2) ·POAD
E) )
!2.~ - (2),PBAO Q~.imJ!!.ifY
-
(2),PBAB
-
Figure 6-91
Example 6.92 (0.001, 1, 3) Let C be the midpoint of the arc AB of circle (0) . D is a point on the circle. E = AB n CD. Show that CA2 = CE· CD.
Constructive description ( (points A M) (on C (t M M All (inter 0 (I C M) (b A Cll (on E (I AM» (inter D (I C E) (cir 0 C» (eq-product C A C ACE C
The machine proof fuA PECD
12 -
D) )
PACA ·PCFiQ
2POCE ,PCEC
simJ?iify ~ (2),POCE
§..~ - (2)·PA co Q~sim1!!.ifY
-
(2),PA CA
-
Figure 6-92
o.
307
TrlancJes
Example 6.93 (0.050, 3, 7) The distance of a side of a triangle from the circumcenter is equal to half the distance of the opposite vertex from the orthocenter.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (orthocenter H A B C) (foot DAB C) (inter A, (1 B C) (p 0 (.4lL. = OA,
AD))
2»
The eliminants
The machine proof
.4lL.~~ OA, - SBCO H PACB · PABC-16S~BC SABHC (-16) .SABC S 0 PacB , PsAC BCO (32) ,SABC
~(~) ~~ -
(2) -SBCO
!!..
-PACB -PABC-SABC+l6S!sc (2) -SBCO -(16SiBcl
-
.imJ!!. i fll -
16S~BC=PBAC · PACB+PACA - PABC
-(PACB-PABC-16S~ sc)
PABC=l(PBCB-PACA+PABA)
(32) -SBCO -SABC
PACB=2(PBCB+PACA-PABA)
Q -(PACB - PABC-16S~scl -( 32SABC) -
PBAC= -
~(PBCB-PACA-PABA)
(32) -PBCB -PBAC -SABC
.imJ!!.ify -(PACB - PAB~-16~sc) PBCB -PBAC he!!.on 16PsAc ,PAcs-16P'Gft"PABc±16PACA ,PABC
-
PBCB -PBAC-(16)
fl! (-4P~CR+4PBCB·PACA+4PBCB-PABAH2) PBCB -( PBCB+PACA+PA BA)-« 2»' .imJ!!.ify -
1
Figure 6-93
Example 6.94 (0.067, 3, 9) The ratio of a side of a triangle to the corresponding side of the orthic triangle is equal to the ratio of the circum radius to the distance of the side considered from the circumcenter_
Constructive description ( (POints A B C) (circumcenter 0 A B C) (footQ 0 A B) (foot E B A C) (foot DAB C) (eq-product ABO Q 0 A E
The eliminants P
D PC&C ,PARC+P8fjA ' PAca-PAcB ,P68Q
EDE P
PBCB
E(16HSABC)' BEB PACA
E~ PCEC= PACA D) )
P. Q (16) -(SABO)' OQO PABA P 0 PaCB ,PACrP6BA AOA (64) ,(SABC)' S 0 PACB -PABA ABO (32) .SABC 16~BC=PBAC-PACB+PACA - PABC
PABC= ~(PBCB-PACA +PABA) PBAC= -
Figure 6-94
~(PBCB-PACA-PABA)
308
Chapter 6.. Topics from Geometry
The machine proof PASA ,POQO PAOA,PEDE
Ii
P A S A ,POQo ' PSCS ,(PACA)2 P AOA .(P~cs ,PACA 'PASC-PACS · P~CA ,PAsc+16PACS ,PACA .s~sC>
simJ!!ify
.9.
PASA ,POQO,PSCS,PACA P AOA ·(PACS ' P ASC-PACA . PAsc+16S~sc)'PAcs
PASA ·(16S!BQ)·Pscs·PAcA PAoA · (PAcs'PAsc-PAcA · PAsc+16S~sC>·PACS · PASA
sim!lify
(l6) ,(SASO)"PSCS,PACA PAOA ,(PACS,PASC PAcA · PAsc+16S~sc)' PAcs
Q
(16) . (PAcs · PAsA)2.Pscs·PAcA ·(64S~Bcl
-
PSCS ' p ACA ,PAS A ,(PACS ,PASC-PACA . PAsc+16S~sC> · PAcs ·«32SASC»2
simJ!!ify PACS ,PASC PAcA·PAsc+16S~sc he'.:!.cm ;-;;-n----,-P-,;A'i'C",S,-'P:...A~s~A~·(..:;1:::.6),-;;-_ 16PSAC,PACS+16PACS ,PASC simJ!!.ify PSAC+PASC ~ PABA ·«2))2 simJ!!.ify
4PABA --2
Example 6.95 (0.083,2,6) AH
=-:-2 + -==2 Be = 40A .
Constructive description ( (points A B C) (orthocenter H A B C) (circumcenter 0 A B C) (Alf' +BC'
= 4AO') )
The machine proof
Figure 6-95 The eliminants
PRCR+PAHA
p
(4),PAOA
Q (PBCB+PAHA) · (64S~Bcl (4) ,PBCS ' P ACA ,PASA
!£
(16)'(Pscs·P~Ac+l6Pscs .S!qC) · (SASC)2 PSCS,PACA ,PASA ·(16S ABC)
simJ?lify P&AC+l6S~ BC PACA,PASA he'.:!.cm 16P&AC+16PSAC,PACS+16PACA 'PASC P ACA ,PABA ·(16)
eJl
16PACA "PA8A
PACA ,PASA ·«2»' simJ?lify
0 PBQS , PACA oPABA
AOA P
(64) .(SASC)2
H PSCS'(PBAC)2 AHA (16),(SASC)2
16S~BC=PSAC ' PACB+PACA ,PABC
PABC=!(PBCS-PACA+PABA) PACB=!(PSCS+PACA-PABA) PBAC= -
!(PSCS-PACA-PASA)
309
6.3. Triangles
Example 6.96 (0.116,4,7) The circumdiameters AP. BQ. CR of a triangle ABC meet the sides BC. CA. AB in the points K. L. M . Show that (KP/AK) + (LQ/BL) + (MR/CM) = 1. Constructive description ( (POints A B C) (circumcenter 0 A B C) (iratio PO A-i) (lratio Q 0 B-1) (lratio ROC -1) (inter K (I B C) (I A 0)) (inter L (I C A) (I B 0)) (inter M (I A B) (I C 0)) (*+~+~ = 1» The eliminants
The machine proof
.II.tl£~~
-(~+it+*)
cu- S~BC
~!:-SACq
BL - SABC
!1. -( ¥f -SABC+* -SABC+SABR) -
a!! §.aJ;;.e.
SABC
k
AK- SABC
- ( * -S1 BC- S ACQ -SABC+SABR-SABC) (SABC)'
-
.imJ!!.i/~ -
!i.
SACQg2SACO+SABC
SBcp~2SBCO-SABC
- ( * -SABC-SACQ+SABR) SABC
SBCO-SACO+SABo-tSABC=-tSABC
-(SBCP -SABC-SACq -SABC+SABR-SABC) (SABC)'
-
.imJ!!.i/~
-
!i -
-(SBCP-SACq+SABR) SABC
-(SBCP-SACq+2SABO-SABC) SABC
~ -
(SBCP 2SACO+2SABO 2SABC) SABC
!:..
-(2SBCO-2S AC o+2SABO-3SABC) SABC
-
o
SABR!hsABO-SABC
=
(-2H-1SABC) SA!C
.im!!1i/~
1
Figure 6-96
Example 6.97 (0.066,2,8) The mediators of the sides AC. AB of the triangle ABC meet the sides AB. AC in P and Q. Prove that the points B. C. p. Q lie on a circle_ Constructive description ( (POints A B C) (inter P (I A C) (b A B» (inter Q (I A B) (b A C» (cocircle
B C P Q) )
The machine proof (-SBPq) -Ppcq (-SCPQ) -PPBQ
~ (PBAC-SABP-!PACA-SABP) -PAcr«-PBAc»' - (t PACA -SBCpH-PBAC -PABP+t PACA -PABP)-(2)-(-PBAc) .im~ily
-
SA8P.PACP,PSAC
PACA -SBcrPABP P (_lPABA -SABCH-PBAC-PACA+ 1 PACA-PABA)-PBAC -(2H -PBAC) 2 PACA -( -PBAC-SABC+¥PAB:-SABc)-PABA -«-PBAC»'
=
.im!!1i/1l
Chapter 6.
310
Topics from c-.etry
The eliminants P PBq
S
q (2PBAC-P"Cd),PABP (2),PBAC
gPACA
o
S8CP
cPq- (2),PBAC
PpCq~!(PACP) S
q(2PB,SC-PACA) ,SABP (2) ,PBAC ~.!.(p ) S P (2PBAC-PABA) ,SABC ABP-2 ABA, BCP (2) ,PBAC P P (2PBAC-PABA) ,PACA S P PARA ,SABa ACP (2) ,PBAC ,ABP (2) ,PBAC BPq
P
Figure 6-97
Example 6.98 (0.083, 4, 13) The two tangents to the circumcircle of ABC at A and C
meet at E. The mediator of BC meet AB at D. Show that DE
II BC.
Constructive description ( (points A B C) (circumcenter 0 A B C) (midpoint H B C) (inter D (I A B) (10 H» (inter E (t C C 0) (t A A 0» (parallel DEB C) ) Figure 6-98
The machine proof
The eliminants S
~ SBCE
12 ~
tPCAO,PBCO (-16H -SBOn-SABel,SACO PCAO,PBCO ,SAOBH
(16) ·( -tSBCO),SABC ,SACO PCAO ,PBCO '( -SABO+tSABC)
g
(-16),SACO D -SROH,SABa
BCP
~ SBCP·(-4S AC o)
-
E PCAQ-PaCQ
BCE S
SAOBH
SAOBH!!, SBOH!!, -
s
!(2SABO-SABC) !(SBCO)
0 PACS - PABA
ABO
(32),SABC
PBCOgt(PBCB)
g
PCAO 2(PACA) (16) ' PBCB·PBAC·SABC·PAC~ · PABd32SABC)·((2))2 OPdCA"Pd8C PACA ,PBCB ·(2PACB ' PABA -32S: Bc H -32SABC )·(32SABC) ACO (-32) .SABC si~ifY -PRAC ,PARC S QPacs ' PSAC BCO- (32) .SABC PACB·PABA-16S~BC
s
heg;on
(-PBAC ,P A BCH16) -16PBAC,PACB+16PACB,PABA-16PACA ,P A BC
!!li (-PBCB+PACA+P"BA),(PBCB-PACA+PABA) ·((2))3 -
(-2P~CB+4PBCB ' PACA -2P~CA+2P~BA) ·«2»2
16~BC=PBAC ' PACB+PACA · PABC PACB=t(PBCB+PACA-PABA) PABC=2(PBCB-PACA+PABA) PBAC= -
!(PBCB-PACA-PABA)
sim!!4i!y 1
Example 6.99 (0.016, 1, 12) The lines tangent to the circumcircle of a triangle at the
vertices meet opposite sides in three collinear points. (The Lemoine axis of the given triangle.)
6.3.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (inter A, (I B C) (t A A 0)) (inter B, (I A C) (t B B 0)) (inter c, (I A B) (t C C 0)) (inter Z~B, Ad (I A B» (~ BC I
311
Triangles
The eliminants
The machine proof
AZC, Z£, SAA,B, BZC, - SBA,B,
(~)/(AZc, ) BC l BZ c •
z,g,
~~~ BC, - PBCO 1b PABO ,SACA,
-SBA,B, • ~ -SA-AtBI
S
BC I
AA.B) -
~ PACO ,SBA,B,
BA,d, -
SAA, B, ,PBCO
= AZC,»
~
BZc.
PACO,(-PCBO ,SABA,H-PCBAO) (-PABO ,SACA,),PBCO,(-PCBAO)
.im1!! i fll -
PACO ,PCBO ,SABA, PABO,SACA"PBCO
PABO ,PCAO ,SABC ,PBCO ,( -PCABO)
6im.J!!.ify -
A
PACQ -PCBO -PBAQ PABO ,PCAO ,PBCO
-
PCABO
~ PSAO , SARC
ABA, -
PCABO
PCAOg~(PACA) PABog~(PABA) PBAOg~(PABA) PCBOg¥(PBCB) PACO g 2(PACA)
Q PACA ,PBCB ,PABA ·«2))'
Figure 6-99
S
PCBAO
~ PCAO,SARC
ACA, -
PBCOgj(PBCB)
~ PACO ,PCBO ,PB .SO ,SABq-{-PCABO)
-
S
PCBAO
1b PCBO ,SA BA,
S
PABA ,PACA ,PBCB'«2))'
Bim!!:.ifll 1
Definition. The symmetric of a median of a triangle with respect to the internal bisector issued from the same vertex is called a symmedian.
Example 6.100 (0.050, 1, 6) The distances from a point on a symmedian of a triangle to the two including sides are proportional to these sides. Constructive description ( (points A B C) (circumcenter 0 A B C) (midpoint A, B C) (inter T, (I B C) (t A A 0)) (foot x T, A C) (foot Y T, A B) (eq-product T, Y ACT. X )
Figure 6-100
The eliminants
The machine proof Pr,YT"PACA Pr.XT"PABA
r.
(16S~BT,)'PACA
-
Pr.XT. ,(PABA)'
~ (l6),(SABT.)',(PACA)' (16S~CT.)·(PABA)'
A B)
~ (PBAO ,SABC)' ,(PACA)',«-PCABO))' -
(PCAO,SABC)',(PABA)"« PCABO))'
.imJ!!ifll (PBAO)' ,(PACA)' (PCAO)',(PABA)'
Q (PA BA)' ,(PACA)' ·«2»' .imJ!!ifll -
(PACA)' ,(PABA)' ·«2))'
-
312
Chapter 6. Topics from c-netry
Example 6.101 (0.033,1,8) Show that the distances of the vertices ofa triangle from the Lemoine axis are proportional to the squares of the respective altitudes.
Constructive description ( (points A B C) (circumcenter 0 A B C) (foot E B A C) (foot DAB C) (inter Bl (I A C) (t B B 0» (inter C l (I A (foot K A Bl Cd (foot J B Bl Cd
B) (t C C 0»
(AK'IiE'IiE' = EiJ'WW) ) The machine proof (PBEiBl',PAKA PBJB ,( PADA )'
.:!... (PBEiB)"PAKA ,PB,C,B, -
Ii -
(l6StB1Cl ),(PADA)' (PBEiB)'·(16S~B,Cl ),PB1C1B l (16),(SBB 1Cl )',(PADA)"PB1C1Bl
.iml!!ify (PBEiB)',(SAB,C, )' (SBB1Cl )',(PADA)'
G
(PBEiB)'·« -PACO,SABB, »',«-PBC,W»' PBCAO»'
- « PBCO,SABB1»'·(P,WA)' ·«
.iml!!iflJ (PBEiB)',(PACO)' (PBCO)',(PADA)'
Q (PBEiB)' ,( PACO)' ,(PBCB)' (PBco)'·«16S~BC»'
K -
«
16S~ BO»' ·( PACO)'·( PBC B)' (256),(PBCO)' ,( SABC)' ,(PACA)'
.iml!!.iflJ (PACO)',(PBCB)' (PBCO)' ,(PACA)'
Q (PACA)' ,( PBCB)' ·«2))' -
~~--------~~B
C,
(PBCB),,(PACA),·«2»'
.imJ!!.ify
C
0
Figure 6-101
Definition The tangents to the circumcircle at the vertices of a given triangle form a triangle called the tangential triangle of the given triangle. Example 6.102 (0.067, 3, 10) Show that the vertices of the tangential triangle of ABC are the isogonal conjugates of the vertices of the anticomplementary triangle of ABC.
6.3.
313
TrlancJes
Constructive description ( (POints A B C) (circumcenter 0 A B C) (pratio A) B A C I) (inter A, (t B B 0) (t C C 0 )) (eqangle B A A, A) A C) )
The eliminants p S
~PBCQ ·SA8Q±PA8A · SBCQ
BAA, -
SBCO
~ PRCO ,PABO
(16),SBCO
ABA, -
SACA) ~ PCAA)
s
(SAB C )
~ PBAC+PACA
0 PaOB ·Pade
BCO
S
(32) ,SABC 0 PAGB , PABA
ABO
(32) ,SABC
PABOgt(PABA) PBcogl(PBCB) PACB=2(PBCB+PACA-PABA)
Figure 6-102
PBAC= -
~(PBCB-PACA-PAB"')
The machine proof (-SABA.) ,PCAA) SACA) ,PBAA,
12
(-}PBCO ,PABO) ,PCAA, ·(4SBCO) SACA)·(4PBCO ,SABO+4PA BA ,SBCO) ·(4SBCO)
-
.im!lifll
-PBCO ,PABO ,PCAA, (16) ,SACA) ,(PBCO ,SABO+PABA ,SBCO)
~
-PBCO ,PABO ,(PB6 C+PACA) ( 16) ,(-SABCHPBCO ,SABO+PABA ,SBCO)
Q
PBCB ,PAB,,,{PBAC+PACA)·«32SABc»' ·(2) (16) ,SABC ·(64PBCB ,PBAC ,PABA ,SABC+32PBCB ,PACB ,PABA ,SABC) ·«2»2
-
eim1!! i/1I
-
PRActPACA
2PBAC+PACB
ell (-PBCB+3PACA+PABA)·«2»' -
( 2PBCB+6PACA+2PABA)·(2)
.imE!ifll
Example 6.103 (1.300 3 38) If two lines are antiparallel with respect to an angle, the perpendiculars dropped upon them from the vertex are isogonal in the angle considered.
Constructive description ( (circle B C R S) (circumcenter 0 B C R) (inter A (I B R) (l C S)) (foot NAB C) (foot MAR S) (eqangle BAM N A C) )
Figure 6-103
314
Cbapter 6. Topics from Geometry
Example 6.104 (3.083,9, 60) Show that the four perpendiculars to the sides of an angle at four cyclic points form a parallelogram whose opposite vertices lie on isogonal conjugate lines with respect to the given angle.
Constructive description ( (circle ABC D) (circumcenter 0 A B C) (inter I (I A B) (I C D)) (midpoint x C D) (midpoint Y A B) (inter E (P D 0 X) (P A 0
,~ A
B
F
Figure 6-104 V))
(inter
F
(P COX) (P BOY)) (eqangle
A I E F I C) )
Example 6.105 (0.001,1,3) The perpendicular at the orthocenter H to the altitude HC of the triangle ABC meets the circumcircle of HBC in P. Show that ABPH is a parallelogram. A
Constructive description «points A B C) (orthocenter H A B C) (circumcenter 0 B C H) (pratio Pp H A B 1) (inter P (I H Pp) (cir 0 H))
Themachi neproo f :
-H
g
BP
= P
(if = 1) )
SAHPp SBHPp
·· ."pSAHPp The e I1ffilnants: 'ftf:= SBHP • p
SBHPp
= ~ simplify =
Pp
-.SHDU
-SABH
Pp
=-
(
)
1.
SABH • SAHPp
Pp
=-
Figure 6-105 (
) SABH •
Example 6.106 (1.533, 10,38) The tangential and the orthic triangles of a given triangle are homothetic. (also see Example 6.90)
Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot DAB C) (foot E B A C) (foot F C A B) (inter c, (t BOB) (t A 0 A)) (inter A, (I c, B) (t C 0 C)) (inter B, (I C A,) (I A C,)) (inter I (I B, E) (I A, D)) (inter JJLA, D) (I c, F»
(4fT = 1tr»
Figure 6-106
6.3.
315
TrlaDcJes
Example 6.107 (0.083,4,11) Let P be the midpoint AH. Show that the segment OP is bisected by the median AA\.
Constructive description ( (POints A B C) (circumcenter 0 A B C ) (midpoint A, B C) (orthocenter H A B C) (midpoint P A H) (inter I (loP) (l A Ad) (midpoint lOP) ) The machine proof
P1- SAA,P
SAA,P~~(SAA,H )
SAOA) SAA,P
-
~ lAOA) .SAA,H
!f
The eliminants ~!. -SAOA)
-~ PI
L
Figure 6-107
(2) ,SAOA) · (16S~Ba> -PBAC ,PACB ,SABA,-PBAC,PABC ,SACA,
.im1!!. i f~
-
(-32) ,SAOA) ,(SABC)' (PACB ,SABA, +PABC ,SACA,) ,PBAC
~ (-32H-tSACO-tSABO),(SABC)' - (tPACB ,SABC-tPABC ,SABC) ,PBAC
.im1!!.i/v -
(32)-(SA CO+SABQ) ,SABC (PACB-PABC) ,PBAC
Q (32)-(-32PACB ,PABA ,SABC+32PACA ,PABC ,SABC) ,SABC 6imJ!! i fu -
(PACB-PA BC),PBAC ·(32SABCH-32SABC)
SAA,H
H(PACB ,SABA) +PABC,SACA) ),PBAC ( \6),(SABo)'
SACA,
~
SABA, ~
-
~(SABC)
~(SABC)
SAQA, ~ - ~(SACQ+SABQ) S 0 PACS,PABA ABQ (32) .SABC S 0 PACA , PAAC ACQ (-32),SABC PBAC= - ~(PBCB-PACA-PABA) PABC=~( PBCB-PACA+PABA) PACB=~(PBCB+PACA-PABA)
PAca ' PABA-PACA,PAAC
(PACB-PABcl ,PBAC
~ (-2PBCB ,PACA +2PBCB' P ABA +2P!C A -2P! BA)·«2))' (4PACA-4PABAH-PBCB+PACA+PABA) ·«2))'
sim!!li/v \
Example 6.108 (4.866, 12, 15) Prove that H A\ passes through the diametric opposite of A on the circumcircle.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (midpoint A, B C) (orthocenter H A B C) (inter I (I A 0) (I HAd) (perp-biesct 0 B 1) ) Figure 6-108
316
Chapter 6. Topics from Geometry
Example 6.109 (0.250, 2, 31) Show that the product of the distances of a point of the circumcircle of a triangle from the sides of the triangle is equal to the product of the distances of the same point from the sides of the tangential triangle of the given triangle. Constructive description ( (circle ABC D) (circumcenter 0 A B C) (foot E D B C) (foot FDA C) (foot G D A B) (inter E, (P D 0 A) (t A 0 A)) (inter F, (P D 0 B) (t BOB)) (inter G, (P DOC) (t C 0 C))
(DE'DP75G' = DE.'DF;"75G.') )
Figure 6-109
Example 6.110 (0.200, 12, 16) If 0 is the circumcenter the triangle ABC, and G is the centroid of ABC. we have
3M = GA2 + GB2 + GC2 + 3m:f. Constructive description ( (points A B C) (circumcenter 0 A B C) (midpoint E A C) (midpoint F A B) (midpoint D B C) (inter G (I A D) (I C F)) (3oG' +ACT +GC' +1R'f = 3M)
B
o Figure 6-110 )
Example 6.111 (1.033, 28,24) The internal and external bisectors ofan angle ofa triangle pass through the ends of the circumdiameters which is perpendicular to the side opposite the vertex considered. Constructive description ( (points B E P) (on L (t E E B)) (lratio C E B-1) (inter 0 (I L E) (b B L» (inter A (I B P) (cir 0 B»
(eqangle
B ALL A C) )
Figure 6-111
Example 6.112 (0.866, 12, 15) The segment of the altitude extended between the orthocenter and the second point of intersection with the circumcircle is bisected by the corresponding side of the triangle.
317
6.3. TriancJes
Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot DCA B) (foot E B A C) (inter H (I C D) (I B Ell (inter K (I C D) (cir 0 Cll (midpoint D H K) ) Figure 6-112
Example 6.113 (1.650, 5, 18) The circumcircle of the triangle formed by two vertices and the ortlwcenter of a given triangle is equal to the circumcircle of the given triangle. Constructive description ( (points A B C) (circumcenter 0 A B C) (foot DCA B) (foot E B A C) (inter H (I C D) (I B Ell (circumcenter 0, A B H) (eqdistance 0 A 0, H) ) Figure 6-113
Example 6.114 (0.933,17,18) A vertex of a triangle is the midpoint of the arc determined on its circumcircle by the two altitudes. produced. issued from the other two vertices.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (orthocenter H A B C) (inter c, (I e H) (cir 0 e)) (inter A, (I A H) (cir 0 A)) (perp-biesct B c, A,) ) Figure 6-114
Example 6.115 (0.833,17,27) /fO is the circumcenter and H the ortlwcenter ofa triangle ABC. and AH. BH. C H meet the circumcircle in D I • E I • Fl. prove that parallels through D I • E}, FI to ~A. OB. ~C. respectively. meet in a point. Constructive description ( (POints A B e) (orthocenter H A B C) (circumcenter 0 A B C) (inter e, (1 C H) (cir 0 e)) (inter A, (I A H) (cir 0 A))
Figure 6-11 S
318
Chapter 6. Topics from Geometry
(inter B, (I B H) (cir 0 B)) (inter I (P A, 0 A) (P B, 0 (parallel I e, 0 e) )
Bll
Example 6.116 (0.133, 7, 15) Show that the foot of the altitude to the base of a triangle and the projections of the ends of the base upon the circunuiiameter passing through the opposite vertex of the triangle determine a circle having for center the midpoint the base.
Constructive description «points A B e) (circumcenter 0 A B e) (midpoint M A B) (foot ve A B) (foot E A 0 e) (perp-biesct MEV) ) Figure 6-116
Example 6.117 (0.333, 12, 16) Show that the symmetric of the orthocenter of a triangle with respect to a vertex, and the symmetric of that vertex with respect to the midpoint of the opposite side, are collinear with the circumcenter of the triangle.
Constructive description ( (points A B e) (orthocenter H A B e) (circumcenter 0 A B e) (midpoint Al B e) (lratio S A H -I) (lratio L A, A -I) (collinear 0 L S) )
Figure 6-117
Example 6.118 (0.817, 26, 25) If DI is the second point of intersection of the altitude ADDI of the triangle ABC with the circumcircle, center 0, and P is the trace on BC of the perpendicular from DI to AC, show that the lines AP, AO make equal angles with the bisector of the angle DAC.
Constructive description ( (points A B e) (circumcenter 0 A B e) (foot v A B e) (inter v, (l A V) (cir 0 All (inter p (I B e) (t v, A e)) (eqangle 0 A V e A P) )
p
Figure 6-118
6.3.
T...... pa
319
Example 6.119 (3.900, 14,40) Show that the triangle formed by the foot of the altitude to the base of a triangle and the midpoints of the altitudes to the lateral sides is similar to the given triangle; its circumcircle passes through the orthocenter of the given triangle and through the midpoint of its base. Constructive description ( (POints A B C) (midpoint A, B C) (foot DAB C) (foot E B A C) (foot F C A B) (inter H (I B E) (I A D» (midpoint P B E) (midpoint Q C F) (cocircle H P Q D) )
A
Figure 6-119
Example 6.120 (0.717,44,38) The sides of the anticomplementary triangle of the triangle ABC meet the circumcircle of ABC in the points p. Q. R. Show that the area of the triangle PQ R is equal to four times the area of the orthic triangle of ABC. Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot DAB C) (foot E B A C) (foot F C A B) (inter P (p A B C) (cir 0 A» (inter Q (P B A C) (cir 0 B» (inter R (P C A B) (cir 0 C» (4SOEF
= SPQR) ) Figure 6-120
Example 6.121 (0.450,5,17) Through the orthocenter of the triangle ABC parallels are drawn to the sides AB. AC. meeting BC in D. E . The perpendiculars to BC at D. E meet AB. AC in two points D i • Ei which are collinear with the diametric opposites of B. C on the circumcircle of ABC. Constructive description ( (POints A B C) (circumcenter 0 A B C) (orthocenter H A B C) (inter D (I B C) (P H A B» (inter D, (I A B) (P D A H» (inter p (I B 0) (cir 0 B» (inter Q (I C 0) (cir 0 C» (collinear P Q D,) ) Figure 6-121
Chapter 6. Topics from Geometry
320
Example 6.122 (0.150, 6, 19) If the altitudes AD. BE. C F of the triangle ABC meet the circumcircle of ABC again in p. Q. R. show that we have (APIAD) + (BQIBE) + (CRICF) = 4. Constructive description ( (points A B C) (circumcenter 0 A B C) (foot DAB C) (foot E B A C) (foot F C A B) (inter H (I A D) (I C F» (inter P (I A D) (cir 0 A)) (inter Q (I B E) (cir 0 B)) (inter R (I C F) (cir 0 C» (.1l:+~+~ = 4) ) AD
BE
Figure 6-122
OF
Example 6.123 (0.433, 9, 21) Through the point of intersection of the tangents DB. DC to the circumcircle (0) of the triangle ABC a parallel is drawn to the line touching (0) at A. If this parallel meets AB. AC in E. F. show that D bisects EF. Constructive description ( (points A B C) (circumcenter 0 A B C) (inter D (t C 0 C) (t BOB» (inter F (I A B) (t D 0 A» (inter E (I A C) (I F D)) (midpoint D E F) )
D
Figure 6-123
Example 6.124 (0.466, 13, 28) In a triangle ABC. let p and q be the radii of two circles through A. touching side BC at B and C. respectively. Then pq = R2. Constructive description ( (points A B C) (circumcenter 0 A B C) (midpoint M B C) (inter Q (P B 0 M) (b A B)) (inter p (P COM) (b A C)) (eq-product Q B P COB 0
B) )
Figure 6-124
Example 6.125 (0.400, 7, 18) The parallel to the side AC through the vertex B of the triangle ABC meets the tangent to the circumcircle (0) of ABC at C in B l • and the parallel through C to AB meets the tangent to (0) at B in Cl. Prove that BC 2 BCl • Bl C.
=
Constructive description ( (points A B C ) (circumcentero A B C ) (inter BI (p B A C ) (t c o c ) ) (inter c, (p c A B ) (t B o B ) ) (eq-product B c B c B cl B, c ) ) Figure 6-125
Example 6.126 (5.733,22,23) Show that thefoot of theperpendicularfrom the orthocenter of a triangle upon the line joining a vertex to the point of intersection of the opposite side with the corresponding side of the orthic triangle lies on the circumcirck of the triangle. X
Constructive description ( (points A B C ) (circumcentero A B C ) (foot D A B C ) (foot E B A C ) (inter H (1 A D ) (1 B E ) ) (inter x (1 E D ) (1 A B ) ) (foot S H C X) (perp-biesct o s A) ) Figure 6-126
6.3.4
The Euler Line
Definition. The circumcenter 0, orthocenter H, and the centroid G of a given tnivigk are collinear and this line is called the Eukr line of the triangle. For the machine proof of Euler's theorem, see Example 3.71 on page 137. Example 6.127 (0.033,4, 2) Let 0 and H be the circumcenter and the orthocenter of a triangle ABC. Show that OH2 = 9R2 - a2 - b2 - 2.
Constructive description ( (points A B C ) (circumcenter o A B C ) (orthocenter H A B C ) (OH' = 09- -AB'-AC'-BC')) Figure 6-127
Example 6.128 (0.167, 6, 9) With the usual notations for the triangle ABC, we have * 4 3 = 4 A B ' + 4 2 + 4 @ + ~ ~ .
322
Chapter 6. Topics from Geometry
Constructive description ( (points A 8 e) (circumcenter 0 A 8 e) (centroid G A 8 e) (orthocenter H A 8 e)
(4AO'
=
4:A'F+4AC'+4BC'+GJr»
Figure 6-128
Example 6.129 (0.816, 9, 9) With the usual notations for the triangle ABC, we have
9m =AlT+E+BC2 +OH
2 •
Constructive description ( (points A 8 e) (orthocenter H A 8 e) (circumcenter 0 A 8 e)
(9AO' = :A'F +AC' +BC' +W) ) Figure 6-129
Example 6.130 (0.250, 6, 8) With the usual notations for the triangle ABC, we have 2 2 2 12m =AB +E +Bc + AH2 + BH2 +CH •
Constructive description ( (POints A 8 e) (orthocenter H A B e) (circumcenter 0 A 8 e) (12AO'
=:A'F +AC' +BC' +W +'ifi'f +CH') )
Figure 6-130
Example 6.131 (5.050, 8, 40) The homothetic center of the orthic and the tangential triangles of a given triangle lies on the Euler line of the given triangle.
Constructive description ( (points A 8 e) (orthocenter H A 8 e) (circumcenter 0 A 8 e) (inter D (I 8 e) (I A H)) (inter E (I A e) (I B H)) (inter F (I A 8) (I e H)) (inter 8, (t e 0 e) (t A 0 A)) (inter e, (I A 8,) (t 8 8 0)) (inter A, (I e, 8) (I B, e)) (inter I (I F e,) (I 8, E)) (collinear 0 H
l) )
Figure 6-131
o.
323
TriucJa
Example 6.132 (0.833, 12, 27) The Euler lines of the four triangles of an orthocentric group are concurrent. Constructive description ( (points A B C) (orthocenter H A B C ) (circumcenter 0 A B C) (circumcenter A, B C H ) (circumcenter B, A C H ) (inter I (I B B, ) (I A A, » (collinear 0 H 1) ) Figure 6-132
Example 6.133 (0.950, 4, 40) Show that the perpendiculars from the vertices of a triangle to the lines joining the midpoints of the respectively opposite sides to the orthocenter of the triangle meet these sides in three points of a straight line perpendicular to the Euler line of the triangle. Constructive description ( (POints A B C) (circumcenter 0 A B C) (orthocenter H A B C) (midpoint A, B C) (inter A, (I B C) (t A A, H » (midpoint B, C A ) (inter B, (I A C) (t B H B, » (midpoint c, A B ) (inter c, (I A B ) (t C C , H » (inter c!.l!. A B ) (I A, B, » (~ =~ » Be ]
6.3.5
Be,
Figure 6-133
The Nine-Point Circle
Definition. The midpoints of the segments joining the orthocenter of a triangle to its vertices are called the Euler points of the triangle. The three Euler points determine the Euler triangle.
A
Example 6.134 (The Nine-Point Circle Theorem) (0.050, 2, S) In a triangle the midpoints of the sides, the feet of altitudes, and the Euler points lie on the same circle. This circle is called the nine point circle of the given triangle.
B
Figure 6-134
324
Chapter 6. Topics 'rom C-try
Constructive description ( (POints A B C) (foot M A B C) (midpoint D B C) (midpoint E A C) (midpoint F A B) (midpoint I E F) (midpoint J M D) (parallel I JAM) )
The machine proof ~ SAMJ
The eliminants SAMJ4t(SAMD)
-
tSAMD
SAMI4 z (SAMF+SAME) SAMF~ - t(SABM)
1..
(2)·(tSAMF+tSAME) SAMD
SAMDg -
1. 2.4M.L -
.f.. -
SAME-tSABM SAMD
§..
-8AQM-SA8M
-
Q -
SAME~
-
i(SACM) z(SACM+SABM)
(2)·SAMD -(SACM+S'¥'M) (2)·(-fsACM-:SABM)
.im!!1ifll 1
Since the orthocenter and the three feet of the triangle fonn an orthocentric group, we conclude from the above machine proof that the circle also passes the three Euler points.
Example 6.135 (Feuerbach's Theorem) (55.500 102 46) The nine-point circle of a triangle touches each of the four tritangent circles of the triangle. Constructive description ( (POints I A B) (incenter C I A B) (midpoint Ml A B) (midpoint M2 A (midpoint M3 B C) (midpoint P Ml M3) (midpoint Q Ml M2) (inter N (t PPM.) (t Q Q M.» (foot D I A B) (cc-tangent I D N M.) )
C)
A
Figure 6-135
Example 6.136 (0.683, 3, 19) The radius of the nine-point circle is equal to half the circumradius of the triangle. Constructive description ( (points A B C) (circumcenter 0 A B C) (foot DAB C) (foot E B A C) (foot F C A B) (circumcenter N D E F) (4"
B
= 0;;(') ) Figure 6-136
Example 6.137 (0.067,5,18) The nine-point center lies on the Euler line, midway between the circumcenter and the orthocenter.
325
Trtaacks
6.3.
H
Constructive description ( (POints A B C) (circumcenter 0 A B C) (orthocenter H A B C) (midpoint M, B C) (midpoint M, A B) (midpoint M, A C) (circumcenter N M, M, M,) (midpoint N H 0) )
Figure 6-137
The eliminants
The machine proof
lII:J,NPM,M2M.
ON
-1iJi. ON
!!.. -
11!
M,l (
-tPM,M,M,+POM,M, -(-PBM!H+tPBM,B-PAM,H+tPAM!A-~PABA) -PBM,O+tPBM,B-PAM,O+tPAM,A-fpABA
2PBCO-2PBCB+2PACO-PACA+2PABO -
(2PBCBHPACB PAQA+2PABC) 2PBCO-2PBCB+2PACO-PACA+2PABO
Q (2PBCB-4PACB+PACA 2PABC) ·((2))' -
ell -
2 PBM,O+PAM,O ) M,l ( ) PHM,M, = 2 PBM,H+PAM,H PM,M,M, ~'~(2PBM'B+2PAM,A-PABA) PAM,O~' - ~(PBCB-2PACO-2PABO) PBM,O ~'H2PBCO-PBCB) PAM'A~' - ~(PBCB-2PACA-2PABA) PAM,H M, = - 41 ( PBCB-2PACH-2PABH ) PBM,B~'HpBCB) PBM'H~' ~(2PBCH-PBCB) POM,M, =
iPM,M2M.-PHM)Ma
~ -(2PBCH-2PBCB+2PACH-PAC6+2PABH)
!l.
-2PHM t M a
PM,M,M,-2POM,M,
-8PBCB+8PABA (4PBCBHPABA) (PBCB-PABA) ·«2»'
PABH!!,PABC PACH!!,PACB PBCH!!,PACB
. imgifll
PABOgt(PABA) PACOg¥(PACA) PBCOg¥( PBCB) PABC=¥(PBCB - PACA+PABA) PACB=2(PBCB+PACA -PABA)
Example 6.138 (4.383,51,41) Show that thefoot of the altitude ofa triangle on aside, the midpoint of the segment of the circumdiameter between this side and the opposite vertex, and the nine-point center are collinear.
Constructive description ( (POints A B C) (foot DAB C) (circumcenter 0 A B C) (inter E (I B C) (I A 0» (midpoint M A E) (midpoint (midpoint B, A C) (midpoint C, B A) (circumcenter N A, B, C,) (collinear M D N) )
A, C B)
Figure 6-138
326
Chapter 6.
Topics from G«metry
Example 6.139 (1.667, 44, 38) The center of the nine-point circle is the midpoint of a Euler point and the midpoint of the opposite side.
Constructive description ( (points A B C) (orthocenter H A B C) (midpoint M, B C) (midpoint M, A C) (midpoint M3 A B) (circumcenter N M, M, M3) (inter H, (1 C H) (1 M3 N» (midpoint H, C H) )
Figure 6-139
Example 6.140 (1.183,24,49) The two pairs of points 0 and N, G and H separate each other harmonically.
Constructive description ( (points A B C ) (orthocenter H A B C) (circumcenter 0 A B C) (midpoint M, A C) (midpoint M3 A B) (midpoint M, B C ) (circumcenter N M, M. M3) (inter G (I B M,) (I C M3)) (harmonic 0 N G H) )
Figure 6-140
Example 6.141 (0.850, 23, 38) If P is the symmetric of the vertex A with respect to the opposite side Be, show that H P is equal to four times the distance of the nine-point center from Be.
Constructive description ( (points A B C) (orthocenter H A B C) (foot DAB C) (lratio P D A -1) (midpoint A, B C) (midpoint B, A C) (midpoint c, B A) (circumcenter N A, B, C , ) (foot K NBC) (pif
= 167ii7f) )
c
B
Figure 6-141
327 Example 6.142 (2.450. 56. 42) Show that the sqlUlre of the tangent from a vertex of a triangle to the nine-point circle is equal to the altitude issued from that vertex multiplied by the distance of the opposite side from the circumcenter.
Constructive description ( (points A B
C)
(circumcenter 0
A B C)
(midpoint AI B C) (midpoint BI A C) (midpoint C I B A) (midpoint MI AI BI) (midpoint M. AI Cd (inter N (t MI MI AI) (t (foot DAB C)
c M. M. AI»
= PAIAD-POAD)
(PANA-PAINAI
Figure 6-142 )
Example 6.143 (0.950.50.37) Show that the symmetric of the circumcenter of a triangle with respect to a side coincides with the symmetric of the vertex opposite the side considered with respect to the nine-point center of the triangle.
Constructive description ( (points
A B C)
(circumcenter 0 (midpoint c i (midpoint BI (midpoint AI
A B C)
A B) A C) B C)
(circumcenter
N AI BI C I )
(inter 01 (lOCI) (I C (midpoint N 01 C) )
N»
Figure 6-143
6.3.6
Incircles and the Excircles
Example 6.144 (Theorem ofIncenter) (0.750.14.29) The three internal bisectors of the angels of a triangle meet in a point. the incenter I of the triangle. Constructive description ( (POints A B I) (foot AI B I A) (lratio A. AI B-1) (foot BI A I B)
(iratio B. BI A-I) (inter C (I A A.) (I B B.» (eqangle A C I I C B) ) Figure 6-144
328
Chapter 6. Topics from Geometry
Similarly the internal bisector at vertex A (B, C) and the two external bisectors at vertices B (C, A) and C (A, B) meet in a point Ia (lb, Ie). Definition, Each of I. Ia. h. Ie is the center of a circle tangent to the three sides of the triangle. The four circles with I. Ia. h. Ie as centers are called the inscribed circle and the exscribed circles or the four tritangent circles of the given triangle. Example 6.145 (0.067, 3, 10) Two tritangent centers of a triangle are the ends of a diameter of a circle passing through the two vertices of the triangle which are not collinear with the centers considered.
Constructive description ( (points 1 B C) (incenter A 1 C B) (inter I. (I A I) (t B B I) (midpoint 0 1 I.) (eqdistance 0 B 0 I) )
The eliminants PlOlgHpl1.I)
PBOBg~( 2PBI.B- Pl1.I+2PI BI) P I. PIAdPIBd' 11.1= (PBIA)' I. -(Phc.-PBIA,PIAI-PIAI,PIBA),PIBI PBI.B (PBIA)' p
A
PCCH"PC8C , PCHl
IBA PBCB ,PBIC P A (16),PIBdSIBC)' BIA PBCB ,PBIC 16S1BC=PBIC,PICB+PICI ,PIBC PIBC=i(PBCB-PICI+PIBI) PICB=2( PBCB+PICI-PIBI) PBIC= -
t(PBCB-PICI-PIBI)
The machine proof Eilll11 PIOI
Q tPBI.B-t Pl1.l+tPIBI -
!PllaJ
!s
(2P:' c A ,PI AI ,PI BI+2P~ I . ,PIAl ,PI BA ,PI BI-Phc. ,PIAI,ptR I),Phc. PIAI 'P1BI ,(pLIA)2
-
simE!.i/y 2PS1A±2P'BA-Pcac
-
4:
PIBI -PhCR,Phcc ,PIBI-2PBCB ,PBIC,PICB,PIBC,PIBI+32PBCB ,PBIC ' PIBl ,SIRc PIBdPBCB ,PBIC)'
simgi!y -(PBCB'PBIC+2PICB'PIBc-32SIAd PBCB ,PBIC he~on -16P8ca'P8ca±32PBcc*P'GB
32P'Ca"PCRC±32Prcr,PcRc PBCB ,PBIC ,(16)
eJ{ -( 4P&c A -4PBC B' PICI-4PBC B' PI B I ),(2)
-
PBCB '( PBCB+PICI+PIBI) ,« 2))3
simgify
Figure 145
Example 6.146 (0.300, 8, 9) The four tritangent centers of a triangle lie on sir circles which pass through the pairs of vertices of the triangle and have for their centers the midpoints of the arcs subtended by the respective sides of the triangle on its circumcircle.
bC
Constructive description ( (points B c I ) (incenter A I c B ) (inter I. (t B B I ) ( 1 A I ) ) (inter K (I A I ) (bc B ) ) (midpoint K I I.) )
B
....
1.
Figure 6-146
Example 6.147 (0.067,4,11) Let incircle (with center I ) of A A B C touch the side BC at X and let A1 be the midpoint of this side. Then the line A l l (extended)bisectors A X . Constructive description ( (points B c I ) (incenter A I c B ) (midpoint A, B c ) (foot x I B C ) (inter o (1 A X ) (1 A1 I ) ) (midpointo A x ) )
The eliminants z g
SIAA, -SIA,X x PBA,I.~BIA~ S I A ~ X = p8A18 xo
-~(sBCI)
~81.4,'
p8Al12='f ( ~ P ~ C I - P ~ C B ) ~ 8 , 4 ~ 8 ~ ~ ( ~ 8 C 8 ) sIAA1A=I i(scIA+sBIA)
A-P
The machine proof
S ~ ~ A =
AP
"d,"IC.zcBm .P
.P
-S
.S
s c ~ ~c;.;I:~;c8F =
1
~81c=-(~cIc+pnI,-~~ce)
1 ~ncI=~(~cIc-pBIn+pnc~) pC81=
- Z1 ( ~ C I C - ~ 8 1 B - ~ 8 C 8 )
- -(PCIC~PCBI~P~~~~P~~B~SBCI-PBIC~PBI~.~~~.C~~SBCI)~P A
A
(~P~cI-P~c~)~SBCI'(PBIC'PBCB)~
-
aim&'if~
-(PcIc.Pc~I - P B I B . P B c ~ ~ (~PBcI-PBcB).P~~c
- -(-2P~Ic+2Pc~c~Pnctl+2P~Im-2Pn~n~P~~~)~((~))2
PY -
(~PcIc-~PBIB).(PcIc+%I~-PBcB).((~))~ aimhi f far
-
1
B
AI
X
C
Figure 6-147
Example 6.148 (0.033, 1, 6) The product of the distances of two titangent centers of a triangle from the vertex of the triangle collinear with them is equal to the product of the two sides of the triangle pawing through the vertex considered
330
Chapter 6.
Topics from <>-netry
Constructive description ( (points B C 1) (incenter (inter
I.
A I C B)
(t
B B 1)
(eq-product
(I
A I»
A I A I. A B A C) )
Figure 6-148
Example 6.149 (0.550, 8, 17) Show that an external bisector of an angle of a triangle is parallel to the line joining the points where the circumcircle is met by the external (internal) bisectors of the other two angles of the triangle.
Constructive description ( (POints B C 1) (incenter A I C B) (inter M (II C) (b B A)) (inter L, (t B B I) (b C A)) (parallel M L, A 1) ) Figure 6-149
Example 6.150 (3.933,4,20) The product ofthe four tritangent radii of a triangle is equal to the square of its area.
Constructive description ( (points B C I) (incenter A I B C) (inter I. (I A I) (t B B I)) (inter I, (I C I) (I B I.» (inter 1& (l B 1) (I AI,» (foot x I B C) (foot X. I. B C) (foot x, I, B C) (foot x, I, B C)
(TJr I.X. 21;Jf;'~ = SABCSABCSABCSABC)
)
Figure 6-150
Example 6.151
22 (0.717, 17,28) In triangle ABC, the bisector of angle A meets BC at L and the circumcircle of triangle ABC at N. The feet of the perpendiculars from L to AB and AC are K and M. Show that SABC = SAKNM .
Constructive description: «points I B A) (incenter C I B A) (circumcenter 0 A B C) (inter L (I B C) (I A 1) (foot M L A C) (foot K L A B) (inter N (I A 1) (cir 0 A)) (SABC = SAKNM) ) 22This is a problem from the 1987 International Mathematical Olympiad.
Figure 6-151
332
Chapter 6. Topics from C-Iry
Example 6.156 (0.033, 1, 6) Show that a parallel through a tritangent center to a side of a triangle is equal to the sum, or difference. of the two segments on the other two sides of the triangle between the two parallel lines considered.
Constructive description ( (POints B C J) (incenter A 1 C B ) (inter M (I A C) (P I A B» (inter N (II M) (I B C» (eqdistance M A M J) )
The eliminants P
PBAB·(SCIA)' (SBCA)' P ~ PCAdSBLA)' AMA(SBCA)' s A Pelo ,PaRI ,SSCI CIA PBIC .PBCB P A (PBIB)' ·(PBCrl' BAB (PBIC)' .PBCB M
IMI
s
A -PRra-Pace ,SSCI
BI A PBIC .PBCB Po A (PCIC)' ·(PCBtl' CAC (PBIC)' .PBCB
The machine proof E..ut..4 PIMI
MPCAC ·SV,, ·sVCA PB A B·SCIA ·SBCA
c
N
Figure 6-156
.im~ifY PCAC ·(SBIA)' PBAB ·(SCIA)·
~ P?'G·P?~,,«-PBIB · PBCI"SBCI»' · (PBIC · PBCB)' · Pijrc·PBCB .im1!!.i/lJ P~/B .PBCI .(PCIC .PCBI .SBCI)2 .(PBIC . PBCB)l,pAIC.PBCB 1
Example 6.157 (0.100, 1,9) Prove the formula: AZ . BX . CY
Constructive description ( (POints B C J) (incenter A I B C) (foot XI B C) (foot Y 1 A C) (foot Z 1 A B)
(AZ"RX'cY' = xrSABCSABC)
x
B
)
= r \l ABC.
c
Figure 6-157
Example 6.158 (0.666,10,18) The projection of the vertex B of the triangle ABC upon the internal bisector of the angle A lies on the line joining the points of contact of the incircle with the sides BC and AC.
Constructive description ( (POints B C J) (incenter AlB C) (foot XI B C) (foot Y 1 A C) (foot L B A 1) (inter J (I A 1) (I x Y»
(1t=t.f»
x Figure 6-158
332
Chapter 6. Topics from C-Iry
Example 6.156 (0.033, 1, 6) Show that a parallel through a tritangent center to a side of a triangle is equal to the sum, or difference. of the two segments on the other two sides of the triangle between the two parallel lines considered.
Constructive description ( (POints B C J) (incenter A 1 C B ) (inter M (I A C) (P I A B» (inter N (II M) (I B C» (eqdistance M A M J) )
The eliminants P
PBAB·(SCIA)' (SBCA)' P ~ PCAdSBLA)' AMA(SBCA)' s A Pelo ,PaRI ,SSCI CIA PBIC .PBCB P A (PBIB)' ·(PBCrl' BAB (PBIC)' .PBCB M
IMI
s
A -PRra-Pace ,SSCI
BI A PBIC .PBCB Po A (PCIC)' ·(PCBtl' CAC (PBIC)' .PBCB
The machine proof E..ut..4 PIMI
MPCAC ·SV,, ·sVCA PB A B·SCIA ·SBCA
c
N
Figure 6-156
.im~ifY PCAC ·(SBIA)' PBAB ·(SCIA)·
~ P?'G·P?~,,«-PBIB · PBCI"SBCI»' · (PBIC · PBCB)' · Pijrc·PBCB .im1!!.i/lJ P~/B .PBCI .(PCIC .PCBI .SBCI)2 .(PBIC . PBCB)l,pAIC.PBCB 1
Example 6.157 (0.100, 1,9) Prove the formula: AZ . BX . CY
Constructive description ( (POints B C J) (incenter A I B C) (foot XI B C) (foot Y 1 A C) (foot Z 1 A B)
(AZ"RX'cY' = xrSABCSABC)
x
B
)
= r \l ABC.
c
Figure 6-157
Example 6.158 (0.666,10,18) The projection of the vertex B of the triangle ABC upon the internal bisector of the angle A lies on the line joining the points of contact of the incircle with the sides BC and AC.
Constructive description ( (POints B C J) (incenter AlB C) (foot XI B C) (foot Y 1 A C) (foot L B A 1) (inter J (I A 1) (I x Y»
(1t=t.f»
x Figure 6-158
333 Example 6.159 (8.250, 21, 34) The midpoint of a side of a triangle. the foot of the altitude on this side. and the projections of the ends of this side upon the internal bisector of the opposite angle are four cyclic points.
Constructive description ( (POints B C I) (incenter A I B C) (foot x B A I) (foot Y C A I) (midpoint AI B C) (foot DAB C) (cocircle x Y AI D) )
c y
Figure 6-159
Example 6.160 (0.466, 8, 17) Show that the midpoint of an altitude of a triangle. the point of contact of the corresponding side with the excircle relative to that side. and the incenter of the triangle are collinear.
Constructive description ( (POints B C I) (incenter A I B C) (foot DAB C) (inter I. (t B B I) (II A)) (foot x. I. B C) (inter M (I A D) (II x.)) (midpoint M A D) ) Figure 6-160
Example 6.161 (1.633, 10, 18) The internal bisectors of the angles B. C of the triangle ABC meet the line AXo joining A to the point of contact of BC with the excircle relative to this side in the points L. M. Prove that ALjAM = ABjAC.
Constructive description ( (POints B C I) (incenter A I B C) (inter I. (t B B I) (II A)) (foot x. I. B C) (inter L (I B I) (I A x.» (inter M (I C I) (I A x.»
(eq-product
A M A B A L A C) )
Figure 6-161
Example 6.162 (0.050, 1, 6) Show that the product of the distances of the incenter of a triangle from the three vertices of the triangle is equal to 4Rr2.
Chapter 6. Topics rrom Geometry
334
Constructive description ( (points B C l) (incenter A I B C) (circumcenter 0 A B C) (foot x I B C) (16W·iX' = TA'TifRf»
The elirninants P
x (!6)·(SBcIl' IXI PBCB P 0 PCAC'PBAwPaC8 BOB (64) ,(SBCA)' S A (-2) ,PCB/,PBC/,SBCl BCA PBIC ,PBCB P A (!6),PCIC,PBIB,(SBC,)' IAI (PBIC)' ,PBCB P A (PBIB)',(PBCI)' BAB (PBIC)',PBCB p. A (PCIC)' ,(PCB/)' CAC (PBIC)' ,PBCB
Figure 6-162
The machine proof (!6)-(PlXd ,PBOB PIA/,PCIC ,PBIB
K -
Q -
(16)·« 16Shcc»' ,PBOB PIAJ'PCIC ,PBIB ,(PBCB)' (4096),(SBCIl' ,PCAC,PBAB,PBCB PIAJ'PCIC ,PBIB ,(PBCB)'·(64StCA)
.imglify
.d. -
(64) ,(SBCIl' ,PCAC ,PBAB PIAJ'PCIC ,PBIB ,PBCB ,(SBCA)'
(64) ,(SBCI)4 ' P3ca·P3sc ·PAcs·PAa,.(PBIC,PBCB)' ·PArc,PBCB (16PCIC ,PBI B ·StCI ), PCIC,PBI B,PBCB '« -2PCBJ'PBCJ'SBCI ))'.(P~IC·PBCB)'
.imElify
Example 6.163 (0.433, 6, 10) If the lines AX, BY, C Z joining the vertices of a triangle ABC to the points of contact X, Y, Z of the sides BC, C A, AB with the incircle meet that circle again in the points XI, Y1, ZI, show that: AX · XXI' BC = 4rS, where r, S are the inradius and the area of ABC. Constructive description ( (POints B C l) (incenter A I B C) (foot x I B C) (foot Y I A C) (foot Z I A B) (inter x, (I A X) (cir I X)) (inter Y, (I B Y) (cir I Y)) (inter z, (I C Z) (cir I Z» (16'iX" ·S~BC
= IT Xx,'BC')
x )
c
Figure 6-163
Example 6.164 (2.783, 16, 13) If h, m, t are the altitude, the median, and the internal bisector issued from the same vertex of a triangle whose circumradius is R , show that 4R 2 h2 (t2 _ h2 ) = t 4 (m 2 _ h2 ) .
335
O. TrtucJes
Constructive description ( (POints B G I) (incenter A / B G) (circumcenter 0 A B C) (foot DAB G) (midpoint A, B C) (inter T (I / A) (I B G»
(4TK.08' .AD"-4AD".08'.AD" = AT,'.TA'-AD" .TA'» Figure 6-164
Example 6.165 (0.050, 1, 10) The external bisectors of the angles of a triangle meet the opposite sides in three collinear points. Constructive description ( (POints B G /) (incenter A / B G) (inter A, (I B G) (t A A /» (inter B, (I A G) (t B B /» (inter G , (I A B) (t C G /» (inter Gu!. A, B,) (I A B» (~=~» BC.
Be2
The machine proof
~ ~ B
A,
c
Figure 6-165
The eliminants
(~)/(~) BC. Be]
~ SS"'S' • ~ SAA1B 1
QI
BC.
PIC .. ·SS"IS, S .... ,S, ·PSCI
~
PIC .. ·PCSI·SS .... , ·PICS .. PIS .. ·SC .... , ·PSC/'PICS ..
-
•iml!!i/u PIC .. ·PCSI ·SS .... , PIS .. ·SC .... , ·PSCI
-
~ PIC6 ·Pcsd-Ps,,/,SSC")·PCIS ..
-
PIS .. ·( PCA/ ,Ssc,,) ,PSC/,PCIS ..
aim~i/SI -
.d. -
P,eA ,PaRI ' PA" PIS" ,PC"I'PSCI
(-PCIC,PCSI,PSCI ),Pcsd16Psl S ,PSCI '51.C!).P' w ' PSCS,PSIC'PSCS ( PCSI ·PSls ,PSCI)-(16PClc,PCSI ·51,cl)·PSCI ·P slc ·PSCS·PSIC·Pscs
Example 6.166 (1.850, 14,28) Prove tlult the triangle formed by the points of contact of the sides of a given triangle with the excircles corresponding to these sides Iuls the same area with the triangle formed by the points of contact of the sides of the triangle with the inscribed circle.
Chapter 6. Topics from Geometry
336 Constructive description ( (POints B C J) (incenter A / C B) (inter /. (t B B /) (I A /» (foot x / B C) (foot y / A C) (foot Z / A B) (foot x. I. B C) (inter r. (t C C /) (I B I)) (inter /, (t B B /) (I C I)) (foot Y. /. A C) (foot z, /, B A) (SX Y Z = SX.Y.z,) )
I.
Figure 6-166
Example 6.167 (Gergonne Point) (0.050, 1,7) The lines joining the vertices of a triangle to the points of contact of the opposite sides with the inscribed circle are concurrent (The Gergonne point).
Constructive description ( (points B C /) (incenter A / C B) (foot X/ B C) (foot Y / A C) (foot Z / A B) (~~ti=l»
The eliminants ~!....l:JuL BZ--PIBA
~r....ful.L Ay--PCAT
u.~~
C X --PBCI Po A (l6) ·PCIC ·PCBdSBCd' cAi (PBIC)'.PBCB p A -PCR,.PalB ·Pncc IBA PBIC .PBCB P A Pare·PeRc ·PaCI ICA PBIC .PBCB P A (16)·PBIB ·PBCdSBCI)' BAT (PBlc)' .PBCB
x
c
Figure 6-167
The machine proof _~.~ . u. ~ =.EB...u. • ~ . u. ~ BZ AY CX
.d. -
-PIBA
AY cx
PaM·PCCA • U. ~ -PRM·PWd ·PCRI PIBA ·(-PCAT) cx PIBA·PCAI"(-PBCI)
(16PBCB · PBCI · shcIH-PCIC·PCBI"PBCI) · PCBI·P~lQ.PBCB o PBIC · PBCB .im1!!i/y (-PCBI"PBIB oPBcIl .(16PcIC oPcBl ostcl)oPBCI"PBlc oPBcB ·pZICoPBCB 1
Example 6.168 (Nagel Point) (0.650, 3, 31) The lines joining the vertices of a triangle to the points of contact of the opposite sides with the excircles relative to those sides are concurrent (The Nagel point). Constructive description ( (points B C J) (incenter A / C B) (inter /. (t B / B) (I A I)) (inter /. (I B J) (1/. C)) (inter /, (I I. B) (l C I)) (foot X. I. B C) (foot Y. /.
A C)
I.
Figure 6-168
6.3.
337
TrtucJes (foot Z. I. A B) (inter J (I A x.) (I BY.» (inter K (I C Z.) (I BY.»
(t7 = ~»
Example 6.169 (3.583, 12,30) Show that the line joining the incenter of the triangle ABC to the midpoint of the segment joining A to the Nagel point of ABC is bisected by the median issued from A. Constructive description ( (points B C I) (incenter A I B C) (inter I. (I A I) (t B B I» (inter I. (I C I.) (I B I» (foot X. I. B C) (foot Y. I. A C) (inter N (I BY.) (I A x.» (midpoint A, B C) (midpoint S N (inter p (II S) (I A, A» (midpoint PIS) )
8
A) I.
Figure 6-169
Example 6.170 (0.816, 12, 13) The sides AB. AC intercept the segments DE. FG on the parallels to the side BC through the tritangent centers [ and [a. Show that: 21BC = IIDE + I/FG. Constructive description ( (POints B C I) (incenter A I B C) (inter I. (t B B I) (I A I» (inter D (P I B C) (I A B» (inter E (I D I) (I A C» (inter F (P I. B C) (I A B» (inter G (I F I.) (I A C» (~+~
=2»
Figure 6-170
Example 6.171 (1.817, 27, 20) Show that the perpendiculars to the internal bisectors of a triangle at the incenter meet the respective sides in three points lying on a line perpendicular to the line joining the incenter to the circumcenter of the triangle. Constructive description ( (POints B C I) (incenter A I B C) (circumcenter 0 A B C) (inter x (t I I A) (I B C» (inter Y (t lIB) (I A C» (inter z (t I I C) (I A B» (perpendicular 0 I X Z) ) Figure 6-171
338
Chapter 6.
Topics from c - I r y
Example 6.172 (0.416,4,19) The side BC of the triangle ABC touches the incircle (I) in X and the excircle (10) relative to BC in Xo. Show that the line AXo passes through the diametric opposite Xl of X on (I). A
Constructive description ( (POints B C 1) (incenter A 1 C B) (inter 1. (t BIB) (I A 1» (foot x 1 B C) (foot X. 1. B C) (inter XI (11 X) (I A X.» (midpoint 1 X X.) )
Figure 6-172
Example 6.173 (0.816, 6, 15) With the notations of Example 6.172, show that if the line All meets the altitude AD of ABC in P, then AP is equal to the inradius of ABC. Constructive description ( (POints B C 1) (incenter A 1 C B) (foot XI B C) (foot DAB C) (midpoint AI B C) (inter P (11 Ad (I A D» (eqdistance API X) ) Figure 6-173
Example 6.174 (0.267, 10, 16) With the notations of Example 6.172, if the parallels to AXo through B , C meet the bisectors CI, BI in L, M show that the line LM is parallel toBC. Constructive description ( (POints B C 1) (incenter A 1 C B) (inter 1. (t BIB) (I A 1» (foot XI B C) (foot X. 1. B C) (inter L (P B A X.) (I C 1» (inter M (P C A X.) (I B 1» (parallel L M B C) )
I.
Figure 6-174
Example 6.175 (0.150, 2, 10) Show that the trilinear polar (see Example 6.203) of the incenter of a triangle passes through the feet of the external bisectors, and this line is perpendicular to the line joining the incenter to the circumcenter, and this lines is perpendicular to the line joining the incenter to circumcenter.
339
U. Triuctes
Constructive description ( (POints B C /) (incenter A / B C) (circumcenter 0 A B C) (inter E (I B J) (I A C» (inter F (I C J) (I B A» (inter D (I A /) (I B C» (inter x (I E F) (I B C» (inter z (I D E) (I A B» (perpendicular x A A /) )
c
Figure 6-175
The eliminants
The machine proof
x PCA,.SSfjp-PSAI ,SCfif
p IIiX
S8ECF
S
="
S8CAI·PCAI -S8I1E ·S8CI-S8CAI -P8I1rSCIE-S8CII
-S8CIII -PCAI -S8111 -S8CII- S 8CI-S8CIII -P 8I1rS CIII -S8CII -S 8CI
Bimg1ifll -(PCAI -SBIII+P8I1r S CIII) -S8CII -SBCI
=" -(0)-(
F-SsAg -Sec r 8EF
S
•• mg1ifll PCAI -SBIIE -S8cl-P8I1rSCIE -S8CII
= "
-S8C.4,J
S
Bimg1ifll PCAI -S 8EF- P 8I1rSCEF
= "
FSalg -SSCA
CEF
PCllrS8EF- P 8I1r S CEF
CIES8CIII S E-SalA ,SscA 811E S8CIII 5 II (-2) -PCB,.PBC,.S8CI 8CII P8IC -P8C8 S ~ Perc -PaAI ·Ssc r CIIIP8IC -P8C8 P II (l6)-P818 · P BCdSBCrl' 8A1 (P8IC)' -P8C8
s
-2PCB/ -PBC/ -S8CI )-S8CI
S8CAI
!SCIA·Sacr
A -P"a · Pscr ,Sscl 8111
P8IC -P8C8
R
Bimg1ifll 0
II (l6)·PCIC -PC8dS8CIl' CIII (P8IC)' -P8C8
The second result has also been proved by the program.
Example 6.176 (0.633, 3, 27) Show that the mediators of the internal bisectors of the angles of a triangle meet the respective sides of the triangle in three collinear points. Constructive description ( (POints B C J) (incenter A / B C) (inter u (I A J) (I B C» (inter v (I B J) (I A C» (inter w (I C /) (I A B» (inter P (b A U) (I B C» (inter Q (b B V) (I A C» (inter 5 (b C w) (I A B» (inter hll A B) (I P Q» (~=~» 85
u figure 6-176
85,
Example 6.177 (0.450,6,25) Show that the lines joining the vertices of a triangle to the projections of the incenter upon the mediators of the respectively opposite sides meet in a point· the isotomic conjugate of the Gergonne point of the triangle.
340
Chapter 6. Topics from Geometry
Constructive description ( (points B C I) (incenter AlB C) (foot x I B C) (midpoint A, B C) (midpoint B, C A) (midpoint c , A B) (inter D (t A, A, C) (P 1 B C» (inter E (t B, B. C) (P 1 A C» (inter F (t C, C, B) (P 1 A B» (inter J (I A D) (I B E» (inter Y (I A D) (I B C» (midpoint A, x
~ F
B
Y) )
· ·.
X
AI
... ... . Y
C
Figure 6-177
Example 6.178 (0.516,3, 19) Show that the line AI meets the sides XY, XZ in two points P, Q inverse with respect to the incircle (1) = XY Z, and the perpendiculars to AI at P, Q pass through the vertices B, C of the given triangle ABC. Constructive description ( (points B C I) (incenter AlB C) (foot XI B C) (foot Y I A C) (foot Z I A B) (inter P (II A) (I x Y» (inter Q (II A) (I x Z» (inversion I x P Q) )
Figure 6-178
Definition The symmetric of a median of a triangle with respect to the internal bisector issued from the same vertex is a symmedian of the triangle Example 6.179 (1.250, 6,18) The three symmedians of a triangle are concurrent (The Lemoine Point or the symmedian point). Constructive description ( (POints A B C) (midpoint M, B C) (midpoint M. A C) (midpoint M. A B) (on B, (a B C M. B A» (on c, (a CAM. G B)) (inter I (I C G,) (I B B,» (eqangle B A I M, A C) )
M, Figure 6-179
B
Example 6.180 (0.566, 5. 19) The three symmetrics of the three lines joining a point and the three vertices of a triangle with respect to the internal bisectors issued from the same vertices are concurrent. (The isogonal conjugate point).
341
Constructive description ( (POints ABC M) (on AI (a A C M A B» (on BI (a BAM B C» (inter N (1 A AI) (1 B BI» (eqangle A C M NCB) ) Figure 6-180
Intercept Triangles
6.3.7
Definition. Let L, M, N be three points on the sides BC, C A, AB of triangle ABC. Then triangle LMN and the triangle determined by lines AL,BM, and CN are called the intercept triangles of triangle ABC for points L, M, N .
Example 6.181 (0.033,2,5) Let Ab B I . CI be points on the sides BC. CA. AB of a triangle ABC such that BAd BC = CBdCA = ACdAB = 1/3. Show that the area of the triangle determined by lines AA I • BBI and CCI is one seventh of the area of triangle ABC. We only need to show
SsABAa ABC
= 2/7.
Constructive description: ( (POints (inter A. (I
A A.)
(1
B B.»
(2SABC
A B C)
(lratio
AI B C 1/3)
(lratio
BI C A 1/3)
= 7SABA,) ).
The machine proof
The eliminants
(2) ·SABC (7)·SABA, ~ (2) ·SABC·SABA)B)
-
(7) ·SABBI ·S ABAI
~ (2)·SABC·( -tSACA) +SABA)
-
(7)·(t S ABC)·SABAI
.i"'1!!.ifll -(2SACA) -3SABA) (7)·SABA I
Figure 6-181
~ -(-iSABc) .im!!iifl/ I (7)·( • S ABC)
Example 6.182 (0.017, 3, 5) Let AI. Bb C I be points on the sides BC. CA. AB of a triangle ABC such that BAdBC = CBdCA = ACdAB = T. The intercept triangle determined by lines AA I • BBI and CCI is A2B2C2 (Figure 6-181). Show that SA,B,C, _ SABC -
(2r-I)'
r2-r+l·
342
Chapter 6. Topics from Geometry
Constructive description: ( (POints A B C) (lratio A, B C r) (lratio (inter A, (I A A.) (I B B,» «r'-r+1) ,SA2AB = (-"+r) ,SABC».
The eliminants
The machine proof
S
("-r+1) ,SABA, (r 1 ).r·S ABC ~
!:!! -
11 -
~ SABBt ,SABA, ABAl -
SABAIB.
SABA,B, ~ SACA, ·r-SACA, +SABA,
(r'-r+l),SABBt .SABAt (r 1).r,SABc ,SABA,B,
SABB, ~ -
-(r' - r+l) ,(-SABc ·r+SABc),SABA, (r-I) .r,SABc ,(SACA,·r-SACA, +SABA,)
.im:g1i/y
B, CAr)
(r-l) ,SABC)
SACA, ~(r-l).SABC SABA, ~SABc·r
(r'-r+l) ,SABA t r,(SACA, .r-SACA, +SABA,)
(r'-r+1) ,SABC·r .i"'!!!i/y r,(SABc·r 2 -SABc·r+SABC) -
Example 6.183 (0.466, 5, 16) Use the same notations as 6.182. If BAt/AIC SA.B,C. ( ..... r'-1)2 CB 1 /B I A = T2, AC1 /CI B = T3 then SABC = (.. r'+r'+I)(r'r'+ ... +l)( ..... + .. +l) · Constructive description: (Formula derivation) «POints A B C) (mratio A, B C r.) (mratio B, C A .. ) (mratio c, A B r.) (inter A, (I A A,) (I B Bd) (inter B, (I B B.) (I C Cd) (inter c, (I C Cd (l
= TI,
A A,» (5 ~~;~' ) )
Example 6.184 (0.033,2,4) Let AI, Bb CI be points on the sides BC, CA, AB of a triangle ABC such that BAt/A 1C = CBt/B1A = ACt/C1B = k. Furthermore, let A 2 , B 2 , C2 be points on the sides BICb C 1 A b A I C 1 of a triangle A 1B 1 C 1 such that C1A2/A2B1 = A I B 2 /B 2 C 1 = B I C 2 /C2 A 1 = k. Show that triangles ABC and A 2 B 2 C 2 are homothetic.
Constructive description «POints A B C) (lratio c, A B YT) (lratio A, B C YT) (lratio B, C A YT) (lratio A, c, B, YT) (lratio B2 A, c, YT) (lratio c, B, A, YT) (inter 0 (I A A2) (I B B2» (inter D (I C C2) (I A B» (parallel A2 B, A B) )
The machine proof ~ --;;-_""SA::;B7'A"1"-;;-_ SABA,'YT+SABA,
.img1i/y ~
The eliminants SABB, ~ - (YT-1) ,SABA,)
SABA, SABB,
-SABA. (YT-1)'SABA,
SABA,~SABB"YT
SABB, ~ - (YT-1) ,SABC) SABA, ~SABC'YT
-SABBt'YT (YT-1),SABA,
-
!:!!
-(-SABC'YT+SABC)'YT (YT l),SABA,
-
.im1!ii/1/ -
Wa.:r:z: SABA,
11 Wa.:r:z: -
SABC'YT
6i"'!li/1I
Figure 6-184
343 Example 6.185 (0.083,4, 11) Let M, N, and P be three points on the sides AB, BC and AC of a triangle ABC such that AM/ M B = B N / N C = C P / P A. Show that the point of intersection of the medians of t:::.M N P coincides with the point of intersection of the medians of t:::.ABC.
Constructive description «POints
A B C)
(lratio M A B Yr) (lratio NBC Yr) (lratio PC A Yr) (centroid GAB C)
(inter L (I
N P)
(£t = -I»
(I
F
M G»
The eliminants
The machine proof
I:.J.f~ NL-SHNO
-(£t) k -
SMNOg~(SCMN+SBMN+SAMN )
SHea
SMPog3(SCMP+SBMP+SAMP)
-(-SMNO)
Q -(SCHP+SBMP+SAHP)·(3) -
(SCHN+SBMN+SAMN) ·(3)
~ -(SBcwYr-SBCM+2SAcwYr-SACM)
-
!:! -
M -
M
Figure 6-185
SCHN+SBMN+SAMN
SAMP~(Yr-I).SACM SBMP~(Yr-I)·SBCM SCMP~SACM'Yr SAMNt:. -
-(SBcwYr-SBCM+2SAcwYr-SACM) 2SBCM 'Yr+SBCM SACM 'Yr
-3SABC· ~~+3SABc·Yr-SABC 3SABC·Yr +3SA Bc·Yr- SABC
(SBCM 'Yr)
SCMNt:. -
«Yr-I),SBCM)
SACM~ SBCM~
. i"'!1i/ll
(SACM'Yr)
SBMNt:. -
-
(SABC 'Yr)
-
«Yr-I),SABC)
Example 6.186 (0.266,3,13) Let M, N , and P be the same as in Example 6.185. Show that the point of intersection of the medians of the triangle formed by lines AN, B P and C M coincides with the point of intersection of the medians of t:::.ABC.
Constructive description «POints (lratio (lratio
A B C) M A B Yr)
(lratio
NBC Yr)
PC A Yr) (centroid GAB C) x (l B P) (I A N» Y (I B P) (I C M» z (I A N) (I C M»
(inter (inter (inter (inter L (I
(""=-1» ZL
Y Z)
(I x
G»
F
Figure 6-186
344
Chapter 6. Topics from Geometry
The machine proof -(~)
.f.. -
-Saxy (sGxz)
~ -SqXy,SACNM
-
1::. -
SCMX ,SANG -SQMx oSapo -SACNM
SCMX ,SANG ,SBCPM
simJ!!.ify -
-SBPO-SACNM
SANG ,SBCPM
Q -(-SBCP+SABP),SACN W(3) -
(-SACN-SABN),SBCpw(3)
E.
-(2SABC'YT-SABC),SACNM (SACN+SABN),(SBCM SACM 'YT)
-
sim!1:ify
!:!.. -
-(2YT-l) ,SABC,SAQNM (SACN+SABN) ,(SBCM-SACM'YT)
The eliminants ~!!&=:. ZL - SGXZ S !SCMX-SANG GXZ- SACNM S y SCM ",SRPG GXY SBCPM SANGg -
t(SACN+SABN)
SBPGg -
a(SBCP-SABP)
SBCPM!:,SBCM-SACM 'YT SABP!:, -
(YT-I),SABC)
SBCP!:,SABC'YT
SABN~SABC'YT SACN~(YT-I).SABC
SACNM~SBCM 'YT-SBCM+SACM SACM~ - (SABC 'YT) SBCM~ - (YT-I),SABC)
-(2YT-I),SABC,(SBCM'YT-SBCM+SACM) (2SABC 'YT-SABC) ,(SBCM-SACM 'YT)
simJ!!.ify -(SBCWYT-SBQM+SAQM) SBCM-SACM 'YT
M -
-(-SABC'Y,f+SABC'YT-SABC) SABC ·Yj. SABC'YT+SABC
sim!1:ify I
Example 6.187 (0.533, 2, 32) Through each of the vertices of a triangle ABC we draw two lines dividing the opposite side into three equal parts. These six lines determine a hexagon. Prove that the diagonals joining opposite vertices of this hexagon meet in a point. Constructive description ( (POints A B C) (lratio AI B C 1/3) (!ratio A2 B C 2/3) (!ratio BI C A 1/3) (!ratio B2 C A 2/3) (lratio CI A B 1/3) (!ratio C2 A B 2/3) (inter x (I A All (I B B2» (inter Y (I C CIl (I A A2» (inter z (I B BIl (I C C.» (inter u (I A A2) (I B Bd) (inter v (I C C.) (I A Ad) (inter w (I B B2) (I C Cd) (inter
C.
c,
Figure 6-187
Example 6.188 (0.216, 4, 11) Let M. N. P be points on the sides AB. BC and AC of a triangle ABC. Show that if MI. NI and PI are points on sides AC. BA. and BC of a triangle ABC such that M MI II BC. N NI II C A and P PI II AB. then triangles M N P and MINI PI have equal areas.
U
345
Trluctes
Constructive description ( (POints A B e) (on M (I A B» (on N (I Be» (on p (I A e» (inter M, (I A e) (P M B C» (inter N, (I A B) (P N A C» (inter p, (I B e) (P p A B» (SMNP
=
SM,N,P,»
The eliminants ~ SOM,N, ,SABP+SBM,N, ,SBOP
S
M,N,~-
SABO
N, SABM, ,SABN N, SBM,N, SABO SOM,N,
=
SABM,
=-
~'-SAOM SONM,~'
-
(
) SONM,
(SOMN)
«~-l).SABO) SABP!:'SABO' ~ AO
SBOP!:, -
SMNP!:'SONN· ~-SAMN· ~+SAMN SABN~SABO' ~ SANN~ - (SAON ' ~) SOMN~ - «~-l).SBOM)
The machine proof ..bui..L SM 1 NI PI
~
SMNpo(-SABO) SOM,N"SABP
SBM,N"SBOP
~
SMNpo(SABO)' -SONM"SABP ,SABo+SBorSABM"SABN
'1J
SMNP ,(S6BO)' -( -SOM N · SABP ' ~BO+SBOP · SAOM ·SABN ·SABO)
-
.im~iJII
SMNp .(S,
SBOP,SAOM ,SABN
A
(SOMN · ~-SAMN · ~+SAMNHSABO)' de de
p :; So M
N'S~BO ' ~+S AOM ' S ABN ' S ABO' ~-SAO M ' S ABN'SABO
M
Figure 6-188
(SOM N · ~-SAM N · ~+SAM N),SABO "'"
dC
SOMN · SABO · ~+SAON ·SABN·~-SAOM ·SABN N :;
(-SBOM·U · ~+SBOM,~+SAOM·U · U)·SABO ~Aa Ae BC · ~-SAOM de 8C
.im=pliJII
-SBOM ·SABO · ~·~+SBOM·SABO·~+SAOM ·SABO · ~ · ~-SAOM ·SABO ' ~
Example 6.189 (0.050, 2,3) Three parallel lines drawn through the vertices of a triangle ABC meet the respectively opposite sides in the points X. Y. Z. Show t/wt area XY Z 1 area ABC = 211.
Constructive description ( (POints A B e) (on x (I Be» (inter Y (I A e) (P B A X» (inter z (I A B) (P e A X» (2SBAO
=
SXYZ) )
Figure 6-189
346
Chapter 6. Topics from G_try
Example 6.190 (0.066, 1,6) Two doubly perspective triangles are infact triply perspective_
Constructive description ( (POints ABC 0 0,) (inter A, (1 C 0,) (I A 0» (inter B, (I A 0,) (1 B 0» (inter C, (1 B 0,) (1 C 0» (inter 0, (I B A,) (I A C,» (inter Zo~, C) (I B A,» (~~ BZOa» .4.02
A1Z02
The eliminants
The machine proof
BZo, z£, -SBCB, A,ZO, - SCA,B, ~~ SABC,
(~)/( BZOa ) A 10 2
Z.£, -
AIZ0 2
SCA, B, _~ -S8CB!
~
.4102 -
AA1C 1 -
(-SABC,)-SCA,B, -SBCB, -( -SAA,C,)
-
ABC, -
SBCB,-(-SBOO,-SACA, )-SBCO,O -
SBCO,O
'!! SBCO -SABO,
S
BCB, -
SABO,O
'!! -SBOO, -SACA,
S
-SOA) Bl
S800 0 1
<::J SBCO -SABO,
S
~ -SBCO -SABO, -SCA,B,-(-SBCO,O)
aim1!!i!y -SaGo ,SABO)
SAA,C J
<::J SBOO, -SACA,
S
.4 1 °2
CA,B, -
SABO,O
SBCB, -SBOO,-SACA,
~ -SBCO-SABO, -(-SBOO, -SACA, H-SABO,o)
-
(-SBCO-SABO,)-SBOO,-SACA,-SABO,O
A
Figure 6-190
6.3.8
Equilateral Triangles
Example 6.191 (The Napoleon triangle) (0.433,8,22) Ifequilateral triangles are erected externally (or internally) on the sides of any triangle, their centers form an equilateral triangle_ The machine proof POaO,O. po,o.o, ~ ~____~~PO~'LO~,~O~.~~____ PO,GO,+!PBGB -r·-tSBO.G-r
g
(9) -Po,o, o. JPBO.B+JPAO.A+f PABA-r'-¥PABA-12SABo,-r
~
(36) -(PO,FO,+}PcFc -r'-t SCO.F-r) lSPBO.B+lSPAo.A+PABA-r' 9PABA-4SSABo, -r
E.
(4)-(!PcO.C+!PBOaB+tPBcB-r'-tPBcB-12SBco.-r) lSPBO.B+1SPAO.A+PABA-r' 9PABA 4SSABo,-r
-
~ 4PCAR,, -r'tlSPCI>ctlSPBEB+PRGR-r'-9PRcB+4PgA -r' 4SSRcr;;-r 4SSAB,,-r lSPBEB+4PBAE-r'+4PAEA-r2+1SPAEA+PABA-r' 9PABA 96S A BE-r K 3Pscs ·r 2tP'CA ·,-2 2PABc·r2t9PABA 48SAAC'r - 9PBCB+2PBAc-r'+PACA-r'+PABA -r'-4SSABc-r ~. 9PaCB±3PACA-6PARC±9PABA-48SA8c,r 9PBC B+6PBAC+3PACA +3PABA -4S8 ABC-r ~ (4PBCBHPACAHPABr32SABC-r)-(2) .imJ!!.illl - (4PBCB+4PACA+4PABA-32SABC-r)-(2) -
6.3.
347
Triucla
Constructive description ( (points A 8 C) (constant .'-3). (midpoint E A C) (tratio 0, E A (midpoint F 8 C) (tratio 0, F C (midpoint G A 8) (tratio o. G 8 (eqdistance 0, 0,0, 0.) )
The eliminants po,o.o, 0.1( = 9 9Po,OO,+PB08o.'-24SBo,o or ) SBO,Og~(SABO,). P808 g H PA8A)
tr) tr)
Po,oo,gi( 2P80,8+ 2PAO,A-PABA)
tr)
Po,o,o, ~ ~(9PO,FO,+PCFC ·.'-24Sco,F ·r) SCO,F!;~(S8CO,). PCFc!;H p8C8) Po,Fo,!;i(2Pco,C+ 2P80,8- P8CB) SABO, '2
-
f2(PBAs ·r-12SABS)
PAO,A '2~(.'+9). PAEA) 1 ( S8CO, 0, = - 12 PCABE·r-12SBCE ) P80,B 0,1 = 9 ( 9PBEB+PAEA ·.'-24SA8E or )
PCo,C '2 ~(9PCEC+PAEA ·.')
P8AS~t(PBAC)' SA8E~t(SA8C)'
S8CE~1(SA8C)' PAEA~4(PACA) P8EB~t(2PBCB-PACA+2PA8A)
PCEC~4(PACA)' PCA8E~~(PBCB-PABC) P8AC= - ~(P8CB-PACA-PA8A) PA8C=~(PBCB-PACA+PABA) Figure 6-191
Example 6.192 (0.883, 15, 31) Continuing from the above example. the lines A01• B02• C0 3 are concurrent.
Constructive description ( (constant .'-3) (POints A 8 C) (midpoint E A C) (tratio 0, (midpoint F 8 C) (tratio 0, (midpoint G A 8) (tratio o. (inter N (I 0, 8) (I 0, A» (inter M (I 0, 8) (I o. C» (~
E A tr)
F C tr) G 8
H
= -H!f;» Figure 6-192
Example 6.193 (0.900,18,31) In Example 6.192. if the three equilateral triangles become similar isosceles triangles. the conclusion is still true. Constructive description ( (POints A 8 C) (midpoint E A C) (tratio 0, EAr) (midpoint F 8 C) (tratio 0, F C r) (midpoint G A 8) (tratio o. G 8 r) (inter N (I 0, 8) (inter M (10,8) (10. C» (~ = -H!f;) )
(10, A»
348
Chapter 6. Topics from c-try
Example 6.194 (0.367 26 54) Let the circumcenters of the equilateral triangles erected externally (or internally) on the sides of any triangle be 0 11 0 2 and 0 3 (Xl. X 2 • and X 3 ) _ Show that So,o.o. + Sx,x.x. = SABCConstructive description ( (POints A B C) (constant r 2 -3) (midpoint E A C) (tratio o. E A tr) (lratio x. EO. -1) (midpoint F B C) (tratio 0, F C tr) (lratio x, F 0, -1) (midpoint G A B) (tratio o. G B tr) (lratio x. GO. -1) (So,o.o.+Sx,x.x. =
SABC»
Example 6.195 (0.050, 5, 8) Let equilaterals BCD. ABF. and ACE are erected externally on the sides of triangle ABC_Show that AD = CF = BE_ Constructive description ( (POints A B C) (constant r'-3) (midpoint M A C) (tratio E MAr) (midpoint NBC) (tratio D NCr) (eqdistance B E A D) ) The machine proof .E.a.E..a. PADA
!:!.. -
!d. -
M. -
The eliminants PADAgPCNC -r'+PANA+SSACN -r SACN!:!, -
t(SABC)
PCNC -r'+PANA+SSACN-r
PANA!:!, -
i(PBCB-2PACA-2PABA)
PUR f P BCB -r'-f PBCB+t PACA+t PABA-4SABc-r
PCNc!:!'i( PBCB)
Q -
Figure 6-195
PlffjR
(4)-(PBMB+PAMA -r' SSABWrl PBcB -r'-PBCB+2PAcA+2PABA-16SABc-r (4)-(tPBCB+tPACA -r'-tPACA+tPABA-4SABc-r) PBCB-r'-PBCB+2PACA+2PABA-16SABc-r
PBEB~PBMB+PAMA-"-SSABM-r
SABM~t(SABC) PAMA:i(PACA) PBMB=i(2PBCB-PACA+2PABA)
~. 2PRCR+2PACA+2PABA 16Suc-r .im~ifll 1 2PBCB+2PACA+2PABA-16SABc-r
6.3.9
Pedal Triangles
Definition. From a point P three perpendicular lines are drawn to the three sides of a triangle. The triangle whose vertices are the three feet of the three perpendicular lines is called the pedal triangle of point P with respect to the given triangle.
Example 6.196 (0.783, 6,23) The orthogonal projections from point D to BC. AC, and AB are E. F. and G respectively_ Let 0 be the circumcenter of triangle ABC. Show that
Dd = MT(l -
~)_ SABC
349
c
Constructive description ( (POints ABC D ) (circumcenter 0 A B C) (foot E D B C) (foot FDA Cl (footG D A B) (-W .SASC = -oK ,S A scHo:4' ,SEFO»
B
Figure 6-196
Example 6.197 (0.083, 2, 5) Let K be the area of the pedal triangle of the orthocenter of triangle ABC with respect to triangle ABC. Show that SK = P~~~a. ASC
4AB BC AC
Constructive description «POints A B C) (foot E A B C) (foot F B A C) (foot G C A B) G
(PASC ,PSAC ,PACS ,SASC = 4A!f .7il'?Al'?,SEFO) )
Figure 6-197
The eliminants
The machine proof P8AC · P'CB , P'BC ,S'BC
(¥) ,PSCS ,PACA ,PASA ,SEFO
Q -
PSAC , P'CB ,P'BC ,S'RC ,P'RA
(tl,PSCS ,PACA ,PASA ,(PSAC ,SSEF+PASC ,SAEF)
.'m~i/"
PSAO ,P'CS ' P'SQ ,S'BC
(t) ,PSCS ,PACA ,(PSAC ,SSEF+PASC ,SAEF)
!'.. -
PSAC ,PACS ,PASC ,SASd PAQ6l' (t) ,PSCS ,PACA ,(PSAC ,PACS ,PACA ,SASE-PSAC,PACA ,PASC ,SACE)
.im}!!i/ll PACS ,PABO ,SABG (¥l ' PSCS ,(PACS ,SASE-PASC,SACE) ~
-
PACS ,PASC,SASdPscsl' (fj' Pscs ·(2Pscs ,PACS ,PA SC ,SASC)
. imglifll 1
Example 6.198 (0.233,14,16) Let K be the area of the pedal triangle of the centroid of triangle ABC with respect to triangle ABC. Show that ..JL A!f ±7il'? ±Al'? • _ -__ c SASC 36Jl2
=
Constructive description «POints A B C) (circumcenter 0 A B C) (centroid GAB C)
~-------L----~B
(foot EG B C) (foot F G A C) (footGG A Bl (Alf,S ASC ±7il'?,SASC±Al'?,SASC =
360K SUC»
Figure 6-198
Chapter 6. Topics from Geometry
350
Example 6.199 (0.083,2,11) Let K be the area of the pedal triangle of the circumcenter of triangle ABC with respect to triangle ABC. Show that SABC = 4K.
The eliminants
Constructive description «points A B C) (circumcenter 0 A B C) (foot EO B C) (foot FO A C) (foot GOA B) (SASC
s S S
G P8AQ , SAfif+P6SQ , SAfjf
EFO
PASA F
POIQ ,SAGE
AEF SEF
S
PACA F P'CO ,SARli
PACA
E PORO , SASO
ASE
= 4SEFC) )
S
PSCS E
PReO,SARO
ACE
The machine proof ~
Figure 6-199
(4) ,SEFC
Q
PSCS
PASOgt(PASA) Pscogl(PsCS) PCAogl(PACA)
SABe'PABA
-
(4),(PSAO,SSEF+PASO ,SAEF)
PACOg¥(PACA)
.f..
SASC ,PASA,(PACA)' (4) ,(-PCAO ,PACA ,PASO,SACE+PSAO ,PACO ,PACA ,SASE)
PBAOg¥(PASA)
-
sim~ifY
Pcsog2(PSCS)
SARC , PARA , PACA
(-4) ,(PCAO ,PASO,SACE-PSAO ,PACO,SASE)
K -
SASC ,PASA ,PACA '(PSCS)' (-4) ·( -PCSO 'PSCS ,PSAO ,PACO ,SASC-PCAO,PSCO,PSCS ,PASO ,SASC)
simg1ify
PABA , P'CA,PRCB
(4)-(PCSO ,PSAO ,PACO+PCAO ,PSCO ,PASO)
Q PAU ,PACd ,PScs ·«2))• • imJ!!.ify -
(4)·(l6PscS ·PACA ,PASA)
-
Example 6.200 (0.300, 2, 14) Let K be the area of the pedal triangle of the incenter of triangle ABC with respect to triangle ABC. Show that SA~C =
;R'
Constructive description «points A B I) (incenter CIA B) (circumcenter 0 A B C) (foot E 1 B C) (foot F 1 A C) (foot G 1 A B) (Tif .s~SC
6.3.10
= 4o:::1"Shc) )
Miscellaneous
Figure 6-200
Example 6.201 (0.016, 1,4) AD, AAl are the altitude and the median of the triangle ABC; the parallels through Al to AB. AC meet AD in P, Q; show that (ADPQ) = -1.
351
Constructive description ( (POints A B C)
(foot
0 A B C)
(midpoint A, B C) (inter p (1 A 0) (P A, A (inter Q (l A 0) (P A, A (harmonic A 0 P Q) )
;1S.~ SACA, ~!. SABA, DQ- SADA, DP- SADA,
(-~)/(~)
g
-SADA! • _~ -SACA, DP
!..
-(-SABA, )·SADA, SACA, ,(-SADA,)
B»
C»
TIle elirninants
TIle machine proof
-
SACA,~ - !(SABC) SABA,~!(SABC)
.im1!!i/1I -SABA, -
SACA,
~ -(iSABC)
p
-.SABC
.im1!!i/1i -
1
B
Figure 6-201
Example 6.202 (0.067, 3, 13) If A l • B l • C l are the midpoint of the sides of the triangle ABC. prove that AlA is the harmonic conjugate of A1C with respect to A1Bl> A1Cl . Constructive description «POints A B C) (midpoint A, C B) (midpoint B, A C) (midpoint c, B A) (centroid GAB C) (inter 0 (1 c, C) (I A, B,» (harmonic c, 0 G C) )
AI
Figure 6-202
TIle machine proof Q (-SA,C,B,O) ,SCA,B, -
SCA,C,B"SA,B,O
Q -( -3SA, B,C, +SCA, B, +SBA, B, +SAA,B, ),SCA, B, ·(3) -
SCA,C, B, ,(SCA,
B, +SBA,B, +SAA,B,
)·(3)
Chapter 6. Topics from G«metry
352
Example 6.203 (0.250, 1, 12) Let P be a point in the plane of the triangle ABC. Let Al = BC n AP,BI = AC n BP,CI = AB n CP, A2 = BC n B ICI, B 2 = AC n A I CI ,C2 = AB n AIBI. Show that A 2,B2 ,C2 are collinear, and is called the trilinear polar of P for triangle ABC. Constructive description ( (POints A B G P) (inter A, (I B G) (I A P)) (inter B, (I A G) (I B P)) (inter G, (I A B) (I G P)) (inter A, (I B G) (I B, GIl) (inter B, (I A G) (I A, GIl) (inter G, (I A B) (I A, B,)) (inter Zc~, A,) (I A, B,)) (~=A,ZCa)) Siel
Figure 6-203
BIZC 2
The machine proof (~)/(A,ZC3 81G2
Z.E, -
)
B.ZC'l
-SB)AaB3 •
~
-5"'IA282
8 1 °2
<2a
SABA]
-
SA1A'JB'l ,SABB I
·58, A'18 7
~ SASA"SAtStO, , SACA],(-SAA)CC 1 -
)
(-SA,C,A"SACA,),SABB"SAA,CC,
simJ!!ify ~
SABA) ,SA) 8, 0, ,SACA]
SA,C,A"SACA"SABB, SABA),SA,B,C"SCB,Ct ,SABC,SSB)CC)
-
(-SA,B, C, , SBCC, ),SACA, ,SABB"SBB,CC,
8imJ!!ify -
~
SABA, ,SCB)C) ,SABC -SBCC, ,SACA, ,SABB,
SABA) .( -SCP B) ·SABC ),SABC'SAC B P -(-SBCP-SABC),SACA"SABB, ,(-SACBP)
sim1!!.ify SABA) ,SCPB) ,SABC -
SBCP ,SACA, ,SABB,
~ SABA) ,SBCP-SACP,SABC, SABCP
-
SBCP ,SACA"SABP ,SABC,SABCP
.imJ!!ify SABA) ,SACP -
1! -
SACA"SABP (-S6BP-SABC),S6CP-(-SABPC) (-SACP-SABC),SABP,(-SABPC)
sim!1 ify I
Example 6.204 (0.083, 1, 6) Let P be a point in the plane of the triangle ABC. Let Al = BC n AP,BI = AC n BP,CI = AB n CP, A2 = BC n BICI. Show that AI, A 2 , B, C form a harmonic sequence.
353
6.3. Trtu&la
Constructive description ( (POints ABC P) (inter A, (I B C) (I A P» (inter B, (I A C) (I B P» (inter c, (I A B) (I C P» (inter A. (I B C) (I B, C,» (harmonic
1be machine proof (-~)/(~) CAl
CAl
~ SCB,C, • _~ SBB,C,
@
B C A, A.) )
CA,
-(-SCPB,)·SABC ·SACBP • ~ (SBCP"SABB,H-SACBP) CA,
e,m!!1'!11
SCPB, ·SABC • ~ SBCP"SABB, CA,
~ Sacp ·$,cP ,S'RC ,S'8CP . ~ SBCP"SABP"SABC ·SABCP
e'' '!!:.'!11
CA,
§...M;;z..~ SABP
CA,
~ (-SABP) ·SACP SABP"(-SACP)
"m!!1 ifll
1
Example 6.205 (0.033,1,6) With the usual notations for the triangle ABC, if EF meets BC in M, show that (BCDM) = -1.
Constructive description ( (POints A B C) (foot (foot (foot
DAB C) E B A C) FC A B) (inter H (I A D) (inter K (I D F)
1be eliminants
1be machine proof
1IJIl.~5..a.s.L S
(-~)/(~)
CM - SCSF'
S
M~.-u. SBSF
(I C
.!.
(I (I E
-
F» B E»
(inter M (I
8 C)
(harmonic
BCD M) )
F»
CEF ABE
-PRIG -SaCE 'P'AA . PABC·SABs ·PABA
o''''!!:.''11
III
PABC ·PBAC ·SABC · PACA
o,m!!1"11
BCS
PABA PAC./I SPACS·SUC· PACA CD - -PACB
u.!!.h.ac....
C
-PUC· Ssm; • U. PABC ·SABE CD
~ -Pa,c ·P,qa ·S,ac,P.4af -
S
CD
F PUC ·SUE PABA
EP".sC ,S ' 8 C
S
CD
BSF
F P'",C,S'CF
. U. CD
=l?Ma. U. PABC
CD
Q -P"ac·P",c a -
PABC ·(-PACB)
"m!!1'!11
1
M
Figwe6-20S
Example 6.206 (0.467, 13, 34) The circle having for diameter the median AAI of the triangle ABC meets the circumcircle in L: show that A(LDBC) = -1, where AD is the altitude.
Chapter 6.
354 Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot DAB C) (midpoint A. B C) (midpoint M A. A) (inter L (cir M A) (cir 0 A» (inter K (I B C) (I A L» (harmonic K D B C) )
Toplca from Geometry
Figure 6-206
Example 6.207 (0.883, 10,38) If L, M, N are tire traces of tire lines AP, B P, C P on the sides BC, C A, AB of tire triangle ABC, and L 1, M 1, Nl tire traces, on tire same sides, of the trilinear polar of P for ABC, show that tire midpoints of tire segments LLI> M Mlo N Nl are collinear. Constructive description ( (POints ABC P) (inter L (I A P) (I B C» (inter M (I B P) (I A C» (inter N (I C P) (I A B» (inter L. (I B C) (I M N» (inter M. (I A C) (I L N» (inter N. (I A B) (I L M» (midpoint L. L Ld (midpoint M. M M.) (midpoint N. N Nd (collinear L. M. N.) )
Figure 6-W7
Example 6.208 (0.550,2,25) Let L, M, N be tlrefeet of tire cevians AP, BP, CP of the triangle ABC and let PI be a point on tire trilinear polar of P for ABC. If tire lines API BPI> CP1 meet MN, NL, LM in X, Y, Z, show that tire triangle XYZ is circumscribed about tire triangle ABC. Constructive description ( (POints ABC P) (inter L (I A P) (I B C» (inter M (I B P) (I A C» (inter N (I C P) (I A B» (inter L. (I B C) (I M N» (inter N. (I A B) (I L M» (on p. (I N. Ld) (inter Y (I N L) (I B Pd) (inter ZJll..M) (I c I'J» (inter Zz (I Y A) (I N. L» (~=~» LZ LZz
Figure 6-208
Example 6.209 (0.466, 4, 26) If Al is tire point of intersection of tire side BC of the triangle ABC with tire trilinear polar p of a point P on tire circumcircle of ABC, show that tire circle APA 1 passes through tire midpoint of BC.
355
Constructive description ( (circle ABC P) (circumcenter 0 A B C) (inter M (I B P) (I A C» (inter N (I C P) (I A B» (inter A, (I B C) (I M N» (midpoint G B C) (cocircle A P A, G) )
Figure 6-209
Example 6.210 (0.050, 1,8) Let ABC be a triangle with LB altitude on C B and M the midpoint of B and C. Show that AB
Constructive description ( (POints B C) (midpoint M B C) (tratio x
The eliminants u£~ BC -
M B H)
5
(foot Y C B X) (iratio Z Y C -1) (inter A (I B Z) (i C X» (inter D (I B C) (P A M X»
SBCX
ASCHy ,SRKZ
MXA
-SBCZX
u~.bua.... BZ-SBCZX
SBXZ~SBCX
(fi=2'H»
.im!ii/ll
u ~ g (-2). ~
-(SSex)' (2),SCMX 'SSXZ
.i"'!ii/" ..=.lXJJ.M....!!!. (2) ,PSMC
-
.£. -
M
Figure 6-210
SCMX~
-
HPBMC ·H)
Sscx~
-
HpCBM ·H)
PSMC~ -
TIle machine proof:
= 2LC. D the foot of the = 2D M.
Hpscs)
Same . U (-2H-SMXA) SZ
PCSM~~(Pscs)
4
(Sscx)' ·(-Ssczx) (2) ,SCMX 'SSXZ'Ssczx
-(Sscx)' .i"'!ii/ll ~ l!. -(-tpcSM ·H) (2)·SCMX'SSCX (2),SCMX - (2H-fpSMC ·H)
-
.im!!!.i/" -
1 .
Example 6.211 (1.117, 33, 38) The three adjoint circles of the direct group have a point in common. Constructive description ( (POints A B C) (inter A, (t C C A) (b B C» (inter B, (t A A B) (b A C» (inter c, (t B B C) (b A B» (inter N (cir c, B) (cir A, B» (perp-biesct B, C N) )
Figure 6-211
Example 6.212 (2.383, 68, 47) The three adjoint circles of the indirect group have a point
in common.
356
Chapter 6.
Constructive description ( (POints A B C) (inter A, (t B B A) (b C B)) (inter B, (t C C B) (b A C)) (inter c, (t A A C) (b B A)) (inter M (cir c, B) (cir A, B)) (perp-biesct B, C M) )
Topics from G--,.
Figure 6-212
Definition. The points N. M in Examples 6.211 and 6.212 are called the Brocard points of the triangle. Example 6.213 (1.000, 12, 50) The Brocard points are a pair of isogonal points of the triangle. Constructive description ( (POints A B C) (inter A, (t C C A) (b B C)) (inter c, (t B B C) (b A B)) (inter A, (t B B A) (b C B)) (inter B, (t C C B) (b A C)) (inter M (cir c, B) (cir A, B)) (inter N (cir B, C) (cir A, C))
(eqangle
A B M NBC) )
Figure 6-213
Example 6.214 (1.600,27,52) The two Brocard points of a triangle are equidistant from the circumcenter of the triangle. Constructive description ( (POints A B C) (circumcenter 0 A B C) (inter A, (t C C A) (b B C)) (inter c, (t B B C) (b A B)) (inter A. (t B B A) (b C B)) (inter B, (t C C B) (b A C)) (inter M (cir CI B ) (cir Al B)) (inter N (cir B, C) (cir A, C)) (perp-biesct 0 N M) )
c
B
Figure 6-214
Example 6.215 (0.683, 12,34) For the Brocard point in Example 6.211 we have LN AB LN BC = LNCA. Similar for the Brocard point in Example 6.212.
=
6.4.
357
QuadrIlaterals
c•
.~
Constructive description ( (POints A B C) (inter A, (t C C A) (b B C)) (inter c, (t B B C) (b A B)) (inter N (cir c, B) (cir A, B)) (eqangle NAB NBC) )
A.
Figure 6-215
Example 6.216 (0.016. 1. 6) Let D and E be two points on two sides AC and BC of triangle ABC such that AD = BE. F = DEn AB. Show that F D . AC = E F . BC.
Constructive description ( (points A B C) (on D (I A C )) (inter E (I C B) (cir B iiD')) (inter F (I E D) (l A B)) (eq-product FDA C E F B C)
The machine proof
The eliminants
POfO , PACA
P
~
PDED ' StRD ,PACA ' StDRg
PDED 'S ABE,PBCB 'S ADBE )
.im1!li/y (SABD)' ,PACA (SABE)',PBCB
!l. -
!. PDED ,(SABE)' (SADBE)'
EFE-
PEFE ,PBCB
(S,Up)' ,PAQA,PBCB
PADA '~BC'PBCB
.im1!li/1I (SABpl' ,PACA PADA ,(SABC)'
P
F PPEp ,(SABpl' DFD (SADBE)'
SABE~
-
(~ 'SABC)
EMu..
BE2 _
Be -
PBCB
PADAgPACA '(~)'
SABDgSABC'~
g (SABC'~)"PACA .im~i/y PACA ' ~ ,(SABC)' E
6.4 6.4.1
Quadrilaterals General Quadrilaterals
Example 6.217 (0.050. 2. 4) The figure formed when the midpoints of the sides of a quadrilateral are joined in order is a parallelogram.
Constructive description «POints ABC D) (midpoint E A B) (midpoint (midpoint G C D) (midpoint
F B C) H D A)
I~='))~:
The machine proof
The eliminants
u.!!..~
u.!!~
FG
Q -
E
Figure 6-217
B
SAFDG
FG- SAFDG
-SA~g
SAFDGg -
-SADF-;SACD
E.~
!l. A
-
-SABD (-2J.(-tSABD) SABD
.im£!i/y
SADFf, -
SADE~
-
~(2SADF+SACD) ~(SACD+SABD) ~(SABD)
Chapter 6.
358
Topics from Geometry
Example 6.218 (0.083, 5, 13) The area of the parallelogram whose vertices are the midpoints of the sides of a quadrilateral is equal to half the area of the given quadrilateral (Figure 6·217).
Constructive descrip· The machine proof (2) ·SEFGH tion SABCD ( (points ABC D) Ii (2) .(SEFG+tSDEG+tSAEG) (midpoint E A B) SA BCD (midpoint F B C) Q SDEF+SCEF+tSCDE-tSADE-tSACE SABCD (midpoint G C D) K 2SCDEtSsog+Sacg-SAog-SAGE (midpoint H D A) (2SEFGH
= SABCD) )
-
(2)·SABCD
~ Sacp+SAQp+SA8Q+SABC
-
(2) ·SABCD
are.!!...-co 2SAQO±2SARQ (2) ·(SACD+SABC)
.imgi/y
The eliminants SEFGH~t(2SEFG+SDEG+SAEG) SAEGg-t(SADE+SACE) SDEGgt(SCDE) SEFGgt(SDEF+SCEF)
SCEF~t(SBCE) SDEF~t(SCDE+SBDE) SACE~-t(SABC) SADE~-t(SABD) SBCE~!(SABC) SBDE~t(SABD) SCDE~t(SBCD+SACD) SABCD=SACD+SABC SBCD=SACD-SABD+SABC
Example 6.219 (0.067,3,13) The lines joining the midpoints of the two pairs of opposite sides of a quadrilateral and the line joining the midpoints of the diagonals are concurrent and are bisected by their common point.
Constructive description ( (points ABC D) (midpoint E A B) (midpoint G C D) (midpoint N A C) (midpoint M B D) (inter 0 (I E G) (1 N M» (midpoint 0 N M) )
The machine proof
I:!!l.£~
-~ MO
MO-SEGM
Q=hi= -
SEGM
M.
-SEON
-
tSDEG+tSBEG
t!..
(-2)·(tSCEG+tSAEG) SDEG+SBEG
-
Q -(-tSCDE-tSADE-tSACE) -
fSCDE-fsBDE-fsBCE
~ tSBCD+tSACD-1SABD-tSABC - tSBCD+ t SACD-tSABD-tSABC
.imEii/y Q
The eliminants SEGM~t(SDEG+SBEG) SEGN~t(SCEG+SAEG) SBEGg-t(SBDE+SBCE) SDEGgt(SCDE) SAEGg-t(SADE+SACE) SCEGg-t(SCDE)
SBCE~t(SABC) SBDE~t<SABD) SACE~-t(SABC) SADE~-!(SABD) SCDE~t(SBCD+SACD)
Figure 6-219
Definition. The intersection of the lines joining the midpoints of two pairs of opposite sides of a quadrilateral is called the centroid of the quadrilateral.
Example 6320 (0.050, 5 6 ) The four lines obtained by joining each vertex of a quadridetermined by the remaining three vem'ces are lateral to the centroid of the
concurrent at the centroid of the given quadrilateral.
Constructive description ((points~~c~) (centroid A, B c D ) (midpoint N A c ) (midpoint M (inter J (I A A I ) (1 N M ) ) (midpoint J N M ) )
The machine proof B D)
The eliminants N
-E r
SAA,M M I S A A ~ M-=1 ( ~ ~ ~ ~ 1 + N 1 S A A ~ N-=I(SACA~) sABAl5 f (SABD+SABC) MJ
- -SAA,N
S A AM ~
M -
-SAA~N
-4 S A D A ~ - ~ S A B A ~
N (~).(-$SACA, ) SADA~+SABA~
sADA1 A=' - ~(SACD+SABD) 4 -(SACD-SABC).(3)= sAcAI3 $(sAcD-sABc)
-
-
(-~SACD+~SABC)'(~) simplif r
-
R
1
Figure 6-220
Example 6.221 (0.083,4,3) The sum of the squares of the sides of a quadrilateral is equal
to the sum of the squares of the diagonals increased byfour times the square of the segment joining the midpoints of the diagonals. Constructive description ( (points A B c D ) (midpoint E A c ) (midpoint F B D )
The machine uroof
(AB'+CB'+CD'+DA' = I
---.I
AC +BD
The eliminants
p
+4sl)
~PEFE+PBDB+PACA
F -
Pc~c+P~PDED+~PBEB+PACA
Ep,,,+p,,,+p,,,+p,,,
PEFE~?(~PDED+~PBEB-PBDB)
pBEBEf ( ~ P B C B - P A C A + ~ P A B A ) pDEDg?(2pCDC+2pADA-PACA)
PCDC+PBCB+PADA+PABA
Example 6.222 (0.033,7,6) The sum of the squares of the diagonals of a quadrilateral is equal to twice the sum of the squares of the two lines joining the midpoints of the two pairs of opposite sides of the quadrilateral.
Constructive description ((points~~c~) (midpoint P A B ) (midpoint Q B c) (midpoint s D A) (midpoint R c D )
(AC.+BD.= 2 $ + 2 r n )
)
Figure 6222
360
Chapter 6. Topics from Geometry
The machine proof
The eliminants
(t )-(PBDB+PACA)
PPRP~t<2PDPD+2PCPc-PCDC ) PQSQ~!(2PDQD+2PAQA-PADA) PAQA~Lt(PBCB-2PACA-2PABA)
PQSQ+PPRP R
(l) _(PBDB+PACA)
= PQSQ+tPDPD+tPCpc-fpCDC
PDQD g t<2PCDC+2PBDB-PBCB)
§...
PCPc~!(2PBCB+2PACA-PABA) PDPD~t(2PBDB+2PADA-PABA)
-
2. -
J:.. -
(2)-(PBDB+PAC6) 2PDQD+2PDPD+2PcPc-PCDC+2PAQA-PADA (2)-(PBDB+PACA) 2PDPD+2PcPC+PBDB-PBCB-PADA+PACA+PABA (2) -(PBDB+PACA) simJ!!ify 1 2PBDB+2PACA -
Example 6.223 (0.050, 4, 10) If a quadrilateral ABC D has its opposite sides AD and BC (extended) meeting at W , while X and Yare the midpoints of the diagonals AC and BD, then 4SWXY = SABCDConstructive description ( (POints BCD A) (midpoint x C A) (midpoint Y B D) (inter w (I B C) (I D A»
c
(SBCDA = 4Sxyw) )
The machine proof 2.B.=.L (4) -Sxyw
!:!:: -
SReDA,SROCA
(4)-(SCXy -SBDA-SCDA -SBXY)
r
SRODA ,SaOCA
-
(4)-(- tSCDX-SBDA+tSCDA -SBDX+tSBDA -SBCX)
2£
SReDA,SRoCA
w
Figure 6-223
The eliminants
s
W Scxy ·SaoA-SCorSsxy
XYW
SBDCA
SBXY~
-
t(SBDX)
SCXY~
-
2(SCDX-SBCX)
SBcx~t(SBCA)
SBDX=~( SBDA-SBCD) SCDX~2(SCDA )
- (2Hf S CDA -SBCD-!SBDA -SBCA)
SCDA=SBDA-SBCA+SBCD
are!!..-co
SBDCA=SBCA -SBCD
-(SBDA+SBCD)-(SBCA-SBCD) -SBDA - SBCA+SBDA - SBCD-SBCA - SBCD+S~CD
.im~ifY 1
SBCDA=SBDA+SBCD
Example 6.224 (0.216, 6, 12) Let P be a point on the line joining the midpoints of the diagonals of the quadrilateral ABCD_ Show that SPAB + SPCD = SPDA + SPBC. Constructive description ( (points ABC D) (midpoint N A C) (midpoint M B D) (lratio P N M r)
A~' Figure 6-224
361
6A. Qu8drilatenls
The eliminants
The machine proof
SADP~SADM or-SADN or+SADN SBCP~SBcM or-SBcN or+SBcN SABP~SABM or-SABNor+SABN SCDP~ScDMor-ScDN or+ScDN
ScnptSABP
SBCP-SADP
E.
Saow r-5CQN rtSgoNtSAAM r
-
SBCM or-SBCN or+SBCN-SADM or+SADNor-SADN
o
o
o
SABN-rtSA8N
M -SCDN or+SCDN+jSBCDor-SABNor+SABN+iSABDor
SADM~ - !(SABD), SBCM~!(SBCD) SABM~!(SABD)' SCDM~!(SBCD) SADN!i, - !(SACD), SBCN~!(SABC) SABN~!(SABC), SCDN~!(SACD)
-SBCWr+SBCN+, SBCD or+SADN or-SADN+, SABD or
!!... -
-Saco ortS'Quor-S,co SA8o rtSABc ·r SABe -SBCDor+SACD or-SACD-SABD or+SABc or-SABC o
.im!!fY~ J
Example 6.225 (0.083, 2, 10) Let ABC D be a quadrilateral. F and E be the midpoints of AD and BC respectively. M = BA n EF. N = CD n EFo Show that ~~ =
gZ
Constructive description ( (POints ABC D) (midpoint E B C) (midpoint F A D) (inter M (I A B) (I E F» (inter N (I C D) (I E F» (~=~»
The machine proof
The eliminants
(~)/(~)
1ZJi.!i.~ CN- SCEF .w~~ BM- SBEF
Ii.. i.c..u..
0
.w BM
-
SDEF
M.
SASE ,SORF
-
SDEF"SBEF
F (-lSADEH-1SCDE+1SACE) (f'sADEH-f'sBDE+f'sABE)
=
.im1!!i!1I -
-(SCDE-SACE) SBDE SABE
SACE~ - 2(SABC)
~ -(iSBCD+isABC)
SCDE~!(SBCD)
-,SBCD-,SABC
E
Figure 6-225
SBEF~- !(SBDE-SABE) SDEF~HsADE) SCEF~- i(SCDE-SACE) SAEF~ - 2(SADE) SABE~!(SABC) SBDE~ - t(SBCD)
C
Example 6.226 (1.333,20,37) Let ABCD be a quadrilateral with AB = CD. F and E be the midpoints of AD and BC respectivelyo M = BA n EF. N = CD n EFo Show that LBM E = LENC (Figure 6-225)0 Constructive description ( (POints ABC X) (inter D (I C (inter M (I A B) (I E F» (inter
X) (cir C N (I C D)
Aif» (midpoint (I
E F))
(eqangle
F A D) (midpoint E B C) B M E E N C) )
Example 6.227 (0.300, 3, 18) Let ABC D be a quadrilateral such that AB = C Do F and E are the midpoints of AD and BC respectivelyo M = BA n EF. N = CD n EF. Show that AM = DN.
362
Chapter 6. Topics from <>-try
Constructive description ( (points ABC X) (inter D (1 C X) (cir C Jilll» (midpoint F A D) (midpoint E B C) (inter M (I A B) (I E F» (inter N (I C D) (I E F» (eqdistance A M D N) )
The eliminants NPCDC ' SbFE
PDND
SbFDE
MPABAoS1 PE
,PAMA=
S~FBE
~(2SABF-SABC) SDFEf:~(SCDF+SBDF ) SCFDEf: - ~(2SCDF-SBCD) SAFEf: - ~(SACF+SABF) SAFBEf: -
SBDF;l(SABD),SCDF;t(SACD) SABF=2(SABD), SACF=2(SACD) SABDg -
(is.JJ. oSABC-f
SACDg - (~ OSACX)' SBCDg CD 2 _ L.u.. CD 2 _ L.u..
(~oSBCX)
C'X - Pcxc' C'X - Pcxc SABC=SABC,SBCX=SACX-SABX+SABC The machine proof ~ PDND
!%: PAMA oS~FD"
M PABAoS~f"o~fQ"
PCDcoSDFE
PCDcoSDFE oSAFBE
§.. PABAO( -tSACF-tSABF)' o( -SCDF+jSBCD)' -
E. -
PCDc o(tSCDF+tSBDF)'o( -SABF+. SABC)' PABA o(tSACD+tSABD)' o(-SBCD+SACD)' PCDco(f S ACD+f S ABD)2 (SABD-SABC)' 0
.imJ!!ify PABA o(SBCD SACD)' PCDc o(SABD-SABC)2
!2. -
E
Figure 6-227
(PABA o S~CI!:-2PABA o SBCx o SACX+PABAo~QK)oPABA o Pt,KQ
(PCXc o P1BA oSABX-2Pcxc o p1BAoSABx o SABC+Pcxc o p1BA oS~BC)o PCXC
. imJ!!ify (SBCX-SACX )2 (SAB X -SABC)2
are!!..-co (-SABX+SABC)' .im.1!!.iflJ (SABX-SABC)' 1
Example 6.228 (0.416,12,36) Let ABCD be a quadrilateral such that AB = CD. P and Q are the midpoints of AD and BC; Nand M are the midpoints of AC and BD. Show that PQl.NMo Constructive description ( (points ABC X) (inter D (I C X) (cir c Jilll» (midpoint P A D) (midpoint Q B C) (midpoint N A C) (midpoint M B D) (perpendicular N M P Q) )
Q
c
Figure 6-228
Example 6.229 (0.866,10,20) The sides BA, CD of the quadrilateral ABCD meet in 0, and the sides DA. CB meet 0 1 Along OA. ~C. OIA. OIC are measured off, respectively. 0
363
Constructive description ( (points ABC D) (inter 0 (I A B ) (I C D» (inter 0, (I A D) (I B C» (lratio EO B -4&) JJ.fl. (lratio F 0 C-~ (lratio E, 0, A - ~) AO, (lratio
F, 0, B -J!£.)
(SEFE,
= SEFF,)
80,
o
)
Figure 6-229
Example 6.230 (0.617,13,70) ABCD is a quadrikueraI. p. Q. R. S the midpoints of its sides taken in order. U. V. the midpoints of the diagonals. 0 any point; OP. OQ. OR. OS. OU. OV are divided in the same ratio in PI. QI. R I• Sj, UI. VI. Prove that PIR Io QISI. UIVj, are concurrent. Constructive description ( (POints ABC D 0) (midpoint P A B) (midpoint Q B C) (midpoint R C D) (midpoint S D A) (midpoint U A C) (midpoint v D B) (lratio P, 0 P YT) (lratio Q, 0 Q YT) (lratio R, 0 R YT) (lratio s, 0 S YT) B (lratio U, 0 U YT) (lratio v, 0 v YT) (inter J (I P, R,) (I Q, S,) (inter z, ll.fl R,) (1 U, v,) 'ter z,( I u, v,) (1 Q, S,) (v, ~ (m z, = ~ v,z,»
o
c
Figure 6-230
Example 6.231 (Theorem ofPratt-Wu) (5.133,72,123) Given a quadrilateral ABDC. let HE. EF. FC. CH be the tangents of circles CAB. ABD. BDC. DCA at A. B. D andC. respectively. Then HA· EB · FD· CC = AE· BF · DC · CH. Constructive description ( (POints ABC D) (midpoint L A B) (midpoint M B D) (midpoint N C D) (midpoint PC A) (inter x (t L L A) (t P P A» (inter Y (t L L B) (t M M B» (inter z (t M M D) (t N N D» (inter w (t N N C) (t P P C)) (tratio 0 A X /) (inter H (t C C W) (I 0 All (tratio U D z J) (inter F (t B B Y) (I U D)) (inter E (I B F) (I A 0)) (inter G (I D U) (I C H» (lH·* = %.~)
)
Figure 6-231
Chapter 6. Topics rrom Geometry
364
Example 6.232 (0.00, 1, 3) Let P, Q be the midpoints of the diagonals of a trapezoid ABC D . Then PQ is parallel to the two parallel sides of ABC D.
Constructive description «POints A B C) (on D (P CAB)) (midpoint Q A C) (midpoint P B D) (parallel P Q A B) )
The eliminants
The machine proof
SABP~!(SABD) SABQ~HSABC)
~
SABQ
E.
tSABD SABQ
-
2.
SABDgSABC
SARQ (2) ·(t S ABcl
-
;::z\
Q~ - SABC .implify
=
~
I
B
A
Figure 6-232
Example 6.233 (0.050, 2, 3) Let P, Q be the midpoints of the diagonals of a trapezoid ABC D. Then PQ is half of the difference of the two parallel sides of ABC D.
Constructive description ( (points A B C) (on D (P ABC)) (midpoint P B D) (midpoint Q A C) (~-~ PQ PQ
=2»
The machine proof
The elirninants
1(- -)
4,g~~ PQ- SACP
9..
I!!i.~~ PQ- SACP
2 ~-~
SAce ,SAGu-SAGP ,SARG
E.. -
SACP~!(SACD-SABC)
(2) . S~cp
lJimJ!l.i/Ji -
SAGO
SARG
T'Z\
(2) ,SACP
S1Re (2)'(f S ACD-;SABC) SAeD
~c
.imglify 1
B
Figure 6-233
Example 6.234 (0.050, 3, 7) Let ABC D be a trapezoid and 0 be the intersection of its diagonals AC and BD. The line passing through 0 and parallel to AB meet AD and BC at F and E . Show that 0 is the midpoint of EF.
Constructive description «POints A B C) (on D (P CAB)) (inter 0 (I A C) (I B D)) (inter E (I B C) (P 0 A B)) (inter F (I A D) (P 0 A B)) (midpoint 0 E F) )
D
C
P<]' A
B
Figure 6-234
365
604. Quadrilaterals
The machine proof -2& OF f..~ -
SAOO
~ S a c Q ,SABn±SAQQ ,SAAC+SACP ,SABQ SAOO ,(-SABC )
Q
-S!RCO ,SBCO ·S.UO ,SABC ,SABCO ( -SACO ,SABO) ,SABC ,(SABCO)'
-
,im1!ii I ~ §.a.c.u. -
SACO
Q -SABC· H -
-SABC' ¥Il. AB
'im1!ii/~ -
1
Other properties of the general quadrilateral can be found in Examples 2.62, 6.2, 6.3, and 6.17.
6.4.2
Complete Quadrilaterals
Definition. Bya complete quadrilateral, we mean the figure consisting offour points (any three of them are not collinear) and the six lines joining any two of them. As in Figure 6-235, A , B,C, D are the vertices of the complete quadrilateral. AB and CD, AC and BD, AD and BC are called opposite sides of the complete quadrilateral respectively.
~
P
R
Q
S
Figure 6-235
Example 6.235 (0.083, 1, 10) The line joining the intersections of two pairs of opposite sides of a complete quadrilateral is divided harmonically by the remaining pair of opposite sides of the complete quadrilateral. In Figure 6-235, this means (PQRS) = -1. For a machine proof of this example, see Example 6.17 on page 269. Example 6.235 can be used to define the concept of harmonic sequence: four collinear points P , Q, R , S are said to be a harmonic sequence if there exists a complete quadrilateral ABCD such that P = BC n AD ,Q = AB n CD , R = BD n PQ , and S = ACn PQ. By the following example, the above definition is independent of the choices of ABCD. Example 6.236 (0.466, 1, 14) Let ABCD and EFGH be two complete quadrilaterals such that P = ABnCD = EFnHG, Q = ADnBC = EHnFG, and S = ACnPQ = EG n PQ are collinear. Then BD, F H . PQ at:e concurrent.
Chapter 6. Topics from Geometry
366
Constructive description ( (points ABC D E) (inter P (1 A B) (I C D»
-
-
(on G (I
!!..
(inter (inter H (1 E (inter R (1 B (inter 0 (I F (*=~»
(1 Q G»
D)
(1 P G» (1 P Q»
H)
(1
Q)
P
Q»
-)
(~)/(~
(inter Q (1 B C) (1 A D» (inter S (1 A C) (1 P Q» E S» F (I P E)
0 SqFH = SPFH
I:.IJ.Q§.£nL
0
~
qO-SqFH
us.!! §..a.D.e. QR-SBDQ
~ SBDP"SqFH SPFWSBDq
S SBDposP9G oSEqr!-SEP9G) SPGrSEPQ oSBDQoSEPqG
-
E.
-
-SBDpoSEQG
.imJ!!.ify -
§..
SEQF
F -SEqGoSEPq SEQPG
SEqGgSEQSo~ S
-SBDpoSEqS
-
5 SEPq oSACq -SAPCQ
BDQ
Q SaGP-SABo -SABDC
ACQ
Q SACO-SARC SABDC
S
SBDe-( SAcpoSABcH-SABDc) SACP-SBCD -SABD-(-SABDC)
-
EQS
S
-SBDP"SACq SACP-SBDQ
,imJ!!.ify
5 SEPqoSACP -SAPCQ
S
SEPQoSACP"SBDQ -(-SAPCQ)
.imJ!lify
EPS S
SEPsoSBDQ -SBDP"SEPq-SACq-( -SAPCq)
-
Figure 6-236
F -SPqGoSEPG SEQPG
SEPGgSEPSo~
SEPG oSBDQ
SEPs o ~oSBDQ
-
-
SPGF
(-SPQGoSEPG) oSEPQoSBDQ oSEQPG
_G -SBDP o SEQSo~ ES
.f..
HSPqGoSEqF SEPQG
QFH
SBDp oSPqG o(-SEqGoSEPq) oSEqPG
-
!!.SPGF"SEPq PFH- -SEPQG
S
SBDPoSPqGoSEqF SPGF oSEPQoSBDQ
.imJ!!ify Q
The eliminants
The machine proof
p -SACO -SABG
ACP SBDp
SAnp,SAco,SABG
SACBD P
SaGP ,SARn SACBD
SACpoSBCD -SABD (-SBCD -SABD) -SACD -SABc oSACBD (-SACD oSABc)-SBCD oSABD oSACBD
.im~ifY
Example 6.237 (0.216, 2, 24) If the intersections of five corresponding sides of two complete quadrilaterals are on the same line 10 Then the remaining sides also meet in '0
Constructive description ( (POints ABC D P) (lratio U A D r.) (lratio v
B C r2)
(inter I (l D B) (1 U v» (inter J (1 C D) (1 U V» (inter K (1 A B) (I U V» (lratio sUP r3) (inter Q (1 I S) (1 K P» (inter R (1 V Q) (I S J» (inter L (1 P R) (1 u V» (inter N (1 A C) (1 u V» (~ = ~»
Figure 6-237
367
604. QudrlIaterals
The machine proof
The eliminants
(i*)/(~)
~!!.&cJL VN- SACV
SJ.J.!:.~
!i..~ . SJ.J.
-
SACU
f.
(-SPUR) ,SACV SACU ,(-SPVR)
-
11
VL - SPVR S !SVJs ' Spvo PVRSVJQS
VL
S
sVsQ ,SpurSAcv ,SVJos SACU ,SVJs ·SPVQ ·(-SVJQS)
-
"mg'/II ~
S
Sy cs.SeK s .5pur5AQY SACU ,SVJS ' SP/S ,SPVK
SACU ' SPV r ... ·(Spurr.
SpuIl ,SpVK
,im,eilll Seyr ,SpI/K ·Be" "SAGY -
!i.
SACU ,SPVJ ,SpurSpVK SpvrSpuv ,SABu-SpurSAcv-<-S,WBV) SAcu ,SpvrSpurSpuv ,SABV ,(-SAUBV)
-
,im1!ii fJl -
:!..
Seyc ,SABIl ·$Plll ,SACY SACU ,SPVJ,SPUI,SABV
5 eyrS .sawS euy ,ScpwS Acd -Scupy) SACU ,SPUv ,Scov ,SpurSABv '(-SCuov)
-
.im~i/Jl $eyc ,SAAr,.SCQu 'S'oy SACU ,SCov ,SpurSABv
L
SPUV ,SBOV ,SABU ,SCOU'SACV ,(-SBUOV) SACU ,Scov ,SPUV ,SBOU ,SABV ,(-SBUOV)
-
,im1!!.i/y -
Ssny ,SABrcScDu ,SACY SACU ,SCOV ,SBOU ,SABV
~ (-SBCp ,,,,),SABU,SCpU'(SABC ,,,,-SABC)
-
SACU ,(-SBCO ,,.,+SBCO),SBOU ,SABc ·r.
"mp!.'/II -
Q. -
QSe,..SeYK SPIKS Q SY1s ·$eKS
VSQ
~ Spyr ... ·(SpuK ·... -SPUK)·SpUrS6Qy
-
S PVQ
svso ,SPUJ ,SACV -SACU ' SV Js ·SPVQ
5ycs ·$PKs ·Sell "SAcy ·SelKS -SACU ' SV JS ' SP/S ' SPV K ' ( -SPI KS)
'im~i/Jl -
!Svso ,SPUJ PURSVJQS
SA8WSCOIl SACU ,SBOU
SABp ·rd-SAco ·r.+SACO) SAco ·r, ,(-SABo·r.+SABO)
-SPIKS
Sp/sg;( ... -I),SpUI SVJsg;SpVJ ·r. SeKsg;( r.-I) ,SPUK SVISg;SPVI'''' S K Selly ,SABY PVK -SAUBV
S
!!Seuy ,SA811 PUK-
S
J
PV)
5
-SAUBV Sefly ·Sony -Scuov
J Serry ·SeQU
PUJ
S
Scuov !...$euy ,SsDU
PUI-
S
SBUOV
I Selly ,SHQY
PVI
SBUOV
SABV);;SABC'''' Scov);; -
(,.,-I),SBCO)
SACV);;(,.,-I) ,SABC SBOV);; -
(SBCO '''')
SBOUg -
(r.-I) ,SABO)
S ACUgSAC 0 ·r. Scoug -
(r.-I) ,SAco)
SABugSABo ·r.
"mg'/II Definition. The line joining the midpoints of a pair of opposite sides of a complete quadrilateral bisects the segment between the intersections of the other two pairs of opposite sides of the quadrilateral. For a machine proof of this theorem see Example 2.36 on page 74. Let us call this line the Gauss line for the given complete quadrilateral. Example 6.238 (Gauss Point Theorem) (0.933, 26, 56) Given five points. we have five Gauss' lines. These five Gauss' lines are concurrent. Let us call this point of concurrency the Gauss point for the given five points.
368
Chapter 6. Topics from Geometry
Constructive description ( (points Ao A, A2 A3 A.) (inter x (I A3 A,) (l A, Ao» (inter y (l A2 A3) (I Ao A.» (inter z (I A. A3) (I A, A2» (midpoint M, A, A 3 ) (midpoint (midpoint M3 Ao A2) (midpoint (midpoint M. A, A.) (midpoint (inter J (l M3 M.) (l M, M 2» (inter zLJ.!..¥. M.) (I M, M2»
M2 x A 2 ) M. Y A,) M. Ao Z)
A,
(~= M,ZI» M21
Ao
Figure 6-238
M2Z/
Example 6.239 (0.033, 3, 6) The centroids of the four triangles determined by the vertices of a quadrilateral taken three at a time form a quadrilateral homothetic to the given quadrilateral in the ratio 1/3.
Constructive description ( (points ABC D) (midpoint Q B C) (midpoint P B A) (midpoint R C D) (inter A, (l D Q) (l B R» (inter D, (l A Q) (l C Pi) (inter J (l A A,) (I D D,»
The machine proof
The eliminants
-!(*')
.41-4
L -
o -
SADD, (-J) ·(-SDA,D,) SADQ·SACp ·SACQP (J)·SCQP·SADA, ·SACQP
.imJ!!.ify
(,4.1.=-3» A,)
:!J -
SADQ ·SACP (3) ·SCQP·SADA,
SADQ·SACp ·( -SBDRQ) (J)·sCQP ·(-SBDR ·SADQ)
.im.Ef.ify SACp·SBDRQ (J) ·SCQP ·SBDR
l! -
1:.. -
Figure 6-239
~ -
SACP·(-tSCDQ+SBDQ) (J) .sCQP ·(-tSBCD) (-tSABC)·(SCDQ-2SBDQ) (3) ·(t SACQ)·SBCD
A,)
SADD, -SDA,D,
l2.! SCQP·SADA,
S
DA,D, -
SACQP
l2.! SADQ·SACP S ADD, SACQP ~ SBDR ·SADQ S ADA, SBDRQ
SBDR~ - ~(SBCD) SBDRQ~- ~(SCDQ-2SBDQ) SCQP~~(SACQ) SACP~ - ~(SABC) SACQ~
-
~(SABC)
SBDQ~
-
~(SBCD)
SCDQ~~(SBCD )
-SABdtSBCD) (J)·(-tSABC)·SBCD
.im~ifY 1
Example 6.240 (5.866, 51, 69) The orthocenters of the four triangles formed by four lines taken three at a time are collinear.
369 Constructive description ( (POints A B e) (on D (I A B)) (on E (I A e» (inter F (I E D) (I C B» (foot G. C A B) (foot FH. A B C) (inter H. (I C G.) (I A FH.)) (foot G, E A B) (foot FHl A E D) (inter H, (I E G,) (I A FH,)) (foot G. F A B) (foot FH. BED) (inter H. (I F G.) (I B FH.)) _ _ (inter H. (1 F G.) (I H. H,)) (~ = ~» G,H G3 H •
F
A
Figure 6-240
3
6.4.3
Parallelograms
The parallelogram has a special position in the area method. The reason is that when eliminating points from length ratios, we need to add auxiliary parallelograms. Thus some properties (2.11, 2.12, 3.6, 3.7, 3.11) of parallelograms are often used in the proofs produced by the area method.
Example 6.241 (0.016,1,3) Let 0 be the intersection of the two diagoTUlls AC and BD of a parallelogram ABC D. Show that 0 is the midpoint of AC. Constructive description ( (POints A B C) (pratio DAB
(inter 0 (I B
C 1)
D)
(I A
e))
(midpoint 0
The machine proof
The eliminants
-~
~~~
co
CO--SBCD
Q~ -
SBCDgSABC
-SBCD SABDgSABC
!2.~ -
Figure 6-241
A e) )
SABC
.im!!f.ify 1
Example 6.242 (0.066, 3, 10) Let I be a line passing through the vertex of M of a parallelogram M N PQ and intersecting the lines N P. PQ. NQ in points R. S. T . Show that ~~ = :~ (or l/MR + l/MS = l/MT). Constructive description ( (POints M N P) (pratio Q M N P 1) (lratio (inter S (I M R) (I P Q))
R N P r)
M
N
Figure 6-242
370
Chapter 6.
(inter T (I
M R)
(I
N
(5 = -Hi»
Q»
The machine proof
The elirninants
!!!J:.
,iI.:!:~
~
MS- SMNSQ
.ll.
4U;:!:
SMNq MR-SMNRQ
- MS
1:.
SMNQ,SMNSQ
S SNPq ,SMqR
-SNQS,SMNRQ
§...
SNQS
SMNQ ' ( SMPRQ,SMNQ+SNPQ,SMqRj-{-SMPRQ) -(-SNPQ ,SMQR),SMNRQ,(-SMPRQ)
.imglify
SMNRQ~SNPQ ·r+SMNQ
-SMNq,(SMPRq,SMNq-SNPq,SMqR)
!!:.
SMPRQ
S SMPRq,SMNq-SNPq ,SMqR SMNSQ SMPRQ
SNPQ,SMQWSMNRQ
SMQR~
-SMNq,(SNPq,SMPq·r+SMPq,SMNq)
-
Topics from Geometry
SNPQ,(-SMPQ·r+SMNQ·r-SMNQ),(SNPQ ·r+SMNQ)
-
(SMPQ ·r-SMNQ ·r+SMNQ)
SMPRQ~SNPQ·r-SNPQ+SMPQ
SMNq ,SMPq
SNPQ~SMNP
SNPQ ,(SMPQ·r-SMNQ ·r+SMNQ)
SMPq~SMNP SMNq~SMNP
.imglify
= s!:; Q S'
.imJ?!.ify
Example 6.243 (0.066,2,9) In the parallelogram ABCD, AE is drawn parallel to BD; show that A(ECBD) -l.
=
Constructive description ( (POints A B e) (pratio DAB e 1) (pratio E A B D 1) (inter F (I B E) (I A D» (inter G (I E B) (I A e» (harmonic E G F B) ) The machine proof
-
SABDE,SACE+SACD,SABE),SABC ,SABDE SABCE,SACD ,SABE,SABDE
.imJ?!.ify
(SABDE,SACE-SACD ,SABE),SABC
-
.!f.
SABCE,SACD,SABE (SACD,SABD+2SA BD ,SABcl ,SABC (SACD+2S A BC) ,SACD,SABD
-
.imJ?!.ify -
SACD
Q &.a.c. -
&.a.c.
SABC
3imglify
B
The elirninants BG- SABC
SAECd( SABC) (-SABCE),SACF (
A
Figure 6-243
u.Q~
(~~)/(~) Q (
~ u;,Q~ FG-
s ACF S
SACF F SACP ,SA8fj
SABDE
F -(SABPE ,SACE SACp,SABE) AECF SABDE
SABCE~SACD+2SABC SABE~SABD
SACE~SACD+SABC SABDE~(2SABD) SACDgSABC
6.4.
371
Quadri"terals
Example 6.244 (0.116,2, 12) The diagonals of a parallelogram and those of its inscribed parallelogram are concurrent.
Constructive description ( (points AB C E ) (pratio 0 C B A 1) (inter F (I A E ) (P 0 BE» (inter G ( \ B E ) (P CA E )) (inter H (I 0 F ) ( \ C G» (inter 0 (I A C ) (I B 0 » (inter z. (\ B D) (\ E H» (inter z, (\ A C) (\ E H » ( E.l."~ HZ EZ, 1
The machine proof
The eliminants
E.l."~ HZ E Z 'l
~~~
~~ . ~
!!.!6~~
~
S
l
H Z , - SAOH
SA C H
EZ.
S ACW ( -SBED) -SAC E;:-(-SCDO ,SBDF) "SCDOF SC DF "SACC"SBED "( -SC DCF )
= 1»
S
S C DF "(-SB C E "SA C E) "SBED "SABE
aimJ!iiJy -
f:. E
-
Figure 6-244
-Sacg,SAcr;;
G
SAGEn ·SaaE
SO DOF SABE
CDO
5
SABE F SRGEP ,SAED
ODF
S
SABE
F5BBo oSAgO BDF
SACED "(-SBED "SAEDH SABE) ( -SB C ED "SAED)"SBED "( -SABE )
-
G
SAGED , SRDE
SCDF "SBED
. imJ!!.ify
Sana -SHOE
AOO
S
SACg oSACEP "Sacg "SRO[ OSAB6
-
SODGF
H
BDH
S
SAGE ·SOOO ,SROE SCDF "SACC "SBED
Q
!!.500E oSACQ
AOH-
-
!!.. -
-SBED
E Z. -
SAC & "SSDH
~
SABE
SBOEDgSABE SA o EDgSABE
SBCED
Q~ -
SABE
.im~ if1J 1
Examp\e 6.245 (0.066, 2, 8) Let ABC D be a parallelogram. Then the feet from A, B I C , D to the diagonals of the parallelogram form a parallelogram.
Constructive description ( (points A B C) (pratio DAB C 1) (foot F A B D ) (foot E B A C) (foot H C B D ) (foot G D A C) (~ FC
A
=1 »
~
B
E
F
H
G
Figure 6-245
The machine proof
The eliminants
~
~Q~
FC
s
-
S
- ( PO BD "SA C D-PBDC ,SAB O ) SAO F "PBDB
f:.
-(PC BP "SAOP-PBPO "SABO)"PBPB ( -PADB "SABO+PABD "SACD ) ,PBDB
-
aimJ!!.i/lI -
PaRO,SAGO
H PCSO -SACp-PSDC ,SABG
AON
SACF
~
0
C
FC- SAOF
Q ~
PAPa ,SABa
PBDB
F -(PADB"SABO-PABp·S,lCD) AOF PBDB
PABDgPABC+PABA PADBgPBOB+PABO PBDOgPABO+PABA
PADB "S A BO-PABD "SAOD SACDgSABC
Q
PBGS -SABC-PARA "S6AC
-
PBCB ,SAB O -PABA "SABC
""'2!_if1J
PCBDgPBOB+PABC
372
Chapter 6. Topics from Geometry
Example 6.246 (0.083,10,8) Ifsquares are erected externally (or internally) on the sides of any parallelogram, their centers form a square. s .'
Constructive description «points A B C) (pratio DAB C 1) (constant i' -1) (midpoint L A D) (pe-square S L D) (midpoint NBC) (pe-square Q N B) (midpoint T A B) (pe-square PTA) --+
--+
(PS-i.PQ
=
.
····.9····
0»
Figure 6-246
The eliminants
The machine proof -+
sP;' - (r-.i-r-+-; - A.i) QP;' - (r-.i-r-+Q - A .i) r-~!(s+ +A)
-+
QP ·i-SP
~
- r-.i'+2r-.i-r- - Q.i+ -; +A .i'-A .;
~ -(Q.i- -; +t s+.i'- s+.i+! s+ -! A .i'+! A) ~
~
- (N.i- N - S+.i)
Q~
N~Kc+S+)
(1).(2-;+ C.i'- C .i-2s+.i'+ s+.i- s+ +A.i'- A)
-+
-+
-+-+
S ~ - ( L ·i- L - D --+ --+ --+ --+ 2 --+ -+ 2 --+ --+ -+ 2 --+
= --+ --+ --+ --+ -+ --+ --+ =n - --+ L .i+ L + D .i-! C .i-t C +t B .i+t B - A n
.'
.i)
A)
2
DgC _s+ +A
(-!).(-D .i- D +C.i+ C - s+.i-s+ +A .i+ A)
.implify
=
(t)'(i+l)'(
-+ -+ -+ -+ .im1!f.ify D - C
+
B - A)
-
0
Example 6.247 (0.250, 7, 8) If squares are erected externally on two opposite sides and internally on the other two sides of any parallelogram, their centers form a parallelogram.
s Constructive description: «POints A B C) (pratio DAB C 1) (midpoint L A D) (constant i' -1) (pe-square S L D) (midpoint NBC) (pe-square Q N B) (midpoint T A B) (ne-square PTA) (midpoint M C D) (ne-square R M C)
(ps=QR»
B
'..
,
Q
Figure 6-247
C
6.4.
373
Quadrilaterals
The machine proof
The eliminants
-sF Ok
Ok~-;:;·;+ -;:; - -q -C.;
!J:
-sP
--;:;-:-.-;+--;:;::':-"-!.-q--:---c-:-.; -5P
--q +! p ~
T _
M~!(P + c)
-+
~
.;+! P -! C .;+! C
sP~T . i+ 7
~ --+ --+ --+ (2)·( T ·H T - 5 - A .j) --+ --+ --+ --+ --+ 2 Q - D .,- D + C .;- C
T'I!(11 +""1)
(2).(-7+1 I1'H1 11-1 ""1 .H1""1) 2
2
1
g J:!..
27-11.;-11+ ""1 .;-""1
L
=2
D
+
A
27-11.;-11+""1.;-""1
-
-
-+LI(-+ -+)
-+ --+ --+ --+ D ·H D - B .;- B
.;m1!!.;fy
§..
-q';, - Ct:·;- N - 11.;)
1
--+ -+ --+ -+ --+ 2 Q - D .;- D + C .;- C -+ --+ --+ --+ --+ _-:---.>..::(2-=5-:--_B=-:-=. ;-~B:-'-+-:.A'7-..:. --,A-=:-)'--"7"" . .; --+ -+ --+ --+ --+ --+ --+ -2 N ·;+2 N - D .;- D + C .;- C +2 B .;
-
- 7 - A -.
-+ -+
(HI)·( D - B )
-27 ·H2 7 +2 P .;- 11.;-11+""1 .;- ""1 -+ -+
(HI)·( D - B )
1.. -(- p -
.;- p +I1'H 11) . ;mJ!!.;fll -+-+
-
(HI)·( D - B )
I
Example 6.248 (0.450, 5, 26) The diagonals of the square in Example 6.246 pass through the center of the parallelogram (Figure 6-246). Constructive description «points A B C) (pratio DAB C I) (midpoint L A D) (constant ;' -I) (pe-square 5 L D) (midpoint NBC) (pe-square Q N B) (midpoint 0 Q 5) (AO-OC = 0) ) The machine proof.
The eliminants
CO+AO n :::-i --+ --+ --+ = !CIJ+t Cs+t AQ+t A5 ~ (t)'(-2N';+2N +cs- c +211'HAS-""1) ~ (-t)·(-cs+ c .;- 11.;- I1-AS+""1) ~ (!).(-27';+27 +2P.;-C·;-C +I1'H 11-2""1) ~ (-t)·(- p .;- P +C 'H C - 11·;-11+ ""1.;+ ""1)
AOg!(AO+AS) cog!(cQ+cs) -+Q - (-+ AQ= N .;- -+ N - -+ B ·H -+) A
.;mpl;fy
=
-+ -+ -+ -+
(t )·(HI) .( D - C
.;m~;fll 0
+B
- A )
cQ';, - (N .;-N +C - 11.;) -+NI(-+ -+)
N =2 C + B ASg; - (7.;- 7 - p.;+ ""1) csg; - (7.;-7 - p .;+ c) -+LI(-+ -+) L =2 D + A --+ D --+ --+ --+ D=C-B+A
374
Chapter 6.
Topics from G-a-y
Example 6.249 (0.250, 4, 15) If similar rectangles are erected externally on two Op· posite and internally on the other two sides of any parallelogram. their centers form a parallelogramo Al t < : : - - - - - - - - - : 7 T D I
Constructive description «points A B C) (pratio DAB C 1) (tratio Al A D rl) (pratio DI Al AD 1) (inter S (1 A DIl (1 D All) (midpoint NBC) (pratio Q N Al A 1/2) (tratio A, A B r,) (pratio BI A, A B 1) (inter P (1 A BIl (1 B A,» (midpoint M C D)
"i":;...:..-----'.----">t D
Figure 6-249
q~!:.atio R M A, A 1/2) (~! =
1»
The machine proof. QR
~
!:.. -
SAQA'M~)(2SAQA,-SADA,-SACA')
-SASA,P SAQA,M
5
-SASA,P SAQA, -tSADA, -tSACA,
SAA,B,
(-2H-SABB!A,oSASA,-SAA,B!oSABA,) (2S AQA, SADA, SACA,H SABBIA,)
SACA, ~±
3im!!1i/y
t
-(iPBAs·r.-}PABAo,,) PBAQ rl-:PBAo r'J -fPBAc *r2
3im!!f:i/y
9. !i
g,
-(2SAS A ,-SABA,) 2SAQA. SADA, SACA,
o
o
-(2PBAS-PABA) 2PBAQ-PBAD-PBAC
-(2PBAS-PABA) 2PBAN PBAA I -PBAD-PBAC -(2PBAS-PAiJA) PBAA, PBAD+PABA -SADD! A! oPABA+2PBAD!oSADA! (PBAA, +PBAD-PABA)oSADD,A,
Qj -( -2PBAA! oSADA! -2PBADoSADA! +2PABA oSADA!) 3im!!4i/y
p ASA2 P
~ (-2)o(2SAS A, oSABA. -S~BA') - (2SAQA. -SADA.-SACA,)o(2SABA,)
~
u.~ SASA,P
QR- SAQA.M
-§.&.
!!:.
The eliminants
(PBAA, +PBAD PABA) o(2SADA,)
~
ABB)A,} ,SASA,+SAAaB) ,SABAa SABB A 2 1
-
(SABA,)
SABB,A, ~2(SABA,) SADA, ~±
SAQA. ~*(PBAQor,) SABA, A,l = 4( PABA or, ) SASA. A,l = 4( PBAsor. )
PBAQ~~(2PBAN-PBAAI ) PBAN=2(PBAC+PABA) P S PBAD! oSADA! BAS SADD,A, PBAD,
~ PBAA, +PBAD
SADD,A, ~2(SADAJ
375
6.4. QuadriJalenls
Example 6.250 (0.083,3,8) Let AI. Blo C lo DI be points on the sides CD. DA. AB. BC ofaparallelogram ABCD such that CAI/CD = DBdDA = ACI/AB = BDdBC = 1/3. Show that the area of the quadrilateral formed by the lines AAI BBlo CCI• DDI is one thirteenth of the area of parallelogram ABCD.
=
Note that we only need to show 3SABCD
Constructive description ( (POints A B C) (pratio DAB C I) (lratio A, C D 1/3) (lratio B, D A 1/3) (inter A. (I A A,) (I B B,» (3SASCO
= 13SABA.) )
,.,..----".--.,.C
13SABA•• The eliminants
The machine proof ~
-
!!l -
11 -
!2. -
~ SABB, ,SABA,
S
(3) ,SABCO (13) ,SABA.
ABA, -
SABA)Bl
(3) ,SABCO,SABA,B, (13) ,SABB"SABA,
SABB, B'2( = 3 SABO ) B, (2SAOA, -3SABA,) SABA,B, = -3
(3) ,SABCO ·(-t S AOA, +SABA,) (l3) ·(fSABO) ,SABA,
SABA, ~ !(SABO+2SABC)
SAOA, ~ - ~(SACO) (-3) ,SABCO ,(-!SACO-SABO-2SABC) SABDgSABC (26) ,SABO'(f S ABO+isABC) SACOgSABC (3) ·(2S,
SABCDg2(SABc)
.im1!li!y Co
B
Figure 6-250
Example 6.251 (0.466,7,8) Use the same notations as Example 6.250. IfCAI/CD ACI/AB = TI,DBdDA = BDdBC = T2 then SAJ:;~'D' = "2 .r,.?~"r,+2' We only need to show that
:;!~~
=
= 2'("2 . ,I:::- ,+2)' r
r
Constructive description (Formula derivation) «POints A B C) (pratio DAB C I) (lratio A, C D ro) (lratio B, D A "2) (lratio c, A B ro) (lratio D, B C "2) (inter A. (I B Bo) (I A A,» (inter B. (1 C co) (I B B,» (inter c. (I D Do) (I c C,» (inter D. (1 A Ad (I D D,»
(SA.AB
= SABCO)
)
Example 6.252 (0.066, 2, 11) Let P and Q be two points on side BC and AD of a parallelogram such that PQ II AB; M = AP n BQ. N = DP n QC. Show that MN= ~AD. Constructive description ( (POints A B C) (pratio DC B A I) (on P (I B C» (inter Q (I A D) (P P A B» (inter M (1 A P) (1 B Q» (inter N (1 D P) (1 C Q»
Figure 6-252
The eliminants
The machine proof
NSCDQ ,SADP SCDQP NSCPQ ,SADM SAMN SCDQP MSADP ,SABQ SADM SABPQ Q (SADP-SABD) ,SABP SABPQ -SABD
(-2),SAMN SADN
!:!..
sim!!~ify
-
M. -
SADN
(-2) ,(-SCPQ ,SADMH-sCDQP) (-SCDQ ,SADP) ,(-SCDQP)
-
(-2) ,SCPQ ,SADM SCDQ ,SADP
SCDQ~SCDP SABQ~SABP SCPQ~SACP
(-2) ,SCPQ ,SADP,SABQ SCDQ ,SADP ,SABPQ
simg1ify (-2) ,SCPQ ,SABQ SCDQ,SABPQ
Q -
Topics rrom Geometry
Chapter 6.
376
(2) ,SACP ,SABP-( SABD) SCDP,(SADP ,SABP-SABP ,SABD)
simJ!!ify
SADP!:, -
(SACD·*-SABD·:~+SABD)
SCDP!:, -
(~-1).SBCD)
SACP!:,( :~ -1),SABC
(2),SACP,SABD SCDP'(SADP-SABD)
SACDgSABC p
(2) ,(SABC·JJ.I:.-SABC) ,SABD Be
= (-SBCD ' :~ +SBCDH -SACD ' :~ +SABD·*-2SABD) simJ!!ify
SBCDgSABC SABDgSABC
(2l,;§ABC·SABL
SBCD · (SACD·~-SABD' :~ +2SABD)
!2.. -
(2),S~8C
.imJ!!ify
SABC·(2SABC)
Example 6.253 (0.033,2,4) Let ABCD be a rectangular, and EFGH a parallelogram inscribed in ABCD such that the sides of EFGH are parallel to the diagonals of ABCD. Show that the perimeter of EFGH is fixed.
Constructive description ( (points A B) (tratio e BAr) (pratio DAB e 1) (on E (I A B» (inter F (I B e) (P E A e» (inter H (I A D) (P E B D» (~+~ =
1»
The machine proof
.Iili!!.~ H!.~
*+~ _H
-
.f.. -
The eliminants 1fij- -SABD AC- SABC
-1iL,SABD+SADE AC -SABD SaCfj"SA8D
SADe-SARe
SABD,SABC
§.. SARD · SARQ - SABD ,SABC simg1ify 1
SADE~ SBCE~
-
(SABD ' ~)
-
(%-1) ,SABC)
:~:
j2::~:~~;;:t Figure 6-253
Example 6.254 (0.116, 3, 15) Three parallel lines are cut by three parallel transversals in the points A, B, C; AI. B 1, C 1; A 2 , B 2 , C 2• Show that B 2 C, C 1 A 2 , AB1 are concurrent. Constructive description ( (POints A B Ad (on e (1 A B» (pratio B, B A AI (inter e, (l AI Bd (P e A Ad) (on A. (1 A Ad) (inter B2 (1 B Bd (P A2 A B))
1)
6.4.
377
Quadrilaterals
(inter c. (I C (inter z, (I C
c,) (I A. B.)
(I
B.» d!!ter..LJ! C B.) A B,) (~ . ~ = 8)Z)
AZl
I»
(I c,
AZ,
S .. CB • • ~ -SCB,B. "Z.
AZl -
S .. c, .. ,
,s.. c, .. ,
(-S .... 1 C · ~) · (-S.. B1 CJ . ~+S .. B1 C) ),S"BB 1 "'A, AA,
~
SABS
1
S ..CB, ~S..c .. ,
-s..C.."SB,C, .., ,(-S .. BB,) S"B,B .."SBCB,
A B,)
SAC)A2
C8 J B'J -
SCB,B. ·S"C,,,,
~
(l
'!1 SAS, BA, ,SSCB)
S
~ -5"'C8,} ,58)C, .41
-
A.)
~~ S ..CB, B,Z, -SCB,B, ~~SB'C, .. ,
~.~
~
(inter z. (I c,
The eliminants
The machine proof BIZ.
A.»
(-S"BB 1 +S"B" I . ~). SBCB 1 ·(-S .... 1 CI . ~) ..tAl ..tAl
..=, - ( S .... ,C, · ~ AT)
S "B, B", ..=,
-
SB,C,A, ... = -
.., S .. C .. , = S" ..
-
(S"BB, - S "B", ' ~) .... ,
(AT ) (~-I) . S"B,C, ( S .... ,C · ~ AT)
,C, ~S.... ,c
SAB1Cl~
(SACA1B.)
-
SBCB, ~ - (SB .. ,C) S .. BB, '!! S .. B .. , S .. C.. ,B, ~ -
. im!!!i!" -
-(*-I),S .. c .. ,B, ,S .. BB, (S"BB,
-S"B", . ~).SBCB,
(S.... ,C+S.. B.. ,)
SB .. ,Cg -
(~-I).S"B",)
S .... ,cg -
(SAB""~)
-(~-I).(-S .... 1 C-S"B" 1 ),S"B" I ..tA,
~
(-S .. B .. , · ~ .... , +S .. B.. ,)-(-SB .. ,C)
.imJ!!i!lI S .... ,C+S.. B.. , -
g
SB",C -S.. B.. , . ~+S"B", -S.. B.. , . ~+S"B",
.im!!1i!y 1
Figure 6-254
Example 6.255 (0.233, 5, 11) A line passing through the intersection 0 0/ the diagonals a/parallelogram ABeD meets the/our sides at E, F, G, H . Show that EF = GH. Constructive description ( (POints A B C) (pratio D (inter 0 (I A C) (I B D» (on G (I A B» (inter F (I C D) (lOG» (inter E (I A D) (lOG» (inter H (I B C) (loG» (U,=-l» GH
C B AI)
E
Figure 6-255
378
Chapter 6.
Topics from Geometry
Example 6.256 (0.050, 5, 5) Let ABC D be a parallelogram and P, Q, R, S be points on AB, BC,CD,DA such that AP = CR and BQ = DS. Show that PQRS is also a parallelogram Constructive description: ( (POints A B C) (pratio (lratio P A B rJ) (lratio R CD rJ) (lratio Q B C r2)
(lratio
DAB C 1)
(~
S D A r2)
= 1»
The eliminants
The machine proof
-Q
S
R
~=~ SBCR=SBCD ·r,
~
SP
sBSCPt:. -
(SBCS+SABc ·r,-SABC)
9. ....&c:..a...-
-SBSCP
SBCS~ - (SBCD ·ro-SBcD-SABc·r2)
!!:.
-SRGP·rt
SBCDgSABC
-
SBSCP
~
-
-SaCQ -r,
-SBCS-SABc·r,+SABC
§.. -
S80DOr)
SBCD·r2+SBCD+SABC·ro+SABc·r,
g
SABC
-SARc ·n .img1ify 1 -SABC 'Tt
B
c
Q Figure 6-256
Example 6.257 (0.400, 20, 16) Let ABC D be a parallelogram and P, Q , R, S are points in AB,BC,CD,DA such that r1 = ~~ = g~ and r2 = ~ = Show that SPQRS = 2r2r1 - r2 - r1 + l. (Figure 6-256) S
g!.
ABeD
Constructive description ( (POints A B C) (pratio DAB C 1) (lratio S D A ro) (lratio P A B rJ) (lratio R CD rJ) (lratio Q B C r2) (SPQRS = (2ro· ... -ro- .. +l)·SABCD»
Example 6.258 (0.266, 6, 24) Let C and F be any points on the respective sides AE and B D of a parallelogram AE B D. Let M and N denote the points of intersection of C D and FA and of EF and BC. Let the line MN meet DA at P and EB at Q. Then AP = QB. Constructive description ( (POints A E B) (pratio D B E A 1) (on C (1 A E» (on F (1 B D» (inter M (1 D C) (1 A F» (inter N (1 E F) (1 B C» (inter P (1 M N) (1 A D» (inter Q (1 M N) (1 E B» ( ~-~» DP - EQ
D
F
Q
c Figure 6-258
379
Example 6.259 (0.400, 10, 22) ut ABC be any triangle and ABDE, ACFG any parallelograms described on AB and AC. ut DE and FG meet in H and draw BL and C M equal and parallel to H A. Then area( BC M L) = area( AB D E) + area( AC FG). H
Constructive description ( (POints ABC D F) (pratio E A B D I) (pratio G A C F I) (inter H (I D E) (I F G» (inter R (I B C) (I H A» (pratio L B H A I) (SACF+SBAD
= SCBL)
D
Figure 6-259
)
Example 6.260 (0.467,46, 18) The circle through the vertices A, B, C ofa parallelogram ABCD meets DA, DC in the points AI , CI . Prove that AID/AICI = AIC/AIB. Constructive description ( (POints A B C) (circumcenter 0 A B C) (pratio DAB C I) (inter Al (I A D) (cir 0 A» (inter C I (I C D) (cir 0 C» (eq-product Al D Al B Al CI
Al C) )
Figure 6-260
Example 6.261 (0.383,8,16) Given the parallelogram M DOMI, the vertex a is joined to the midpoint C of MM I. If the internal and external bisectors of the angle COD meet M D in A and B, show that M D2 = M A . M B. Constructive description ( (POints / 0 C) (incenter D / 0 C) (midpoint E 0 D) (pratio M DEC I) (pratio MI 0 E C I) (inter A (I 0 I) (I D M» (inter B (I D M) (t 0 0 /» (~ =
tft»
o
M.
Figure 6-261
Example 6.262 (0.633, 46, 34) ABC D is a parallelogram with center O. E is the midpoint ofOA. F is the midpoint ofOB. G is the midpoint ofOC. H is the midpoint 01 aD. P = DEnCF; Q = AFnDG; R = AHn BG; S = BEn CH. Prove that PQRS is a parallelogram.
380
Chapter 6. Topics from c-try
Constructive description ( (POints ABC 0) (pratio DAB C 1) (midpoint E 0 A) (midpoint FOB) (midpoint G 0 C) (midpoint HOD) (inter p (I D E) (I C F)) (inter Q (I A F) (I D G)) (inter R (l A H) (I B G)) (inter S (I B E) (I C H))
Figure 262
(~=1))
Example 6.263 (0.200, 5, 11) ABCD is a parallelogram. 0 = AC n BD. Two perpendicular lines passing through 0 cutthe sides AB, BC, CD, D A in points E , H , F, G. Prove that EHCG is a rhombus. Constructive description ( (points A B C) (pratio DAB C 1) (inter 0 (l A C) (I B D)) (on E (l A B)) (inter F (I C D) (l EO)) (inter G (I A D) (t 0 0 E)) (inter H (I B C) (loG))
D
BJc::....--.........::~~~
H
(~= l))
6.4.4
Figure 6-263
Squares
Example 6.264 (0.016, 1, 2) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that SABC = SAHF. Constructive description ( (points A B C) (tratio FA
B 1)
(tratio
HAC -1) (SABC
The eliminants
~ -SAFH
SAFH!f, -
!L
PCAF~4(SABC)
sra
-(->CAF)
!..~
M
Figure 6-264
B
)
The machine proof
-
C
= SAHF)
4SABC
.imgli!!I
i(PCAF)
381
604. Quadrilatenla
Example 6.265 (0.050, 3, 5) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. M is the midpoint of BC. Show that F H = 2AM (Figure 6-264).
Constructive description The machine proof ( (points A B C) (midpoint M B
~ (4) ·PAMA C)
!!
(tratio
FA B 1)
-
(tratio
HAC -1)
-
(PHFH = 4PMAM) )
The eliminants PFHF~PAFA+PACA+8SACF
PAfAtPACA±8SACf
(4)·PAMA
SACF~HpBAC ) PAFAf"PABA M 1( ) PAMA= - 4 PBCB-2PACA-2PABA
E.
2PUC+PAQA+PABA (4).PAMA
!1.
2PllAc+PfaA+PAfA PBAC= (4H-!PBCB+.PACA+.PABA)
~
-
!(PBCB-PACA-PABA)
-(-2PBCBHPACAHPABAl (PBCB-2PACA-2PABAH2)
.im!!}.i!" 1
Example 6.266 (0.016, 3, 3) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Passing through A a perpendicular to BC is
drawn which cuts F H in N. Show that N is the midpoint of F H. (Figure 6-264)
Constructive description
The machine proof
The eliminants
( (POints A B C) (tratio FA B 1) (tratio HAC -1) (midpoint N F H)
fuJUt.
PCBN~!(PCBH+PCBF )
(perpendicular
PABC
!!!. -
}PCBH+}PCBF PABC
!!..
PCBftPABC-4SA8Q
-
NAB C) )
PCBH~PABC-4SABC PCBFf"PABC+4SABC
(2)·PABC
E. ...1E..i.a.a...
-
(2) ·PA BC
.im!!}.i!1I 1
Example 6.267 (0.033,3,2) On the two sides AB and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that FC = BH. (Figure 6-264)
Constructive description
The machine proof
The eliminants
( (points A B C) (tratio F A B 1) (tratio HAC -1)
f±£I;..
PBHB~PACA+PABA-8SABC
(eqdistance
F C B H) )
PBHB
!!.. -
E. -
Para PACA+PABA-8SABC PACAtPABA-8SARC PACA+PABA-8SABC
.imgli!1I
PCFC~PACA+PABA-8SABC
382
Chapter 6. Topics fmm Geometry
Example 6.268 (0.066,4,6) On the two sides A B and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. Show that F C I B H . (Figure 6-264) The eliminants
Constructive description The machine proof ((pointsA B C ) EE~K (tratio F A B 1)
(tratio H A c -1) (perpendicular F c B H I
PCBH -~SARCF+PARF PABC-~SABC F -p -p +~SABC)
-
=
PCBH~PABC-~SABC H PFBH=
- (ISABCF-PABF)
PABF~PABA
sABCF~~(PBAC+~SABC) PBC~-P.,CA+P& 2 -(-~BCS+~ACA-PABA+~~ABC)~(~) PBc~-P~c~-P& ( P B c B - P A c A + P A B A - ~ ~ A B c ) . ( P~AC=' ~) -2 ( B$~B$~~ABC
Example 6.269 (0.033,2,8) On the two sides A B and AC of triangle ABC, two squares ABDM and ACEN are drawn externally. F and G are the feet of the perpendiculars drawnfrompoints D and E to BC. Show that SABC= SDFB SCGE.
+
The eliminants
Constructive description The machine proof ( (points A B C ) SARC ( t r a t i 0 ~ ~ ~ 1 ) (f00t F D B C ) (trati0 E C A -1) (f00tG E B C ) (SABC = SBDF+SCQE))
-S Ry;CB-
CP SCEC=
-(SCEG-SBDF)
G E -
S
~BcEgi(pAcB)
.P
pBCEE
'7
-
S~ac.Pncn -(PCBD-SBCD-PACB~~ABC)
F
c
B
G
S~ac.Pacn -(-PACB.SABC-PABC.SABC) s i mplify A
Fieure 6-269
-
~(sABC) F -Pcnn.SSBDF= PBCB ~BcDgi(pABc)
S~pc.Pnca -(-PBCB.SBDF-PACB.SABC)
&EDg ~(sABC) p A B C = ~ p ~ ~ ~ - p ~ ~ 2 PBCB+PACA-PAB~ 2
PACB+PABC
Example 6.270 (0.050,6,8) On the two sides A B and AC of triangle ABC, two squares ABEF and ACGH are drawn externally. P is the midpoint of EG. Show that B P = CP.
Constructive description ( (p0intS A B C )
E c B A -1) 1) (tratio G (midpoint P E G)
..
..
E m A G
Hgure6-270
C
383
6.4. Quadrilaterals
(eqdistance
B PCP) )
The machine proof
The eliminants
~
PCPc~-
j(PEGE:-2PCGC-2PCEC)
PcPC
PBPB~-
i(PEGE-2PBGB-2PBEB)
~ -tPE:GE:+iPBGB+iPBE:B
PCGCgPACA
- • PE:GE:+. PCGc+. PCEC
PBOBgPBCB+PACA+SSABC
Q PCiC-2Ps6R-2Paca-PACA-8SAC6-16SABc -
PCE:C
PACA
PEGEgPCE:C+PACA-SSACE:
SSACE:
SACE~
§.. -(-PBCB+2PBAC-PACA-PA BA) PBCB-2PBAC+PACA+PA BA
-
-
~(PBAC+4SABC)
PBEB~PABA PCEC~PBCB+PABA+SSABC
.im~i/ll
Example 6.271 (0.033, 3, 6) Let ABC D be a square and G a point on CD. A square GCEF is erected externally. Show that DEl.BG and DE = BG.
Consb1Jctive description ( (POints D C) (tratio BCD 1) (iratio G CDr) (lratio E C B -r)
(perpendicular
. .'
§.. -
G .'
'.
.
..
PCBGgPCBC
PORQ
PCBG·r+PCBO
DEB G) )
PDBOg -
PQBG (r+1)' PCBG
(PDBc ·r-PDBC-PDBD ·r)
PCBC!PDCD
Q -P08c rtPo8ctPpBo 'r o
F
".
c
B
PGBE~(r+1)'PCBG
PGBE:
AC3JD .... ..
The eliminants
The machine proof f.o..a.a
!!. -
(r+1)' PCBC
PDBD!2(PDCD) PDBC!PDCD
-(-PDcp·r-PDCp) (r+1)'PDCD
E
Figure 6-271
Example 6.272 (0.183,6,17) On the two sides AB and AC o/triangle ABC. two squares ABDE and ACFG are drawn externally. Let P and Q be the centers o/squares ABDE and AC FG. MandL the midpoints 0/ BC and EG. Show that PM Q L is a square.
ConSb1Jctive description ( (POints A B C) (midpoint M B C) (tratio E A B 1) (midpoint P B E) (tratio G A C -1) (midpoint Q C G) (midpoint LEG) (U LQ
= 1»
B
M
Figure 6-272
C
384
Chapter 6.
The eliminants
The machine proof
14.&!:. 5.M..E..=
~
QL -
QL
SECQ
SCECg;J(PACE+4SACE) SMEPCg -
2. 5.M..E..= -
SMEP~2(SBME) PCM AP~ ~(PCAE-2PCAM+PBAC)
Q (2),(-}PCMAP+SMEP+SAMP) tPACE+SACE
!:.. -
1£
(-2)·(t PCAE-PCAM+ipBAC-2SBME-2SAME+2SABM) PACE+4SACE (2PCAM
PBAM+PBAC+PABMHSABC) PBAC+PACA -4SABC
-
SACE~~(PBAC) PACE~PACA-4SABC
SAME~HpBAM) SBME~
M. t PBAC+PACA-tPABC+iPABA-4SABC ~
~(PCMAP-4SMEP-4SAMP)
SAM P~t(SAM E-SABM)
tSCEC
-
SECQ
SECQ~t(SCEC)
.f. 5.M..E..= -
Topics from Geometry
-
HPABM-4SABM)
PCAE~4(SABC)
PBAC+PACA -4SABC ( 4PBcB+l2PAcAHPABr32SABC)·(2) «2))3 ,(-PBCB+3PACA+PABA-8SABcl
PABM~t(PABC) PBAM~?(PBAC+PABA)
simgiify
PCAM~2(PBAC+PACA)
PABC=~(PBCB-PACA+PABA) PBAC= - ~(PBCB-PACA-PABA)
Example 6.273 (0.016, 1, 3) Let ABC D be a square and I be a line passing through B. From A and C perpendiculars are drawn to I and the feet are Al and CI respectively. Show that AAI = BCI .
Constructive description ( (POints B e X) (tratio
ABe I)
(foot Al A B X) (foot e l e B X) (eqdistance A Al
The machine proof PAA]A PBCIB ~
Bel) )
The eliminants P <2! (PCBX )' P
PAA]A,PBXB PtBX
~ ( 16Shu),PBXB (PCBX)',PBXB
BCIB - PBXB ~ (l6HSBXA)' AAIAPBXB
SBXA 4 HpCBX) X D
.imgiify (l6)'( SBX 9)' (PCBX) ~ (16)·« tpCBX )).
-
(PCBX)'
.im~ifY I
B"-------""C
Figure 6-273
Example 6.274 23 (0.067, 5, 13) Starting with any triangle ABC, construct the exterior (or interior) squares BCDE, ACFG, and BAHK; then construct parallelograms FCDQ and E B K P. Show that P AQ is an isosceles. 23This example is from Amer. Math. Mon. 75(1968), p.899.
385 Constructive description ( (POints A B C) (tratio F C A I) (tratio D C B-1) (pratio ED C B I) (tratio K B A -I) (pratio Q F C D I) (pratio P K BEl) (eqdistance A Q A P) )
The machine proof
-
PA9A 2PEBK+PAKA+PAEA-PABA
9..
2Pfcn±PAoAtP'fA-PACA
-
2PEBK+PAKA+PAEA PABA
!S..
2Pecn±P,o,tP'fA-P'CA
-
PAEA+PABA+8SABE
.f -
2Pecn±P,p,tP'fA-P,CA PADA PACA+2PABA+8SABD 8SABC
Q PacstP'fA±8Ssor-8SABC -
K"------4 H
Figure 6-274
The eliminants
PA9A PAPA
1:.
G
PBCB-2PABC+2PABA-8S A BC
L
PRCR-2PAGR+2PACA-8SARC - PBCB-2PABC+2PABA 8SABC eJ1 (2PACA+2PA BA-16SABC)·(2) - (2PACA+2PABA-16SABc) ·(2)
PAPA~2PEBK+PAKA+PAEA-PABA PA9A~2PFCD+PADA+PAFA-PACA PAKA~2(PABA) PEBK~4(SABE) SABE~SABD-SABC PAEA~PADA-PACA+PABA SABDg -
HPABC-4SABC)
PADAgPBCB+PACA -8SABC PFCDg4(SBCF) 8BCFt; -
HPACB)
PA FAt;2(PACA) PABC=l(PBCB-PACA+PABA) PACB=2(PBCB+PACA-PABA)
.im!!f:ify
Example 6.275 (0.516, 9, 32) On the four sides of a quadrilateral ABC D. four squares are drawn externally. Show that the two segments joining the two centers of the squares on two opposite sides are perpendicular and have the same length.
Constructive description ( (POints ABC D) (tratio E A B I) (midpoint L B E) (tratio GAD -I) (midpoint P D G) (tratio F C B-1) (midpoint M B F) (tratio H C D I) (midpoint N D H) (eqdistance M P L
N) )
Figure 6-275
386
Chapter 6. Topics from Geometry
Example 6.276 (0.067 2 14) On the hypotenuse AB of right triangle ABC a square ABF E is erectedo Let P be the intersection of the diagonals AF and BE of ABF Eo Show that LACP = LPCBo
Constructive description «points B C) (tratio A C B r) (tratio F B A-I) (tratio E A B 1) (inter P (1 B E) (1 A F)) (eqangle A C P PCB) )
B
Figure 6-276
The machine proof
The eliminants
( -ScAe) oPBce (-SBce)oPAce
P
.f.. -
S
SCAF oSBAEo( -PBCF oSBAE:)0(SBAEFl 2 ( SBAFoSBCE)oPACEoSBAF o«-SBAEF))'
SSAEF
P PaCE ,SRAE
sce
SSAEF S PSCAE ,SRAfj CAe SSAEF PAC E !!,PCAC -4SSCA
~ SCAF o«-}PBAB))2 0PBCF - (SBAF)2 °(-t PCBA+SSCA) 0(PCAC-4SSCA)
-
SSAEF ~S8Af · S8Cfi
sce-
p
simJ!!.ify SC6dSBAE)2 0PBCF (SBAF)2 0SBCEoPACE
L
PPACg-SBAE
Ace
SSCE!!, -
SSAE~ SSAF~
_(_lPSAC+SSCA)0(PBAB)2 0( pscs-4SsCA) (4)0« -tPSAB))20(PCBA -4SSCA)0(PCAC-4SBCA)
t(PCSA-4SSCA)
-
i(PSAS)
-
i(PSAS)
PscF~Pscs-4SsCA
simJ!!.i!y (PBd c-4SsCAHPBCS-4SSC6) (PCSA -4SBCA)0( PCAC-4SBCA) ~ (PBCB or 2 +PBcsor)0(Pscs or+PscB) - (PBcsor+PBcB)0(PBcsor 2+PBcB or)
SCAF~
-
HPBAC-4SsCA)
p cAc 4Pscso(r)2 PCSA 4 PBCB
s imJ!!ify
S scA 4 -
HPscsor)
p sAc 4Pscso(r)2
Example 6.277 (0.617, 65, 21) ABCD is a squareo G is a point on AD such that AG = ADo E, F are points on AB, BC such that EF II ACo H = EG n F Do Show that AH=BCo
Constructive description ( (points B C) (tratio ABC 1) (pratio DAB C 1) (lratio GAD -1) (lratio
G
F B C r)
(inter E (1 A B) (P F A C)) (inter H (1 G E) (1 F D)) (eqdistance A H B C) )
Figure 6-277
387
Cyclic Quadrilaterals
6.4.5
Definition A quadrilateral whose vertices lie on the same circle is said to be cyclic. To prove theorems about cyclic qUadrilaterals. we use the techniques developed in Section 3.6. Exam~
6.278 (PtolemD Tbeo~m).J!).OSO, 2, 12) Let A. B. C, D be four cyclic points. ThenAB · CD+AD · BC = AC·BD. Constructive description ( (circle A B G D) (AB .co+AD -iiC
= AC-iiD) )
The machine proof
The eliminants
fiJ -AB+iiC-;W Bb -AG ch2!d .in(G D)-ain(AB)-JI +lin(BG)-sin(AD)-JI lin(BD)-d-sin(AG)-d
AC=sin(AG)-d iiD=sin(BD)-d
AD=sin(AD)-d iiC=sin(BG)-d AB=sin(AB)-d CO=lin(GD) -d
.in(AG)=Sc
,im1!!i/ ~ sin( G D)-sin(ABl+sin( BG)-sin( AD) .in(BD)-sin(AG) CD~r
-
sin(BD)=SD -GB-SB -GD .in(AD)=Su
Su -SC -GB-SC-SB -GU (SU -GB-SB-GU)-Sc
sin(BG)=Sc -GB-SB -Gc
.im!!1i/~
sin(AB)=SB
ain(GD)=Su-Gc-Sc-Gu
Example 6.279 (Brabmagupta's Formula) (0.267, s, 16) Let ABC D be a cyclic quadrilateral_ Then
S where p
2
= (p -
-
-
-
-
AB)(p - BC)(P - CD)(P - AD)
1~=_=2 = 16(4AC BD -
2
P ABGD )
= HAB + BC + CD + AD). Constructive description ( (circle A B G D) (l6S?BCD = 4AC' -iiD' -PaBAu»
Example 6.280 (0.033, 1, 10) Show that in a cyclic quadrilateral the distances of the point of intersection of the diagonals from two opposite sides are proportional to these sides.
Constructive description ( (circle A B G D) (inter J (I B D) (I A G» (footQ J B G) (foot S J A D) Figure 6-280
388
Chapter 6.
(eq-product
Topics from Geometry
I S B C I Q A D) )
The machine proof P'S,.PRGR PIQI,PADA
§... (16S~Q,) · PBCB PIQr P1DA
-
g
(16) ' ~~Q,.P¥cR (16S BCI ).pADA
L
(-SACD · SABD)' · P~CR·S~RCQ
SBCD ·S~BC, p1DA · S~BCD
-
.imglify SACD ,SABD,SACD ,SABD ,P qCB SBCD ,SABC ,SBCD,SABC ,PADA co=..cir ( ED.Ab.AC) .( iiD .Ab .;tiJ)-( ED .Ab.AC).(-iiD.Ab .AiI)'(2Bc·» '(2d)4 -
:I
( -C D· BD · BC) ·( - BC · AC ·AB )·( -C D· B D· BC)·( - BC ·AC ·AB)·(2AD )2 ·(2d)4
.imgli fy
Example 6.281 (0.016, 1, 6) In the cyclic quadrilateral ABC D the perpendicular to AB at A meets CD in AI. and the perpendicular to CD at C meets AB in C I . Show that the line Al CI is parallel to the diagonal BD.
Constructive description ( (circle ABC D) (inter A, (I C D) (t A A Bll (inter C, (I A B) (t C CD))
(parallel
A,
A.
........
c,
B D) )
The machine proof SBDA I SBDC,
The eliminants
s
~ PROP ,SABD
BDC, ~
S
BDA] -
~
SBDAI ,(-PBCAD) -PBCD,SABD
-
D
PRAn ·SaGo·PaGAD
PBCD ,SABD'( -PBCAD) ~impJ.iJy
-
-PaA V ,SROO PBCD ,SABD
S
-PBCAD
iiD.Ab.AiI (-2) .d
ABD
PBCD= S
BCD
PBCAD PRAn ·Saon
2(ED.Bc'cos(BD))
ED·iiD ·Bc (-2).d
PBAD=2( Ab .AiI.cos(BD))
co=..cir -(2Ab.AiI'cos(BD».( -ED .iiD.BcH2d)
-
(-2Cb .B'C .cos(BD)).( -iiD.A'b.Ain·(2d)
.imglify
Figure 6-281
Definition. The symmetric of the circumcenter of a cyclic quadrilateral with respect to the centroid is called the anticenter of the cyclic quadrilateral.
Example 6.282 (0.083, 3, 9) The perpendicular from the midpoint of each diagonal upon the other diagonal also passes through the anticenter of a cyclic quadrilateral.
389
U . Quadrilaterals
Constructive description «circle ABC D) (circumcenter 0 A B C) (midpoint T A D) (midpoint P A B) (midpoint Q B C) (midpoint R C D) (!nter M (P PO R) (P Q 0 (IDter I (I P R) (10M » (midpoint I P R) )
The machine proof
The eliminants
-it
u!..~ RI- SORM
l...
~
-
-SORM
SORM~
M.
-(-SOPTQ -SOPR)
-
( SOPR) -( SOTR)
T» .im1!!.i/~ SOPTg SOTR
Ii
SOPTg fSDOT+fScOT
-
Q. (2) -(-SOTP+tSCOT+tSBOT) -
SDOT+SCOT
1:..
-(-SCOT+S,WT) SDOT+SCOT
-
-
(SOPR)
~ SOPTg -SOPR
S
OPM-
SOTR
SOTR~~(SDOT+SCOT) S
OPTQ
Q
(2S0TP-SCOT-SBOT) -2
SOTP~~(SBOT+SAOT) SDOTJ;~(SADO) SAOTJ; - ~(SADO) SCOTJ; - ~(SCDO-SACO)
T _lSCDO+1SADO+1SACO
= -fSCDO+!SADO+!SACO Figure 6-282
Example 6.283 (0.066, 4, 5) Show that the anticenter of a cyclic quadrilateral is the orthocenter of the triangle having for vertices the midpoints of the diagonals and the point of intersection of those two lines_
Constructive description ( (circle ABC D) (circumcenter 0 A B C) (inter I (I A D) (I B C» (midpoint Q B C) (midpoint S A D) (midpoint J S Q) (!ratio M J 0 -1 ) (perpendicular M S B C) )
The machine proof
The eliminants
fr..a.M..
PCBM~2(pCBJ-tPCBO)
PCBS
M. -
!... -
i. -
9. -
2PCB ,-PCBD PCBS (2H}PCBS+iPCBQ-}PCBO) PCBS
PCBJ4t(PCBS+PCBQ) PCBSG;2(PCBD+PABC)
PCBQ~~( PBCB).
PCBOg!( PBCB)
PCBQ-PCBO+tpCBD+}PABC fpCBD+fpABC (2) -(-PCBO+}PCBD+}PBCB+tPABC) PCBD+PABC
Q (-2)-(-PCBD-PABC) -
(PCBD+PABC)-(2)
Figure 6-283
Example 6.284 (0.050, 2, 5) Show that the anticenter of a cyclic quadrilateral is collinear with the two symmetries of the circumcenter of the quadrilateral with respect to a pair of opposite sides_ Constructive description ( (circle ABC D) (circumcenter 0
A B C)
390
Chapter 6. Topics from Geometry
(midpoint Q B C) (midpoint S A D) (midpoint J (iratio Q, Q 0 -1) (iratio s, SO-I) (inter M (I 0
S Q) J) (I Q, S,»
SZJ. M SOQ, JS, JM- SJQ,s,
SZJ.
SJQ,S, ~2SSJQ,-SOJQ,
JM
M. ~
-
J 0 M) )
The eliminants
The machine proof
-
(midpoint
Soq,JS, sJQ,S,
SOQ,JS, ~ SSJQ, 'Q
SOJQ, 'Q -
-SOJQ,-2S0SJ 2SSJQ, -SOJQ,
(SOJQ, +2SoSJ) (SOSJ) 2{SoQJ)
Ii! -( 2SOS J-2S0QJ) Q,
Figure 6-284
-
- 2SOSJ+ 2SOQJ
.im!!1i!u
1
Example 6.285 (0.050, 1, 12) Show that the product of the distances of two opposite sides of a cyclic quadrilateral from a point on the circumcircles is equal to the product of the distances of the other two sides from the same pointo Constructive description ( (circle ABC D E) (foot PEA B) (foot Q E B C) (foot R E C D) (foot SEA D) (eq-product ESE Q E PER)
__
)
(16) oS~PE R(16) 0S1,PE PADA' PERE PCDC (16)0~aE P (16)0~ HE PEQE PBCB' PEPE PABA S
PESE oPE9E PEPE oPE1<E (16S~QE) o PEqE
-
PEPE oPERE oPADA
.!i
(16) o S~DE o PE9EoPCDC
-
PEPE o(16SbDE)oPADA
5l
PEPEosbDE oPADA oPBCB
.f..
(16) o S~PE o s~aEoPCDCoPABA
-
PESE
Q
PBCB=2{Be·). PADA=2{AD') DEofEocv (-2) oJ BEoiEoA"8 SABE (-2) oJ SCDE
S~QE o (16ShcE)oPCDC
-
PABA=2{A"8')
(16S~BE)oSbDE o PADA OPBCB
co~cir ( DEoiEoAD)'o( fE oBEoBe)' 0(2CV')0(2A"8')0(2d)' -
2
PCDc=2{cv') fE oBEoBe SBCE (-2) oJ
j
(-BE oAEoAB)'0(-DEoCE oCD)'0(2AD )0(2BC )0(2d)'
.im!!1i/u
E
The eliminants
The machine proof §.
~~
Figure 6-285
SADE
DEoiEoAD (-2)oJ
1
Example 6.286 (0.050, 1, 12) Show that the projections of a point of the circumcircle of a cyclic quadrilateral upon the sides divide the sides into eight segments such that the product offour nonconsecutive segments is equal to the product of the remaining fouro
391
Consb'Uctive description ( (circle ABC D P) (circumcenter 0 A B C) (foot A, P A B) (foot B, P B C) (foot C, P C D) (foot...!2. P D & (.!!I.~~~ = AlB SIC OlD DIA
1»
Figure 6-286
The machine proof
The eliminants
~ . ~ .!.!.J. ..!!I.
~ ~..E.iIlJ!... ~ ~ ...fD.CE.... AD, - -PDAP' DO, - -POOP
AD, DO, OB, BA,
!.!.J. l!! ..£raL.
~ ...EA.D.L • ~.!.!.J. ..!!I. -POAP
@ ~
PApe ·Pac e
PDAp ·(-PODP)
•
-
BA, - -PABP PABP= - 2{iiP.As'coo(AP»)
!.!.J. ..!!I. OB, BA,
PAne ' POGP ' POle
PDAP ,PODP ,(-PBOP) ~
OB, - -PBOP
.!!I. ~...f.B..A.tL.
DO, OB, BA,
• ~
PBOP= - 2{Ci>.iiC.coo(BP»)
BA,
PODP= - 2{.5P.Cb.cos(CP») PDAP=2{AP.Ai>.coo(DP»)
P,pp ·PnCP · PCBp ,PA.p
PDAP"PODP"PBOP ,(-PABP)
PBAP=2{ AP.As.coo(BP)}
",,=..cir (-2.5P.Ai> .coo(AP)H2Ci>.Cb'coo(DP» .
-
(2AP.AD.coo(DP».( -2fjp.Cb .coo(c P» (2iiP .iiC.coo(cP».(2AP.As.coo(BP» (-2CP.iiC.coo(BP)H -2BP.AB.coo(AP»
POBP=2(iiP.iiC.coo(cp») PDop=2( Ci>.Cb'coo(DP)} PADP= - 2(.5P.Ai>'cos(AP»)
.im~ifJl 1
Example 6.287 (0.866, 4, 29) The four lines obtained by joining each vertex of a cyclic quadriklteral to the orthocenter of the triangle formed by the remaining three vertices bisect each other. v
o ~-----=:::ilIjIB
Figure 6-287
Figure 6-288
ConSb'Uctive description ( (circle ABC D) (orthocenter D, A B C) (orthocenter A, BCD) (inter M (I D D,J (I A A,J) (midpoint M A Ad
)
Example 6.288 (1.750, 11,35) A line AD through the vertex A meets the circumcircle of the triangle ABC in D. If U. V are the orthocenters of the triangle ABD. ACD. respectively. prove thllt UV is equal and parallel to BC.
392
Cbapter 6. Topics from Geometry
Constructive description ( (circle ABC D) (orthocenter U
A B D)
(orthocenter v
A C D)
(eqdistance v
U B C) )
Example 6.289 (1.633, 12, 35) The sum of the squares of the distances of the anticenter of a cyclic quadrilateral from the four vertices is equal to the square of the circumdiameter of the quadrilateral. Constructive description ( (circle ABC D) (circumcenter 0 A B C) (orthocenter A, BCD) (midpoint M A A,)
,
__---.D
(MA' +M7f +MC' +w = 4W» Figure 6-289
Example 6.290 (0.750,6,41) The line joining the centroid of a triangle to a point P on the circumcircle bisects the line joining the diametric opposite of P to the orthocenter. Constructive description ( (circle ABC P) (circumcenter 0 A B C) (orthocenter H A B C) (centroid GAB C) (lratio Q 0 P -1) (inter I (I H Q) (I P G)) (midpoint I Q H) ) Figure 6-290
Example 6.291 (0.616,9,48) Show that the perpendicular from the point of intersection of two opposite sides. produced. of a cyclic quadrilateral upon the line joining the midpoints of the two sides considered passes through the anticenter of the quadrilateral. Constructive description ( (circle ABC D) (circumcenter 0 A B C) (inter I (I A D) (I B C)) (midpoint Q B C) (midpoint S A D) (midpoint J S Q) (lratio M J 0 -1) (perpendicular I M S Q) )
Figure 6-291
Example 6.292 (0.450, 3, 31) If Ha. Hb• He. Hd are the orthocenters of the four triangles determined by the vertices of the cyclic quadrilateral ABC D. show that the vertices of ABeD are the orthocenters of the four triangles determined by the points Ha. Hb• He> Hd•
Constructive description ( (circle A B c D ) (orthocenter HD A B C ) (orthocenter HA B c D ) (orthocenter HC A B D ) (perpendicular HC HA B H D ) ) D
6.4.6
@
'
H.
Figure 6292
Orthodiagonal Quadrilaterals
Defmition A quadrilateral is said to be orthodiagonal if its diagonals are perpendicular to each other. Example 6.293 (Theorem o f Brahrnagupta) (0.133, 4, 15) In a quadrilateral which is both orthodiagonal and cyclic the perpendicular from the point of intersection of the diagonals to a side bisects the side opposite.
Constructive description ( (points A B C ) (circumcenter o A B C ) (foot E B A C ) (inter D (1 B E ) (cir o B ) ) (foot FF E C D ) (inter F (1A B ) (1E FF)) (midpoint F A B ) )
The eliminants ~ ~ S A E F
.s
FJ-P
-
SBEF~
BF
PCDEg -(~~oEE-~cBE)~(~oEo-~BoE~ PBEB -(POEO-PBOB).SABE PBEB ~EcDgpBcs EP .S D
SAED=
~
B
C
E
=
~
E~PEAC+P& PACA ~Pcso.Paro+P~on.P= POBE= PACA &BE=
The machine proof
Figure 6-293
394
Chapter 6.
Topics from Geometry
PaCe ,SASf: ( 2POBE-PCBE) ,SBCE
E
P:CB , PSAC ,SABC,P!CA
= (2PCBO,PBAC ,PACA-PBCB ,PBAC,PACA+2PACB , PACA ,PABO Jim1!!.i!lJ
PAcB'PA cA,PABc) , PAcB ' SABc , plcA
PAQ8,PBAG
2PCBO,PBAC-PBCB , PBAC+2PACB ,PABO-PACB , PABC
Q -
PACB , PBAC ,(2)' -4PACB , PABC+4PACB , PABA
.im!!!.i!y
-PRAQ PABC-PABA
~ -(-PBCB+PACA+PABA) ,(2) Jim!!!.i!y 1 -
(PBCB-PACA-PABA) ,(2)
-
Example 6.294 (0.333,4,34) Let E be the intersection of the two diagonals AC and BD of cyclic quadrilateral ABCD, Let I be the center of circumcircle of ABE. Show the IE.L DC.
Constructive description ( (circle ABC D) (inter E (1 B D) (1 A C» (circumcenter I A B E) (perpendicular I E C D) ) Figure 6-294
Example 6.295 (0.600,7,16) In an orthodiagonal quadrilateral the two lines joining the midpoints of the pairs of opposite sides are equal. Constructive description ( (POints A B C) (foot FD B A C) (on D (1 B (midpoint P A B) (midpoint Q B C) (midpoint R C D) (midpoint S D A) (eqdistance S Q P R) )
FD»
c Figure 6-295
Example 6.296 (0.850,7,30) In an orthodiagona/ quadrilateral the midpoints of the sides lie on a circle having for center the centroid of the quadrilateral. Constructive description ( (POints A B C) (foot FD B A C) (on D (1 B FD» (midpoint P A B) (midpoint Q B C) (midpoint R C D)
c
A
D
Figure 6-296
395
6.A. Quadrilaterals
(midpoint 5 D A) (inter 0 (I Q 5) (I P R» (perp-biesct 0 5 R) )
Example 6.297 (3.300, 36, 29) If an orthodiagonal quadrilateral is cyclic. the anticenter coincides with the point of intersection of its diagonals. Constructive description ( (POints A 8 C) (circumcenter 0 A 8 C) (foot MBA C) (inter D (18M) (cir 0 8» (midpoint J 0 M) (midpoint R C D) (inter P (1 A 8) (I J R» (midpoint P A 8) )
Figure 6-297
Example 6.298 (0.967, 14, 15) In a cyclic orthodiagonal quadrilateral the distance of a side from the circumcenter of the quadrilateral is equal to half the opposite side. Constructive description ( (POints A 8 C) (circumcenter 0 A B C) (midpoint P A 8) (foot Fo 8 A C) (inter D (1 8 Fo) (cir 0 8» (Pcoc = 4Popo) )
Figure 6-298
Example 6.299 (0.967, 15, 14) If a quadrilateral is both cyclic and orthodiagonal. the sum of the squares of two opposite sides is equal to the square of the circumdiameter of the quadrilateral. Constructive description ( (POints A 8 C) (circumcenter 0 A 8 C) (foot Fo 8 A C ) (inter D (I 8 Fo) (cir 0 8ll
(liB' +OC' = 4oK)
)
Figure 6-299
Example 6300 (2.117,26,20) Show that the line joining the midpoints of the diagonals of a cyclic orthodiagonal quadrilateral is equal to the distance of the point of intersection of the diagonals from the circumcenter of the quadrilateral .
396
Chapter 6.
Topics from Geometry
Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot E B A C) (inter D (I B E) (cir 0 B» (midpoint U A C) (midpoint v B D) (eqdistance U v 0 E) ) Figure 6-300
Example 6.301 (1.050, 16, 16) If the diagonals 0/ a cyclic quadrilateral ABG Dare orthogonal. and E is the diametric opposite 0/ D on its circumcircle. show tlult AE = G B.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot FD B A C) (inter D (I B FD) (cir 0 B» (inter E (I D 0) (cir 0 D» (eqdistance A E C B) ) Figure 6-301
Example 6.302 (1.250, 4, 84) AlBl GlDl is a quadrilateral with an inscribed circle O. Then 0 is on the line joining the midpoints 0/ AlGI and BID l . Constructive description ( (circle ABC D) (circumcenter 0 (on BT (t B B 0» (on AT (t A A 0» (on CT (t C C 0» (on DT (t D DO» (inter A, (I B BT) (I A AT» (inter B, (I C CT) (I B BT» (inter c, (I DDT) (I C CT» (inter D, (I A AT) (I DDT» (midpoint M A, C,) (inter N (I B, D.) (l 0 M» (midpoint N B, D,) )
6.4.7
A B C)
4<------':::-.+-=--~c. 0,
o
Figure 6-302
The Butterfly Theorems
For the machine proof of the general butterfly theorem for a circle. See Example 3.81 on page 143 Example 6.303 (0.516,4,31) The ButteTjly theorem/or a circle.
397 F
Constructive description ( (circle
A 8 C D)
(circumcenter 0 A 8 C) (inter E (I A 8) (l C D» (inter G (I 8 C) (t E E 0» (inter F (I A D) (t E E 0»
2»
(~ =
Figure 6-303
Example 6.304 (Butterfly Theorem for Quadrilaterals) (0.166, 3, 19) Let ABC D be a qUiJdrilllteral such that AB = BC and AD = CD. M is the intersection of AC and BD. Passing through M two lines are drawn which meet the sides of ABC D at P, Q, S, R. Let G = PRnAC. H = SQnAC. Show thot GM = MH.
Constructive description ( (POints
(on
8
A C)
(b A
C»
(midpoint
(on (on
M A C) D (18M)) p (I A 8))
(inter Q (I
C D)
(on S (I 8
C»
(I
P M»
(inter R (I A D) (I S M» (inter G (I A C) (I P R» (inter H (I A C) (I S Q» (~=¥C»
The machine proof !!. ( ~)/{14.11.) AG CN -
!!
-SCQS • ~ -SMQS AG
Q -
SMPR ,SCgS SMQS ,SAPR
(-SIIIPS ,SAMD),SCgS ,SAMDS
-
Aof---+--ll:.!!!.+---l...
SMQS ,(-SADP'SAMSH-SAMDS)
,i"'1!!.i/" -
-SMPS ,SAMD ,SCgS SMQS ,SADP'SAIIIS
= (-SBMQ·~+SBMQ+SCMQ JM2 JM2 JM2 · ~)·SADP ·(SABM · ~-SABM) S
-(-SBMp · Il.+SBMP+SCMp·Il.),SAMD,(SCBQ · Il.-SCBQ)
D
Figure 6-304 ,im_.Ji/1I
:c
-(SBMP'Il.-SBMP-SCMp · Il.),SAMD ,SCBQ
JW
JW
(-SCMDP · SBMP · SCMD · ~+SCMDP'SBMP · SCMD+SCMDP ·SCMP'SCMD ' ~) ' SADP , SABM , SCMDP .im~i/J/
-SAND ,SCAlP ,SCBa SCIIID ,SADP'SABIII
398
Chapter 6. Topics from Geometry o
o(-SCBM 4f.) oSCBD =P -SAMD ...41' SCMDo(-SABDo*)OSABM
sim1!!.i/lI -
-SAUO,SCSM ,SCRP SCMD oSABD oSABM
=
D -(SABM °!J.IL-SABM) oSCBMoSCBM"Jl2.
JUt
BM
(SCBM °f:S- - SCBM) oSABM °ts- °SABM
simJ!lify -
=.&...a.M.. M. SABM
-
-(1SACB)
simJ!lify I
-
- 2 SACB
Example 6.305 (Butterfly Theorem for Quadrilaterals) (0.066, 2, 14) Let ABC D be a quadrilateral such that the intersection M of its diagonals is the midpoint of ACo Passing through M two lines are drawn which meet the sides of ABC D at P, Q, S, R. Let G = PRnAC. H = SQnACoShow that GM = MHo
Constructive description ( (points A C B) (midpoint M A C) (lratio D M B r.) (lratio P A B r2) (lratio R A D
The eliminants
(~)/(~)
I:UJJ!. S MqS
~ -SCQS
(I C D» (1 B C» (I P R» (I S Q»
(~~=~»
0
-SMQS
Q
r3)
(inter Q (1 P M) (inter S (I R M) (inter G (I A C) (inter H (I A C )
The machine proof
CH - SCQS
I:!.Si AG
SMPR oSCQS SMQs oSAPR
-
§..
SMPRo(-SCMRoSCBq)o(-SCMBR) ( SMRQ oSCBM)oSAPRoSCMBR
-
simJ!lify -
-SMPRoSCMR oSCBq SMRQ oSCBM oSAPR
-SMPR"SCMR oSCMp oSCBD o(-SCMDP) SMPR oSCMD oSCBM oSAPR oSCMDP
~im1!.!.illl Saws,SOMp-SCRP -
!!
SCMDoSCBM oSAPR
SQMvora oScMpoSCBD SCMD oSCBM o(-SADp or3)
-
£. -
CQS
S -SCMR oSCBq SCMBR
S
Q SMPRoSQMQ
MRQ
SOB
-SCMDP
g,SCMP·SqRQ SCMDP (SADPo,,)
SCMR~SCMD or3 SADPt': -
(SABD o,,) (SCBM o,,)
-(-SCBM",,)oSCBD SCBM o(-SABDo,,)
SCBDg -
«r,-I) oSCBM)
=:.&J..a.rJ. SABD
Q -( -SCBWr, +SCBM) SABM or,+SABM
.imJ!li f y -
=
-
«r,-I)oSABM)
-
-
SAPR~
SCMPt': -
.imJ!l.if y
M
S
SMRq oSCBM SCMBR
SABDg -
-
Figure 6-305
! MQS-
-SCMpoSQBD SCBM oSADP
sim2!.ify
D
AG - SAPR
S
Q-
£: B
1:!.Sif!~
=.&...a.M.. SABM
-(lS ACB ) -tSACB
.imJ!l.ify
SABMM -
!(SACB)
SCBMM!(SACB)
399 Example 6.306 (0.100,2, 15) Let ABCD be a quadrilaJeral and M the intersection -of its diagonals_ Passing through M two lines are drawn which meet the sides of ABC D at P,Q,S,R. LetG = PRnAC. H = SQnAC. Show that MG AG
MHCM
= CH MA '
Constructive description 1be machine proof .wi ( (POints A B e) ...MJ
(lratio MAe ro)
-1Ul-SlM. CH AM
(lratio 0 M B rl)
~
g
(lratio R A 0,..)
(inter Q (I P M) (1 eo» (inter S (I R M) (I Be» (inter G (I A e) (I P R» (inter (~
B
(I
A e)
(I S
Q»
= *ff'iff» 8
•
.wi AO
-
(-SMRO -SBCM)· ~ ·SAPR-SBMCR
.im!4i/ll
SMPR -SCMR -SBcq
SMRO -SBCM - ~ -SAPR
£:
SMPR-SCMR -SCMP -§"S;P-( -SCMpP)
SMPR -SCMD -SBCM - ~-SAPR-SCMDP
.'m~'ltI
-SaMB.Sa~8CQ SCAID-SBCM - ~-SAPR
-SCMoon0§.gtpoSaQn
.im!4i/ll
SCMe.:§.ACO
SBCM - ~ -SADP ~
SR~",-SRCD
SBCM - ~-(-SABD-f'2) D .i"'!4i/ll
-SRCO
~-SABD
g
-=J-SBCWrl+SBCM)
~ -(-SABM -rl+SABM) .im!4i/ll
-SRCM
~ -SABM
M -
-(-SABC -ro+SABcl-ro
(ro-l)-SABC-ro
.im!4i/ll
S SMRq -SBCM SBMCR
MOS SCOS
S -SCMR-SBCq SBMCR
S
§.. -SMPR,(-SCMR-SBCq)-(-SBMCR)
SCMD -SBCM - ~-(-SADr,..)
Figure 6-306
AO - SAPR
S
-SMPR,SCqS
SAlOS · ~ ·SAPR
!! A''-~f---;;~--t~7'C
IUlt!.SMqS CH - SCQS
.wi!!~
-SCQS
-(-SMQS)'~
(lratio P A B ,..)
1be eliminants
OSMPR-SCMD
-SCMDP
MRO
S
OSCMP-SRcn
BCO
SAPR~
SCMDP -
(SADr,..)
SCAI R~SCM D-'"
SADP~
-
(SABD -f'2)
SCMP~SBCM -f'2 SABDg -
«rt-1)-SABM)
SBCDg -
«rl-l)-SBCM)
SABM~SABC-ro ~~!A=l AM
ro
SBCM~
-
«ro-l)-SABC)
400
Chapter 6. Topics from Geometry
Circles
6.5 6.5.1
Chords, Secants, and Tangents
Example 6.307 (The Secant Theorem) (0.016, 1, 8) The product of the distances of a given point, from any two points which are collinear with the given point and lie on the circle, is a constant. This constant is called the power of the point with respect to the circle.
The eliminants
Constructive description The machine proof ( (circle ABC D) E..s.=. PSED (inter E (I B D) (I A C» (eq-product K (-PACA · SSCD ·SASD) · S~8CD -
A E C E BED E) ) .
(-PSDS ·SACD ·SAScl·S ASCD
simJ!l.ify PACA oSSGooSABQ PSDS ·SACD·SASC _
-
(2A'C
2 ) -(
2
-ED·fib·fjC)·(-iiD·AboAiJ)
(2BD )-(-CD·AD.AC)·(-BC ·AC·AB)
simg1ifll
Figure 6-307
Example 6.308 (0.050, 4, 3) The power of a point with respect to a circle is equal, both in magnitude and in sign, to the square of the distance of the point from the center of the circle diminished by the square of the radius of the circle. Constructive description ( (points A B j (on 0 (b A B» (on E (I A B» (tPAES
The elirninants
= OE'-OA'»
The machine proof POEO-PAOA E
PASA ·
_
M 2 AB
E
-PASA · ~ AB
-
2
PSOS ·*-PAOA ·*+PASA· * simEJ.ify
(*-l) .PASA
-PASA"*
Q PASA · H-PAS A simEJ.ify
Psos-PAoA+PASA·if-PASA -
PASA·ii-PASA
Figure 6-308
-
Example 6.309 (The Tangent Theorem) (0.017 3 4) The product of the distances of a given point, from any two points which are collinear with the given point and lie on the circle, is equal to the square of the tangent line from the given point to the circle.
6.5
401
Circles
Constructive description «POints AT) (on 0 (b AT» (tratio P TOr) (inler B (I A P) (cir 0 All (eq-producl P T P T PAP B)
The machine proof
The eliminants
fr..ex.
PAPB~POPO-PAOA
PAPB
!i -
)
Popo~(rl+l)' PrOT PrPT~PrOT·(r)l
FrPT , PAPA
POPO ,PAPA-PAPA ,PAOA
sim~ifY
!:...
ProT ·r
-
PrOTgPAOA
PrPT POPo PAO A 2
PrOT'r'+PrOT-PAOA
Q~ P"'OA'r 2
-
p
s imJ!!.ify 1
T
Figure 6-309
Example 6.310 (0.066, 3, 5) From the ends D and C of a diameter of circle (0) perpendiculars are drawn to chord AB. Let E and F be the feet of the perpendiculars. Show that OE = OF,
Constructive description ( (POints A B D) (circumcenler 0 A B D) (iratio C OD - 1) ( fOOl E C A B ) ( fOOl FDA B)
(midpoinl (midpoinl
The machine proof
The eliminants
_a
-M4&-t
a=.;i4...2.
FM
4&_1 =~
M
-*+t
F -(4&-l) AR 2 ·PABA
M A B)
_
M E F) )
-
PBAD-fpABA
§.. -(PBAc-tPABA),PABA (PBAD-tPABA),PABA Bim1!! i fll
-
£. -
FM
~_1
AB
•
~!:.bJ..w. AB- PABA 4&!.~ AB- PABA PBAC
g2( PBAO- t PBAD)
PBAOg~(PABA)
-(PBAc-tPABA) PBAD-tPABA
-(2PBAO-PBAD-tPABA) PBAD-tPABA
Q (-2),(-PBAD+tPABA) -
c ~---
Figure 6-3 \0
(PBAD-t PABA)·(2)
.im~ifll 1
Example 6.311 Let A, B, C, D, E and P be six cyclic points, From P perpendicular lines are drawn to AB. BC. CD. DE. and EA respectively, Let thefoot be L, M, N,. K. and AL BM CN DK ES 1 S , Sho wthatTB'M'C"ND"KE''SA--'
402
Chapter 6. Topics from c - t r y
Constructive description «circle ABC D E P) (foot L P A B) (foot M P B C) (foot N PC D) (foot K P D E) (foot S PEA) (~~rH~~Hf
=
-1»
Figure 6-311
The eliminants
The machine proof
u.~~ 1lJS.!!h.D.L AS - -PEAP' EK - -PDEP !i.I:f.!!.~ Ul.~...!±.B.L DN--PCDP' CM--PBCP 4!..f....EJuL BL --PABP PABP= - 2(iiP.AiJ.cos(AP»)
M, .tu& .g.lU4 .Al!. AS EK DN eM BL
~ ~ . ~ . g . IJ:I. . 4!. -PEAP
~
t!. -
M.
-
EK DN CM BL
-Pu;p·P,mp
PEArt PDEP)
•
!i.I:f..IIJ4..4!.
PBCP= - 2(&.iiC .cos(BP»)
DN CM BL
PAF;P'PF:Dp'Pacp
PEAP,PDEP '( -PCDP)
•
PCDP= - 2(.i5P.CiJ.cos(CP»)
IIJ4. 4!. CM' BL
-PABe-PgoP'POCp,PCBP
PEArPDEP ,PCDP ,(-PBCP)
PDEP= - 2(EP.jjE'C08(DP»)
• ~
PEAP=2( AP.XE'cos(EP»)
BL
k
PA gP ·PgDP·PnCP·PQ8p ,PRA P
-
PEAP ,PDEP,PCDP,PBcr( PABP)
PBAP=2( AP.AiJ.cos(BP») PCBP=2(iiP.iiC.cos(cp»)
co~r -( -2EP.XE'cos(AP)).(2.i5P.jjE'cos(EP)) (2A"P.A'E'cos(EP)).(-2EP.iiE'cos(DP))
PDcp=2( & .CiJ'cos(DP») PEDP=2( .i5P.jjE'cos(EP»)
(2& .CiJ'cos(DP»)-(2iiP.iiC'cos(CP»)-(2AP.AiJ'cos(BP)) (-2M.ED.cos(C P))·( -2CP·BC,cos(BP)).( -2BP.::iB.cos(AP))
PAEP= - 2(EP.XE .cos(AP»)
sim!!J.ify 1
Example 6.312 (0.416,4,22) If ABCD is a rectangle inscribed in a circle, center 0, and if P X, P Xl, PY PYi. are the perpendiculars from any point P upon the sides AB. CD, AD, BC, prove that P X . P Xl + PY . PYI is equal to the power of P with respect to the circle O. Constructive description ( (POints A B P) (on D (t A A B)) (inter C (P B A D) (P DAB)) (inter 0 (I B D) (I A C)) (foot x P A B) (inter X, (I C D) (I P X)) (foot Y P A D) (inter Y, (I B C) (I Y P))
(or -"('5;(' = tPxPX , +tPyPY,) )
A
X
B
Figure 6-312
Example 6.313 (0.150, 3, 15) Let D be a point on the side C B of a right triangle ABC such that the circle (0) with diameter CD touches the hypotenuse AB at E. Let F = AC n DE. Show that AF = AE.
6.5
403
Circles
The eliminants
Constructive description ( (POints C E) (on D (t E E C)) (midpoint 0 D C) (inter A (t E E 0) (t C C 0)) (inter B (1 D C) (1 A E)) (inter F (1 D E) (1 A C)) (eqdistance A F A E) )
P AFA
F P9AdSEPA)2 (SOEAP )2
S
A POEQ ,PficQ-16SCEQ ·SCEQ
OUP ( 16).SOEO P A PEOE·(PEOOt EAE (16).(SOEO)
s
EDA
Po
A POfjo ·PEen
(16).SOEO
~POOdPCE;0)2
OAO- (16).(SOEO)2 SOEOgt(SOED)
F
PEOOg~(PEOD) PEOEg1( 2PEDE- PODO+ 2POEO)
PDEog~(PEDE) POEOg~(POEO) PooOgi(PODO) D
SOED= B
i1( POEO · -) ~
PEODgPOEO
PEDEgPOEO ·(~)2 D -2
Figure 6-313
PODO=(~ H)·POEO
The machine proof fAu. PEAI'
!:. -
POAo ·S1 p A PEAE ·S'bEAD
~ Pooo · P~EQ ·«-tPDEO . PEOO»2 . «-4S0EO»' ·(16ShEQ) - PeoE·Pioo ·(: PDEO ·PEOO-4S0EO ·SOED )2.« -4S0EO))2 .(165&EO) .im!!1i/~
Q -
POOdPOEO)2 .(ppEQ)2 PEOE ·(PDEO ·PEOO- 16SOEO ·SOED)2
(lPODoHct POEO))2 ·«}PEDEW (¥PEDE-!PODO+f p OEOH!PEDE ·PEOD-85&ED)2 -2
-2
Q (POEO · ~ +POEO) ·(POEO)2 .(POEO · ~ )2 .im1!!.i/1I - ! i
j
(POEO · ~ +POEc) ·«-P~EO · ~ ))2
-
1
Example 6.314 (3.250, 21, 23) Show that the sum of the powers. with respect to the circumcircle of a triangle. of the symmetries of the orthocenter with respect to the vertices of the triangle is equal to the sum of the squares of the sides of the triangle. c, Constructive description ( (points A B C) (circumcenter 0 A B C) (foot DAB C) (foot F C A B) (foot E B A C) Figure 6-314
404
Chapter 6.
(inter H (I B E) (i (!ratio A, A H -1) (!ratio B, B H -1) (!ratio c, C H -1)
Topics from Geometry
A D»
(A,02_oB'+B.O'-oB'+c.o'-oB'
= AF+BC'+CA'»
Example 6.315 (0.700, 17, 17) Two unequal circles are tangent internally at A. The tangent to the smaller circle at a point B meets the larger circle in C, D. Show thot AB bisects the angle CAD.
Constructive description ( (points A B C) (inter 0, (t B B C) (b A B)) (inter 02 (I A 0.) (b A C» (inter D (I B C) (cir 02 C))
(eqangle CA
B B A D) )
Figure 6-315
Example 6.316 (1.850, 65, 19) If G is the centroid of a triangle ABC, show that the powers of the vertices A. B, C for the circles GBC, GCA, GAB, respectively, are equal.
Constructive description ( (points A B C) (centroid GAB C) (circumcenter 0, B C G) (circumcenter 02 A C G) (::w,' -o.G' = 00;" -M) ) Figure 6-316
Example 6.317 (0.083,4,10) Let C be a point on a chord AB of circle O. Let D and E be the intersections of perpendicular of OC through C with the two tangents of the circle at A and B, respectively. Show that CE = CD.
Constructive description «points A B) (on 0 (b A B)) (!ratio CA B r) (tratio x B 0 s) (inter E (I B X) (t C C 0)) (inter D (I C E) (t A A 0» (midpoint C D E) ) Figure 6-317
6.5
Circles
405
The machine proof
~!!.~ CE--POCAE
-~ CE
POCAE! -(PeRO K ' PoAa+P~gB"O~8AQ-PQ A y ' Paca)
Q~ - -POCAE
PO AX?£PBAO-4SABO " POCX?£PBCo-4SBOC "
~
-
The eliminants
PQA'o · PCAQK
-PCBOX ,POAC-POCX ,PBAO+POAX ,PBCO
PCBox?£4(SBOC") PBCOg -
«r-1)'(PBAO-PABA ·r») POAC'!4SBOC" ) (4POAC ,SBOC·.+4PBCO ,SABO ·' 4PBAO ,SBOC " ) SBOCg - «r-1),SABO) POAcgPBAo·r, PBOBgPAOA .im~iJJI -POAe ,SAOO PBAO= - !(PBOB-PAOA-PABA) POAC ,SBOC+PBCO ,SABO-PBAO ,SBOC ~
-
f. -
-PBAo·r·( -SABO·r+SABO) -PBAO ,SABo ·r'+PBAo ,SABo ·r+PABA ,SABO ·r2-PABA,SABo·r
~i m~i/Jl
-PAAQ
PBAO
PABA
~ -(-PBOB+PAOA+PABAH2)
-
(-PBOB+PAOA-PABAH2)
QE.!..a..s, -
PABA
Example 6.318 (0.083,5,14) Let G be a point on the circle (0) with diameter BC, A be the midpoint of the arc BG. AD .1 BC. E = AD n BG and F = AC n BG. Show that AE = BE( = EF). Constructive description
( (POints G
B)
(midpoint M B G) (tratio A M B r) (inter 0 (I A M) (b A B» (!ratio COB -1) (foot DAB C) (inter E (I A D) (I B G» (inter F (I A C) (I B G» (eqdistance A E B E) )
Figure 6-318
Example 6.319 (0.050, 2, 6) Let D be the intersection of one of the bisectors of LA of triangle ABC with side BC, E be the intersection of AD with the circumcircle of ABC. Show that AB . AC = AD . AE.
406
Chapter 6. Topics from Geometry
Constructive description ( (circle A B C) (circumcenter 0 A B C) (lratio DB C~) (inter E(I A Drtc1¥ 0 A» (PDAAE
The machine proof
PDAE~(2). POAD
P~f..... (2)·AC·AB ~
= 2)iB.A'C) )
P. DA'C ,PR,4g+AB ,PCAO AC+AB DAD
2P~g·PAQA (2) ·AC·AB,PADA
.im~if1J
PCAOg!
.J:B..4g..
PBAOg!
AC·AB
g A£:rijePt1B'~AQ AC·AB·(AC+AB)
g
The eliminants
PACA=2(A'C') PABA=2()iB')
2A'C' ~R!l.t2-1£'PAGA AC·AB·(AC+AB).(2)' .
-
2 --
----2
co-cor 2AC ·AB+2AC·AB
=
(2)'AC.iB.(Ac+AB)
.im~if!J 1
Figure 6-319
Example 6.320 (0.067,5,12) Let ABC be a triangle. Through A a line is drawn tangent to the circle with diameter BC at D. Let E be a point on AB such that AD = AE. The
perpendicular to AB at E meets AC at F. Show that AE/ AB
Constructive description ( (POints E D) (on A (b E D» (on B (l A E» (tratio 00 D A 1) (inter 0 (I 00 D) (b B D» (lratio COB -1) (inter F (l A C) (t E E A» (eq-product A B A CAE A
= AC/ AF.
The eliminants 4f:~~ AC- PEAC PEACg2PEAO-P AB o 2PDBO q ·{iEAD-PDBD,PEAOq -PDBD ,PEAD PEAO ( 2) , PBDOO PEAOO oJ> PEAD+4SEDA PDBOO oJ> PDBD-4SDAB F) )
PBDOO °J>4(SDAB)
SDAB~SEDA ' * PEAB~PEAE' * PDBD~ - (pDAD · a-PDAD-PEAE· ~' +PEAE·~-PEDE ' ~) AE AE AE AE ~~ EA
-
(~) AE
PEAD=!(PDAD+PEAE-PEDE) PDAD 4PEAE
The machine proof
(-*)/(*) ~~.-a PEAE
g
g
EA
a
-(2PEAO-PEABl • PEAE EA -(PBDOq,PEAB+2PDBOq,PEAD-PDBD ,PEAOq-PDBD ,PEAD) • ~ PEAE,(-PBDOo) EA
Figure 6-320
6.5
qg
4PORO ,S,.,OAt4PUR,SOAR 8Pr;;AQ,SpAR • .u& P£,Ud4SDAB) EA
B -
_
407
Circles
44·(-PDAD ,S£DA · .u&A" +PDAD ,S£DA +2PEAD ,SEDA · .u&-PEAE ,SEDA · .u&+P£DE ,SEDA . .u&) AS Afi A~
_!::
AS
-
PEAE ,S£DA ·i f ·(-l)
.im..Ji/~ -(PDAD·44-PDAD-2P£AD · .u&+P£A£· .u&-P£D£· .u&) 'J::. A~ '6 Ali AI} -
P£A£
eJI -( -2PDApl -
P£A£ ·(2)
~~ -
P£A£
.im~i/~ 1
Example 6.321 (0.600, 2, 40) The cross ratio of four points on a circles with respect to any points on the circle is constant. Consttuctive description (circle A E E, BCD) (inter G (I A B) (I C E» (inter F (I D E) (I A B» (inter G, (I E, C) (I A B» (inter F, (1...& 12l.JI A B» (4L . ~ SF AG
=~ . ~) ) BF. AGt
Figure 6-321
The machine proof ~ . 4L
~
~.~ AD, BF,
~
-S£,BD ~S AO. · AEID
•
/lSi. .4L AD BF
~
-S£,BD,SA£,C • /lSi. .4L (-S£,BC) ,SA£,D AD BF
t;
(-SAED) ' S£, BD ,SA£, C • /lSi. S£, BC,SA£, D ,S£BD AD
Example 6.322 (0.083,1,17) The tangent to a circle at the point C meets the diameter AB. produced. in T; Prove that the other tangent from T to the circle is divided harmonically by CA. CB. CT and its point of contact.
408
Chapter 6. Topics from Geometry
Constructive description «POints A C) (on B (t C C A)) (midpoint 0 A B) (foot EC A 0) (lratio DEC -1) (inter T (I A 0) (t C C 0)) (inter A, (l T D) (I C A)) (inter B, (I T D) (I C B)) (hannonic T D A, B,) )
The elirninants
The machine proof
~l!!&:.az.
(-~)/(~) DAI DB}
DB, - SCBD
~~~
DA, - SACD S T PRoo ,SAGa CBT PAOC S r..PACO-SACQ ACT- -PAOC
~~ . -~ SCBT
DA,
~ -(-SACT)·SCBD
SCBT '( SACD)
-
1:..
-PACO ·SACO,SCBD ,(-PAOC) ( PBCO·SACO)·SACD ·( PAOC)
-
sim1!1 i fy
PAQQ ·Scap
SCBDg2(sCBE)
s
E PCAo oSACQ
Q PAco·(2ScBE)
ACE PAOA S E PARO -SAgB CBE PABA
-
PBCO,SACD
-
PBco ·(2SACE)
SACOgt(SACB)
K
PACQ-PA8Q"SACS , P6QA
-
PBCO ,PCAO ,SACO,PABA
Q
(tPACA) ·PABC,SACB ·(tPABA) (fpCBcl·(fpCABHtSACB) ·PABA
PCAOg¥(PCAB) PBCo g PCBC)
-
sinlJ!!.ify -
PACA,P68G
PCBC ,PCAB ell PACA ,(PCBC+PABA-PACAH2) - PCBC,(-PCBC+PAB A+PACA)·(2)
Figure 6-322
SACDg2(SACE)
!}. -2PCBC , PACA - -2PCBC ,PACA
1(
PAOAgl(PABA) PACog2(PACA) (PCBC-PABA-PACA) PCAB 2 PABC=l(PCBC+PABA-PACA)
PABA~PCBC+PACA
sim!}j,ifu 1
Example 6.323 (0.650, 10, 19) Let R be a point on the circle with diameter AB. At a point P of AB a perpendicular is drawn meeting BR at N. AR at M. and meeting the circle at Q. Show that PQ2 = PM · P N. Constructive description ( (POints A Q R) (circumcenter 0 A Q R) (lratio BOA -1) (foot P Q A 0) (inter M (I P Q) (I A R)) (inter N (I P Q) (l R B)) (eq-product P M P N P Q
P Q) )
Figure 6-323
Example 6.324 (0.066, 3, 12) Through point F on the circle with diameter AB a tangent to the circle is drawn meeqng the two lines. perpendicular to AB at A and B. at D and E. Show the OA 2 = DF . EF.
6.5
409
Circles
Constructive description
The machine proof
The eliminants
( (POints
~ PAOA
P
(on
A F)
B (t F F A»
(midpoint 0
(on
~ -PWfO , PABf
A B)
-
M (t F FO»
(inter (inter
PAOA ,PSFAM
D
(l
F M) (t A A B»
Q
E
(I
F M) (t B B A»
-
(eq-product
D F F E 0 A 0 A)
)
E
PEME ,PEAR , PARE
PAOA ,PSFAM ,( -PSFAM)
M PPOF ' ML'J ,PFAB ,PASF fO -
PSFAM
PAOA·« -4SAFS ·* » '
D PEME ,PEA
MFD
PSFAMt:!, -
4(SAFS '* )
PFMFt:!,PFOF '(I!J5)' PAOAgHpASA) PFOF e t(2PFSF-PASA+2PAF A) SAFS= -
Hu,.u.)
PASF=2(u, )')
e(iPFSrlPAfA+iPAFA),PFAS ,PASF ( •• ).( PASA) ,(SAFS)'
PAFA=2(u,j')
-
(16).(2u,+2ufl·« -u, ·u.»'
.i~i/lI
In the above machine proof, F
8
-PSFAM
.imJ!!.i/1I PfOf ,PfAR ,PARf (16),PAOA,(SAFS)2
~ (2u~+2up '(2un·(2u~) '«2»'
Figure 6-324
E PUCQ ,PARE
DFE P
PFAB=2(u,j') PABA=2( u~+u~) PFBF=2(u,)')
1
= (0,0), A = (Ul' 0), B = (0, U2),
Example 6.325 (0.066, 2, 12) Let AD be the altitude on the hypotenuse BC of right triangle ABC, A circle passing through C and D meets AC at E, BE meets the circle at another point F. Show that AF 1.. BE,
Constructive description
The machine proof
The eliminants
( (POints A B) (tratio CAB H)
fu..E.
(foot
E.
PEBF~-(POEO-PBOB) POEO~PCOC
PEBF
DAB C)
(midpoint M C D) (tratio 0 M C I) (inter E (I C A) (cir 0 C» (inter F (I E 8) (cir 0 E»
(perpendicular
A F 8 E) )
C
-
PABs ·PaEe
-POEO ,PSES+PSES ' PSOS
.imgi/lI ~
PAB" -(POEO-PBOB) PABe
-
-(PCOC-PBOB)
Q
-PARC PCMC-PSMS
-
~
PARC -1JSi.' ,PCDC-IJSi. ,PCDC CD CD
.imJ!!.i/1I
P~
(~+1)' ~ 'PCDC
Q B
Figure 6-325
PABC,PSCB ,(PACS)' (-PSCS+PACBH-PsCB)· pl cB
.imgi/II
PARC £ !?uu. .imJ!!.i/1I PSCS-PACS - PASA -
PASE~PASC PBOSgPCMC ·(J+PSMS Pcocg«(J+l),PCMC Ml( (2~+l)'.PCDC ' ) PBMB=4 PCMCt:!,t(PCDC)
D~
PCDC= PSCS IJSi.E=l:.aJa. CD- PACS PACS£PASA,(r)' Pscs£(r'+l)' PASA PASC£PASA
Chapter 6. Topics from Geometry
410
Example 6.326 (0.067, 5, 9) Let M be the midpoint of the arc AB of circle (0), D be the midpoint of AB. The perpendicular through M is drawn to the tangent of the circle at A meeting that tangent at E. Show ME = MD. Constructive description ( (points A D) (on M (t D D A)) (lratio B D A-I) (inter 0 (I M D) (b A M)) (inter E (P M 0 A) (t A 0 A)) (perp-biesct M D E) ) Figure 6-326
Example 6.327 (0.200, 4, 11) The circle with the altitude AD of triangle ABC as a diameter meets AB and AC at E and F, respectively. Show that B , C, E and F are on the same circle. Constructive description ( (POints A B C) (foot DAB C) (midpoint 0 A D) (inter E (I A B) (cir 0 A)) (inter F (I A C) (cir 0 A)) (cocircle E F C B) )
The eliminants
The machine proof
s imJ!!.ify -
§..
-SBcE'(2PCdO-PBAC) SABC,PBBC
A
-(PBOB ,SABC-PAOd ,SdBC) ·(2PCdO- PBdC) ,PdBd SdBC·(-2PBOB ,PBdO+PBOB,PBdC+2PBdO ,PdOd-PBdC,PAOA) ,PABA
s i mJ!lify -
2PQAQ
PBAQ
2PBdO- P BAC
Q PaAO-Pade -
.g
PBAD-PBdC (-PBCB ,PBdC+PBdC ,PACB+PdCA ,PABC) ,PBCB ( PBCB ,PBAC+PBAC ,PABC+PACB ,PABA),PBCB
simJ!!.ify -
PaGS -Pade PBCB ,PBdC
P8AC-PAGR PAGA-PARG PBAC,PABC-PACB,PABA
ell (-2P9CR+2P~c. -4PACA 'PABA+2P~ R .)·«2))' aim.!!if" -
(2PBCB+2PACA-4PACA ,PABA+2PABA)·«2))'
-
Figure 6-327
6.5
Circles
411
Example 6.328 (0.033,1, 11) Let A, B, C, D be/our points on circle (0). E = CDnAB. CB meets the line passing through E and parallel to AD at F . GF is tangent to circle (0) at G. Show that FG = FE. By Example 6.309, we only need to prove the following statement. Constructive description ( (circle A B D C) (inter E (I A B) (I C D» (inter F (I B C) (P E A D» (eq-product F B F C E F E
F) )
The machine proof f.tw:;. PEFE f. PBCB ,SBDE,SACE'S'"snc PADA 'S1cE 'S!BDC l,.m~'I'II Paes-S80B -SACE
-
PADA ,( SBCE)2
~ PBCB '( SBPC ,SABDl-!
SAPC ,SABCl-!SApBC)' PADA ,«-SBDC ,SABC»' ,(SADBc)'
-
tlim1!!.'/JI ro~cir
-.
PRes ,SARo ,SAPG
PADA ,SBDC,SABC
- - -
- --
(2BC ,).(-BD .AD.ABl-! -DC·AC·AD)'«2d)' (2AD H-DC ·BC·BDH-BC·AC·AB)·«2d»'
aim!!:i!"
Example 6.329 (1.783 25 21) The bisector 0/ triangle ABC at vertex C bisects the arc AB a/the circumcircle a/triangle ABC.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (midpoint M A B) (inter N (10M) (cir 0 A» (eqangle A C N NCB) ) Figure 6-329
Example 6.330 (0.066, 2, 6) Let PT and P B be two tangents to a circle, AB the diorneter through B, and T H the perpendicular from T to AB. Then AP bisects T H.
412
Chapter 6.
Constructive description The machine proof ( (POints B P) -li (on 0 (t B B P» HI (inter T (cir 0 B) (cir P B» L~ - -SPAH (lratio A 0 B-1) (inter H (I B 0) (p T B P)) !i SPTA ,(-SBPO) - -SBTPA ,SBPO (inter I (I T H) (l A P)) simJ!!.ify hLL (midpoint I T H) ) -
SBTPA
Topics from Geometry
The eliminants
li!..2zLL HI--SPAH SPAH~SBrpA SBTPA 4
-
(SBPr- 2S BPO)
SPTA 4 - (2Spor+ S BPT) S r(2)'PBPO,SBPO BPT PPOP Spor;;' -
(SBPO)
~ -2SPOT-SSPT
-
:r. -
-SBPT+2SBPO (2PPOP ,SBPO 2PBPO ,SBPO)'PpQP (-2PPOP,SBPO+2PBPO ,SBPO)'Ppop '( -I)
Example 6.331 24 (0.083 3 15) Let M be a point on line AB. Two squares AM C D and BM EF are drawn on the same side of AB. Let U and V be the center of the squares AMCD and BMEF. Line BC and circle VB meet in N. Show that when point M moving on line AB. the line M N passes through a fixed point.
Constructive description «POints A B) (on M (I A B» (tratio C MAl) (midpoint U A C) (tratio E M B-1) (midpoint v E B) (midpoint 0 A B) (tratio R 0 A-I) (inter N (I B C) (cir v B» (inter T (l B C) (I M R» (~z
= ~»
The machine proof
u.!~ cr--SMCR
(~)/(~) I:~ . lJ.J:t. SBMR
CN
!:!... PCBy·SUr;8 - SBMR ,(PCBv-tPBCB)
11 -
PCBvoSMC'f (tPMABO),(PCBV-:PBCB)
Q (-4),PCBv-
y. -
(-tPABM) ,(PCBv-tPBCB) (4) ·(tpCBE) ,(SBMC+SAMC) PABM·(tPcBE-tPBCB)
~ (4),(PMBCHSBMC) ,(SBMC+SAMC)
-
N D
A
g
PABM ,(PMBC-PBCB+4SBMC) (4) ,(PBMB-PAMBH-i PAMB-iPAMA) PABM"(-PAMB-PAMA)
simJ!!.ify -
The eliminants
PBMB-P AMH PABM
M -PABA ' ~+PABA
= -PABA ' ..ul ~+PABA
Figure 6-331 24This is a problem from the 1959 Inlemational Mathematical Olympia.
liNN CN
Pcsy
SBMR~HpMABO) SMCR~SMCO
PMABO~
-
!(PABM)
SMCO~!(SBMC+SAMC) PCBV~~(PCBE) PCBE~PMBC+4SBMC PBCBgPBMB+PAMA SAMCg -
HPAMA)
SBMCg -
HPAMB)
PMBcgPBMB
PABM~
-
(~-I)'PABA)
PAM B ~(4i4L-l),PABA .4i4L AB AB
PBMB~(~-1)2 .PABA
6.5
Circles
413
Example 6.332 (0.616, 22, 31) Let M be the midpoint of the hypotenuse of the right triangle ABC. A circle passing through A and M meet AB at E . F is the point on the circle such that EF II BC. Show that BC = 2EF.
c Constructive description ( (POints A 8 ) (on C (t A A 8)) (midpoint M 8 C) (midpoint x A M) (on 0 (t x x A» (inter E(I A 8 ) (cir 0 A» (inter F (P E 8 C) (cir 0 E)) (u, = 1/2» Be
B
Figure 6-332
Example 6.333 (0.767,14,21) Let P A tangent to circle (0) at point A. M is the midpoint of P A. C is a point on the circle. PC and MC meet the circle at points E and B . respectively. P B meets the circle at D. Show that ED is parallel to AP.
Constructive description ( (points A 8 C) (circumcenter 0 A 8 C) (inter M (I 8 C) (t A A 0» (lratio P M A-I) (inter D (I 8 P) (cir 0 8)) (inter E (I C P) (cir 0 C)) (parallel ED A M) )
p
Figure 6-333
Example 6.334 (0.366, 6, 23) If P is any point on a semicircle. diameter AB. and BC. CD are two equal arcs. then if E = CAn P B. F = AD n PC. prove that AD is perpendicular to EF.
Constructive description ( (POints A P C) (circumcenter 0 A PC) (lratio 80 A-I) (inter D (cir C B) (cir 0 8)) (inter E (I P 8) (I C A» (inter F (I P C) (I A D)) (perpendicular E F A D ) )
Figure 6-334
Example 6.335 (1.900,49,18) From the point S the two tangents SA. SB and the secant SPQ are drawn to the same circle. Prove that API AQ = BPI BQ.
414
CIuIpter 6.
Constructive description ( (POints A B P) (circumcenter 0 A B P) (midpoint M A B) (inter s (10M) (t A A 0)) (inter Q (I s P) (cir 0 P» (eq-product A Q B PAP B
Topics from GeooIetry
s Q) )
Figure 6-335
Example 6.336 (0.466, 6, 19) Show that the lines joining a point of a circle to the ends of a chord divide harmonically the diameter perpendicular to the chord. Constructive description ( (POints A D P) (circumcenter 0 A D P) (lratio BOA -1) (foot F D 0 A) (lratio E F D -1) (inter Q (I A 0) (I P E)) (inter s (I A 0) (I P D)) (harmonic A B Q S»
D
Figure 6-336
Example 6.337 (0.133,3,16) From a point A two lines are drawn tangent to circle (0) at B and G. From a point P on the circle perpendiculars are drawn to BG, AB, and AG. Let D, F , E be thefeet. Show that PD 2 = PE· PF. Constructive description ( (POints B C P) (circumcenter 0 B C P) (inter A (t B B 0) (t C CO» (foot D P B C) (foot EPA C) (foot F P A B) (eq-product PDP D PEP
F) )
B
F
Figure 6-337
Example 6.338 (1.113, 14, 24) Through P a tangent P E and a secant P AB of circle (0) are drawn. The bisector of angle APE meets AE and BE at G and D. Show that EG = ED.
6.5
Circles
415
Constructive description ( (POints P E C) (on y (a P C E P C» (inter A (I E C) (I P Y» (midpoint M A E) (tratio Z (inter 0 (t M M E) (I E Z» (inter B (I P A) (cir 0 A» (inter D (l P C) (I B E» (eqdistance E DEC) )
E P H)
E
p
Figure 6-338
Example 6.339 (0.216, 3, 18) Let N be the traces of the internal bisectors of the triangle
ABC on the circumscribed circle (0). Show that the Simson line of N is the external bisector of the medial triangle of ABC.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (midpoint D B C) (midpoint E A C) (midpoint (inter N (l 0 F) (cir 0 A» (foot K N A C) (eqangle E F K K F D) )
F B A)
Figure 6-339
6.5.2
Intersectional Circles
Example 6.340 (0.216, 3, 18) Through the two comnwn points A. B of two circles (0) and (0 1 ) two lines are drawn meeting the circles at C and D. E and F . respectively. Show thatCE II DF.
For a proof of this theorem based on full-angles. see Example 1.115 on page 47. Constructive description ( (circle B A C E) (midpoint x A B) (tratio 0 x B H) (foot M 0 A C) (foot NOB E) (lratio D M A-i) (lratio F N B-1) (SCED
=
SCEF·) )
Figure 6-340
416
Chapter 6.
The eliminants
The machine proof
SCEF!,2SCEN-SBCE
~
SCEF
£.. -
SCEDg2ScEM-SACE S !!P8EQ ,S8CFi CENPBEB S M PACQ,SACFi CEM PACA
SeED
2SCEN-SBCE
Q 2SCgM SACE -
2SCEN-SBCE
!:!.
(2SCEM-SACE) ,PBEB 2PBEO ,SBCE-PBEB ,SBCE
-
PBEogPBEx-4SBEx ·r PAcogPAcx-4SBcx ·r
SBEX~
simJ!!ify (2SCEM SIlCE) ,PBEB (2PBEO-PBEB) ,SBCE
M. -
(2PACO ,SACE PIlCd ,SACE),PBEB (2PBEO-PBEB) ,SBCE ,PACA
-
(2PBEX-PBEB-8SBEX ·r) ,SBCE ,P ACA (PBCIl+4SBAC·r) ,SIlCE ,PBEB (PBEA+4SBIlE ·r),SBCE ,PIl CIl
~(SBAE)
PACA=2(Ac)') EE·iiE·iiC SBCE (-2).d Ai;. iiE· iiA SBAE (-2) .d
Q (2PllcX PACIl 8SBcx ·r),SACE ,PBEB
K
-
PBEX~~(PBEA+PBEB) SBCX~ - ~(SBAC) PACX~~(PACA+PBCA)
simJ!!ify (2P,WO-PIlCd),SIlCE ,PBEB (2PBEO-PBEB) ,SBCE,PACA
-
Topics from Geometry
PBEA=2(Ai;.iiE.cos(BA»)
co~r ( 4Ac.BC .B;t ·r+4AC .BC.cos(BA).d) .( CE.AE;·Ac).(2BE ).«2d)2 PBEB=2(iiE)') l
(-4AE·BE ·BA·rHAE·BE·cos(BA)·d)·( -c E .BE.BC).(2AC') .«2d)'
SACE SBAC
EE·Ai;·Ac (-2)·d Ac·iiC·iiA (-2).d
PBCA=2( Ac.iiC.cos(BA»)
Example 6.341 (1.583, 39, 25) Let A and B the two common poilus of two circles (0) and (0 1 ) , Through A a line is drawn meeting the circles at C and D respectively. G is the midpoint of CD. Line BG intersects circles (0) and (Od at E and F , respectively. Show that EG = GP. Constructive description ( (points A B C) (midpoint x A B) (on 0, (t x x A» (inter 0 (I x 0,) (b A C» (inter D (I C A) (cir 0, A» (midpoint G C D) (inter E (I B G) (cir 0 B» (inter F (I B G) (cir 0, B» (midpoint G F E) )
Figure 6-341
Example 6.342 (0.133, 6, 14) Three circles, centers A, B, C, have a point D in common and intersect two-by-two in the points AI, B 1, C 1• The common chord DCI of the first two circles meets the third in C 2 . Let A 2, B2 be the analogous points on the other two circles. Prove that the segments A 1 A 2 , B 1 B 2 , C1 C2 are twice as long as the altitudes of the triangle ABC.
6.5
Circles
417
Constructive description ( (POints ABC D) (foot F C A B) (inter c, (cir B D) (cir A D» (inter C. (1 D C,) (cir C D))
The elirninants ~~-(PCC,C-PCDC) _ ~ CF' PDC,D CF' P ~ (4)-(PBDB -P B6 p-PBAD -PABD+PAPA -PABD) DC,D PABA
~~~ CF' -
(~=2» CF'
SABC
PCc. c';J -(PCDC -PABA -2PBDB -PBAD-2PBCB -PBAD +4PBAD-PABD-2PADA-PABD-2PACA -P ABD)/ PABA SABC=¥(II. -Z• ) SABD=2(Y,-z.) PABD=( -2(:I:,-z.)-z.) PACA=2(1I~+z~) PADA=2(1I~+z~) PBAD=2(
z. -Z.)
PBCB=2(1I~+z~-2z, -z.+z~) PBDB=2(y~+z~-2z. -z.+z~) PABA=2(z~) PCDc=2(1I~-2\1l -II' +1I~+z~-2z.-z, +z~)
Figure 6-342
The machine proof l(_~) 2 CF' ~ - ( - ~D Cf -rcc,c+ ~D) Cf -rCDC
-
(2) -PDC,D
,im!1i/1I PCC,C - PCDC _ ~ (2)-PDC,D ~ -2P
~
~
CF' -p
-p
~
~
-p
~
-p
-p
~
-p
2 • 4PSDB , PSAD-4PSAD ' P ABD+"PADA ' P ABO ·5ABC'PABA
,im!1ifll -(PCpC-PABA-PBPB -PB6p-PBCB-PBAp+2PBAp-PABD-PAPA -PABp-PACA -PABp) -SABD (2) -(PBDB -PBAD PBAD-PABD+PADA -PABD) -SABC ~
-
-( -8\1l -I/1"z~) -Y,-z. -(2) .im1!li fll (2)-( 411fzg )-11' -z. -(2)
-
1
In the above proof, we have A
= (0,0), B = (XO, 0), C = (Xl> yd, and D = (X2' Y2)-
Example 6.343 (0.050, 3, 8) Show that an altitude of a triangle is the radical axis of the two circles having for diameters the medians issued from the other two vertices. A
Constructive description ( (POints A B C) (midpoint B. A C) (midpoint c, A B) (midpoint M B B,) (midpoint N C c,) (on-radical A M B N C) ) Figure 6-343
418
Chapter 6. Topics rrom Geometry
The eliminants
The machine proof
PANA{;f, -
PSMRtPAMA
~(Pcc,c-2PAC,A-2PACA)
-PCNC+PANA
PCNc{;f,Hpcc,c)
~
PAMA~
-
PRM8±PAMA
1:1. (2) '(k PBB,B-t PAB I A-t PABA) ~
-
-
~(PBB,B-2PAB,A-2PABA)
PBMB~HpBB,B) PAC,A ~ ~(PABA)
-tPcC,C+tPAC,A+tPACA PCC,C-PAC,A-PACA
PCc,C ~ H2PBCB+2PACA-PABA)
PBBIB-PABIA-PABA tPBCB-tPACA-fpABA
PAB,A~HpACA)
~ (2)·(t PBCB-k PACA-k PABA) PBCB-PACA-PABA simJ!!ify
PBB,B ~ ~(2PBCB-PACA+2PABA)
Example 6.344 (0.800, 12, 18) Let A and B be the two common points of two circles (0) and (Od. Through B a line is drawn meeting the circles at C and D respectively. Show ACIAD = OAI01A. Constructive description ( (POints A B C) (midpoint x A B) (on a, (t x x A» (inter a (I x 0.) (b A C» (inter D (1 C B) (cir a, B» (eq-product A C a, A A D a
A) )
Figure 6-344
Example 6.345 (0.783, 16,25) From a point P on the line joining the two common points A and B of two circles (0) and (0 1 ) two secants PCE and PF D are drawn to the circles respectively. Show that pc· PE = PF· PD. Constructive description ( (points A C F X) (on y (t x x A» (inter a (l x Y) (b A C» (inter a, (1 x Y) (b A F» (lratio B x A -1) (on P (l A x» (inter E (1 P C) (cir a C» (inter D (1 P F) (cir a, F» (eq-product PCP E PDP F) )
p
Figure 6-345
6.5
Circles
419
Example 6.346 (1.283,60,67) Ifthree chords drawn through apoint of a circle art! taken for diameters of three circles, these circles intersect, in pairs, in three new points, which are collimar. Constructive description ( (circle A B c D) (midpoint M I A D) (midpoint M~ B D) (midpoint ,usc D) (inter E (cir Ma D ) (cir M I D ) ) (inter F (cir MS D) (cir M I D)) (inter G (cir M~ D) (cir M~ D)) (colliinear E F G ) )
R~gm 6-36
Example 6.347 (0.733, 11, 38) I f three circles pass through the sume point of the circumircle of the triangle of their centers, these circles intersect, in pairs, in three collinear points. Constructive description ( (circle A B c P) (inter A1 (cir B P) (cir c P)) (inter BI (cir c P) (cir A P)) (inter cl(cir B P) (cir A P)) (collinear A1 BI c l ))
Figure. 6-347
Example 6348 (1.083,24,66) Ifthree circles having apoint in common intersect in pairs in three collinear points, their common point is cyclic with their centers. Constructive description ( (points AI BI P) (on cl(1B, A,)) (midpoint M I P BI) (on TI(t M I M I P) 1) (midpoint Ma P C I ) (on TZ(t MZ M? P) 1) (midpoint MS P AI) (on TS(t MS M S P) 1) (inter A (1 M I TI)(1 Ma Ta)) (inter B (1 M S Ts)(1 Ma Ta)) (inter c (1M I TI)(I MS TS)) (cocircle A B c P) )
I
R ~ g m6-348
Example 6.349 (0333, 7, 15) Show that the radical axis of the two circles having for diameters the diagonals AC, B D of a irapezoid ABCD passes through the point 4 intersection E of the nonparallel sides BC, AD.
420
Chapter 6. Topics from Geometry E
Constructive description ( (points A B C) (on D (P CAB» (inter E (I A D) (I B C» (midpoint N, A C) (midpoint No B D) (on-radical E N, A No B) ) Figure 6-349
Example 6.350 (0.250,8,13) Given two circles (A). (B) intersecting in E. F. slww that the clwrd E1Fl determined in (A) by the lines MEE1• MFFl joining E. F to any point M of (B) is perpendicular to M B. Constructive description ( (POints E F M) (circumcenter B E F M) (midpoint D E F) (lratio A D (inter E, (I M E) (cir A E» (inter F, (I M F) (cir A F» (perpendicular E, F, M B) )
B r)
Figure 6-350
Example 6.351 (0.667, 4, 44) From the midpoint C of arc AB of a circle. two secants are drawn meeting line AB at F. G. and the circle at D and E. Slww that F . D. E. and G are on the same circle. Constructive description ( (circle A C D E) (circumcenter 0 A C D) (foot MAO C) (lratio B M (inter F (I A M) (I D C» (inter G (I A M) (I C E)) (cocircle D E F G»
A -1)
Figure 6-451
Example 6.352 (1.067 39 18) From point P two tangent lines P A and P B of a circle are drawn. D is the middle point of segment AB. Through D a secant EF is drawn. Then LEPD = LFPD. Constructive description ( (POints A B E) (circumcenter 0 A B E) (midpoint D A B) (inter p (I 0 D) (t A A 0)) (inter F (I E D) (cir 0 E)) (eqangle E P 0 0 P F) ) Figure 6-352
6.5
421
Circles
Example 6.353 (0.100 4 13 ) Let D and E be two points on sides AB and AC of triangle ABC such that DE II BC. Show that the circumcircles of triangle ABC and ADE are tangent.
Constructive description ( (POints A B e) (circumcenter 0
(on
D
(I
A B e)
A B»
(inter E (I A e) (P D B e» (circumcenter N A D E) (SANO
=
0» Figure 6-353
6.5.3
The Inversion
Dermition Suppose that we have given a circle whose center is 0 and the radius has the length r =I O. Let P and Q be any two points collinear with 0 such that
OP . OQ
= r2 .
Then P is said to be the inverse of Q with regard to the circle. and the transformation from P to Q is called an inversion. The point 0 is called the center of inversion, the given circle the circle of inversion, and its radius the radius of inversion.
Example 6.354 (0.001, 1, 3) Two inverse points divide the corresponding diameter harmonically.
Constructive description ( (POints B A) (midpoint 0 A B) (lratio PO A r) (lratio Q 0 A ~)
(harmonic A
B P Q) )
The eliminants
The machine proof
~g~
(-*)/(~)
BQ
!1. -~.r+l -~.-*
AO
BP
-(~-r)
~g -
p -(r-I)·(~ ·r-I) ~
=
~ ' r-l
4tt:,~
(I)
(r-I) .(-~+r)
.im1!!i _ /ll
~ 'r-I ...w..~-r AO
Q (- tr-J. ).( 1) .im1!! i /ll
- (-ir-I)·(l)
-
1
Figure 6-354
Example 6.355 From a point P outside a given circle, center 0, the tangents are drawn to the circle. Show that P is the inverse of the point of the intersection of 0 P with the chord of contact.
422
Chapter 6. Topics from C - t r y
Constructive description «POints A C) (on B (t C C A» (midpoint 0 A B) (inter P (I A 0) (t C C 0» (foot EC A 0) (inversion 0 A P E) )
The eliminants
The machine proof
~!.=E.w.A
(-%)/(-~)
OE- PAOC 2I!f.~ AO--PAOC
~ ~.-21! PAOA
!:.. -
AO
PAOAgt(PABA)
-Pono,PAPG
Pcocgi(2PCBC-PABA+2PACA)
PAOA·(-PAOC)
PABA~PCBC+PACA
8im2f.ify ~ PAOA
Q -
tPcBc-iPABA+tPACA ;PABA
~ PCBC+PACA 8im2f.ify
-
PCBC+PACA
-
Figure 6-355
Example 6.356 (0.433, 10, 11) Prove that two pairs of inverse points with respect to the same circle are cyclic. or collinear.
Constructive description ( (POints A B) (on 0 (b A B» (lratio P 0 A r,) (lratio Q 0 A f-) (lratio ROB ~) (lratio SOB f.-) (cocircle P Q R S) )
The eliminants SPQS~SBPq
...
PRSR
s
PRS
SQRS
S pORo ·1-PORO· ... +PBRB· ... -PBOB ·... +PBOB
'1
S SOPR 'ra-SoPRtSBPR
... S SoqR· ... -SoqR+SBqR r,
SPQR/!,SBPQ· ... PBRB/!,(r,-I)' .PBOB
PORO/!'PBOB·r~ SBPR/!,( ... -I).SBOP SOPR/!,SBOP"'" SBQR/!,( ... -I).SBOQ SOQR/!,SBOQ· ... Q -(SBop·r,-SBOP-SABP) S BPQ
PPQP
rl Q
Popo ·r~-PoPO ·'" +PAPA·r'-PAOA·r, +PAOA ~
SBOQ~~
SABP!:, -
«r,-I).SABO)
SBOP!:,SABo·r, PAPA!:,(r,-I)'.PAOA
Figure 6-356
POPO!:'PAOA·r~ PBOBgPAOA
The machine proof PPqp·( -SqRS)·SPRS PRSR"SPQS·( -SPQR)
§.. -
Pp p·(-So R·r,+SO R-SB R ·(SOPR ·... -SOPR+SBPR)· ... ·~ (PoRo ·ri-PoRo· ... +PBRB ·r'-PBOB· ... +PBOB)·SBPQ ·( -SPQR)· ,
8im.E!.ify PpqP·(SOqR·... -SOqR+SBqR)·(SOPR· ... -SOPR+SBPR)· ... (PoRo·r~ PORO· ... +PBRB·... PBOB· ... +PBOB)·SBPQ·SpQR -
!!:
PpqP"(SBoq·r~-SBoq)·(SBoP"r~-SBoP)· ...
(PBoB·r;-2PBOB·r~+PBOB) ·SBPQ·SBPQ·...
6.5
Circles .imz!i/II -
9.
423
PPqp · SSO~ ·SSOp
PSOS ·S SPQ (POPO ·~-POPO ·rl+PAPA ·rl-PAOA ·",+PAOA) · SASO · SSOr~ PSOS ·( Ssop · rl+SSOP+SAsp)' ·rl ·~
-
.imz!i/II
( POPO ·~-POPO ·'" +PAPA ·rl-PAOA ·rl +PAOA)·SASO ·SSOP
-
E.
PSOs ·(Ssor ... -SSOP-SAsp)' ·rl (PAOA ·rt- 2PAOA ·rf+ PAOA)·SASO·SASO ·...
PSOS ·(SASO ·~-SAso)' ·rl
-
.im!li/II
~
Q
~
PSOS
-
PAOA
.imz!i/II -
Example 6.357 (0.067. 6. 9) The inverse of a line not passing through the center of inversion is a circle through that point.
Constructive description « POints 0 A) ~) r.) r) (midpoint u P 0) (lratio G 0 R !Jl.4D.) . PORO (eqdlStance G U U 0) )
(lratio Q 0 A (tratio R Q 0 (lratio PO A
The eliminants PuGug PRUR·POAO+POUO ·PORO-POUO ·POAO-PORO ·POAO+PbAO/PORO Pouogl(PoPo) PRURgi(2PRPR-POpo+2PORO) POPO!':POAo ·(r)' PRPR!':PARA ·r-PoRo ·r+PoRo+POAO·r2- POAo ·r
PORO~(rf+l)· POQO PARA ~PAQA+POQO ·rf Q~
POQO= (r)2 Q(r- l)' · PoAO P AQ A (r)'
The machine proof fwJJJ. POUO _ -
PRUR ·PORO ·POAO+POUO ·PJ.RO-POUO·PORO ·POAO-PJ.80 ·POAO+PORO .PJ.AO POUO ·PbRO
. im1!f.i/1I PRUR ·POAO+POUO ·PORO-POUO ·POAO-PORO ·POAO+PJ.AO -
!L -
E.
POUO ·PORO t PRPR ·PO AO+t p oPO ·PORO-t p OPO ·POAO-tpORO ·POAO+P4AO (!POPO) ·PORO 2PARA ·POAO·r+PORO·POAO ·r'-2PORO ·POAO ·r-2PJ.AO ·r+4PJ.AO POAO ·,-2 ·PORO
-
.im1!f.i/1I 2PA8A .r+P080 .r2-2P080 .r-2POAQ .rt4POAO -
11
p
(r)' ·PORO 2PAqA ·r+Poqo ·rf ·r2+PoqO ·r2-2Poqo·r-2POAo ·r+4POAO (r)' .(POQO ·'1+ POQO)
-
.imz!ih -
2PAqA ·r+Poqo ·rf ·r2+POQO ·r2-2Poqo ·r-2POAo ·r+4POAO (r)' .(f1+l).POQO
Figure 6-357
Chapter 6. Topics from Geometry
424
Example 6.358 (0.133, 5, 11) If the circle U passes trough two inverse points of circle O. Then the inverse of any point on circle U with regard to circle 0 is on circle U. In other words. the inverse of circle U with regard to circle 0 is itself.
Constructive description ( (POints A 0 X) (lratio PO A r) (iratio Q 0 (midpoint Mu P Q) (tratio
A ~)
U Mu P ru)
(inter E (I P X) (cir UP» (inter F (I E 0) (cir U E» (inversion 0 A E F) )
PEOF
-
sim~iJY
PUEU ,POEO+POEO 'POUO ~
PAOA
-(PUEU-POUO) -
!!.. -
II!!!
PAQA
PPMuP-POMuO
sim!!1ify
-
PuEUfgppuP
PAQA
-2.!:."PPQP-2.!:. ·PPQP PQ PQ
PAQ A
(- ¥o 'r)'(POPO 'r' -POPO'r+PAPA ·r-PAOA ·r+PAOA)
.f..
POUOgPPMUP 'rf,+POMuO PpuPg(rf,+1) ,PPMu P
PPQP~ POPQ ·,.2-PO PQ ·r+P A
P'
sim_plify
PEOF~-(PuEU-POUO)
PAPA
-(Ppup-POUO)
(~+1)'~ ' PPQP
~
Figure 6-358
The eliminants
The machine proof ~ .f.. PAOA·POgO
(r)
A ·r-P AOA . r+P AO A
POMUO MJI ~(2~+1)2 ' PPQP) PPMUP
r.;v ~(pPQP)
PA O A·r ·(~·r+l)' AD
lU:~~.~ PQ % 'r+l AO
- ~~ ·(POPo·r2 - POPo·r+PAPA·r-PAOA·r+PAOA)
PAPA~(r-l)"PAOA
PAOA ·r-(r 2 -1)' .(-I) sim£lify r·(PAOA ·r 4 -2PAOA ·r 2+PAOA)·«-I»2 -
r
POPO~PAOA ·(r)'. %~
-
(r)
Example 6.359 (10.817,267,29) The inverse of a circle not passing through the center of inversion is a circle.
Constructive description ( (points 0 A X) (iratio PO A r,) (lratio Q 0 A r2) (midpoint U P Q) (lratio E 0 A .;.) (iratio F 0 A'./-) (midpoint I E'F) (inter R (I P X) (cir U P)) (lratio G 0 R fu.Aa.) • PORO (eqdlstance G I I E) )
x
Figure 6-359
6.5
425
Circles
Example 6.360 (0.033, 4, 24) The harmonic conjugate of a fixed point with respect to a variable pair of points which lie on a given circle and are collinear with the fixed point, describes a straight line. This line is called the polar of the fixed point with the circle, and the fixed point is said to be the pole of the line. Constructive description ( (circle E F A) (circumcenter (inter P (lOA) (I E F» (!ratio Q 0 A 24) (inter M (I E 7{' (t Q Q A» (harmonic E F M P) )
0 E F A)
Figure 6-360
Example 6.361 (0.216, 5, 6) Let P and Q be inverse points with regard to circle OA. Then for any point C on circle 0, we have C P . AQ = CQ . AP. Constructive description ( (POints A B C) (circumcenter 0 A B C) (lratio PO A r)
... Q A~------~--+--
(lratiOQOA ~)
(eq-product CPA Q C Q A
P) )
Figure 6-361
Example 6.362 (0.400,8,16) If the circle (B) passes through the center A of the circle (A), and a diameter of (A) meets the common chord of the two circles in F and the circle (B) again in G, show that the points F, G are inverse for the circle (A) . Constructive description ( (points DEC) (circumcenter A DEC) (midpoint / (inter B (I A /) (b A E» (inter F (1 E C) (1 A D» (inter G (I A D) (cir B A» (inversion A D G F) )
E C)
Figure 6-362
Example 6.363 (0.466, 6, 15) Show that the two lines joining any point of a circle to the ends of a given chord meet the diameter perpendicular to that chord in two inverse points. Constructive description ( (points A E D) (circumcenter 0 A E D) (foot / E A 0) (lratio FIE (inter P (lOA) (I D E» (inter Q (lOA) (I D F)) (inversion 0 A P Q) )
-1)
Figure 6-363
Chapter 6. Topks from c - t r y
426
Example 6.364 (0.700, 8, 25) T p. TQ are the tangents at the extremities of a coord PQ of a circle. The tangent at any point R of the circle meets PQ in S; prove that T R is the polarofS. Constructive description ( (points P Q R) (circumcenter 0 P Q R) (on A (t P PO» (inter T (I A P) (t Q Q 0)) (inter 5 (I P Q) (t R R 0» (inter 5, (I T R) (I 0 5)) (inversion 0 P 5 SIl )
T
Figure 6-364
6.5.4
Orthogonal Circles
DermitioD. Two circles 0 1 and O2 with a common point A are said to be ortOOgonal 01 A .L02A.
if
Example 6.365 (0.050, 4, 7) If two circles are ortOOgonal. any two diametrically opposite points of one circle are conjugate with respect to the other circle. Constructive description ( (POints D E) (on 0, (b D Ell (lratio F 0, E -1) (on 0 (t D DOll) (footG FO E)
(inversion 0
D G E) )
The eliminants PEOGgPEoF
o -1 PEOF=PEDF+PDO'D ' ~ -4SDO,F"~+4SDEOl ' ~
PDODgPDO, D'( -'U!.)' DO,
SDO,F~SDEO, PEDF~2PEDOI-PDED PEDO, PEO, E
= - ~(PEO'E-PDOID-PDED) ~ PDO, D
The machine proof EIuuL Q EIuuL PEOG
-
PEOF
Figure 6-365
6.5
427
Circles
Example 6.366 (0.016, I, 3) Show that the two poles of the comnwn chord of two orthogonal circles with respect to these circles coincide with the centers of the given circles.
Constructive description ( (POints A R) (on 0, (b A R»
(
RO, ~-PAO,R PAO,A
~)/( ~) RO.
~ PAO,A. _~
(inter O. (I 0, (inversion 0,
Q
PAO,R
-
-
ODO PAO,R ~ -PRO,R' PRO,R - PAO,A
il;7=
RO,
-PAO,RoPAO,A
PAO,R ,(- PRO,R)
.im!!!if~ -
<2J
0,0. -
0.02
(foot D A ROd (lratio B D A -I) D) (t A A 0,» R DO.) )
The eliminants
The machine proof
PAO, A PRO,R
PAO,A PAO,A
.im~if~ 1
Figure 6-366
Example 6.367 (0.001, 1, 5) A circle orthogonal to two given circles has its center on the radical axis of the two circles.
Constructive description: ( (POints D, D.) (on 0 (b D, (on o. (t D. D. 0» (on-radical 0 0, D, o. D.) )
D.»
(on 0, (t D,
The machine proof POO,O-PD.O,D,
poo.o-po.o.o.
P0.0. o. o'P = 0.00. '(~)' 0.0 02 "'0'0"2
po.oo.
+1).Po.oo.
PD.O.D] 0, =PD.OD." (~)' D.O
0.00 2 Poo,o =(~ +1)'Po,oo,
<2J Po, 00, -
po.oo. Q Po, 00, - Po,oo, Figure 6-367
The eliminants Poo.o=(~
~ P001 0 - PD ,O)D.
-
D, 0»
po.oo. gpo, 00,
.im~if~
Example 6.368 (0.283, 8, 12) If two circles are orthogonal, any two points of one of them collinear with the center of the second circle are inverse for that second circle.
Constructive description ( (POints M E F) (circumcenter B M E F) (inter A (I E F) (t M M B» (inversion A M E F) ) Figure 6-368
428
Chapter 6. Topics from Geometry
Example 6.369 (0.350, 4, 18) The two lines joining the points of intersection of two orthogonal circles to a point on one of the circles meet the other circle in two diametrically opposite points. Constructive description ( (points C E F) (circumcenter Q C E F) (iratio D Q C -1) (midpoint I E F) (inter p (i Q I) (t E E Q)) (inter A (i E C) (cir P E» (inter &.1.1 F D) (I C E» (~=~» EA
Figure 6-369
EAI
Example 6.370 (0.850, 11, 19) Show that in a triangle ABC the circles on AH and BC as diameters are orthogonal. Constructive description ( (points A B C) (orthocenter H A B C) (midpoint M B C) (midpoint N A H) (inter I (I A B) (cir N A» (perpendicular MIN I) )
B
M
Figure 6-370
Example 6.371 (1.183, 10, 12) The circle I BC is orthogonal to the circle on hIe as diameter.
Constructive description ( (points B C I) (incenter A I B C) (inter 18 (t C C I) (I B I) (inter Ie (t B B I) (I C I) (circumcenter 0 B C I) (midpoint M 18 Ie) (perpendicular M BOB)
)
Figure 6-371
Example 6.372 (0.533, 16, 32) Show that given two perpendicular diameters of two orthogonal circles, the lines joining an end of one of these diameters to the ends of the other pass through the points common to the two circles.
6.5
Circles Constructive description ( (points p D I) (circumcenter A P D I ) (lratio E A D -1) (midpoint x P I) (inter B (1A X ) (t P P A ) ) (inter c (1I E ) (cir B I ) ) (lratio F B G -1) (perpendicular F G D E ) )
figure 6-372
Example 6373 (0.083,4,15) Show that if AB is a diameter and M any point of a circle, center 0 , the ~o circles AMO, BMO are orthogonal. Constructive description (points A M ) (on B (t M M A)) (midpoint 0 A B ) (circumcenter I A M O ) (circumcenter J B M O ) (perpendicular I M M J ) )
.___...----
(
B
@A
0
Figure 6373
Example 6374 (0.783, 11, 23) If the line joining the en& A, B of a diameter A B of a given circle ( 0 )to a given point P meets (0)again in AlpEl, show that the circle PAl El iS 0rthogo~ltO(0). Constructive description ( (points A1 B1 A) (circumcenter o A] BI A) (lratio B o A -1) (inter P (1A A ] ) (1B B,)) (circumcenter ol A] BI P ) (perpendicular ol AI AI o) )
6.5.5
The Simson Line
Figure 6-374
Let D be a point on the circumscribed circle of triangle ABC. From D three perpendiculars are drawn to the three sides BC, AC, and AB of triangle ABC. Let E. F,and G be the three feet respectively. Then E,F and G are coIlinear. The line E F G is called the Simson line of the point D with respect to the triangle ABC, and D is called the pole of the simson line. For machine proofs of Simson's theorem see Example 3.79 on page 141. Example 3.106 on page 160 and Example 5.5 5 on page 245.
Chapter 6. Topics from Geometry
430
Example 6.375 (0.933, 3, 37) The Simson line bisects the line joining its pole to the orthocenter of the triangle. Constructive description ( (circle ABC D) (orthocenter H A B C) (foot FDA C) (foot G D A B) (inter N (I G F) (I D H» (midpoint N D H) ) Figure 6-375
Example 6.376 (0.050, 1, 8) Let D be a point on the circumcircle of triangle ABC. If line DA is parallel to BC, show that the Simson line D(ABC) is parallel to the circumrtuiius OA.
Constructive description The machine proof ( (POints A B C) ~ -SAOF (circumcenter 0 A B C) (pratio PD ABC 1) Q PSAO , SABO (inter D (I A PD) (cir 0 A» - (-SAOF) , PABA ~ -PBAO,SA8Q , PACA (foot FDA C) (-PCAD ,SACO) , PABA (foot G D A B) Q (2POAPp ,PBAPp),SABO,PACA ' PAPpA (parallel G F 0 A) ) ( 2POAPD ,PCAPD) ,SACO , PABA ,PAPDA
-
.im1!li/" -
I!B -
Q -
PBAPp,SABO , PACA PCAPD,SACO ,PABA
The eliminants S
G -PBAQ , SA8Q
AOG
S
PABA F -PCAQ ,SAGQ
AOF
PACA
D (2) ,POAPp , PCAPp PCAD PAPDA D (2),POAPp ,PBAPp
PBAD
PAPDA
PCAPD Pg PACB PBAPD Pll -
s
PABe ,SABo ,PAQA
PACB,SACO , PABA ,(-1)
S
-PABC ,PACB,PABA , PACA ' ( 32SABC) PACB ,PACA , PABC , PABA ,(32SABC)
(PABC)
OPACA , PABQ
ACO ABO
(-32) ,SABC 0
PAC8,PABA
(32),SABC
.im!1i/y 1 Figure 6-376
Example 6.377 (0.033, 1, 8) If E, F, G are the feet of the perpendiculars from a point D of the circumcircle ofa triangle ABC upon its sides BC, CA, AB, prove that the triangle DFG, DBC are similar,
Constructive description ( (circle ABC D ) (foot FDA C) (footG D A B) (eq-product D B D F D C
D
G) ) Figure 6-377
6.5
Circles
431
The machine proof PaDs -PocO
PCDC ,PDOD
Q -
f. -
PAos ,Papa -PABA
PCDC·(16S~BD)
PBDB ·(16S'4fo) ,PABA (16) ,PCDC 'S ABD ,PACA
oo~r (2BD' H -CD.AD.AC)2.(2Ai3').(U)' (2Cif).( -BO.AO.AB)2 .(2::iC') .(2d)2 .im!!1.if~ I
Example 6.378 (0.050, 3,5) Simson lines corresponding to pairs of diametrically opposite points on the circumcircle of a triangle meet a side of the triangle in two isotomic points.
Consttuctive description ( (circle ABC 0) (circumcenter 0 ABO) (lratio 0, 00-1) (footG 0 A B) (foot G, 0, A B) (midpoint C, A B) (midpoint c, G Gd )
The machine proof -~ GIC)
~_1
C ~
-4L-L -~+t
G _(~_l) . PABA ...J AA 2 -
PBAD,-fpABA
G -(PBAD- 1 PABA) ,PABA (PBAD,-lpABA) ,PABA
=
B
The eliminants -e
~_1
o,e,
~_1
~d..4.ll,....1...
AB
,
~~PBAD, AS -
PABA
~QEaAJJ. AB- PABA
PBAD, ~2(PBAo-tPBAD)
PBAOg~(PABA)
.i~i/~ -(PBAD-1PABA) PBAD,-, PABA
Qj -(PBAD-tPABA) - 2PBAO-PBAD-f p ABA
A
o
-(PBAD- 1 PABA)·(2)
=(2) ,(-PBA!+f p ABA) o Figure 6-378
Example 6.379 (1.817,4,35) If the perpendicular from a point D of the circumcircle (0) of a triangle ABC to the sides BC. CA. AB meet (0) again in the points N. M. L. tM three lines AN. BM. CL are parallel to the simson of D for ABC. Consttuctive description ( (circle ABC 0) (circumcenter 0 A B C) (foot F (footG
(inter
0 A C)
0 A B) L (I D G)
(parallel C
(cir 0
0))
L F G) )
Figure 6-379
432
Chapter 6. Topics from GeometrY
Example 6.380 (4.283, 26, 25) Show that the simson of the point where an altitude cuts the circumcircle again passes through the foot of the altitude and is antiparallel to the corresponding side of the triangle with respect to the other two sides.
Constructive description ( (POints A B C) (circumcenter 0 A B C) (foot GC A B) (inter D (I C G) (cir 0 C» (foot E D A C) (foot F D B C) (inter G, (I A B) (I E F» (~=~» BG BG 1
Figure 6-380
Example 6.381 (1.633,4,39) If the Simson line D(ABC) meets BC in E and the altitude from A in K , show that the line DK is parallel to EH, where H is the orthocenter of ABC.
Constructive description ( (circle ABC D) (foot E DB C) (foot G D A B) (orthocenter H A B C) (inter K (I E G) (I A H» (parallel D K E H) ) Figure 6-381
Example 6.382 (1.483, 7,44) Show that the symmetries, with respect to the sides of a triangle, of a point on its circumcircle lie on a line passing through the orthocenter of the triangle.
Constructive description ( (circle ABC D) (orthocenter H A B C) (foot FDA C) (foot G D A B) (iratio G, G D -1) (lratio F, F D -1) (collinear F, G, H)
0,:
Figure 6-382
Example 6.383 (4.133,4,53) Thefour Simson lines offour points of a cirl:le, each taken for the triangle formed by the remaining three points, are concurrent.
6.5
Circles
433 o
Consttuctive description ( (circle A 8 C D) (circumcenter 0 A 8 C) (foot E D A 8) (foot FDA C) (foot G C A 8) (foot H C A D) (foot I A D 8) (foot J A 8 C) (inter K (I H G) (I E F)) (inter M (I E F) (II J» = ~) )
(*
Figure 6-383
Example 6.384 (5.333, 58, 76) If two triangles are inscribed in the same circle and are symmetrical with respect to the center of that circle, show that the two simsons of any point of the circle for these triangles are rectangular. Consttuctive description ( (circle A 8 C D) (circumcenter 0 A 8 C) (lratio A, 0 A-I) (lratio 8, 08-1) (lratio C, 0 (foot G D A 8) (foot FDA C) (foot G, D A, 8d (foot F, D A, Cd (perpendicular G 1 F, G F) )
C -I)
Figure 6-384
Example 6.385 (1.683, 15, 48) Show that the Simson lines of the three points where the altitudes of a triangle cut the circumcircle again form a triangle homothetic to the orthic triangle, and its circumcenter coincides with orthocenter of the orthic triangle.
Consttuctive description ( (POints A 8 C) (circumcenter 0 A 8 C) (orthocenter (foot D A 8 C) (foot E 8 A C) (foot F C A 8) (orthocenter G D E F) (lratio D, D H -I) (lratio E, E H -1) (lratio F, F H -I) (foot D. D, A 8) (foot Eo E, A 8) (foot F. F, A C) (inter A, (I E Eo) (I D D.)) (inter 8, (I D D.) (I F F.» .i!!!ter ~ (I (inter I (I A, F) (18, E» =
H A 8 C)
E Eo)
(7t Tt) )
(I
F F.»
Figure 6-385
Example 6.386 (0.633, 3, 40) The perpendiculars dropped upon the sides BC, C A, AB of the triangle from a point P on its circumcircle meet these sides in L, M, N and the circle in AI, B I , C 1• The Simson line LM N meets Bl C I , CIA I , AIBI in L 1, M I , N I . Prove that the lines ALl, BM\. CNI are concurrent.
434
Chapter 6. Topics from Geometry
Constructive description ( (circle ABC P) (circumcenter 0 A B C) (midpoint x A B) (foot L P B C) (foot M P A C) (foot N P A B) (foot DO P L) (foot E 0 P M) (foot FOP N) (lratio A, D P -I) (lratio B, E P -1) (lratio c, F P -I) (inter L, (I B, c,) (I M N» (inter M, (I c, A,) (I M N» (inter N, (I A, B,) (I M N» (inter J (I B M,) (I (inter J (I A L,) (I C N,) = ~»
(th
A L,)
B.
Figure 6-386
Example 6.387 (0.467,5,38) The circumradius OP of the triangle ABC meets the sides of the triangle in the points AI. BI> CI. Show that the projections A 2• B 2• C2 of the points AI. B I . C I upon the lines AP. BP. CP lie on the Simson line of P for ABC.
Constructive description ( (circle ABC P) (circumcenter 0 A B C) (inter A, (I B C) (I P 0» (foot A. A, A P) (foot G P A B) (foot F P A C) (inter F, (I A C) (I G A.» (~=g» CF)
CF
Figure6-387
Example 6.388 (0.950, 3, 37) Let L. M. N be the projections of the point P of the circumcircle of the triangle ABC upon the sides BC. CA. AB. and let the Simson line LM N meet the altitudes AD. BE. CF in the points L1> MI. N l . Show that the segments LM. Ll Ml are equal to the projection of the sides upon the Simson line.
Constructive description ( (circle ABC P) (circumcenter 0 A B C) (orthocenter (foot L P B C) (foot M P A C) (foot N P A B) (inter L, (l A H) (I N M» (inter M, (I B H) (I N M» (inter N, (I C H) (I N M» (foot A, A M N) (foot B, B M N) (LJ.!!l =1» ML
H A B C)
Figure 6-388
6.5
435
Circles
The Pascal Configuration
6.5.6
For a machine proof of Pascal's theorem, see Example 3.80 on page 142. Example 6.389 (The Converse of Pascal's Theorem On a Circle) (0.933, 7,44) s
Constructive description ( (circle A E B D C) (circumcenter 0 A B C) (inter P (I A B) (I E D» (lratio Q B C r)
(inter S (I D C) (I P Q» (inter F (I E Q) (1 A S» (cocircle
FA B C) )
Figure 6-389
Example 6.390 (pascal's Theorem: The General Case) (0.666, 3, 23) Let A, B, C, D, F and E be six points with P = AB n DE, Q = BC n EF and 8 = CD n FA collinear. Then Pl = AC n DE, Q1 = BE n CF and 8 1 = AB n F D are collinear. Constructive description «POints BAD E S) (on C (I D S» (inter P (I B A) (1 D E» (inter Q (I B C) (I S P» (inter F (I E Q) (I A S» (inter 5, (I D F) (I B A» (inter P, (I A C) (I D E» (inter Z2 (I B E) (I s, p,» (inter z, (I C F) (I s, P,» (~ . ~=1» P.Zl S.Z2
The machine proof ~.~ ~ SCFS, • ~ ~ SCFS, ,SBEP,
p,z,
S,Z2
!1
5CFS, ' ( -SAEC ,SBDE),SADCE
-
(
-
-
S,Z2
SDEC ,SACF) ,SBES,'(
.im.!!!.i/y ~
SCFP,
SCFP, ,SBES,
SADCE)
-SCFS, ,SAEC ,SBDE SDEc ,SAcrSBEs,
-(-SDcrSuF)·Sgc·5BPE ,SBPAF SDEc ,SAcr(-SBDrSBAEH SBDAF)
Cbapter 6. Topics rrom Geometry
436 sim1!!iJlI
SncPOSBAf "SA6C,SRPF; SDEC,SACF,SBDF ,SBAE
L -
SESq,SADC ,SAEQ,SBAS ,SAEC,SBDE'( -SAESq),SAEsq SDEC ' ( -SASC,SAEQ
Bim!1ify
H -SASQ,SBDE-SAES,SBDQ) ,SBAE,(SAESQ)2
-SESq ,SADC ,SBAS,SAEC,SBDE SDEC ,SASC,(SASQ,SBDE+SAES ,SBDQ),SBAE
2. -
-( -SESe-SBSC),SADC,SB6S,SAEC ,SBDE,SBSCP'( -SBSCP) SDEC ,SASC ' (
sim~pfY
SBSCP,SASP,SBSC,SBDE
SBSCP,SAES ,SBSP ,SBDC),SBAE'(
SBSCP)
-SeSp-SaSC-SA DC ,SRA S,SAEC,SBDE SDEC,SASC ,(SASP ,SBSC ,SBDE+SAES,SBSP,SBDC) ,SBAE
.f.. -
(
SDES,SBM;) ,SBSC·S,WC,SB6S,SAEC,SBDE ,(SBDAE)' SDEC,SASC'( -SBDAE,SAES,SBDC ' S B DE'S BAS SBDAE ,SADE ,SBSC ,SBDE ,SBAS ) ,SBAE'(
8im~ifY
SBDAE)
SUES ,SBSC"SAOC-SAEC SDEC ,SASC ,(SAES,SBDC+SADE ,SBSC)
c
SDES·(SBDS·~-SBDS) · SADS·~·(SAES · ~+SADE , ~-SADE)
ns os = SDES · ~ · (SADS , ~-SADSHSAES·SBDS · ~+SADE · SBDS·*-SADE·SBDS) ~
DS
=
simplify
Example 6.391 (Brianchon's Theorem) (1.750,2,137) The dual of Pascal theorem. Constructive description ( (circle ABC D E F) (circumcenter 0 A B C) (on Br (t B B 0» (on Ar (t A A 0» (on Cr (t C C 0» (on Dr (t D DO» (On Er (t E EO» (on Fr (t F F 0» (inter Al (I B Br) (I A Ar)) (inter DI (l E Er) (I D Dr)) (inter BI (I C Cr) (I B Br» (inter EI (I F Fr) (I E Er)) (inter C I (I D Dr) (I C Cr)) (inter FI (I A Ar) (I F Fr)) (inter I J!.pl Ed (I Al Dd) (inter J (I Al Dd (I CI Fd) (~=~» Dll D } I
Figure 6-391
Example 6.392 (Kirkman's theorem) (0.650,2,50) Given six points A. B. C. D. E. and F on a circle (ora conic). the three Pascal lines [BAECDFj. [CDBFEAj. [FECABDj are concurrent. There are 60 Kirkman points for one Pascal configuration. Constructive description ( (circle ABC D E F) (inter p (I C D) (I A B» (inter Q (l D F) (I A E» (inter S (I A C) (I B F» (inter T (I B D) (I A E)) (inter x (I A B) (I E F» (inter Y (I D F) (I A C)) (inter I (I S T) (I P Q)) (inter J (I P Q) (I x V))
(5t = ~» Figure 6-392
6.5
Circles
437
Example 6.393 (Steiner's theorem) (0.700,2,63) Given six points A, B. C. D. E . and F on a circle (ora conic). the three Pascallines [ABEDCFj. [CDAFEBj. [EFCBADj are concurrent. There are 20 steiner points for one Pascal configuration. Constructive description ( (circle ABC D E F) (inter P (I C D) (I (inter Q (1 F A) (I D E» (inter s (I B C) (I F A» (inter T (1 A D) (I B E» (inter x (I A B) (I E F» (inter y (I C F) (I A D» (inter J (I S T) (I P Q» (inter J (I P Q) (1 x Y» (~= ~»
A B»
T
Figun: 6-393
Example 6.394 (0.967, 2, 30) Given five points .40. Ai> A 2• Aa and AoAI
~.
then points
n A2Aa. AoAI n A 2A 4• AoA2 n AIAa. AoA2 n AIA4. AoAa n A I A 2• Ao~ n AIA2
are on the same conic. (There are 60 such conics for five points.)
Constructive description ( (points Ao AI A, A. A.) (inter Po (I A, A.) (I Ao (inter PI (I A, A.) (I Ao Ad) (inter P:. (I AI A.) (I Ao A,» (inter p. (I AI A.) (I Ao A,» (inter p. (I AI A,) (I Ao A.» (inter Po (I AI A,) (I Ao A.» (inter P (I P:. p.) (I Po Pd) (inter Q (1 p. P.) (I PI P:.» (inter s (I p. Po) (I P:. P.» (collinear P Q S) )
Ad)
Example 6.395 (4.383,3,84) Given six points Ao. AI. A 2• Aa. ~ and As on one conic. then points AoAI n A2Aa. AoAI n A4AS. AoA2 n AIAa. AoAa n A I A 2• Ao~ n AlAs. AoAs n AIA4 are on the same conic. (There are 45 such conics for one Pascal configuration. ) Constructive description ( (circle Ao AI A, A. A. A.) (inter Po (I A, A.) (I Ao AI» (inter PI (I A. A.) (I Ao Ad) (inter P:. (I AI A,) (I Ao A,» (inter Ps (I Al A,) (I Ao A,» (inter p. (I Al A,) (I Ao A.» (inter Ps (I AI A.) (I Ao .40»
p,
Figun: 6-395
Chapter 6. Topies from
w
m
(inter x (15 p4) (1~g PI)) (inter Y (1p4 %) (1PI pi)) @r ~(1 PO) % (1fi 5)) (inter zl (1p2 5) (1x Y))($$ = Example 6396 (4.566,4,75) Given six points Ao. Alp A2, A3. A4 and A5 on one conic, then points AoAl n A2A3, AOAl n A4As, AoA2 n AIAa, AoA3 n A1A5, A o A ~n A1A2, AoA5 n A1A3 are on the same conic. There are 90 such conics for one Pascal configuration. Constructive description ( (circle AO A, AZ AS AS) (inter PO (1A* AS) (1AO A])) (inter PI (1A, AS) (1AO A])) (inter P, (1AI A ~ )(1AO A*)) (inter ~s (1A] AS) (1AO AS)) (inter P4 (1A1 A21 (1Ao A,)) (inter P, (1AI AS) (1AO AS)) (inter Y (1p4 ~ s )(1PI P,)) (inter z (1f i PO) (1f i 5)) (inter x (15 p4) (1PO PI)) - (interw(1~0 ~ 1 () 1 z)) ~ =
(s s))
Figure 6-396
Example 6397 25 (0.866,2,55) Given skpoints Ao, Al, A2, At, Aq and As on one conic, then points AoAl n -42-43, AoAl n A4A5. AoA2 n A3A4, AOA3n d2A5r AIAq n d2A5, AlA5 n A3A4 are on the same conic. There are 60 such conics for one Pascal configuration. Constructive description ( (circle ao A, AS) (inter PO (1AZ AS) (1AO A])) (inter PI (1A. 4) (1ao A])) (inter PZ(1A$ A,) (1AO AZ)) (inter 5 (1AZ Aa) (1AO AS)) (inter p4 (1AZ A ~ )(1AI A,)) (inter % (1AS 4) (1 AS)) (inter x (1p4 P,) (1AO A])) (inter Y (1AS A,). (1 . ~i fi)) (inter z (1 ~s ~ g (1 ) A* Aaii (inter w (1%pb) (1x Y))
(s s)) =
25Example 6.397 was a new theorem found by Wu in 1980 [36]. Examples 6.395 and 6.396 were found by us.
6.5
439
Circles
Cantor's Theorems
6.5.7
Example 6.398 (0.083, 3, 8) The perpendiculars from the midpoints of the sides of a triangle to the tangent lines of the circumcircle at the third vertex of the triangle are concurrent. and this concurrent point is the center of the nine point circle of the triangle
Constructive description ( (POints A B C) (circumcenter 0 A B C) (midpoint L B C) (midpoint M A C) (inter N (P LOA) (P MOB» (eqdistance N L N M) )
Figure 6-398
The eliminants
The machine proof .l:JJi..J....
t:f:
MNM
-
SALOM~ - (SAOL+tSACO) SBLOM~ - (SBOL+tSBco-tSABO)
(-SSOL-tSBCO+tSABO)',PAOA (-SAOL-!SACO)' ,PBOB
SAOL!;, -
L (-.lSABO)' ,PAOA
!(SACO+SABO)
SBOL!;,- !(SBCO)
= (-!SABO)' ,PBOB
0 PBa8 ' PA~A ' PABA
p
BOB
.e..uu.
.im.!!!i!1I PBOB
(64)·S ABC
P 0 P8aB ' P~A.PA8A ,00,0 (64)· ABC
Q PBCB . PACA ' PABA .(64~RC) -
SABO
N51. COt' ,PAOA PLNL SASO
?CQM ,PAOA,sf RQ S ALOM ' PSOS ' s ASO
.imJ!i/1I ~IQM ' PAOA ALOM' PSOS
!!!.
NS~CQt"PBOB
P
PMNM
PBCB ,PACA ,PABA ·(64 ABC)
.im~i!1I 1
Example 6.399 (0.017,5, S) Let A, B, C, D be four points on a circle O. The perpendiculars from the centroids of the four triangles ABC, ABD. ACD. and BCD to the tangent lines of circle 0 at points D , C, B , A are concurrent.
Constructive description ( (circle ABC D) (circumcenter 0 A B C) (centroid M A B C) (centroid NAB D) (centroid L BCD) (inter P (P MOD) (P N 0 (parallel P LOA) )
The eliminants S
~SaI.lON ·S .. no±SCQQ ·SAQM
AOP-
SCDO
SAOL!;,- i(SADO+SACO+SABO) SCMON!;!, -
SAOM~ C»
(SCOM+tSCDO-tSBCO-tSACO)
-i(SACO+SABO)
SCOM~l(SBCO+SACO)
Chapter 6. Topics from C - t r y
440
The machine proof ~ -SAOL
E.
SOMON oSAQO±SQQQoSAQM
-
1:.. -
(-SAOL)-(-SCDO) (SCMON"S,WO+ S CDO ,SAOM)·(3) (-SADO-SACO-SABO),SCDO
!:!. ( M. -
Figure 6-399
3)-( 3SCOWSADO+3SCDO ,SAOM SCDO ,SADO+SBCO ,SADO+S,WO,SACO) (SADO+SACO+SABO) ,SCDO ·(3)
(3) ·(3SCDO ·S .WO+ 3SCDO ,SACO+3SCDO ,SABO) (SADO+SACO+SABO) ,SCDO ·(3)'
8img1ify
Example 6.400 (1.900, 42, 81) Let A, B, C, D, E be five points on a circle O. The perpendiculars from the centroids of the triangles whose vertices arefrom A,B,C,D,E to the lines joining the remaining two points are concurrent.
Constructive description ( (circle ABC D E) (circumcenter 0 A B C) (centroid A. BCD) (centroid B. A C D) (centroid c. A B D) (midpoint A2 A E) (midpoint B. B E) (midpoint c. C E) (inter N (P A. A. 0) (P B. B. (perpendicular N C. C E) )
0»
Figure 6-400
For more results related to Cantor's theorems, see Example 3.82 on page 143, Example 5.50 on page 242, Example 5.51 on page 242, and Example 5.56 on page 245.
6.6
A Summary
We have given 400 machine proved geometry theorems in Sections 6.2-6.5 and 78 machine solved geometry problems in Part I of the book. Thus totally there are 478 machine solved geometry problems in this book, including 280 proofs produced automatically by a computer program.
In order to access the overall perfonnance of the algorithm/program, we will first list the machine computation times and proof lengths of the examples in Part I.
441 No. 2.35 2.36 2.37 2.41 2.42 2.46 2.47 2.52 2.53 2.54 2.55 2.56 2.58 2.59 2.62 2.65 2.66 3.36 3.40 3.41 3.42 3.43 3.44 3.45 3.51 3.52 3.53 3.68 3.69 3.70 3.71 3.79 3.80 3.81 3.82 3.102 3.105 3.106 3.107
page 73 74 75 78 80 83 84 86 87 88 89 90 92 93 94 95 97 118 121 121 122 122 123 124 127 128 129 136 137 137 137 141 142 143 143 152 159 160 161
time 0.017 0.117 0.067 0.050 0.133 0.067 0.033 0.083 0.250 0.300 0.300 0.167 0.517 1.033 0.050 0.117 0.117 0.001 0.050 0.017 0.750 0.083 0.033 0.250 0.067 0.050 1086.8 0.050 0.067 0.033 0.033 0.033 0.083 0.050 0.067 1.050 0.867 0.117 0.033
maxt 1 2 1 1 1 2 1 4 6 9 10 2 17 15 1 1 1 1 3 1 3 2 1 6 1 3 3125 2 2 1 3 1 1 1 4 48 5 2 2
lerns 3 9 7 5 14 4 3 5 8 13 7 17 18 18 4 10 10 2 3 4 15 10 7 16 6 6 65 2 5 5 3 12 14 14 5 15 8 7 4
No. 3.108 3.109 3.110 3.111 3.112 4.23 4.24 4.25 4.40 4.41 4.47 4 .~3
4.54 4.61 4.62 4.63 4.64 4.65 4.87 4.88 4.89 4.90 4.91 4.92 5.38 5.46 5.47 5.48 5.49 5.55 5.56 5.57 5.58 5.59 5.61 5.62 5.63 5.64 5.65
page 161 162 162 163 164 176 176 176 185 185 188 190 190 194 194 195 196 196 206 206 207 208 209 210 234 240 240 241 241 245 245 247 249 250 250 251 252 252 253
time 0.200 0.350 0.050 1.383 1.100 0.033 0.017 0.033 0.067 0.883 0.083 0.017 0.033 0.083 0.133 0.083 0.067 0.083 0.100 0.117 0.067 0.133 0.167 99.200 0.083 0.017 0.100 0.117 0.083 0.017 0.717 0.083 0.083 0.117 0.067 0.133 0.017 0.050 0.267
Table 1. Statistics for the Examples in Part I
maxt 5 4 3 4 5 1 1 2 2 9 3 1 1 4 3 2 3 3 4 5 3 4 4 140 4 2 3 4 5 1 4 5 6 9 4 5 3 3 7
lerns 8 7 7 11 10 4 2 7 6 10 7 4 3 8 8 5 6 6 7 6 6 6 10 78 8 4 7 5 5 9 5 7 5 6 3 3 2 3 9
442
Chapter 6. Topics from Geometry
We use a triple (time,maxt,lems) to measure how difficult a machine proof is:
1. time is the time needed to complete the machine proof. The program is implemented on a NexT Turbo workstation (25 MIPS) using AKCL (Austin-Kyoto Common Lisp). 2. maxt is the number of terms of the maximal polynomial occurring in the machine proof. Thus maxt measures the amount of computation needed in the proof.
3. lems is the number of elimination lemmas used to eliminate points from geometry quantities. In other words, lems is the number of deduction steps in the proof. The following table contains some statistics for the timings and proof lengths of the 478 machine solved problems in this book.
Proving Time Time (secs) %ofThm. t < 0.1 45.3% t < 0.5 68.8% 85.5% t<1 t<5 97.45% 98.9% t < 10 t < 1087 100%
Proof Length Maxterm %ofThm. 16.9% m= 1 m<2 33.0% m<5 66.9% 81.7% m < 10 98.7% m < 100 100% m < 3125
Deduction Step Lemmas %ofThm. 7.1% 1< 3 1<5 16.7% 1<10 42.6% 73.2% I < 20 1 < 50 95.1% 100% 1 < 137
Table 2. Statistics for the 478 Theorems
Remark. 1. We can see that our program is very fast and can produce short proofs for many difficult geometry theorems. Eighty-five percent of the 478 proofs can be completed within one second, and the average maximal term and deduction step for the 478 examples are 14.8726 seconds and 17.35 steps respectively. 2. If we set a standard that a machine proof is readable if one of the following conditions holds
(1) the maximal term in the proof is less than or equal to 5;
(2) the deduction step of the proof is less than or equal to 10; or (3) the maximal term in the proof is less than or equal to 10, and the deduction step is less than or equal to 20, then 76.9 percent (or 368) of the proofs of the 478 theorems produced by our prover are readable. The machine proofs for 59.2% or 283 of the 478 theorems are actually presented in this book. 8 2 H not considering
Morley's theorem (see 3 of the this relllllrlt), the average maximal term is 8.37.
6.6.
443
A Summary
3. In spite of this success, there are still some geometry theorems for which the method/program performs badly. For instance, for Morley's theorem (on page 129), we have time= 1086.8, maxt = 3125. and /ems = 65. There are two main factors in the description of the geometry statements that affect the machine proof: the number of points in the statement and the type of constructions needed to describe it. Generally speaking, the number of points in a geometry statement is fixed and reflects the difficulty level of the statement in nature. On the other hand. the difficulty related to the type of constructions is somehow artificial. According to our experience. the geometry relations can be listed in ascending order of difficulties as follows: collinear, parallel. ratios. perpendicular. circles. angles. Morley's theorem involves information mainly about angles, and most of the other "difficult geometry theorems" involve perpendiculars, circles. or angles. We have two ideas for further improvement of the method/program. First, we can put elimination results for more constructions into the program instead of dividing these constructions into other simple constructions. Second, we can use new geometry quantities to produce short and readable proofs. In our method. we mainly use areas and Pythagoras differences. which deal perfectly with geometry statements about collinear and parallel, but do not always work well for a geometry statement about perpendicular. circles. and angles. For instance, to express the fact that the sum of two angles is equal to another angle, we have to use a complicated equation of areas and Pythagoras differences. Besides area and Pythagoras difference. we also discussed how to use other geometry quantities such as the vector, the complex number. and the full-angle, to produce short and readable proofs. The approach based on full-angles presented in Section 3.8 is quite promising. It uses the angle as the basic geometry quantity, and hence might produce short proofs for difficult geometry problems about angles, circles, and perpendiculars.
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List of Symbols
~ABCD
5 5 266 37
LIAB,CD]
44
6ABC 'VABC (ABCD)
triangle ABC the area of triangle ABC the cross-ratio the oriented angle from AB to CD the full-angle from AB to CD
--+ lAB , CD)
the exterior product
--+ (AB,CD)
the inner product the exterior product the inner product
227 227 223 Iz,y] (z , y) 219 -+ A 225 AB 225 AB 3 AB 39 AB 138 ABnCD 9 16 ABIICD ABl.CD 30 Ca 61 CL 107 27 CABC O(ABCD) 267 PABC 27 PABCD 29 6 SABC 7 SABCD 178 Sa VABCD 167 169 VABCDE 147 Klx] 147 class(P) 147 i nit(P) 147 Id(P) 147 Iv(P)
the vector from the origin to A
the vector from A to B the directed line segment from A to B the oriented chord the cochord of AB the intersection of AB and CD AB is parallel to CD AB is perpendicular to CD the Hilbert intersection point statements in the plane the linear constructive statements in the plane the co-area of triangle ABC the cross-ratio for five points the Pythagoras difference of triangle ABC for point B the Pythagoras difference of quadrilateral ABC D the signed area of triangle ABC the signed area of quadrilateral ABCD the Hilbert intersection point statements in the space the signed volume of tetrahedron ABC D the signed volume of polyhedron A-BCD-E the ring of polynomials the class of polynomial P the initial of polynomial P the leading degree of polynomial P the leading variable of polynomial P
455
456
LIst of Symbols
prem(P,Q) a~b a sim~liry b a~b
a a a
co~s
a a a
21~es
b
ar~-co
b
72
a is simplified
132 Pythagoras difference is expanded 121 constant is substituted 132 using Herron-Qin's formula
b
heg,on
147 the pseudo division of P for Q point A is eliminated from a
72
b
68
using the area coordinates
79 using the two-line configuration co=cir b 140 using the co-circle theorem volu~e-co b 195 using the volume coordinates
Index ALINE,126 ARATIO, 133,177 Artin,81 affine geometry, 82, 83, 174 affine space, 225 algebraic approach, 81 algebraic extension, 149 algorithm, 70 AFFINE,70 PLANE,116 SOLID,205 VECroR,236 angle, 51 , 129 anticenter, 388 anticomplementary triangle, 288 area coordinate, 68, 115 area coordinate system, 130 distance formula, 131 line equation, 131 associated field, 81 auxiliary para1lelograms. 58
CIR, 144, 182 COCIRCLE, 127 COLLINEAR, 112 COLL,63,178 CONG,111 CONSTANT, 121 COPL,178 CPLANE,180 Cantor's theorem, 143,245,439 Cantor's first theorem, 242 Cantor's second theorem, 242 Cartesian product, 81 Cayley-Menger formula, 202 Ceva's theorem, 10,61,62,262 polygons, 93.94,95 skew quadrilaterals, 190 Collins,53 center of inversion, 421 centroid, 242, 358 tetrahedra, 190 centroid theorem, 12,240, 281 tetrahedra, 188,231 cevian,262 cevian triangle. 262 circle of inversion. 421 circumcenter, 303 circumcircle, 303 class, 147 co-angle inequality, 23 co-angle theorem, 20, 24 co-angle triangles, 20 co-area, 27 co-circle theorem, 39, 42 co-face theorem, 170 co-side theorem, 8, 57, 259
BLINE,I07 BPLANE,180 Brahmagupta's formula, 40 Brocard points, 356 basis. 218 butterfly theorem. 41, 396 quadrilaterals, 397 general form for circles, 41 , 143 CENTROID,135 CIRCLE,I40 CIRCUMCENTER,135 CIRC, 161
457
458 co-side triangles, 8 co-trihedral theorem, 175 co-vertex theorem, 169 cochord, 138 collinear, 4 complete quadrilateral, 365 complex number, 246 configuration, 90 congruent, 222 conjugate rays, 267 consistency of proofs, 52 construction, 60, 108, 177 constructive configuration, 180 constructive geometry statement, 9, 107, 178 linear, 109 coordinates, 218 coplanar, 167 cosine law, 35 cross-ratio, 266, 267 cyclic quadrilateral, 387 DPLANE,180 Desargues' axiom, 18,82,83 Desargues' theorem, 272 dimension, 168,218 dimension axioms, 55 directed line segment, 3 directed line, 3 EQ-PRODUcr,112 EQANGLE, 127 EQDISTANCE, 112 Echols' theorem, 249, 250 Erdos' ineqUality, 25 Euclid's parnllel axiom, 58 Euclidean plane, 3 Euclidean space, 221 negative, 222 Euclid,81 Euler line, 321 Euler points, 323 Euler triangle, 323 Euler's theorem, 137
elementary geometry, 52 elimination, 9 area,65,68 area ratio, 65, 68 cyclic points, 140 length ratio, 66, 188 Pythagoras difference, 112, 203 volume, 183 equilateral triangle, 124, 248 exscribed circle, 328 extension, 149 exterior product, 223, 227 FOOT2L~, 178,230 FOOT2PLANE,181,230 FOOT,110 Fano plane, 91 Feuerbach's theorem, 324 field associated with, 81, 83 field, 217 final remainder, 148 finite geometry, 85 flat full-angle, 44 formula derivation, 86 full-angle, 44, 125, 154
Gauss line, 367 Gauss-line theorem, 74 Gauss point, 367 Gergonne point, 336 Griinbaum, 94 generally false, 152 generic point, 150 geometric approach, 81 geometric information base, 156 geometric object, 107 geometric quantity, 60, 107, 125,230, 177, linear, 112 quadratic, 113 geometry knowledge base, 156 Hanson, 52 HARMONIC, 112, 120
459 Herron-Qin's fonnula, 36, 106, 105, 132,228 quadrilaterals, 37, 38,106 Tetrahedra, 201, 202 in Miokowskian geometry, 243 Hilbert, 59, SI Hilbert intersection point statement, 61, 178 Hilbert's mechanization theorem, 59 harmonic pencil, 267 harmonic sequence, 120,267,365, homothetic, 291 homothetic center, 292 hyperbolic trigonometric, 244 INCENTER, 135 INTER,IOS INVERSION, 120 International Mathematical Olympiad, 122,196,209,330,412 incenter theorem, 327 incidence, S2 initial, 147 inner product, 219, 227 inscribed angle theorem, 45, 128 inscribed circle, 32S intercept triangle, 341 inverse, 421 inversion, 120,421 irreducible, 148 isogonal conjugate point, 340 isogonal conjugate, 301 isotomic, 261 isotropic, 220 isotropic line, 226 LINE,I07 LRATIO, 120 Lagrange identity, 225 Lemoine Point, 340 Lemoine axis, 310 Lesening's theorem, 272 Loveland, 52 leading degree, 147 leading variable, 147
25,
light cone, 220 MIDPOINT,112 MRATIO,I20 Menelaus' theorem, 73, 11, 176,259 Miokowskian plane geometry, 243 Miokowskian space, 222 oegative, 222 Miquel Point, 161 Monge's theorem, 210 Morley's theorem, 129, 442 machine proof, 72 at lemma level, 72 at proposition level, 72 at axiom level, 72 medial triangle, 288 median, 281 metric vector space, 219 NE-SQUARE, 252 NE-TRIANGLE, 249 Nagel point, 336 Nehring's theorem, 274 nine point circle, 48, 323 non-degenerate condition, 52, 62, 109 noosingular, 219 OLINE,1S0 ON,108 ORTHOCENTER,135 oriented angle, 36, 37 oriented chord, 39, 138 oriented quadrilateral, 7 oriented triangle, 6, 55 ortbic triangle, 295 orthocenter theorem, 31 tetrahedron, 199 Miokowskiao Geometry, 244 orthocenter-dual, 31 orthocentric, 298 orthocentric quadrilateral, 297, 393 orthogonal circle, 426
460 PARALLEL, 112 PARA,63 PARTIO,108 PE-SQUARE, 252 PE-TRIANGLE, 249 PERPENDICULAR,112 PERP, Ill, 178 PLANE,180 PLINE,107 POINT,108 PPLANE,180 PRATIO,I77 PRLL,178 PRLP, 178 Pratt-Wu's theorem, 363 Pappus point theorem, 271 Pappus' theorem, 14,80,270 Pappus' line, 271 Pappus-dual,271 parameter set, 147 Pascal's configuration, 59 Pascal's theorem, 142,435 Pascalian axiom, 18, 82, 83, 84 Pascal's theorem, 40 Ptolemy's theorem, 40, 139 Pythagoras difference, 27, WI Pythagoras difference theorem, 28, 29 Pythagorean theorem, 23, 27, 221, 227 paradox, 52 parallel translation, 103 parallelogram, 16, 17, 19,51,58 parallel, 16, 171, 226, 228 pedal triangle, 348 perpendicular, 219 plane, 168 polar, 425 pole, 425, 429 polygon, II polygrams, 94 position coordinates, 4, 56 position ratios, 4, 56 predicates, 63
Index
predicate form, 63, III, 178 prover, 51 pseudo division], 147 pseudo remainder, 147 pure constructive, 238 pure point of intersection theorem, 59 radius of inversion, 421 ratio constructions, 119 ratio of the directed segments, 55, 16,58 rectangular coordinate system, 221 reduced, 147 reducible, 149 regular n-polygons, 250 remainder formula, 148, 149 right full-angle, 44 Shephard, 94 SPHERE,182 SRATIO,230 SYMMETRY, 120 Seidenberg, 52 Simson line, 429 Simson's theorem, 42, 141, 160, 245 Steiner's theorem, 185 Steiner-Lehmus' theorem, 26 segment arithmetic, 83 signed area, 6, 55, 228 similar, 21 sine law, 34 skew area coordinate system, 130 skew volume coordinate system, 211 solid metric geometry, 225 square distance, 227 square, 121, 220 subspace, 218 successive pseudo division, 148 symmedian, 340 symmedian point, 340 symmetric matrix, 219 TANGENT, 112 TUNE,l07
461 TPLANE.I80 TRATIO.I09
tritangent circles. 328
Tarski.52 tangent function. 125 tangential triangle. 312 theory of geometry. 81 transcendental over. 149 translation. 172 transversal for polygrams. 95 three perpendiculars. 199 triangular form. 147 trilinear polar. 352 triple scalar product, 224 trisector. 129
type of constructions. 443
~o-lineconfiguration.
79
unordered geometry. 53 vector. 218. 225 vector space. 217 visual angle theorem. 34 volume. 228 volume coordinates. 192
VVu. 53. 59. 147.363.438 Zhang. 2. 81
