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PRODUCED FOR PRE-COLLEGE PH!LOMATHS
1983 - 1984 by
Professor Samuel L. Greitzer 1906 - 1988
Volume 2
Number 1· 5
Third Edition Original Copyright © 1983
Second Edition Copyright © 1995 Third Edition Copyright © 1996
The American Mathematics Competitions
ProCessor Samuel L. Greitzer in Memoria During the interval of time between 1982 - 1987 Professor Samuel L. Greitzer served as the editor and author of essentially all of the articles which appear in the Arbelos. His untimely death, on February 22, 1988, was indeed a sad day for all those who knew him. Professor Greitzer emigrated to the United States from Odessa, Russia in 1906. He graduated from the City College of New York in 1927 aOO earned his Ph.D. degree at Yeshiva University. He had more than 27 years experience as a junior and senior high school teacher. He taught at Yeshiva University, the Polytechnic Institute of Brooklyn, Teachers College and the School of General Studies of Columbia University. His last academic teaching position was at Rutgers University. He was the author or co-author of several books including Geometry Revisited with H.S.M. Coxeter. I was extremely pleased that Professor Greitzer agreed to write and edit the Arbelos, since I frequently receive requests for references to publications which are appropriate for superior students, and for material which will help students prepare for the USA Mathematical Olympiad. Professor Greitzer served as a coach of the summer Mathematical Olympiad training program from 1974 to 1983. Consequently, many of the articles in the Arbelos are a reflection of his lectures and thus appropriate for talented and gifted students. Professors Greitzer and Murray S. Klamkin accompanied the USA team to the International Mathematical Olympiad from 1974 [the first year the USA participated] to 1983. Their success in coaching the team is indicated by the fact that it usually placed among the top three [out of 30-35 participating countries]. The contributions of Professor Greitzer to the development of students of mathematics and teachers from many nations will be lasting. We shall miss his humor, words of wisdom, mathematical insight aOO friendship.
Dr. Walter E. Mient1«J Executive Director American Mathematics Competitions University of Nebraska-Lincoln
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Preface To The Third Edition 1983 . 1984 During the five year period from 1982 through 1987 Professor Greitzer authored 25 regular issues of the ARBELOS. This Volume 2 contains the issues published from 1983 - 1984. The tradition of Professor Greitzer to present a geometry problem on the cover of each of the issues of the ARBELOS led to a challenge to the readers for their solution. I was extremely pleased that he agreed to prepare a solution manuscript and they appear in Volume 6 of the ARBELOS series. The Second Edition of this publication was completed by Andre' W. Mientka and it reflected his elegant use of computer technology and software. The Third Edition essentially remains the same as the Second, except that page headings were included as well as the inclusion of one Table of Contents. These changes were made in a noteworthy manner by Mrs. Amy Fisher, a member of the American Mathematics Competitions Lincoln office staff. Your comments and suggestions are welcomed. Dr. Walter E. Mientka Executive Director American Mathematics Competitions 1740 Vine Street University of Nebraska @ Lincoln Lincoln, NE USA 68588-0658
e-mail:
[email protected]
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Preface To The Second Edition 1983 - 1984 At the time Dr. Greitzer created and wrote the Arbelos he did not have access to desktop publishing. In fact, he used a typewriter with key inserts for the mathematical symbols. Using current computer technology and software, this Second Edition was created in order to produce a publication which was easier to read and contained diagrams which were more definable. This was accomplished by using current computer technology and software. It was completed by Andre' W. Mientka whose academic training and technological expertise is reflected throughout the manuscript. Except for some minor editorial and mathematical corrections he reproduced the mathematical content and discourse as was originally written by Professor Greitzer. Your comments and suggestions are welcomed.
Dr. Walter E. Mientka Executive Director American Mathematical Competitions 1740 Vine Street University of Nebraska, Lincoln 68588-0658
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CONTENTS OF THE ARBELOS, VOLUME 2
Volume 2, Chapter t· Cover - Spherical Law Of Cosines Preface 1983 - 1984 Which Is Correct? Kiirschak Komer Something About Vectors Answers and Questions Parity Assorted Problems
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3
8
23-44
23
Cover - Fibonacci From Pascal!
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16
17
21
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Volume 2, Chapter 2 t
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1-22
1
2
Preface
24
Generating Functions Assorted Problems Topological Considerations An Intriguing Problem To Improve Problem Solving Just for Fun
25
29
30
36
39
43
Volume 2, Chapter 3*
45-68
45
46
47
54
Cover - An IMO Problem
Preface Transformations Many Cheerful Facts
1t •••.•.•••••••.•.•.••.•.••••••••••.•....•..........•.. ...•..•.•.•.•••••.••....•.•.••••••••••••
Week-Enders The Euler Line Assorted Problems Kilrschak Komer From Our Readers
57
62
63
65
66
67
• Originally published as the Arbelos, Volume 2, Issue I, September, 1983. t Originally published as the Arbelos, Volume 2, Issue 2, November, 1983. Originally published as the Arbelos, Volume 2, Issue 3, January, 1984.
*
iii
Volume 2, Chapter 4'
69-90
69
70
71 77
83
84
85 89
90
Cover - Outer Napoleon Triangle
Preface
What are the Odds?
Napoleon, Miquel, Simson
How To Catch A Lion
Week-Enders
Many Cheerful Facts
Assorted Problems
Ktirschak Komer
Volume 2, Chapter 5··
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91-116
91
Cover - Our Monthly Problem
Preface Topics in Number Theory Just For Fun Ramsey's Theorm & The Pigeon Hole Principle Projective & Affine Planes Ktirschak Komer Answers
92 93
100
l0l 104
115
116
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, Origimdly published as the Arbelos, Volume 2, Issue 4, March, 1984. •• Originally published as the Arbelos, Volume 2, Issue 5, May, 1984. Iv
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1983-1984: No.1
September, 1983
Copyright © 1983 1
ARBELOS
Volume 2, Chapter 1
1983 - PREFACE -1984
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With this issue, we start our second year of trying to provide mathematics beyond the usual for students who, in our ~ opinion, would like to know more than the classroom provides. ~ Also, we are trying to provide problem materials that are out of the ordinary in form and content, and that could be solved by ingenious ] methods. For the first year, our selection of students was quite small, since there was no effort made to reach too many readers. Nevertheless, this number has grown, due mainly to word of mouth "advertising". Even with so small a clientele, there have managed to be a number (small, of course) of readers who have shown extraordinary interest by solving the problems provided and by sending in their solutions. Now it does take time to prepare an issue. For example, this issue is being prepared in March of 1983. Therefore, where solutions were sent in more than two months after each issue was mailed it was impossible to acknowledge such solutions. ]. At this time, therefore, we wish to acknowledge the excellent contributions made by David Mo~ws, Mark Kantrowitz, Dougles Davidson and Glenn Ellison. In the future, please send your solutions to me early - say at least a month before the next issue is due, so that it can be read, evaluated, and due credit given.
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Very truly yours, Dr. Samuel L. Greitzer
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Volume 2, Chapter 1
WHICHISCORRECT? In order to check on the relative accuracy of two important formulas of world significance, we find it necessary to go into spherical geometry for a spell. In any case, in view of the present state of solid geometry in general and spherical geometry in particular, better students should be aware of this area. Referring to the figure at the A right, we see a sphere with center at O. On this sphere, there are no "straight" lines. We have, instead,
great circles - that is , circles whose
centers are at O. Two such "lines" are
AMA and ANA. First, they ~
intersect. Hence, on a sphere, no lines
are parallel. Next, for the "line", we have no A
"length" as we know it. MN is an arc Figure 1 of a circle, and is measured by the central angle, LM 0 N. Again, a triangle will consist of three arcs, measured in degrees, and three angles, also measured in degrees. See, for example, MMN. Additionally, if arcs AM and AN are 900 each (called quadrants), the angles at M and N will be right angles, so that the sum of the angles of MMN exceeds 1800 • We have also drawn A M and AN in planes A MA and ANA respectively. These will be parallel to OM and ON respectively, so that LA of MMN will equal LMAN which equals LMON which, in turn equals arch MN. That is, an angle on a sphere is equal in degrees to the arc cut off by its sides at a quadrants distance from the vertex. Now look at MXY, where AX and AYare quadrants, and LA is equal to one degree. We agree to call this a "spherical degree". Still referring to Figure 1, we call attention to the following: first, there are 360 spherical degrees about the point A (the pole), 3
ARBELOS
Volume 2, Chapter 1
and 360 spherical degrees about pole A. These cover the entire sphere, so that the area of the sphere is 720 spherical degrees. Next, examine Figure ANAMA. This has two sides, but a definite area. We call it a "lune". We have already called attention to the fact that the angle at A (or A) is equal in degrees to the arc MN. Since AM and AN are quadrants, the area of spherical MMN equals A spherical degrees. Therefore, the area of the entire lune is equal to 2A spherical degrees. There is a great temptation to continue and develop more of spherical geometry. For example, since the area of a triangle (or any figure) cannot exceed that of the sphere, there is a limit to the area of any triangle ! Next, we cannot have similar triangles. And finally, two triangles can have equal sides and angles Figure 2 and not be congruent. In Figure 2, we have isosceles MBC with median AM. MBM and MCM cannot be made to coincide. Such triangles are called symmetric. However, right now, our main goal involves the development of some spherical trigonometry. i Our intention, naturally, is to work on the sphere, not within it. If we wished to draw a circle on a sphere, our center would have to lie on the sphere. We could then have a radius which would be an arc of a great circle. The circumference would not equal 21tf and the area not 1tf2.
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Volume 2, Chapter 1
With this in mind, let us examine MBC with right angle at C. (Figure 3). From B, we drop perpendicular BB to meet OC, then BA perpendicular to OA, and finally draw line AB. First, LBAB and LA are equal. (the tangents to the circles at A are parallel to lines AB and AB.) Now _ BB
sinA = sin A = _ .
Figure 3
AB '
BB = r sin BOC = rsin 0,
AB = r sin BOA = r sine.
Therefore, we can remain on the surface of the sphere and say that sin A = sin alsin c. Of course, sin B = sin b/sin c. This does look a lot like what we have in plane trigonometry. However, we must be careful not to draw too strong an analogy. For example, you can use the same diagram to show that cos A = tan b/tan c ! Now we can expand our work to any spherical triangle. In Figure 4, CH is drawn perpendicular to side AB. Now sin A = sin hlsin b c sin B = sin hlsin a, and, on eliminating sin h, we derive a Law of Sines for spherical triangles: sina sin b ( sinc)
A - - - - . J B sinA = sinB = sinC H
and we need not go within the sphere in this case either. For our next derivation, we request that the reader refer to the diagram on the frontispiece. In this figure, MBC is drawn on a sphere with center 0 and radius r.
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Volume 2, Chapter 1
We draw tangentCA and line OA, which, since they are coplanar, intersect at A. Similarly, we draw tangent CB and line OB, and finally, AB. Note that LOCA and LOCB are right angles, since a tangent is perpendicular to a radius at their point of contact. Note also that LACB = C while LAOB= c. Now (AB)2 = (OA)2 + (OB)2 -2 (OA)(OB)cos c (AB)2 = (CA)2 + (CB)2 -2 (CA)(CB)cos C. Subtracting (and remembering that dOCA and dOCB are..!!gh! triangles, we have: 0= r 2 + r 2 + 2 (CA)( CB)cos C - 2 (OA)(OB) cos c. (OA)(OB)cos c = ~ + (CA)(CB)cos C Divide by (OA)(OB) and substitute for the ratios that appear, and we finally find: cos c = cos a.cos b + sin a.sin b.cos C and we have a Law of Cosines for spherical triangles such that it is unnecessary to go inside the sphere. Of course, we may also write: cos a = cos b.cos c + sin a.sin b.cos C, cos b = cos c.cos a + sin c.sin a.cos B. In the special situation where LC is a right angle, our first formula becomes cos c = cos a . cos b Note: As a special exercise, derive this last from the diagram of Figure 3 directly. It is obvious that this last formula takes the place of the Pythagorean Theorem in Euclidean Geometry. Why, then, can't we find out which law holds for our world? Certainly, there is lots of difference between
cos c = cos a . cos b and In fact, there were many mathematicians, among them Gauss, who suggested that, given a sufficiently large triangle 6
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Volume 2, Chapter 1
(hopefully right) and sufficiently accurate measurements, one could check. Now there are many ways to derive the series expression for cos x, say. One of the very cutest is that by Euler. Or, one can use calculus, etc. Any way, we have,
a 2 a 4 a 6
cos a = 1- 21+ 41- 61 +
b 2 b 4 b6
cos b = 1- 2! + 4! - 6!'+
c2
cos
C=
c4
c 6
1- 21+41- 61+
However, with the world so large, the arcs a, b, c are very small indeed. In fact, we find that all powers above the second are too small to be included. This gives us 1-
c2
21+
b2
a2
=
(1- 21+ )(1-21= )
c2 a2 b2 1 - - = 1 - - - - + terms of order above 4. 2! 2! 2!
2 2 or finally, c = a + b 2 , whether we use the Pythagorean theorems for Euclidean or spherical geometry. Sorry about that !
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Volume 2, Chapter 1
KORscHAK KORNER To the problems of this column the following students submitted most credible solutions during the past school year: Barbara Baethge (TX), Andrew Chin (TX), Constantine Coster (AL), Douglas Davidson (VA), Glenn Ellison (CT), Douglas Jungreis (NY), Jeremy Kahn (NY), Mark Kantorowitz (MA), David Moews (CT), Michael Reid (NY), Mark Reider (PA); David Steinsaltz (NY) and David Zuckerman (NY). Many thanks are hereby expressed for their splendid efforts. This year we will feature some more problems posed in this famous competition since 1928 (for the years 1894-1928 see Hungarian Problem Books I and II in the NML series). Students are invited to set aside an uninterrupted 4-hour period to compose complete well-written solutions to these problems and to submit their work to the address given below for a critical evaluation. 2 1/1931. Let p be a prime greater than 2. Prove that - can be p 2 1 1 expressed exactly one way in the form -= -+ - in terms of p x y distinct positive integers x and y. 2/1931. Assume that a., a 2 , a 3 , a 4 , as and b are integers such that ai + a; + ai + a ~ + a~ = b
2
.
Show that not all of these
numbers can be odd.
3/1931. A and B are two points on a given straight line.
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Determine a point P on this line so that 1+ AP + 1+ BP is
maximal. Dr. George Berzsenyi
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Volume 2, Chapter 1
SOMETlDNG ABOUT VECTORS We define a vector as being an ordered n-tuple of real numbers, and shall represent one either by means of a Greek letter of by (xl' ••• ,x n ). For our purposes, we let n = 2 or n = 3. We need some means of operating with vectors, and agree to use the familiar "parallelogram Law". That is, we shall be dealing with "free" vectors. Using the number pair definition, this means that if a = (ai' a 2 ) and ~ = (bl , b2 ) , then the sum is found by locating the points (ai' a2 ) and (hi' h2 ) on a grid, joining each to the origin, and then constructing the parallelogram with those lines as adjacent sides, and letting the diagonal emanating from the origin represent the sum. As the diagram shows, one can also add two such vectors by (bl' b [ ~(al+bl' ii?j2) a 2 +b 2 )
~ a+~
a
a
(ai' a 2 )
attaching the tail of one to the head of the other. One reaches the same point P in either case. Obviously, one can add any number of vectors in the same way - just attach the head of one to the tail of the one before it. We also invent the vector 0 = (0, 0, 0) and the vector -a = (-ai' - a 2 , - a 3 ), and we can now add or subtract. To multiply a vector by an element not a vector (a scalar)
we multiply each term, thus:
ta =(ta l , ta 2 , ta 3 ).
Again, given coplanar vectors a., ~, 'Y, all coplanar, then, if we select y to be the diagonal of a parallelogram whose sides lie in the directions of a. and J3, it is easy to multiply these by scalars so as to make xa. and yj3 sides of that parallelogram - that is, so that 'Y = xa+ ~. In this case, a. and J3 form a "basis" for all 9
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Volume 2, Chapter 1
vectors in that plane. In the same manner, any three vectors not in the same plane may constitute a basis for threespace. The length (magnitude, norm) of a vector is found by using the Pythagorean Theorem - that is, if a = (a p a 2' a 3 ), then the length of ex equals
Ial = Ja}2 + a 2 2 + a 32 , with suitable changes
for two, four or more dimensions. The direction of a vector, however, must be expressed in terms of some fixed coordinate system or its equivalent. It is usual to select three mutually perpendicular unit vectors, as a coordinate system. In the plane, we use i, j, with j 90° counterclockwise from i. In three dimensions, we use i, j, k. These are obviously bases for vectors in two and three dimensions respectively. Thus, a vector A in two dimensions (in a plane) can be written as A = xi + yj. In three dimensions, we could write A = xi + yj + z k. Now consider two standard vectors a = (xPy}) and ~ = (x 2 ,y 2)' The endpoints determine the line AB. What conditions must the vector
A= (x,y) follow if its endpoint is to
lie on this line? First, the vector AB = ~- a. Also, PB = ~ - A. Then ~- A = t(~ -a), which simplifies to A= ta + (1- t)~.
~
Noting that the sum of the coefficients on the right equals the coefficient on the left, we may conclude generally, that, for P to lie in line AB, xa+y~+ zA. =0 x+y+ z=o Similarly, for the end point of a vector 0 to lie on the plane determined by the endpoints of the vectors a, ~, y, we must have xa+y~+ zy+w8=O 10
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Volume 2, Chapter 1
x+y+z+w=O. With what we have so far, we can o quite a bit in the way of solving
roblems. Thus, suppose we have a
ircle with radius 4r with a circle with 4r
ldius r rolling on its interior. We seek
le locus of a point Y as the circle rolls.
Directly, we know that the X M A ector OY is equal to ix + jy. [owever, we can get to point Y from 0 by following the path 1M + MP + PR + RY. In terms of vectors, we have 3r cos e) + j(3r sin e) - i(r cos(3e -1t» + j(r sin(3e -1t». This mplifies, first to ir(3 cos e + cos3e) + jr(sin e - sin3e) = ix + jy. ld finally to x = 4r cos3 e y = 4r sin3 e. A useful formula is obtained in case, in our second figure, e point P divides the segment AB in the ratio m:n. Then AP = m PB, or n(A-a) = m(~-A) which, solved for A., elds
A=na+m~. m+n ate the relative position of m and n in this formula. Of course,
P is the midpoint of segment AB, the formula becomes
= (a+ ~)2.
We pause here to present a very pretty problem. Assume a
,lygon (not necessary planar) in threespace. Let its n vertices
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labeled Ao' AI' •••, A n_ I in any order. Suppose further It one travels from vertex A o to AI' then halfway from I
to A2 , from which one travels one-third of the way to A 3 ,
11
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Volume 2, Chapter 1
thence one-fourth of the way to A 4' and so on, until one travels one-n-th of the way from the point then reached toward A o' Prove that, however the vertices are labeled, one arrives at the same point, the centroid of the figure. Notice that for a triangle, one does reach the centroid, the intersection of the medians. We tum now to products of vectors. We have already been introduced to the product of a scalar and a vector. We now continue with the scalar product of two vectors, ex and ~. It is defined to be equal to 1a11~1 cos e, where e is the angle between the two vectors. It is also called the dot product, and written ex • ~. We see at once that, if ex and ~ are not zero, but ex • ~ = 0, the two vectors must be perpendicular. Also, for the unit vectors i, j, k, we have i. i = j . j = k. k = 1, while, i. j = j. i = j . k = k. j = i . k = k •i = O. Now,if ex=a l i+a 2j+a 3k ~=bli+b2j+b3k
and we multiply these, we find that ex • ~ = alb l +a 2b 2 +a 3b 3. As a very simple application of all this, note that we also have ex • ~ = 1aI1~1 cos e. When we equate these two expressions, a little algebra (and recalling that cos e ~ 1) yields
(a~ +a~ +a;)(b~+b~ +b;r~~(albl +a 2b 2 +a 3b 3)2 thewell known Cauchy inequality. We continue with a definition of the vector product of two vectors. This is written ex x ~, is usually called the "cross product", and is defined as equal to v p.11~1 sin e, where V is a unit vector perpendicular to the plane of vectors ex, ~, and e is the angle between the given vectors ex and ~. 12
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Volume 2, Chapter 1
One can see that the modulus of a vector product of two
vectors equals the area of the parallelogram determined with a, ~
as adjacent sides. Half of this, of course, equals the area of the
triangle two of whose sides are a and ~.
As simple applications of these definitions, one can easily derive the Law of Cosines by using the dot product and the Law of Sines by using the cross product. More interesting, in view of the present situation with solid geometry, is the use of vectors to derive "solid" results.
a For example, suppose that, in the diagram at the left, we are given v.a = 0, v. ~ = 0, and let 'Y be a vector in the plane of a and ~.
Then we may write 'Y = xa +Y~. Take the dot product with V and we have v. 'Y = x v. a + Y v. ~ = 0, which means that v and 'Yare perpendicular. We have an old text on solid geometry in which this takes over thirty steps to prove. Again, suppose that a, ~ and 'Yare all perpendicular to V, as in the same diagram. Must a, ~ and 'Y be coplanar? We are given a. v = 0, ~. v = 0, 'Y' v = 0. Suppose that, in addition, we take a, ~,v as a three-dimensional basis, and write 'Y = xa + ~+ zv. The dot product of this with V will be v''Y=x v.a +Y v.~ + z v.v = z v.v = and, since v.v is not zero, z must equal zero, 'Y = xa + ~ , a, ~, 'Y coplanar. We still need a simple way to represent a vector product. Once again, we use i, j, k as a basis, so a = a l i + a 2j + a 3k
°
~=bli+b2j+b3k.
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Volume 2, Chapter 1
Note that i x j = k, j x k = i, k x i = j. Note also that i x i = j x j = k x k = O. Ifwe multiply, remembering that a x ~ = - ~ x a, then a x ~ = i(a 2 b 3 - a 3b 2 ) + j(a 3b I - a Ib 3 )+ k(a Ib 2 - a 2 b I ) We may write this product in determinant form:
ax
~=
j
k
al
a2
a3
b
b
b
I
2
3
al
a2
a3
bI
b2
b3
c1
c2
c3
We can use the determinant at the top of the page to evaluate the triple vector product y x ( a x ~). If we multiply and simplify our product (which requires some careful algebra), we derive the result y x (ax ~) = a (y. ~)- ~ (y. a) The relative position of the vectors is important. Changing the position of the parentheses will change the value of the product. When working on problems involving figures on a plane, it is often useful to select a point not in that plane and use vectors to the figure from that point. There is usually a gain in symmetry. 14
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Now we introduce the triple scalar product. This is a scalar, since it is the dot product of two vectors. As before, a x ~ = 't a ~ sin e, and t a unit vector perpendicular to a and ~,as in the diagram on the previous page. Then 't. Y can be considered as the projection of y along the altitude of the parallelepiped formed by a, ~, y. That is, a x ~, y yields the volume of the parallelepiped. Modifications of the cross product and the triple scalar product that will yield the area of a triangle or a tetrahedron are obvious. It is interesting to note that, in determinant form, ax~.y= a.~xy=
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Volume 2, Chapter 1
We include a few problems that illustrate application of the material covered here. a) ABCD is a square and 0 a point not in the plane of the square. Prove: (OA)2 + (OC)2 = (OB)2 + (OD)2
b) In the diagram at the right, prove that the midpoints P, Q, R of BD, CF, AE lie on a line.
A
B~
C
D
E
c) In trapezoid ABCD, AB and CD are parallel, and M is the midpoint of AD. Find the area of triangle MBC.
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A
B
d) In the quadrilateral ABCD, P, Q, R, S are the midpoints of
AB, BC, CD, DA. Prove that PR, QS bisect each other.
R
e) Prove that, in any parallelogram, the sum S 1 - - - - + - - - \ of the squares of the four sides is equal to the sum of the squares of the two A ' - - - - - 1 " ' - - - -.... B P diagonals.
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Volume 2, Chapter 1
ANSWERS AND QUESTIONS a) The square of a natural number can end only with a 1, 4, 6, 9, 5 or O. We are given a number which consists of !l ones followed by !l twos and ending in 25. What is the square root (if any) of the number? 2 1 1 1 1t = What is the sum of b) We know that 2"+-2+2"+
1
2
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T.
3
111
-+-+-+ ... ? 12 32 52 . c) Among the numbers whose digits in the decimal system are all ones, some are prime numbers and some are not. What are the factors of 1,111,111 ? d) The sum of the squares of the sides of a parallelogram equals the sum of the squares of the diagonals (easy using vectors I). Find the sum of the squares of those diagonals of a parallelepiped that are not diagonals of faces in terms of the edges. e) We know that n! is divisible by 1, ••• , n. If 100 < n! < x < (n + I)!, show that there is a value of x that is divisible by n 3 .
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PARITY
I Sometimes, very simple concepts are enough to solve complicated problems. One such concept is that of parity. We aware of this when a neighbor said he had been trying, I onwereandmade off, for some time to find five odd natural numbers whose sum would equal 100. took some time to convince him of the I rules for parity - namely:
the sum of two odd numbers is even,
sum of two even numbers is even,
I thethe sum of an odd number and an even number is odd. We met with another problem in a math journal: If p and p + 2 are odd primes, then p + 1 is divisible by 6, I given p greater than 3. Since consecutive integers are alternately odd and even, p + 1 is even by 2. Moreover, one of three consecutive I integersandisdivisible divisible by 3. Since neither p nor p + 2 is divisible must be. I by 3, PThe+ 1following problem appeared in an official state examination: If the discriminant of a quadratic equation whose coefficients are I integers is equal to 23, what is the nature of the roots? Here, if the discriminant is b 4ac, then 4ac is always divisible I by 4. Now, if b is even, then b is divisible by 4, so the discriminant is divisible by 4. However, if b is odd, then b leaves a remainder of 1 when divided by 4. In this case, the entire I discriminant leaves a remainder of 1 . Therefore, whatever the discriminant, the remainder on division by 4 must be either 0 or when 23 is divided by 4, the remainder is 3. Hence I it isHowever, impossible to have a discriminant equal to 23. A very pretty problem is the following: Assume that, on a I blackboard (a very large one!), the numbers from 1 through 1983 have been written. One selects any two of these numbers, erases I them and replaces the two numbers thus erased by their difference. I It
2
-
2
2
1.
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Volume 2, Chapter 1
This reduces the totality of numbers by one. One continues this operation until just one number is left. Is that number odd or even? If the two numbers are both even, the operation reduces the number of even numbers by one, but leaves the number of odd numbers unchanged. If one number is even and the other odd, the operation reduces the number of even numbers by one, but leaves the number of odd numbers unchanged. If both numbers are odd, the operation reduces the number of odd numbers by two, and increases the number of even numbers by one. That is, the operation will reduce the number of odd numbers by zero or two. Now the number of odd numbers in 1983 is 992 - an even number. Hence the operation ultimately leaves no odd numbers, and the one number left is even. We should add the following rules to those we have already given: The product of two odd numbers is odd, The product of two even numbers is even,
The product of an odd number and an even number is even.
Of course, a problem can be so stated as to hide the fact that it can be solved by considerations of parity. For example, Solve x Y + 1 = z (x, y, z prime numbers) First, y cannot be odd, for in that case, one could factor x Y + 1, one factor being x + 1. In this case, z would be composite. Hence y = 2. Next, in x 2 + 1 = z, x cannot be odd, for then z would be even and therefore composite. Hence 22 + 1 = 5 is the solution. A nice problem is the following: Ifthe ten's digit of the square of a number is equal to 7, what is the unit's digit? Let the number be lOt + u. Then its square is 2 100t + 20tu+ u 2 = OOt 2 +2tu)xlO+u 2 • The term at the left here ends with a zero. Now of all possible values of u 2 , 01, 04, 09, 16, 25,36,49,64,81, all but two have an even number in the ten's spot, which would lead to an even number when added. Only 16 18
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Volume 2, Chapter 1
and 36 do not. Now since (10t 2 + 2tu) is~, only 16 and 36 could produce a 7 in the tens place. In both cases, the units digit is 6. We look next at the following problem: Solve in natural numbers a 3 - b 3 _c 3 = 3abc a 2 =2(b+c) Let us resist our natural impulse to eliminate something, and look at the problem parity-wise. First, a is even. Next, a is larger than b or c. From a> b and a> c, we get 2a > (b + c). That . IS,
4a> a 2 .
This means that a < 4, so a = 3 or 2 or 1. Since a is even, a =2. Hence b=c= 1. Parity considerations play quite a part in Number theory. They help in finding the expressions
a= m 2 _n 2 , b =2mn, c = m 2 +n 2 that yield all Pythagorean triples. And they also help to prove the impossibility of solving
xn+yn =zn
where n equals or exceeds 3 in many cases. From an Olympiad, we present the following: Find the largest number which is the product of positive integers whose sum equals 1983. We have. a t +a 2 + ••• +a n =1983, at' a 2 • a 3
• •••• an
Suppose there is an a.I
~
=
tnaX. lffiUffi.
4. Replace it with two terms - namely
a.I - 2, 2. This obviously leaves the sum, 1983, unchanged. As for the product, we get 2a.I -4. Now, from a.I -4 ~ 0, we get, by
adding a., 2a.I - 4 ~ a.. Therefore, this sort of substitution I I increases the product. Since we can thus replace all a.I 19
~
4, our
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Volume 2, Chapter 1
product will consist of only 3'sand 2' s; thus 3x2 Y = our desired product. However, 23 = 8 can be replaced by 32 = 9. Therefore y = 0, 1, or 2 at most. Now 1983 is divisible by 3, and equal to 3 x 661. Hence x = 661, y = 0, and our maximum product equals 3 661 . We present for your amusement the following problems: a) Find all solutions (if any) in natural numbers of x 2 + y3 = 3(a 2 + b2 ). b) Find all solutions (if any) of the following expression, composed of integers: + +nj + ••• + = 1599.
nt ni
nt4
c) Find all solutions (if any) of the following expression, composed of integers: a 2 + b 2 + c 2 = a 2 b 2 . We complete this short article with a problem that you may find fun and challenging. The number 1210 is remarkable. The leftmost digit tells how many zeros the number has, the next digit tells how many ones it has, the tens digit tells how many two's it has, and the units digit how many threes it has. Similarly, the number 21200 is remarkable. The leftmost digit tells how many zeros the number has, the next digit how many ones the number has, the next digit how many twos the number has, and the last digits how many threes and how many fours the number has. Can you construct a ten-digit number that has the same property ? (If you can, send it to me. I lost my answer.)
And I remember nothing more That I can clearly fix,
Till I was sitting on the floor,
Repeating, "Two and five are four,
But five and two are six."
C.L.Dodgson
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Volume 2, Chapter 1
ASSORTED PROBLEMS 1) Given the recursion formula x n+ I =
1
X
+ -X n
(n = 0, 1,
... ).
n
Given also
X =
o
5. Show that 45 < x lOOO < 45.1 . (Yugoslavia)
2) Let M be the set of all natural numbers (in the base 10), which do not contain the digit 9. If xl' x 2' ••• , X n is a subset of n 1 n distinct members of M, show that < 80.
Ij=l X j
(Sweden)
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3) Construct a cubic equation with integer coefficients whose 1t 51t -31t roots will be sin 14' sin 14' sin 14 . (VietNam) These problems are among those that have been presented for possible use in International Mathematical Olympiads by the nations listed.
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Volume 2, Chapter 1
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Volume 2, Chapter 2
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1983-1984: No.2 23
ARBELOS
Volume 2, Chapter 2
PREFACE
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With recognition at last that mathematics education is in " very poor condition finally accepted, even by the most recalcitrant, ~ there are many efforts under way to improve this situation. One of the attempts to improve mathematics results involves teaching "problem solving". Now there questions whether this skill can be taught, but there is little question that it can be improved in those who already are good at it. This can be done, for example, by seeing examples of problem solving by others. For those who wish to do so, this means reading mathematics. One never knows when or where an idea may occur that the reader will find interesting and useful. Often, it means looking at something familiar in a different way. We saw something of this when we met with e i8 = cos
e+ i sin e.
Of course, we can all derive e in: = -1. What
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was new was the idea of factoring cos2 e+ sin 2 e= 1 to obtain (cos 8 + i sin 8)(cos 8 - i sin 8) = 1, from which cos
e-i sin e = e -ie.
We suggest a lot of reading, perhaps with membership in ]' some group that issues periodicals. We also suggest trying to solve problems. This can only help the would-be problem solver. Dr. Samuel L. Greitzer Math. Department Rutgers University New Brunswick, NJ 08903
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Volume 2, Chapter 2
GENERATING FUNCTIONS
If a function f(t) is expressed as a series f(t) = a o + a,t + a 2t 2 + a 3 t 3 + ••• , then f(t) is said to be the generating function of the sequence ao' al' a 2 , a 3 , . . . . Since
(l+t)"=(8)+(~)t+(~)t2+ ••• ,
the function (1+t)" isthe
generating function of the binomial coefficients. Since 1 1 4 2 3 (1-t) =1+t+t +t +t + ••• ,then (1-t) isthegenerating function of (1, 1, 1, ••• ). There are questions about the role of the !. We agree to use it as a "dummy" symbol and make no assumptions about it. Thus, for example, if we differentiate the generating function just 1 2 3 above, we find that 2 = 1+ 2 t + 3t + 4t + ••• , and we (1- t)
have a generating function for the set (1, 2, 3, 4, ••• ). That is, we agree that, if necessary, we can differentiate or integrate generating functions to derive new generating functions. Thus, from the generating function for binomials, we can, by differentiating, get n(1 +t)n-l =( V+ 2 ( ~)I+ 3 ( ~) t'
+ •...
Now, by giving t some value, we can derive interesting sums. For example from the binomial generating functions, we easily derive, for t = 1,
2" =(%)+(y) + (~) + ••• + (g), or n. 2 "-1 = (y) + 2 (~ ) + 3 ( ~) + ••• + n ( g). From (1-t)"=(%)-(y)t+ (~)t2 - ••• ,wefind,onletting t = 1, that ( %) + (~) + (~) + •••
= ( y) +
( ~) + (~) + •••. By
means of differentiation or integration, many interesting relations 25
ARBELOS
Volume 2, Chapter 2
involving binomial coefficients have been developed. The reader might try to derive the result: n 1 n 1 n + .•. = (1)-2"(2)+ '3(3) 1 1 + ••• + 1 +2 n .
The fascinating thing about generating functions is the process of generating them. This is a matter of skill and imagination. For example, the binomial coefficient (?) is the number of ways of selecting, from n objects, subsets which contain r objects. After all, when one expands (1 + xl t)(1 + x 2 t)(1 + x 3 t) ••• (l + x n t), one is selecting one of two elements from each parenthesis. If we let the x.I = 1, we get the number of elements in sets of 1, 2, ••• ,n things at a time. This is useful in probability theory.
3 5 7 9 1+-+-+-+-+ ••• 2 4 8 16 . We write the series in the form: 357 1+_t 2 +_t 4 +_t 6 + ••• 248 and integrate, getting 111 t + -t 3 + -t 5 + -t 7 248 which equals 26
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Volume 2, Chapter 2 2
t t 4 t6 t ( 1+2"+"'4+ "'8+ ••• ). t2
If we let u = 2"' the expression in parentheses takes the form 1+ u + U2 + u3 + ••• = (1 ~ u) , which makes our integral equal to
2t
2.
(2- t )
Finally, differentiate to get back our original function, and 2 (2 - t 2 ) + 4 t 2 we find that the generating function is 2 2 and, (2- t )
letting t = 1, the sum of the series is 6. The reader is invited to try lx2 2x3 3x4 4x5 this method on - + - + --+ - + ••• . 3 9 27 81
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By expanding the function, we find that the generating function for 1 the Fibonacci sequence is F(t) = 2 . This lends itself to I-t-t
manipulation to find new relations among elements ofthe Fibonacci sequence. We cannot resist applying generating functions to the
following problem from the 1983 USA Mathematical Olympiad. It
goes:
On a given circle, six points A, B, C, D, E, and Fare
chosen at random, independently and uniformly with respect to arc
length. Determine the probability that the two triangles ABC and
DEF and disjoint, I.E., having no common points. Consider the
product (A+B+C)(A+B+C)(A+B+C)(D+E+F)(D+E+F)(D+E+F).
When this generating function is expanded, we cannot get D, E or
F from the first three parentheses, nor A, B, C from the last
three. Hence, when we expand, we will get, for the term ABCDEF,
only terms with ABC and DEF distinct.
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In (A + B+ C)3 (D + E + F)3, the number of terms of the form ABC is, from the multinomial theorem, equal to (1!
~~!! ~1 !) = 6.
Similarly, for terms involving DEF, the number ofthese equals 6. The product equals 36. The number of ways 6 points can be placed on a circle equals 5! = 120. Hence the probability desired is 36 3 120 or 10'
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One can derive interesting results from suitably constructed t t 2 t3 t generating functions. Thus, e = I +T+2T+ 3T+ ••• ,so e t is
1
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the generating function for O! +i!+ 2! + 3! + ••• By differentiation and/or integration, one can derive new results involving inverses of factorials. One very useful application of generating functions is to
1 lit t 3
---+ B - - B - + ••• where the B. are the (e t -1) t 2 1 2! 3 4! I Bemouilli numbers. However, this takes us too far afield, so we must reluctantly close at this point.
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Volume 2, Chapter 2
ASSORTED PROBLEMS
1. Seventeen cities are served by four airlines. It is noted that there is a direct service (without stops) between any two cities and all airline schedules are both ways. Prove that at least one of the airlines can offer a round trip with an odd number of landings.
I
(Australia)
o
is a point outside a circle. Two lines, OAB and OCD through 0 meet the circle at A, B, C, D with A, C the midpoints of OB,OD respectively. Also, the acute angle between the lines is equal to the acute angle at which each line cuts the circle. Find cos e and show that the tangents at A, D to the circle meet on the line Be.
e
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(Great Britain)
3. Prove: In any parallelepiped, the sum of the lengths of the edges is less than or equal to twice the sum of the lengths of the four diagonals. (Netherlands)
t
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Volume 2, Chapter 2
TOPOLOGICAL CONSIDERATIONS With the possible exception of finite geometries, Topology is the most general of all forms of geometry. The concepts of distance, area and volume are omitted, as are those of congruence, similarity and, indeed, shape. Two figures are considered topologically equivalent if one can be deformed by a continuous deformation into the other. Thus, a triangle and a circle are topologically equivalent, as are the surface of a cube and that of a sphere. The question that arises as to what there is left to study with so much discarded. One condition is that of incidence. If point P lies on line L, then, under a topological transformation (as described above), the new point P will still be incident with the new curve L. A second invariant is that of the interior and exterior of a closed curve. That is, if a closed curve lies on a surface, it divides that surface into three sets - the bounding curve C itself, and two other sets, one inside the curve and one outside. Moreover, a curve joining a point inside the curve to one outside the curve must cross the bounding curve C. (This is known as Jordan's Theorem, and is not easy to prove in full generality.) The study of elements of topology in terms of such concepts as concepts as compactness, connectedness, and other set-theoretic concepts makes up point-set topology. We shall restrict ourselves mainly to relations among points, lines, surfaces, etc., which constitute combinatorial topology. Much of graph theory is included. Let us examine a connected graph - that is, one where it is possible to reach any vertex from any other. We agree to call the number of edges on any vertex its valence. Then our graph has vI + V 2 + v 3 + ••• + V n = V vertices. It follows that vI + 2v 2 +3v 3 + ••• + nv n = 2 E ,
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since each edge has two endpoints. Now 2v 1 +2v2 +4v 3 +4v 4 +
= even No.
Hence, by subtracting, we arrive at
v I + v 3 + V5 + ••• = even No. - 2E, which is even. That, in a graph, the number of vertices of odd valence is even.
Now examine a graph all of whose vertices have even valence. Then one begin at A, say, and finally return to A, since, if there is a line to a vertex, there must be a second line. If the path followed does not include all the edges, there must be a vertex with lines on it that have not been used, so we can complete a second circuit (see Diagram at left). Now the diagram at the right shows how we can combine both circuits to form one larger circuit. We can continue until we have traced each line once. That is, a graph all of whose vertices have even valence is unicursal. Next, if a graph has one vertex of odd valence, it has two. If we join these two vertices, the graph now has all vertices of even valence, so it is unicursal. Now remove the line we added, and the original graph is unicursal provided we begin at one odd vertex and end at the other. The graphs below are unicursal.
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Volume 2, Chapter 2
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!!
To add a little difficulty to these, each can be drawn without lifting your pencil or going over any line twice. Try them. Ifwe consider a graph (which consists only of vertices and edges, and assume it drawn on a surface topologically equivalent to a plane, we get what is called a planar graph. Let us assume more that the graph is drawn on a surface topologically equivalent to a sphere. Then we can consider the total figure as consisting of vertices, edges and faces. For such a figure, we assume the reader knows Euler's Formula, V - E + F = 2 . Proofs of this formula (due to Descartes) are too well known to need repeating. Note that such figures include all our usual solid figures (only the vertices, edges and faces, of course). Exercise: Draw a solid with planar faces and edges straight lines having 7 edges. Now a map on a sphere is an example of the figure we have introduced, and the problem of coloring such a map with the fewest colors has just been solved. Four colors are necessary and sufficient. We can always alter a connected map so as to make all its vertices have valence 3. The drawing shows how.
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Now consider such an altered map. The number of faces is f 1 + f 2 + f 3 + ••• = F. Then the number of edges will equal f 1 +2f2 +3f3 + •••
~2E.
In addition, since each vertex has valence 3, 3V=2E.
Using Euler's Formula (multiplied by 6) 12 = 6V - 6E + 6F = 4E - 6E + 6F = 6F - 2E.
That is, 6F = 2E + 12. Substituting for E,
6f1 +6f2 +6f3 +6f4 + ••• ~ 12+f1 +f2 + •••.
5f1 +4f2 +3f3 +2f4 +f5 - f6 - f7
•••
~
12,
so that, if the left side is to be positive, one at least of f 1, f 2 , f 3 , f 4 , f5 must be positive. This proves that, in such a
map, there must be at least country with five boundaries or less.
There are graphs which, when spanned by faces, cannot be planar. Two of these are famous. We call them the Gas, Water, Electricity graph and the Five Castles graph.
I @@@
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In the figure at the left, it is required to supply gas water and electricity to the three houses, all lines to lie in the plane (and no tricks, like laying one electricity line under all houses). This is
easily shown to be impossible. In the figure at the right, it is
required to build a wall joining each of the five castles. This is also impossible. In fact, it has been shown (by Kuratowski) that the
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Volume 2, Chapter 2
necessary and sufficient condition that a graph be non-planar is that it include one or both of these situations. We now derive a fonnula for the sum of the angles of the faces of a polyhedron. That is, we assume all edges straight line segments and all faces planar. The vertices all have valence at least equal to 3, and the faces all have at least three sides. Then v +v 4 +vS + ••• +vm=V, 3
J ~
~
J Now 3v 3 + 4v 4 + 5v S +
=2E,
and 3f3 + 4f 4 + 5f S +
=2E.
J
Recall that the sum of the face angles of a polygon of n sides equals (n - 2)1t . Then the sum of the angles of this polyhedron equals 1tf3 + 21tf4 31tfs + ••• + (n - 2)1tfn =
+
L
Now, from the fonnula above, we have 31tf3 + 41tf4 + 51tfs + ••• + n1tfn = 27tE , 21tf3 + 21tf4 + 21tfs +
+ 21tfn = 21tF .
Subtracting, we arrive at: 1tf3 + 21tf4 + 31tfs + ••• + (n - 2)1tfn = 21t(E - F) . That is,
L
= 21t(E- F) , and since, from V - E + F = 2 , we get E-F=V-2,
L
= 21t(V-2) .
Now you might try the following problem: The sum of all the angles of the faces of a polyhedron, except .Qill:, equals 3,150 degrees. How large is the missing angle? It is remarkable how much can be accomplished by the use of the Euler Fonnula and some simple thought. For example, it is easy to 34
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show that, even if the faces and edges are distorted, there are no more than five "regular" polyhedra.
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Volume 2, Chapter 2
AN IN1RIGUING PROBLEM First - the problem: If each of two students selects an integer at random, and these are compared, what is the probability that the two numbers will be relatively prime? Solution: Let the two integers be A and B. Let p be any prime number. If A is divided. by p, the remainder may be anything from 0 to (P-l) - that is, there are p residues of which only one - 0 - is favorable. The probability that p is a factor of A equals
1 p For the same reason, the probability that p is a factor of B
1
equals -. Hence the probability that p divides both A and B p 1 equals 2 ' Thus, the probability that p does not divide both A p 1 and B equals (1- 2'") .
P Now we consider this last expression for all primes, that is, 2, 3, 5, .7, •••. We can write our desired probability in the form 1 1 1 1 P = (1- 2'2)(1-'32)(1-52)(1-7) ••• or 1
P=
1
1
1
1 1 1 · ••. Each of the fractions (1- 2'2) (1-3'2) (1- 52)
can be expressed as an infinite geometric progression, so that
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Volume 2, Chapter 2
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-=(1+"2"+ 22+-2-3+ ••• ) P 2 (2) (2) x(1+-+--+--+ ••• ) 32 (32)2 (3 2)3
I
x(1+-+--+--+ 52 (5 2)2 (5 2)3 and so on for all primes. On multiplying all these, we find that
I
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1
1
-p = ""2+-2-+2""+2+""2+ •••. Now all we have to do is to 1 2 3 4 5 find the sum of the series above. At this point, we decide to get intuitive. (What's good enough for Euler is good enough for us, I hope!)
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First, we know that using various methods, . . x 3 x5 x 7 sinx = x - - + - - - + ••• However, smce sm x = 0 when 3! 5! 7! x = 0, ± 1t, ± 21t, ± 37[, etc., and since, if we know the roots of a Polynomial equation to be, say rl' r2' r3' etc., we may write the equation as a product (x-r1 )(x-r2)(x-r3) ••• =0, we take a chance and write: x2 x2 x2 sinx= x(1--)(l--)(1--) •••.
2 2 2
1t
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41t
91t
We are taking a chance, because what works for a finite number of roots may not work for an equation with an infinite number of roots. Also, the infinite product may have, as a multiplier, an entire function, such as eX, which does not affect the number of roots but alters the value of the product when some value of x is substituted. 37
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Volume 2, Chapter 2
Fortunately~
this is not the case. The infinite product correctly expresses sin x. let us expand the infinite product~ getting at least the coefficient of the term in x 3 . We find it to be the negative of
J
Now~
1
1
1
1
J
(-2+--2+-2-+ --2+ ... ). In the infinite series~ the
7t 47t 97t 167t
coefficient of x 3 equals -1/3! . We equate the coefficients and
1
1
1
1
1
find that 7t 2 (12+ 22"+'32+ 4 2 +
1
1
1
1
... ) = 3!
~ That is~
7t 2
(}2"+ 22"+ 3"2+ ... ) ="'6 . Now we can finish our probability problem. Since 1 7t 6 p="'6~ p= 7t 2 . 2
It should be noted that the series converges rather rapidly~
and after just a few primes~ has an approximate value of regardless of how large A and B are taken.
6
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Volume 2, Chapter 2
TO IMPROVE PROBLEM SOLVING
There are various ways to improve the ability to solve problems. One way, of course, is to read and see how others have solved problems. However, individual activity is better. One might generalize, say from two dimensions to three or more in geometry or from a quadratic equation to one of higher degree in algebra. We thought of this while working on a problem involving the area of a triangle. To be precise, how many ways of finding the area of a triangle can you invent? Let us look at a few ways.
ah bh ch First we have K = _ a = __ b = _ c where , 2 2 2'
h a ,h b , h C are altitudes to sides a, b , c. We call this set of
formulas set (1). Now h a = b. sinC , and we can substitute in (1) to get (ab.sinC) (be.sinA) (ca.sinB) K= 2 = 2 2 and we label this set of formulas set (2). Next, a. sin C = c. sin A, etc., and we can substitute in (a 2 sin B. sin C) (2 sin A) with similar rules involving set (2) to obtain K= b 2 and c 2
.
Call these set (3).
Using R, the circumradius, we remember that
I
c = 2R. sin C, so K =
(:~)
. This is set (4).
If r is the inradius and s = K= rs. This is rule (5).
39
(a+b+c) 2
we know that
ARBELOS
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Volume 2, Chapter 2
Of course, we all know Hero's Fonnula, 2 K = S(s- a)(s - b)(s - c) . This is rule (6).
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Now, from Set (1), we can multiply and get (abc) 3 K = (abc)(hahbhc)/8=4Rihahbhc)/2)R , and using rule (4), we have K 2 = Rh a hbh c /2.
(7).
This rule gives us the area when we know the three altitudes and the circumradius. In a similar manner, we can use rules (2) to derive K=2R 2 sinA.sinB.sinC. (8). Now we can continue to work with these eight rules to derive more fonnulas. For example, how about K = Rr(sin A + sin B + sinC) ? Or K = s2 tan ~ tan ~ tan ~ ? Now we can spread ourselves and use other data. For example, if r a is the radius of the escribed circle tangent to side a, and rb, rc are similarly defined, show that K 2 = r. r rbr . a e This leads naturally to further extensions. Can one find the area in tenns of the medians? We have a fonnula in tenns of the altitudes. Can one find a fonnula in tenns of the angle bisectors? If we know the altitude, the median and the angle bisector that intersect side a of a triangle, can we find the area? Now we can proceed to generalize. How about finding similar results for quadrilaterals? We could start with the fonnula for the area of an inscribed quadrilateral with sides a, b, c and d (a+b+c+d) where s= 2 ' namely, K 2 =(s-a)(s- b)(s-c)(s-d), or we might strike out on our own, using other data. For instance, if a quadrilateral is inscribed in a circle and circumscribed about another circle, we have K2 = abcd . 40
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Next, we can continue to other polygons. We might get new and surprising results. However, we might also continue into the third dimension, and try for formulas for the volume of a tetrahedron given various sets of data. Our issue of September, 1982 gave one such result. We think this was new. What about generalizing our results to triangles on a sphere, for example? We know that the area of a spherical triangle is measured by its spherical excess, that is by E = A + b +C - 7t . Can we find a formula for the area of such a triangle in terms of its sides? Exercises of this type strengthen ones problem solving ability and incidentally add to the list of results that might be useful in solving many problems we may encounter. We do not have to restrict our efforts to geometry. How many ways can you find to derive at least one root of a cubic equation? There are methods named after mathematicians like Cardan, Newton, Graeffe, for example, and one can use trigonometry or a computer to find approximations. We have even found a method using the Hessian (look up the Hessian!) All great mathematicians have amused themselves by playing with number theory - among them Gauss, Euler, Fermat, Archimedes, and others. One idea, due to Leonardi of Pisa (Fibonacci) has become so popular that there is actually a publication, the Fibonacci Quarterly, containing results derived from the set 0, 1, 1, 2, 3, 5, 8, 13, ••• , where each element from the third on equals the sum of the two preceding it. Try showing that, if f.1 is the i-th Fibonacci number, f l +f2 +f3 + ••• +fn = fn+2 -1 f
n-I
.f =f 2 +(_I)n n+1 n
41
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Volume 2, Chapter 2
There are hundreds of results like these, and one can invent an original fonnula by trying. Finally, try to prove the result shown on the front page of this issue!
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Volume 2, Chapter 2
JUST FORFlJN 1) Three coins are tossed and, after each toss, those coins that come up heads are removed, and the remaining coins tossed again. What is the probability that, in exactly three tosses, all the coins will have been removed? 2) If f(1) = Jf, f(2) = J2-..fi , f(3) =~2- J2-..fi, etc., so that f(n) has n nested two's, find an expression for f(n), where n is any natural number. 3) A triangle with one side equal to 8" is inscribed in a circle whose radius is 6". The sine of the angle opposite the eight inch side is a mean proportional between the sines of the other two angles. Find the area of the triangle.
These are fairly easy problems, the solution of which depends on looking at each in a different way or using some rule or method not usually presented in class. There is nothing, however, that has not appeared in a previous issue of ARBELOS.
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Volume 2, Chapter 3
arbelos PRODUCED FOR PRECOLLEGE PHILOMATHS
A
R
P Prove: PQ = PR and PQ and PR are perpendicular
January, 1984
1983-1984: No.3 45
ARBELOS
Volume 2, Chapter 3
PREFACE Arbelos began as a very small operation in 1982. The first issue went to about thirty students who, we thought, would be interested in its contents. There was no advertising of the publication. As of now, the number of subscribers has grown to about one hundred. It has grown by word of mouth, mainly. However, it has always been a one-man production. Now it seems that this may change. We have received a letter from Dimitrios Vathis, who lives in Chalcis, Greece, proposing a problem for us to solve, and asking for mathematical information. We have been hoping that this would happen. We would like to make this a cooperative effort. Let us have input from our readers! If you have a problem that you think would be interesting to our readers, please send it in. It may be published. If you would like to have some mathematical topic appear in Arbelos, feel free to write and ask for it. Elsewhere in this issue, we print the problem proposed by Mr. Vathis. I hope to receive solutions from other readers. Arbelos is growing. We know it is being read in Australia. And now - Greece! Dr. Samuel L. Greitzer
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lRANSFORMATIONS In synthetic plane geometry, there is one postulate that is used seldom or never after the very beginnings of the development of the subject, even by the earlier geometers, and this is - a geometric figure may be moved (in space) without changing size or shape. It is quite possible that Euclid and his successors were aware that plane geometry was possible only in three dimensions at least. We know that Desargues' Theorem requires three dimensions for its proof and that this theorem or some equivalent theorem based on it is essential for plane geometry. Modern geometers use postulates to hide this fact. Sometime, we will discuss this theorem fully. However, our object now is to show how elementary transformations can be used to solve problems. All that is needed is ingenuity and imagination. Our first transformation is that of "translation". In this transformation, we merely push a figure about parallel to itself from one position to another, thus:
B ,,:::::::::::::::. :::.,.,:.:
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:::::
We see that AA' and BB' are equal in length and parallel. This transformation was used to prove the first few theorems in elementary geometry, and no more. Let us use it for the A ~N Q M following problem: Given MBC p , and line segment MN; required, to .' ·'R C construct segment PQ parallel to B
&
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Volume 2, Chapter 3
and equal to MN, and with P on AB and Q on AC. We construct segment BR parallel to and equal to MN and translate (slide) it as shown. This is, of course, extremely simple to see - and do. Now we offer the following problem for your perusal: Given the circle, center 0, and the line segment MN. Construct a line segment parallel to and equal to MN intersecting the circle at A and B, and such that points A, B trisect the segment.
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Of course, the construction may not always be possible, as, for example, when MN exceeds three times the diameter of the circle in length. Next, we have the rotation shown. The center of the rotation /,..0 .:
o
".
'.
"
...
/Sl.
remains fixed, and corresponding points of the figures lie on circles with 0 as center. Of course, we can have rotations in succession,
each with a different center. This makes for complicated figures,
but these can be used to solve complicated problems.
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We may have special rotations with special names, like the half-tum (180), quarter tum (90°). Here we illustrate with two problems. In the first, we have MBC, with equilateral triangles ABP and ACQ constructed externally on sides AB and AC. Prove that segment PC = BQ.
I I
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I We rotate MPC about A through 60° clockwise, and see that P falls on Band C falls on Q. Finished! I Q PQ I I
I
A H I (b) we try something harder. On sides AC and BC of I MBC,Now squares AP and BQ are constructed. Altitude CH is constructed and extended to meet segment PQ at M. Prove that I PM MQ. (Diagram a). I Q
p---~
B
H
(a)
=
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Volume 2, Chapter 3
Solution: rotate ~CPQ 90° clockwise about C (diagram b). Then PCA is a straight line, and CM is perpendicular to CH and therefore parallel to base AB. A line parallel to the base of a triangle divides the other two sides proportionally. Since CA = CP, BM = MP. This completes the solution. We invite the reader to try the following: Two circles intersect at points M and N. Construct a line segment through M, say AMB, such that M bisects AB. We also call your attention to the diagram on the cover (which appeared in a recent International Math Olympiad). Still another transformation is a reflection. Weare given a diagram and a "mirror". The image of
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!Z S\ A'
A
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M
A in the diagram above lies on the line perpendicular to the mirror M and such that A and A' are equidistant from the mirror. The fact that a translation can be represented by two reflections or that a rotation by reflections with intersecting mirrors is interesting but is immaterial to our purpose of using transformations to solve problems.
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There are examples ofthe use of reflections that are familiar to the reader. For some reason, A
A[\/r p
B
"
C
.
M
···':B'
M
(b)
(a)
there are those who do not quite accept the proof that base angles of an isosceles triangle are equal by reflection of the figure about the angle bisector as mirror (diagram a). It is correct. And diagram b shows how to locate point P on the mirror M such that AP + PB is a minimum. There are other such problems that involve more than one mirror. We add one problem below: On the mirror M, locate the point P such that the sum of the two tangents from P to the circles is a minimum.
o
M
All transformations thus far have been such that lengths (and areas) were unchanged. Another and new transformation is called a dilation. In this, the figure and its image are homothetic. All lines remain parallel under the transformation but all corresponding lines in the image are some constant k times those in the figure. The situation is shown in the diagram below:
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Volume 2, Chapter 3
J ]
rl Finally, if one performs, in either order, a rotation followed by a dilation, one has what is called a spiral similarity. We have found these most helpful. First, in the figure, the two triangles are homothetic, and S is the center of similitude or the homothetic center. Next, this listing of transformations is not exhaustive. There are others - such as point reflection, for example. However, we have enough to be able to solve some problems that are quite complicated. First, the diagram on the cover shows triangle ABC, on the sides of which triangles ABR, ACQ, and PBC are drawn, as shown. Prove that line segments PR and PQ are equal and form a right angle.
tJ t]
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In Figure 2, equilateral triangles are constructed on the sides of the I triangle ABC. P, Q, R are centroids. Prove that PR QR and angle PRQ 60 I I
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Volume 2, Chapter 3
(1)
=
=
0
•
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Volume 2, Chapter 3
MANY CHEERFUL FACTS
ci
The graph ofAx 2 + Bxy+ + Dx + Ey+ F = section (or a special form thereof).
°
is a conic
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Let L\ = B2 -4AC. (a) If L\ < 0, the graph is an ellipse; (b) If L\ = 0, the graph is a parabola; (c) If L\ > 0, the graph is a hyperbola. Let K be the area of a convex quadrilateral with sides a, b, c, d, with two non-adjacent angles A and C, and with semi-perimeter (a+b+c+d) s equalto 2 . Then: (a) K 2 =(s-a)(s- b)(s-c)(s-d)-abcd cos2 (
A+C 2 );
(d) If d = 0, the figure becomes a triangle, and
K 2 = s(s- a)(s - b)(s- c) .
Note: In view of (a), it follows that the quadrilateral with maximum area having the sides a, b, c, d is inscriptible. In view of its age, Geometry has had the opportunity of extension into many interesting vistas.
54
a
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If two lines issuing from a vertex of a
triangle make equal angles with the
sides, the lines are called isogonal lines.
In the diagram, if LBAP = LCAQ B .................._ ......_.aC P Q (or LBAQ= LCAP, of course), then AP and AQ are isogonal lines. If, as in the diagram,
LBAQ= LCAP, LCBQ= LABP and LACQ = LBCP, then B ~;;""':;""'--_"':"""";;~c
points P and Q are isogonally
conjugate points.
Some immediate results are:
(a) Nearly every point in the plane has a definite isogonal
conjugate. Exceptions are
(1) The isogonal conjugate of a point on the circumcircle is at infinity; (2) The isogonal conjugate of any point on a side of a triangle is the opposite vertex. For all points in the plane of a triangle, the isogonal conjugate points of a curve trace out a unique curve. There are a few theorems that are quite easy to prove. We state these (some with hints). (b) If three lines from the vertices of a triangle are concurrent, their isogonal conjugates are concurrent. (Use Ceva).
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(c) The perpendiculars to the sides of a triangle from isogonal conjugate points are inversely proportional. (Easy with similar triangles)
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(d) If AP, AQ are isogonal lines that meet the circumcircle at P and Q, then PQ and BC are parallel. (Note that BP =CQ). (e) If P and Q (in the second diagram on the previous page) are both inside the triangle, then
LBPC+LBQC=180o+LA . (f) The orthocenter and the circumcenter of a triangle are isogonally conjugate points. (To start, MBP and MB Care similar, so AP and AB are isogonal lines. Now draw the altitude and diameter through B and C).
B
(g) The isogonal conjugate of a median is called a symmedian. There are many special theorems that apply to these special lines.
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1t So much has been written about the number 1t that it would appear to be wasting time and space to discuss it further. Our excuse this time is that there are some methods for evaluating
it that may be new or unfamiliar to the reader. We pass rapidly over the values chosen by the ancients, either by measurement or guesswork. The Babylonians used the value 3, and this is also to be found in the Scriptures. Other values were: 256 81
355 113
18(3 -..[2)
It would appear that Archimedes first determined the value of 1t mathematically, by using inscribed and circumscribed polygons. He found
For centuries, this was the only method used for evaluating 1t, and many people spent many months finding its value to more and more decimal places. Not until the seventeenth century did various "modem" mathematicians try other methods. The first would appear to have been John Wallis (1616 1703). Actually, he used interpolation methods that involved crude ideas of the calculus. Because we hope it will be instructive, we give another method. We start with: x2 x2 x2 sill1tx = 1tx(1-12)(1- 2 2 ~(1- 32 ~ (Weare assuming this because both sides have the same roots. In fact, we are assuming a lot more, but so did Wallis, 57
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1 Newton, Euler, etc.). In this equation, let x = 2' Then we find:
]
1t 1t 1 1 1
sin -2 = -2x(1- 22")(1- 2"2)(1- 22') ...
12 22 32
2 2 2 1t 2 - 1 4 - 1 6 - 1
1= -2( 2 2)( 2 2)( 2"T) .•. , from which
12 22 32
1t 2x2 4x4 6x6 8x8
'2= lx3 x 3x5 x 5x7 x 7x9 x •••
and he had expressed 1t as an infinite product. With the help of a computer, one can write a very simple program to find it, but it does take time. It will be recalled that we had, in the previous issue, derived the series:
2 1 1 1 1 - = -2+ -2+ -2+ -2+ 6 1 ' 2 3 4 1t
... ,
and one could use this to evaluate 1t. This is also slow. Ifwe 1 multiply both sides of this series, the right side by (1the
22")'
3
left by
4'
2 1 1 1 1 we get 8= 12+ 3'2+5'2+"72+ •••. This is only a 1t
bit faster. However, we do have at least three ways of finding 1t. Now we look at another approach, due to James Gregory (1638 - 1675) originally, rediscovered by Leibnitz (1646 - 1716), but with the prettiest derivation due to Euler (1707-1783). We have commented before on Euler's methods, and here is another. First, let us accept the series, e Z = l+z+
z2
z3 +
z4 +
2T 3T 4T 58
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as known. Euler got this from the Binomial Theorem. Now, let z = x In a . On substituting in the above, we find: eX In a = aX = I + (x In a) +
(x In a) 2
2!
+
Now let a = 1 + u, and we arrive at: (x In(l + u))3 ( X In(l + u))2 (1 + u) X= 1+ (x In( 1 + u)) + 2! + 3! + From the Binomial Theorem, we know that x x(x-l)u2 x(x-l)(x-2)u 3
(1+u) =l+xu+ 2! + 3! +
In this expansion, the coefficient of x is equal to
u2 u 3 u 4
u - - + - - - + ••• 2 3 4
Comparing the two expansions above, we arrive at:
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u2 u 3 u 4
In(1 +u) = u- - + - - - + •••
2
3
4
(now we can compute logarithms, if we wish to). In this last expansion, replace u by -u, and u2 u 3 u 4 In(1-u)=-u-T-T-4 Subtract the last expression from the one before it, and we have the rule: l+u u 3 uS In(--) = 2(u + - + - + l-u 3 5
...) .
Now along comes Euler. He starts with e iz = cos z + i sin z : e- iz = cos z - i sin z, 59
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which he has already derived. By division, 2iz
e
cos z + 1 sm z I + i tan z = cos z - i sin z = 1 - i tan z '
!1
This last should look familiar to the reader. When we substitute in the formula for the log of the differences just above, we let u = i tanz , and find 2iz 1+ i tan z i tan3 z i tan 5 z i tan 7 z lne = In( l ' ) = 2(i tanz3 + 5 7 + •••), -ltanz
tan 3 z tan 5 z tan 7 z
or 2iz=2i(tanz- 3 +-5-- 7 + •••). We can cancel the multiplier 2i. Now let tanz = a, and we have arrived at the a 3 a 5 a7 result: arctan a = a - T + 5- 7 + •••. This is the formula that seems best suited to computing 1t. Ifwe let a = 1 , we get the well-known 1t 1 1 1 -= 1--+-- -+ ... 4 3 5 7
There is no need, however, to let a = 1 . For 1
111
1
1
1
1
1
a=-, arctan - = - - - - + - - - - - + •••. For a= 2 2 2 3.23 5.2 5 7 .2 7 3' we find arctan
1
1
1,
3' = 3'- --3-+ --5 - --7-+··· . Neither of
3.3 5.3 7.3 these would seem to be much good for evaluating 1t. However, if you notice that 1 1 1t arctan -+ arctan - = 2 3 4 (which is easy to show) we see that we can find and this way is faster. Just add the two series. 60
1t
'4
and hence 1t,
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Other mathematicians have used formulas based on different values of For example, Dase based his evaluation on the formula
e.
7t 1 1 1 - = arctan - + arctan - + arctan 4 2 5 8
Rutherford used the formula 7t
1
1
1
'4 = 4 arctan 5- arc 70 + arctan 99
!!
and Machin used the formula 7t
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1
1
'4= 4arctan 5-arctan 239
!!!
One wonders how these mathematicians derived their initial formulas (hence the exclamation points). Nowadays, there is really no good reason to try to find the value of 1t to an enormous number of decimal places. The processes have their use. For example, one can determine whether a computer is ready to compute accurately by having it compute either e or 1t and comparing the results with stored values. Mathematicians are also computing its value to see whether there is any system to the number of occurrences of the digits. So far, there does not seem to be any such system. In cases where a mathematical or statistical investigation needs "random" numbers, one C&ll use the value of 1t to provide these. Hence its value is useful in statistics or actuarial science. It remains as one case in which an enormous amount of study was expended, sometimes for no useful reason.
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WEEK-ENDERS These are relatively easy problems, which can be solved in a few minutes, as a relief from the harder ones.
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1) When the sum of the first n natural numbers is added to the sum of the cubes of the first n natural numbers, the result is 3080. Find n. 2) On a graph, points A(3, 7), B(9, 1), C(ll, 13) and D(8, 10) are on the successive sides, produced if necessary, of a square. What is the area of the square? 3) Find the area between the X-axis and that portion of the curve x
y = 2 cos 2 '2 between x = -1t and x
= 1t.
4) A five-digit number is a multiple of 41. Ifthe left-most digit is shifted so that it becomes the right-most digit, the new number is a perfect cube. What was the original number? 5) Hypotenuse AB of right triangle ABC is divided into three equal parts by lines drawn from the vertex. If these lines are 7 units long and 9 units long, how long is the hypotenuse?
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THE EULERLINE A C,r--------~r_-------....,.B'
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For those who may have forgotten the Euler Line because it does not arise often in our modem classes, we present the following: In MBC, let 0 be the circumcenter and H, where the altitudes intersect, the orthocenter. M is the midpoint of side BC. Hence AG AM is a median and G, where G M= 2, the centroid. Construct A' B', B' C' and C'A' parallel to AB, BC, CA respectively. The M'B' C' and MBC are similar, and the ratio 2 of corresponding lines is T' However, AH is the perpendicular
bisector of side B' C' ,and H is the circumcenter of M' B' C' .
Hence AH = 2.0 M and AH and OM are parallel. Therefore MGH and ~MGO are similar, and H, G and 0 are collinear (with HG = 2. GO). The line HGO is known as the Euler Line of MBC.
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The Euler Line has many properties - too many for us to list more than a few. Thus, the center of the Nine-point Circle lies on the Euler Line, midway between 0 and H. The radical axis of the circumcircle and the Nine-point Circle is perpendicular to the Euler Line. Now we can present the problem proposed by Mr. Vathis: A
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··•··· H
G
0
L .
B....- - - -......~----~C
J
Let L be the Euler Line of the triangle ABC. Prove that the Euler Line of the triangle ADE is parallel to side BC of triangle ABC.
II ill
M
Note: Don't use similar triangles or ratios of segments. Dim. Vathis Chalcis, Greece
Our Note: You are welcome to send in a solution by any means legal, that is. Let's not disappoint him!
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ASSORTED PROBLEMS Find all integers n with the property: There exists a permutation (al' a 2 ,
••• ,
bn) is also a permutation of
(0,1,2, ••• ,n-I).
I
,an) of n-tuple
(0,1,2, ••• ,n-I) suchthat,if b I ,b 2 , ••• , b n arethe remainders of the division of aI' a Ia 2 , a I a 2 a 3 , a Ia 2 a 3 a 4 , ••• a Ia 2 ••• an by n, then (b I' b2 , b 3'
8)
•••
(Bulgaria)
0 is a point outside a circle. Two lines, OAB and OCD through 0 meet the circle at A, B, C, D with A, C the midpoints of OB, OD respectively. Also the acute angle between the lines is equal to the acute angle at which each line cuts the circle. Find cosS and show that the tangents at A, D to the circle meet on the line BC. (Great Britain)
e
9)
Prove that for all real numbers
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(U.S.S.R.)
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KORsCHAK KORNER This column features those problems of the famous Hungarian Ktirschak Contest which have not yet appeared in English translation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books ! & II in the MAA' s New Mathematical Library series (available from the MAA headquarters); the problems (with brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Csirmaz of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers of this column.) Student readers are also invited to set aside an uninterrupted 4-hour period to compose complete, well - written solutions to the problems below, and to submit their work to the address given below for a critical evaluation. 1/1982. A cube of integral edge-length is positioned in such a manner that all four of the vertices on one of its faces have integral coordinates. Prove that the coordinates of its other four vertices must also be integers. 2/1982. Prove that for each positive integer k>2 there exist infinitely many positive integers n such that the least common multiple of the numbers n, n + 1, ••• , n + k -1 is larger than the least common multiple of the numbers n + 1, n + 2, ••• , n + k . 3/1982. The set of integers is colored with 100 colors so that each color is utilized and that the following condition is satisfied: Whenever two closed intervals [a, b] and [c, d] of equal length are chosen such that a is ofthe same color as c, and b is of the same color as d, then the colors of the intermediate integers also agree (i.e., if x is an integer with 0 ~ x ~ b - a, then a + x and c + x are of the same color). Prove that -1982 and 1982 are of different colors. Dr. George Berzsenyi 66
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Volume 2, Chapter 3
FROM OUR READERS First, we should explain once more why there appears to be so long a time between the receipt by you of the Arbelos and our replies. We do have to prepare each issue at least two months before you receive it. Thus, for example, the November issue is already "finalized" and we are working on the January issue, which is just about complete. You may expect your solutions and comments about the November issue to appear in March, unless you can submit them speedily. And now for the replies we have received. The most complete reply comes from Mark Kantrowitz. He found solutions to most of the problems in the September issue, showing lots of hard work had gone into them. He did give the factors of 1,111,111 an 239x4649. He also had the "self explanatory" number, in which each digit gives the number of times that digit appears in the number. For a ten-digit number, we have 6,210,010,000, where the 6 gives the number of zeros in the number, the 2 the number of ones, etc. Douglas Davidson and Brian Kelly also gave the solution to this problem. Welcome to Brian Kelly! We hope to hear more from him. In giving his correct solution to Problem 3, Joseph Sachs went through a long and complicated procedure to get the result. Remember - the sooner we receive your results, the sooner we can acknowledge them. Keep writing.
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Samuel L. Greitzer
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arbelos PRODUCED FOR PRECOLLEGE PHILOMATHS
p
THE OUTER NAPOLEON TRIANGLE
1983-1984: No.4
March,1984 69
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PREFACE Arbelos has been growing quite rapidly. Between our last issue and this, our membership has tripled, and is now more than three hundred. Naturally, this is very gratifying.
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We try to have articles on subject matter that does not appear in the classroom and problems whose solution require ingenuity rather than mechanical or rote methods. Our Week-enders are
simple problems of this type and our Assorted Problems are
Olympiad type problems actually proposed by the countries for
which they are given credit. Occasionally, we include problems of ~
the type found on the Putnam Examination. All of these require __
imagination and ingenuity to solve.
We ask for cooperation from our readers in two ways. If :" you find a solution to a problem, send it to us. It will help assure us ~ that our whole effort is not a one-way operation. If there is any topic that you would like presented, write us and let us know. If you can prepare an article for us, please send it in. We naturally enjoy receiving complimentary notices. However, if you feel critical, don't hesitate to criticize. All this will help improve our work. Finally, spread the word that Arbelos exists. We can always use more subscribers. And help us by sending us your subscription for the coming School year as welL Let us hear from you! Dr. Samuel L. Greitzer Math. Department Rutgers University New Brunswick, NJ 08903
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WHAT ARE THE ODDS ? We were looking at a problem in a work by the late Rev. Charles Lutwidge Dodgson (1832 - 1898) when we came on the following problem: A bag contains 2 counters, as to which nothing is known except that each is either black or white. Ascertain their colours without taking them out of the bag. His answer was that one had to be black and the other white. Naturally, this led us to reexamine what we knew about probability. In order not to be unnecessarily repetitive, we will merely skim through the elements. As most developments do, we use (six-faced) dice to explain. Anyone die has spots or numbers on each face, and, when a die is cast, one of these six faces lands on top. We call this an elementary event (or possibly a sample space). Since it appears that any face has the same chance of coming out on top, we assign to each event the probability 1/6. Now the case where the number on top of the die is even occurs when that number is either 2, 4 or 6. This we call an event, and its probability of occurring equals 1/6 + 1/6 + 1/6 = 1/2. We may add thus because the elementary events are .
independent. Now we can introduce some theory. Let peE) stand
for the probability that event E occurs. Then E means that the event does not occur, and the probability of this is P(E). Also peE) + peE) = 1 With two dice, an event can consist of the sum of the die faces, or the product or the sum of the squares, etc. Since most texts use the sum of the faces, let us use the product. The probability that a number on one die will show is still 1/6, but the probability of two (distinct) dies showing numbers is 1/36.
Therefore the probability that the product of the numbers on the faces is, say, 6, comes from the events lx6, 2x3, 3x2, 6xl which makes it 1/36 + 1/36 + 1/36 + 1/36 = 1/9. The reader might like to figure out the probability that the product will be odd. Let us now write P(AB) for the probability that both A and B occur, and peA + B) for the probability that A or B or both occur. Then 71
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P(A + B) =P(A) + P(B) - P(AB). In the special case where A and B are mutually exclusive, then P(AB) = 0, and we have P(A + B) = P(A) + P(B). These relations can easily be extended to three or more events, so that, for example, P(A+B+C)=P(A)+P(B)+P(C)-P(AB)-P(BC)-P(AC)+P(ABC). For a finite number of events, or for discrete events, these few rules, properly applied, can usually solve all problems. For example, try this: A bag contains one counter, known to be either black or white. A white counter is put in the bag, and then a counter removed. This counter is white. What is the probability of now removing a white counter? Frequently, one can solve problems in probability by merely determining the number of favorable events, f, and dividing by the total number of events, t. That is, p = fIt. Again, some thought can make one realize that a problem, seen in a different way, is simple. For example: From a complete deck of cards, two cards are drawn in succession. What is the probability that the second one is a king
II II
? We come next to the concept of conditional probability. Let A and B be two events. Define the probability that A occurs, given that B has occurred, written P(AIB) by P(AB)IP(B). Otherwise,
P(AB) = P(B). P(A1B) . Obviously, we can also write:
P(AB) = P(A) .P(BIA). This is the basis for a result that, carefully used, can be quite valuable. Extending it a bit, let B l' B 2 , B 3 be mutually exclusive events. Then
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P(B 2 A) = P(B 2 )· P(AIB 2 )
= P(A). P(B 2 IA)
P(B 3A) = P(B 3 )· P(AIB 3 ) = P(A). P(B 3 IA).
Thus, we have: P(BIIA) = P(BI ).P(AIB I) (peA) , with similar results for P(B 2 IA) and P(B 3 IA).
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If we note that peA) = P(BIA)+ P(B 2 A)+ P(B 3 A) and substitute in any of the formulas above, we have Bayes' Theorem.:
P(BIIA) = P(BI)P(AIB )+ P(B )P(AIB )+ P(B )P(AIB ) 2 2 I 3 3 with similar rules for P(B 2 IA) and P(B 3 IA). One must take care when using Bayes' Theorem not to make unwarranted assumptions about the various probabilities. Let us try the following problem:
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We have three boxes. Box I contains three dimes; box II contains two dimes and a nickel; box III holds two nickels and one dime. From a box selected at random, we abstract one coin and it is a dime. What is the probability that the dime came from Box I? First, P(I) = P(I1) = P(lII) = 1/3. Next, P(dlI) = 1, P(dlII) = 2 / 3, and P(dIIII) = 1/3. If we substitute in our formula, we find that P(lld) = 1/2. In the same way, we find that P(II1d) = 1/3 and P(IIIld) = 1/ 6. In this problem, it was a simple matter to find the various probabilities. However, when Bayes' Theorem is used for quality control, it is not so easy to derive the probabilities. This is because the probabilities depend on sampling. For example, one checks a sample of 100 light bulbs to find how many are bad, and then repeats this procedure many times. There is no assurance that any average one gets by this process is exact. However, this is the best one can do in this case. A few remarks are in order at this point. First, probability theory as a mathematical system is quite as precise as is any other mathematical system. But, just as any mathematical system, when 73
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applied to any other area - say biology, chemistry, physics, etc., may exhibit difficulties due to the area, so it is with probability theory. Where there are precise "a priori" elementary probabilities involved, the results are precise. Often, however", it is impossible to determine a priori probabilities. In that case, one collects data, divides the number of favorable cases by the total number of cases, and uses such data as fundamental probabilities. Such procedures frequently produce results less than accurate. Where even experimental determination of primary probabilities is impossible, one uses intuition or possibly intelligent "reasoning" to find primary probabilities. Let us, for example, consider a satellite of Jupiter that is about as large as the earth. We wish to discuss the presence or absence of animal life on it. Since we have no other way of proceeding, we use the principle of "insufficient knowledge", thus: we assume a 50-50 chance that an animal is on the satellite. Hence the probability that there is an axolotl on the satellite is 1/2 and the probability that there is not is also 1/2. Similarly, the probability that there is not a bandicoot is also 1/2. Our theory tells us that the probability that neither exists is 1/4. Continuing, the probability that neither axolotl, bandicoot nor coati exist on the satellite is 1/8. If we continue with the dingo, echidna and goa, the probability that none exist on the satellite is 1/64. By the time we get to Man, the probability that none of the listed animals including Man are on the satellite is very small. So much for insufficient knowledge. Next, Bayes' Law is special in that we use it to determine a cause given effects. Because people sometimes use primary probabilities derived from experiment rather than a priori probabilities, one sometimes arrives at conclusions that are questionable at least. We can easily invent situations like that of the previous paragraph, where Bayes' Law leads to ridiculous results. Properly used, it is as exact as any mathematical law. Finally, our main purpose was to present Bayes' Law, so we were forced to omit a lot of probability theory that is very pretty as well as productive. Thus, we merely mention continuous probability and geometric probability. Finally, we return to the problem at the start of this article, and give the proposer's solution for your study:
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We know that, if a bag contained 3 counters, 2 being black and one white, the chance of drawing a black one would be 2/3; and that any other state of things would not give this chance. Now the chances, that the given bag contained (a) BB, (b) BW, (c) WW are respectively 114, 112, 114. Add a black counter. Then the chances that it contains (a) BBB, (b) BWB, (c)
WWB, are, as before, 114, 112, 114. Hence the chance, of now drawing a black one, 1
1 2
= 4 xl + '2 x 3' +
1 1
2
'4 x3'= 3' .
Hence the bag now contains BBW (since any other state of things would not give this chance). Hence, before the black counter was added, it contained BW,
i.e. one black counter and one white. Is this conclusion right or wrong?
We conclude this (all too short) article with a few problems :hat the reader will find pleasant. A, Band C are shooting at a target. The probability that A hits the target is 4/5. The probability that B hits the target is 3/4. The probability that C hits the target is 2/3. They shoot and one misses. What is the probability that it was C?
2) A letter arrived with the contents unsigned and the postmark
smudged. It came from either Oneonta or Fredonia, both in New York, but all one could read was the consecutive letters "on". What is the probability that it was mailed from Oneonta ?
3) The sum of the digits of a seven-digit number is 59. What is the probability that it is divisible by 11 ?
4) In a horse race between three horses, the odds on one horse are 3 to 2 and the odds on a second horse are 3 to 1. What are the odds on the third horse ? 5) I have two coins, one normal and one with two heads. I pick one at random and toss it four times. It comes out heads each
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time. What is the probability that I picked the two-headed coin ? Note:
In the American Mathematical Monthly, Volume 45 (year 1938), there is a very fine article by H. Petard, on how to catch a lion. You have a cage in the desert. To catch a lion At any given moment there is a positive probability that there is a lion in the cage. Sit down and wait.
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NAPOLEON, MIQUEL, SIMSON Our start is a problem so simple that it hasn't the status of a "theorem". To present this, let us refer to the diagram on the front. We have d ABC, with triangles ARB, BPC, CQA constructed externally on its sides. The sum of the angles at P, Q and R is equal to 180°. Then the circumcircles about triangles ARB, BPC and CQA are concurrent at O. The proof is simple. Assume that the circles about triangles ARB and CQA drawn, concurrent at O. Then LAOB =1t-R, LCOA =1t-Q, soLBOC =21t- the sum of these angles, or 21t- (21t- Q - R) =(Q + R). But (Q + R) =(1t - P), so the circle on P, B, C passes through point O. There is one simple consequence of this. The line ZY is perpendicular to chord AO and the line ZX is perpendicular to chord BO. Hence LYZX = LR. Similarly, LZXY = LP and LXYZ = LQ. Therefore, if similar triangles be constructed on the sides of a triangle, with a different angle at the "outside" each time, the centers of the circumcircles of these triangles form a fourth similar triangle. Now for our first result. On the frontispiece, we have constructed three eQuilateral triangles. Therefore, the centers X, Y, Z form an eQuilateral triangle. This is known as the outer Napoleon triangle. Incidentally, if the equilateral triangles be constructed inwardly, as reflections of the outer triangles, the centers of these triangles form another equilateral triangle, called the inner Napoleon triangle. It is said that about 1792, when Napoleon was only a general, he discussed these two triangles with noted French mathematicians, which led Lagrange to remark, ''The last thing we need from you, General, is a lesson in geometry."
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B
c..----~---.....;aC
Ha
An interesting special case occurs when the lines through point 0 are perpendicular to the sides. One such special case occurs when we construct the altitudes of a triangle. In this case, point 0 is at the orthocenter of the given triangle, and triangle H a H bH c is the orthic triangle. There are a number of useful results that we can derive from this situation. Note that the angles at H a and H c are right angles, and that points B, H a , H, H c lie on a circle. This makes LHHaH c = LHBHc =90- LA. Similarly,
LHH a H b = 90 - LA. Therefore, the altitude AH a bisects the angle of the orthic triangle, and the same holds for altitudes BH b and HHc ' The inquisitive student can discover many facts from this diagram. For example, since points B, C, H b , H c lie on a circle, AH ABc = AH bAC, so thattriangles AH c H c and ACB are similar. Thus, LAHbHc = LABC. That is, HbH c and BC are anti parallel. Now we invite the reader to prove that
a) The radius AO of the circumcircle of MBC is perpendicular to HbH c ' b) Side HbH c equals
a(AH)
2R .
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A R
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.......-.&.._.a C p
B ....- - - - -.......--~C
p
The point 0 need not coincide with the orthocenter. It may even lie outside the triangle, as shown in the diagram above. Nevertheless, there are many quaint and curious results that can be derived. c) In the diagram at the left, the point 0 happens to be the circumcenter of the triangle. Find the lengths of segments PQ,
QR, RP. d) In either diagram, let h a ,h b ,h c be the altitudes to sides BC, CA, AB. Prove that
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What adjustment must be made in the case where the point 0 is outside the triangle? e) Prove LBOC = LA +LRPQ . Note: This theorem is, in our opinion, easily as important as the Point 0 theorem. It enables us, for example, to determine the area of the pedal triangle.
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We come now to the last name in our heading. The Theorem to be proved is: If the point 0 lies on the circumcircle of the given triangle, then the feet P, Q, R of the perpendiculars to the sides are collinear - and conversely. A
We present a longish but simple proof. First,
LROP =7t- B.
LAOC = 7t-B
Hence LAOC = LROP. And since LROC = LROC,
LAOR = LCOP. Now AOQR and POQC are both inscriptible. Hence
LAOR =LAQR
LCOP=LCQP.
Since A, Q, C are collinear, R, Q, P are collinear. The converse is also true. The line RQP is called the Simson line. However, it seems to have been discovered by William Wallace. This means, for example, that if three points on the sides of a triangle are concurrent, their Miquel point lies on the circumcircle. 81
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Again, there are many theorems involving the Simson line. We close by presenting a problem: e) Prove that the Simson lines of diametric points on the circumcircle of a triangle are perpendicular to each other. R
x We will prove a more general theorem fIrst. In the diagram above, AOXC is inscriptible. The four points lie on the circle. Hence LAXO = LACO. ewe have not drawn CO. Imagine it there.) Next, OQPC is inscriptible. There are right angles at Q and P. Hence LQPO = LQCO = LACO. Therefore, LAXO = LQPO, and AX and PQR are parallel!
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Now select two points P and P' on the circumcircle. The angle between the Simson lines of these two points equals the angle XAY between the lines through A that we found previously. The chords PX and P'Y are perpendicular to Be, so they are parallel and cut off equal arcs XY and PP' on the circumcircle. Therefore, the angle XAY, which equals the angle between the Simson lines of P and P', is half the central angle POP'. Now the problem stated is an obvious corollary of this theorem. Finally, there are several results that follow from what has been developed thus far. For example: f) What is the Simson line of a vertex of a triangle ?
g) What is the Simson line of a point diametrically opposite a vertex of a triangle? h) Try and prove: The Simson line of any point bisects the line joining that point to the orthocenter of the triangle.
How to Catch a Lion We observe that a lion has at least the connectivity of the torus. We transport the desert into four-space. It is then possible to carry out such a deformation that the lion can be returned into three-space in a knotted condition. He is then helpless. H. Petard
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WEEK-ENDERS 1) For those working on the solution of Fennat' s last theorem. If xn + y n = Z n has solutions, then the lesser of x, y equals or exceeds n.
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2) Determine all solutions in integers of x n +(2+x)n+(2-x)n =0, O~n~ 100.
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3) For which integer values of n is n 4 + 4 n a prime number?
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4) Find all solutions in integers of y 3 - x 3 = 91. 5) The diagram below shows a square whose side is equal to 12. Points A and B are selected as shown, and lines drawn to the vertices as indicated, intersecting at P and Q. (a) Find the area of the "lozenge" APBQ: (b) Find the area cut from the square by the line through PQ.
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MANY CHEERFUL FACTS
There are many simple concepts that, for one reason or another, are not stressed or even mentioned in the mathematics class. We have already referred to the concept of parity (see Arbelos for September, 1983). Now we present another concept that is barely mentioned, if at all, in the class. This is the concept of ratio and proportion. Geometry texts of fifty years ago did devote time to the subject, but with the modernization of geometry, this was among the topics removed. Nevertheless, one can find many uses of ratio and proportion, both in and outside of geometry. We start with elementary definitions. The fraction called a ratio and the equality
a
c
b= d
a
b
is
is called a proportion. From
these definitions, we can derive many new forms for proportions. ace Assuming b = d = f ... , we may write: (a+c+e+ ••• ) a a+b -(b-+-d-+-f-+-••-.-) = ~' a - b
c+d a rna d ' b = mb ' etc.
= c-
In geometry, these rules were used, for example, to prove that perimeters of similar polygons were proportional to corresponding sides, that areas of similar polygons were proportional to squares of corresponding sides, etc. We can do better. It should be clear that corresponding lines of similar figures are proportional. Thus, in similar triangles, the ratio of corresponding altitudes equals the ratio of corresponding angle bisectors or the ra~o of corresponding angle bisectors, and so on.
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However, we can also use ratio and proportion to advantage in non-geometric situations. For instance, let us simplify (a+b+c+d)(a-b-c+d) = (a+b-c-d)(a-b+c-d) . The urge is usually to multiply like mad. Let us, however, rewrite this as: (a+b)+(c+d)
(a- b)+(c-d)
(a + b) - (c + d)
(a- b)-(c-d)
In the olden days, this used to be written as: if ad =be, then a:b =c:d and recited thus:
If the product of two quantities equals the product of two other quantities, any pair may be made the means and the other pair the extremes of a proportion. Note the archaic terms "means" and "extremes". At this juncture, our rules tell us we may write: (a+b) (c+d) and finally, alc
(a-b)
=--(c-d)
2a
2c
2b
2d
-=
=bid.
6x+2a+3b+c Similarly, 6x + 2a - 3b - c
16a 6a 2x = -= -= from 3b+c b +3c 3b +9c 8c 3c b which x/b =alc, and, finally, x =ab/c. becomes
6x+2a
2x+6a
2x+6a+ b+3c
2x+6a- b-3c
6x+18a
We don't have to restrict ourselves to one transformation namely, (a+b)/(a-b) = (c+d)/(c-d). Let us examine the following equation: 2x+7 5x+l
2x-3 5x-9·
Subtracting numerator from numerator and denominator from denominator, we arrive at
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2x-3 10 5x _ 9 = 10 = 1, so 2x - 3 = 5x - 9, and x = 2 .
Much faster and easier than "cross-multiplying". In a previous issue, we suggested that the reader try to invent lots of formulas for the area of a triangle. We present two such here. First, we recall that the area of a triangle with sides a, b, c and circumradius R is abc K= 4R . Let us now start with the Law of Sines, thus: abc - - - - - - - - - 2R sinA - sinB - sinC . First, we multiply all these ratios, and find: abc (sin A)(sin B)(sin C) Dividing by 4R and clearing fractions, we have: abc K= 4R =2R 2 (sinA)(sinB)(sinC). Again, recall that the area of a triangle with semiperimeter s and inradius r equals rs. Now: a+b+c (sinA) + (sinB) + (sinC)
2R
s = R(sin A + sin B + sin C). Finally, K = rx = rR(sin A + sin B + sin C).
We have developed two new formulas for the area of a triangle given different information. By equating these two formulas, one can also find an interesting relation between the inradius and circumradius of a triangle in terms of the angles of the triangle, or a relation involving the sum of sines of the angles of a triangle and the product of the sines of the angles. 87
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We hope to have called to your notice the value of knowing and using ratio and proportion wherever possible, provided one can recognize that its use would be helpful.
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ASSORTED PROBLEMS 1) Let Fo (n::;; 1) be the Fibonacci sequence where
FI = F2 =1 , Fo+ 2 = Fo+I + Fo (n ::;; 1) and P(x) a polynomial of degree 990 verifying P(k) = Fk for k = 992 , ••• , 1982 . Prove that P(l983) = FI983 -1 . Rumania
2) A circle y is drawn on AB as diameter. The point C on y is the midpoint of the line segment BD. The line segments AC and DO, where 0 is the center of y, intersect at P. Prove that there is a point E on AB such that P is on the circle drawn on AE as diameter.
Sweden 3) Show that if the sides a, b, c of a triangle satisfy the equation 2(ab 2 + bc 2 + ca 2) = a 2b + b 2c + c 2a + 3abc, then the triangle is equilateral. Great Britain
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,
KURSCHAK KORNER This column features those problems of the famous Hungarian Kiirschak Contest which have not yet appeared in English translation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books I & II in the MAA's New Mathematical Library series (available from the MAA headquarters); the problems (with brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Csirmaz of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers of this column.) Student readers are also invited to set aside an uninterrupted 4-hour period to compose complete, well - written solutions to the problems below, and to submit their work to the address given below for a critical evaluation. 1/1960 Among any four members of a group of travelers, there is one who has met all of the other three before. Prove that among each foursome there is one who has met all of the other travelers before. 2/1960 In the infinite sequence, a 1 ' a 2 ' a 3 ' ••• , of natural numbers, a 1 = 1 and for k> 1, ak
~
1+ a l + a 2 + ••• + a k_ l
.
Prove that every natural number either appears in this sequence or can be expressed as the sum of distinct members thereof. 3/1960 In square ABCD, E is the midpoint of side AB, F is on side BC and G is on side CD so that AG and EF are parallel. Prove that FG is tangent to the circle inscribed in the square. Dr. George Berzsenyi
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arbelos PRODUCED FOR PRECOLLEGE PHILOMATHS
R ,..--...--------.,.c
D
ABCD is a parallelogram.
PQ and AB are parallel;
RS and AD are parallel;
Prove: BP, CT, DS are concurrent at O.
May, 1984
1983-1984: No.5 91
ARBELOS Volume 2 Chapter 5
PREFACE Two years have gone by since the fIrst issue of Arbelos appeared. For your Editor, these two years have been a challenge. Our purpose was to present mathematics not usually presented in our classrooms in a form that would be useful to the interested student. We do hope that our efforts have been of some use and have been moderately successful. However, we have had too few reactions from our readers. As we close out the school year, we have two requests. First, of course, subscribe for the next school year and let others know about our little publication. Second, however, we would like to have your reactions to the publication. Let us know if you are pleased with its contents, what other subject matter you would like to see in it, and especially what you would like to have changed. Thus far, we have avoided such subject matter as matrices, calculus, Would you suggest that we include these in the future ? We have also avoided applications of mathematics to other fIelds physics, chemistry, modeling, and so on. Should we include short articles on some or all of these subjects ? If the postage worries you, you might send your comments with your renewal or subscription check. Professor Mientka will get your letters to me. If you wish, you might send your comments to me.. They will help me immeasurably. Have a Happy Summer! Dr. Samuel L. Greitzer
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TOPICS IN NUMBER THEORY Students who wish to improve their problem-solving ability should spend lots of time in the study of Number Theory. Nowhere will be found as many ingenious concepts and ingenious solutions as will be found in this branch of mathematics. We have already touched on the subject with our article on Diophantine equations in our issue of March, 1983. There, we introduced the Farey sequence, continued fractions, Euler's method for solving Diophantine equations, and so on. We suggest that you reread this article. A) It was Gauss who introduced the concept of arithmetic, modulo some number, without any explanation of just why he was led to the idea. It has been an astoundingly fruitful concept. We start with a definition: If two integers, a and b, are such that (a - b) is divisible by a third integer m, we write a: b(mod m). This means that, for any number a, there are many possible values for b - namely, b + 00. However, all numbers congruent to a (mod m) are said to constitute a "residue class": since 7 : 2(mod 5), the set 2 + 5t constitutes a residue class. It includes
-3, 2, 7, 12, 17, •••. The mathematics of congruences is very like that of ordinary algebra or arithmetic. We list some properties without proof, since the proofs are quite simple. Let a: b(mod m), c: d(mod m); Then (a+c): (b+d)(mod m), (a-c): (b-d)(modm) and ac : bd(mod m) . If a: b(mod m) and b: c(mod m), then a:c(modm). Also,if a:b(modm), then ka:kb(modm). Only when ax: bx(mod m) can one not conclude that a: b(mod m). If a and m are relatively prime, one may cancel otherwise not. Remark:
If x:a(modm), then x 2 :a2 (modm), and, in
general, x n : an (mod m). Further, if f (x)= bOX n + b IX n-I + ••• +b n and f(a)= boa n +bIall- I + ••• +b n , 93
ARBELOS Volume 2 Chapter 5
then f(x) = f(a)(mod m). Now any number is a sum of powers of ten, and, since 10 = l(mod 9), f(lO) = f(l)(mod 9). Now, for any number, f(l) is nothing more than the sum of the digits of f(10), which shows that any number is divisible by 9 if the sum of its digits are divisible by 9. This is the well-known rule for "casting out nines".
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The reader can easily find a rule for division by 11 if he or she notes that 10= -1(mod 11) and compares f(lO) with f(-l) (mod 11).
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Corresponding to linear equations, we may have linear congruences to solve. These have the form ax = b(mod m).
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Suppose we wish to solve 5x = 3(mod 7). There are various ways of doing this. The amateur might wish to rewrite this as 5x - 3 = 7y and solve the Diophantine equation. Or, one might write:
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5x = 3(mod 7) 0= 7(mod 7) and add to get 5x = lO(mod 7), whence x = 2(mod 7). Or one might multiply the given congruence by 3 to get 15x = 9 (mod 7) and use the concept of residue class to get x = 2(mod7). This is where a bit of ingenuity on the part of the student comes in handy. It is this ingenuity that the student should try to develop.
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Aside: When, given ax + by = 1, a and b are relatively prime,
the equation always has a solution. This solution can be obtained in many ways - the Euclidean algorithm, continued fractions, Euler's method, Farey fractions, etc. See the issue of March, 1983. For the same reasons the congruence ax = l(mod m) has solutions if a and m are relatively prime. Of course, simultaneous congruences occur. If a pair has a B) solution, one can get it by means of substitution. First, we can, given ax = b(mod m), always transform into x = c(mod m). We did this in the problem on the previous page. Therefore, suppose this done, and suppose we have: x = 2(mod5) x = 3(mod7) We want an x that satisfies both congruences. From the first
congruence, x = 5t + 2. Then 5t + 2 = 3(mod 7) 5t = 15(mod 7)
5t= 1(mod7) and
t =3(mod7).
Therefore t = 3 + 7u, and x = 5(3 + 7u) + 2. Finally,
x = 17 + 35u, or x = 17(mod 35). Trial shows that the residues
including 17, 52, 69, ••• satisfy the two congruences.
Where there is a solution, this method will always produce it.
C) The Chinese Remainder Theorem gives us a method for solving a series of two or more linear congruences. Various texts ascribe it to the Chinese mathematician Sun Tsu in the first century
A.D. Hence the name. We will prove the theorem only for three simultaneous linear congruences, but the reader will see that it holds for n congruences. Let x == at (mod mt)' x == a 2 (mod m 2 ) and x == a 3 (mod m 3 ) be the congruences, and M = m t m 2 m 3 . Note that m t , m 2 , m 3 are relatively prime in pairs. If they are not, there may be no solution. Consider x == l(mod m t ) , 95
ARBELOS Volume 2 Chapter 5
x == O(mod m 2), x == O(mod m 3 ). Disregarding the fIrst congruence, we see that the general solution for the other two is M x = m2m 3 t, which we write x =- t. m l
M We now seek to solve - t == l(mod m l ). It has a
ml
solution, AI. Then x = A lMa I I ml is a solution of x == a I(mod m l ) x == O(mod m 2) x == O(mod m 3 )
.
Similarly, consider x == O(mod ml ) x == l(mod m 2) x == <Xmod m3 ) ,and in the same manner, we fInd that x = A 2Ma 2 I ~ is a solution of x == O(modm l ) x ==a2(mod~) x == O(modm 3 ) • For x == <xmod m l ), x == O(mod m2 ), x == l(mod m 3 ) , we derive a solution x = A 3Ma a 1m 3 . M M M satisfIes all the Hence x = A 1a l -+A 2a 2 -+A 3 a 3 ml m m 2 3 congruences. In general,
M
M
M +A a -)(modM).
x == (A1a l -+A 2a 2 - + ml m
nnm
2
n
As an example, consider x == l(mod 2),
x == 3(mod 5),
x == 5(mod 7) .
From x == 1(mod2), x == (Xmod5), x == O(mod 7), we fInd fIrst that x == O(mod 35) or x = 35t, and then solve 35t == l(mod 2). Then t == l(mod 2) and we may take Al = 1.
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From x = O(mod 2), x = l(mod 5), x = O(mod 7), we find first that x = 0(mod14), or x = 14u, and then solve 14u = 1(mod5). Then 4u = l(mod 5), or -u = l(mod 5), and we may take A 2 =-1. From x = O(mod 2), x = O(mod 5), x = l(mod 7), we find first that x = O(mod 10), or x lOv, and solve lOv = l(mod 7). Then 3v = l(mod 7), or 3v = 15(mod 7), so v = 5(mod 7) and we take A 3 =5 . Now x= (lxlx35+(-1)x3x14+5x5xlO)(mod70), and simplifying find x = 243(mod 70) and finally, x = 33(mod 70) . Try x =33 in each of the congruences.
=
This theorem is useful because it states that, under proper conditions, a solution exists for these congruences. Our experience has been that it is more useful to use the solution method when there are only two or three congruences to handle. D)
A famous theorem due to Fermat (called his "little" theorem
2»
is to distinguish it from his conjecture about x n + yn = zn (n> the following: Let p be a prime and x any number. Then x P = x (mod p). Incidentally, when x and p are relatively prime, we may divide by x and get xP-l = 1(mod p). There are many proofs of this theorem. The simplest, in our opinion, is the following: Expand (l + 1+ ••• + l)P using the multi-nomial
(x ones)
theorem. Then we get
p! 1P + 1P + ••• + 1P plus a lot ofterms of the form -..;;..... a!b!••• c!
(x terms)
in which each term is an integer. Since p is prime, no element in
the denominator divides p, which means that every such term is
divisible by p. Therefore, we have
(1+1+ ••• +1)P:=(l+1+ ••• +1)(modp), or xP=:x(modp). Among other uses, this theorem will solve many problems involving factoring. For example, the theorem states that 97
ARBELOS Volume 2 Chapter 5
n 13 - n = o(mod 13) for every n. By factoring on the left, one can also show that n 13 - n is divisible by 2, 3, 5 and 7. We leave for the reader the problem of finding the units digit in 3200. (Hint: note that 34 = l(mod 10)). E) As in algebra, we have a factor theorem for congruences. The proof is short. Suppose f(x) = O(mod p), with p prime. Assume that a is a root, so that f(a) =O(mod p). Then f(x) -f(a) = (Xmod p). However, f(x) - f(a) is divisible by (x - a), so f(x) - f(a) = (x-a)Q(x). Hence f(x) = (x - a)Q(x) + f(a)(mod p), from which f(x) = (x - a)Q(x)(mod p). With the help of the factor theorem, we can now prove Lagrange's Theorem, which states: Let f(x) = O(mod p), where p is prime, have incongruent
solutions a 1,a 2' ••• ,a n' then f(x) has no more than n factors.
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We sketch a proof by induction. From our work with linear
congruences, we know that, when f(x) is linear, we have one solution.
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Assume the theorem true up to functions of degree up to (n-1).
From the factor theorem, we may write: f(x) = (x-a)Q(x) (modp) where a is a root of f(x). Now let c be another root. Then: fCc) = (c-a)Q(c) (modp). Since c and a are incongruent, we must have Q(x) = O(mod p). Since, by definition, this has at most (n-l) solutions, f(x) has at most n. We can use the information we now have to prove Wilson's Theorem. This asserts that if p is an odd prime, (p -I)! = -l(mod p) and conversely. When p is prime, we have x p x = 1, 2, ••• ,p - 1. Also, we see that 98
1
= 1(mod p) for
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(x -1)( x - 2) ••• (X - P + 1) = O(mod p) for the same values. We write x p- 1 -l==(x-l)(x-2) ••• (x-p+l)(modp) Now let x = O. Then -1 even number of factors.
= (p-l)!
since we are working with an
When p = 2, we get -1 == 1(mod 2) which is true.
Therefore Wilson's Theorem holds for all primes.
This is not only necessary but also a sufficient condition for
primality. However, it is not very useful in determining whether a
given number is a prime. The arithmetic gets complicated. Problems using what has been developed here pop up in many places. For example, one of the problems in a recent International Olympiad used the Chinese remainder theorem. Moreover, they can lead to new results. For example, from xp-1-l==O(modp) we note that p-l is even, so we may factor and get
x(p-l)/2 ==±l(mod p), which has far results.
We urge students to spend time on Number theory. One very good
reference is The Theory of Numbers, by Niven and Zuckerman.
Another is the Elementary Theory of Numbers by Sierpenski, and a
third is an Introduction to Number Theory, by J.E. Shockley.
Douglas Jungreis used Wilson's theorem to get a correct solution to
a problem by Bulgaria.
We have received a beautiful solution to the problem on the cover of
the March Issue of Arbelos from Ede lijasz Vasquez of Medellin,
Colombia S.A. (better than any I could do).
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ARBELOS Volume 2 Chapter 5
JUST FOR FUN We present these puzzles for your enjoyment - we hope. Try them yourself, try them on others - in your class or your club. For a change, answers are provided. 1) The members of a math club met after school at a drugstore, where each ordered the same item (which cost more than one cent). The combined check came to between $3.00 and $4.00. How many members were there?
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2) A census taker visited a man who had three daughters. He was informed that the product of their ages was 72, and the sum of their ages the same as the house number. Since the census taker still could not determine the ages, the man told him that his oldest daughter liked chocolate sodas. How old were the daughters? 3) Three tramps met and decided to "dine". One had three rolls, another had five rolls, and the third had eight cents. They shared the rolls equally, after which the third tramp left the eight cents to be distributed. How much should each tramp get? 4) The following puzzle is at least 250 years old, but we like the names, so here it is: Three Dutchmen name Hendrick, Elas and Cornelius and their wives Gurtrun, Katrun and Anna bought hogs. Each bought as many hogs as he (or she) paid dollars for one. Each husband spent $63 more than his wife. Hendrick
bought 23 more hogs than Katrun and Elas 11 more than
Gurtrun. What was the name of each man's wife?
5) Six barrels have capacities of 15 gal., 31 gal., 19 gal., 20 gal., 16 gal., and 18 gal. Five contain wine and one champagne. The owner sold some wine to one customer and twice as much to a second customer, but kept the champagne.
Which barrel contained the champagne?
(All sales were by the barrel.)
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RAMSEY'S THEOREM AND THE
PIGEONHOLE PRINCIPLE The Thirteenth Putnam Competition (March, 1953) includes the following problem: Six points are in general position in space ( no three in a line, no four in a plane). The fifteen line segments joining them in pairs are drawn and then painted, some segments red, some blue. Prove that some triangle has all its sides the same color. This is an example applying a theorem of Ramsey on formal logic. In this case, we investigate triangles (3-sets), the edges of which are lines (2-sets) which are either blue or red. We write this condition R(3, 3:2) = N, and wish to find the least value of N for which there exists a triangle all of whose edges are blue or all of which edges are red. One of the troubles with Ramsey's Theorem, which asserts that such an N always exists is the difficulty of stating it simply. It is just not a simple theorem. Hence we will try to use examples to illustrate the theorem. We ask the reader's forbearance. In the problem above, the Ramsey number N is equal to 6. That is, if we start with six points as described, there will be at least one triangle all of whose edges have the same color. We call such triangles monochromatic triangles. If one seeks monochromatic triangles using three colors, it
has been found that the Ramsey number R(3,3,3:2)=N is 17.
That is, if we start with seventeen points, join each point to the other
sixteen, and color these segments with one of three colors, say red,
blue or green, there will be at least one monochromatic triangle.
More generally, consider R(5,6,7:3). Here we have three
colors, as the integer 3 indicates. We are examining pentagons (5),
hexagons (6) and heptagons (7). The segments are all drawn and
colored, and every triangle is colored in one of the three colors. The problem is to determine for what value of R there will be at least one pentagon which is monochromatic, or one hexagon which is monochromatic or one monochromatic heptagon. Finding the different Ramsey numbers has been very difficult thus far. The latest information we have is as follows: 101
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=R(t, 2 ; 2) =t, R(3, 2 ; 2) = R(2, 3 ; 2) = 3, R(3,4 ; 2) = R(4, 3 ; 2) =9, R(4,2 ; 2) =R(2, 4; 2) =4, R(2, t ; 2)
R(3, 3, 3 ; 2)
= 17,
=6, R(4,4 ; 2) = 18, R(3,5 ; 2) = 14. R(3, 3 ; 2)
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R(4,4 ; 2) tells us that, given 18 points as described, each pair of points colored in one of two colors, there exists at least one monochromatic quadrilateral.
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We might mention that anyone extending the list above will be performing an important mathematical activity. For example, it is known that 17 ~ R(3, 6 ; 2) ~ 19. Maybe a reader might derive a more definite result and become famous !
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Let us solve the Putnam problem at any rate. Select one vertex. From this vertex, there will be five segments going to the other five points. At least three of these segments must be of the same color, say red. For these three segments, we examine the other ends. If any segment joining two of these ends is red, we have a red triangle. If none of the segments joining the three ends is red, they are blue and we have a blue triangle. It is easy to show that, for five points, it is possible to join these by segments so as not to have a monochromatic triangle. Thus,R(3, 3 ; 2) = 6 .
The case R(2, 2, ••• , 2; 1) where there are n two's, is simplest. Here, the Ramsey number has the value n + 1. In fact, it is a restatement of the Pigeonhole Principle, which is well known to our readers (or should be). It is usually stated as follows: If one has n pigeonholes in which to place n + 1 or more letters, then one pigeonhole must contain more than one letter. In this form, it is used in many problems, stated in a variety of ways. For example, in a set of 6 people, there are either three who are acquainted with each other, or there are three who are strangers. Sometimes, handshakes are involved, and sometimes languages, etc. We now invite the reader to try the following problems. a) From a set of ten distinct two-digit numbers, it is possible to select two disjoint sets whose members have the same sum.
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b) If five points are selected in a unit square, two of these must be
distant from each other by an amount equal to or less than
cos 45. c) From a row of n integers, one may always select a block of adjacent integers whose sum is divisible by n. d) Given five integers, one can always choose three of these whose sum is divisible by 3. e) Given the numbers 1, 2, ,200, one can pick 100 of these, no one of which divides another Gust pick 101, 102, ,200). However, one selects 101 numbers, there is always one which is a divisor of another.
f) If the arithmetic mean of n positive numbers is equal to A,
then at least one of the n numbers is equal or greater than A,
and at least one is equal to or less than A.
g) In New York City, there are at least two residents who have the same number of hairs on their heads. 000
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We now present a problem that appeared in the U. S. A. Mathematical Olympiad for 1978. Nine mathematicians meet at an international conference and discover that among any three of them, at least two speak a common language. If each of the mathematicians can speak at most three languages, prove that there are at least three of the mathematicians who can speak the same language.
How To Catch a Lion
We place a spherical cage in the desert, enter it and
lock it. We perform an inversion with respect to the
cage. The lion is then in the interior of the cage, and
we are outside. H. Petard
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PROJECTIVE AND AFFINE PLANES Because of considerations of space (no pun intended), much of the content of this article will have to be intuitive in nature. We apologize in advance. We start with the concept of incidence applied to undefined objects called points, lines and planes, and the condition "on". A point is incident with a line when it is "on" the line, and a line is incident with a point when it is "on" the point. The same holds for points or lines on a plane. What is meant by "on" is left to your intuition. We can define a plane as consisting of all the points on all the lines on a point 0 which are incident with all the points not on O. Again you should keep your intuitive concepts of point, line and plane. At any rate, a projective plane is a plane. It is not a set of equations nor a matrix. We can exhibit the beginnings of the concept of duality now. On this plane, line . . Reading the words at the lines pomt top yields one definition and reading the words at the bottom yields another. Note that on the projective plane there is no parallelism. All lines intersect. This should cause no difficulty. Consider what one has to believe on the Euclidean plane. For a line L and a point 0 not on it, when we allow a second line to be on 0 and rotate it conterclockwise, the point of intersection with L travels to the right and then suddenly appears at the left and travels right, and no one seems to ask questions. We now define a perspectivity on the plane. any two distinct
points
are on a
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Why, sometimes I've believed as many as six impossible things before breakfast. C. L. Dodgson 104
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Suppose we have a point 0 on the plane, a line L not on 0 on the plane, and a set of points A, B, C etc., on L. We call this set a range of points. We draw the lines on 0 and on the points of the range. This configuration is called a pencil of lines. Let the pencil be cut by a second line, L not on 0, so that we have A on OA, B on OB, etc. We say that points A'B'C' and ABC are perspective from 0, and write A' B' C' 1't ABC . I
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If we perfonn the same operation on A' B' C', selecting another point 0 and another line L', we get new points A", B" , C". Now we say that A", B" ,C" and A, B, C are projectively related, and write A" B"C" ",ABC. The diagram may clarify this. I
It is interesting to construct the dual figure. We replace "point" by
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0'
"line" and vice versa, and work only with a perspectivity. Try to follow the construction in the figure below. L
We say that the two sets of lines A, B, C and A' B' C' are perspective from the line O. We have projectivities between figures on two planes. Given a figure with points A, B, C, ••• on a plane P and a point not on P, we can join this point to the points A, B, C, ••• to get a pencil of lines. If this pencil be cut by plane P not on P, we get a new figure, A' B'C' ••• on P in perspective with figure I
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ABC •••. If we do the same with points A' B' C' "', we get a new figure A' B' C" ••• which is projective with figure ABC •••. Notice that there are sometimes vast differences between two projectively related figures. Shapes may be different, lengths of line segments and degrees of angles usually change, areas are also changed. One might begin to wonder what use the whole procedure has. However, incidence is preserved, straight lines project into straight lines and things do not seem to be so bad. A circle projects into another conic section, which is also of some help. A triangle remains a triangle, though the image may not look anything like the original. Perhaps we can find more useful properties. First, three points on one line can always be projected into three points on another (or the same) line. In the diagram, draw a line C C' and select a point 0 on it.
o
c' 0'
Now project A, B, C into A, X, C' on line AC' • M ' and XB' meet at 0', and AXC' projects into A 'B'C' . However, if ABCD 7'\ A'B 'C'D' under one projectivity and ABCD 1'\ A'B 'C'D under another projectivity, then 0' and 0" coincide. This is the Fundamental theorem of projective geometry. 106
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Corollary: If, under this operation, a point coincides with itself, the projectivity becomes a perspectivity. The (simple) proof is left for the reader. Next, take two pencils of lines on points P, Q which are projectively but not perspectively related. We assert that the points of intersection of corresponding lines of the pencil define a conic. Again, we just sketch a proof (see Diagram). A
B
T
The two pencils are PA, PB, PC and QA, QB, QC. We also add the lines PX and QY which meet at T. The lines AB, AC cut the pencil in projective ranges. By comparing intersections, we derive. ABUX "AVCY Since A projects into itself, this projectivity is actually a perspectivity. Hence BV, UC, XY are concurrent - obviously at S. Given five points, then, we may find as many more points as we wish. Just draw a line through S. This will intersect AB and AC at points X, Y, and the intersection of PX and QY is a point on the conic.
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Now consider the pencil on 0 cut by line L at A. B. C. D.
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Comparing areas of triangles. we find that h(CD) = (PC)(PD)sin w
h(AB) = (PA)(PB) sin u h(AD)
=(PA)(pD)sin(AOD)
h(CB) = (PC)(PB)sin v.
Combining and dividing we arrive at (AB)(CD)
(AD)(CB)
sinu.sin w sin(AOD)si n v .
However. if we cut the pencil by line L' and get A'. B' • C'. D' and compare areas again. we get the same combination of angle sines. Now ABCD 1'\ A'B C'D'. We call the combination of fractions above a "cross ratio". and we have shown that the value of the cross-ratio of four points is constant under projection. We write a cross-ratio as R(ABCD). and have R(ABCD) = R(A'BC'D) when ABCD 7'\ A'BC'D' . The order of the points is very important!
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D
Now consider the complete quadrangle PQRS. with line L on double points A and B separating C and D. Let Now ABCD 7'\ PRCO 7'\ BACD . Also. R(BACD) =
Therefore ')...2 = 1. and ')... = -1 . In this
situation. we call A. B. C. D. a harmonic set and write it H(AB. CD) =-1 .
Now we can combine what we have to date. On the projective plane. select a line. We may remove it or we may leave it and call it an ideal line. We will call two lines parallel when they intersect on the ideal line. Obviously. if two lines are parallel to a third. they are parallel to each other.
We can always project any line into the ideal line. This usually simplifies diagrams by creating parallel lines. Also. suppose that, given H(AB, CD)=-I, we have D on the ideal line. This .. . (AC)(BD) (AC) gIves us H(AB, CD) = (AD)(BC) = (BC) = -1, SInce, In the limit.
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= 1. Therefore C is the midpoint of AB.
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A - - -....-
...- - - - - - - -.......~p
B
In the diagram above, ABCD is a parallelogram whose opposite sides intersect on the ideal line PQRS. That is (AB)(DC) = P and (AD)(BC) = Q , and P and Q are on the ideal line. AC and BD are two diagonals. Comparing this with the previous diagram, we see that P, Q, R, S form a harmonic set, so that H(PQ, RS) = -1. However, not that: BODS
PRQS
AOCR,
(BO)(DS) Therefore H(BD,OS) = -1 = (BS)(OO)
(BO) (00)' which means
that (BO) = -(DO), and 0 is the midpoint of BD. In the same manner, from H(AC, OR) we derive (AO) =-(BO) , and 0 is the midpoint of AC.
~L:::::~p B
C
In the diagram above, the ideal line is (PR). Let segment (MN) be parallel to side (BC). Then (BC) and (MN) intersect at P. Now we make use of the cross-ratio. We have BMAQ CNAR, so that R(BMAQ) = R(CNAR) = A. .
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ARBELOS Volume 2 Chapter 5
(BM)(AQ) R(BMAQ) = (BQ)(AM) (CN)(AR) R(CNAR) = (CR)(AN)
(BM)
(AM) = A. , and
(CN) (BM) (CN)
(AN) = A., and (AM) = (AN) .
That is, a line parallel to one side of a triangle divides the other two
sides proportionally.
Where there are no "metric" properties involved, it may be
possible to provide a simple proof for a complicated problem. One
might be able to select a line of the figure, make it the ideal line, and
go on from there.
o
So much for the Affine Plane. We return for a moment to the Projective Plane. Since we have no metric properties in Projective geometry,
there can be no circles, only conics. In the diagram above, we have
the ideal line, and three conics. One does not intersect the ideal line
and we call it an ellipse. The second is tangent to the ideal line, and
this is a parabola. The third intersects the ideal line in two points.
We call this an hyperbola. The tangents to the hyperbola at the
intersections with the ideal line are its asymptotes.
We have already presented several theorems from Projective geometry - Desargues' theorem, Pappus' theorem, Pascal's theorem and Brianchon's theorem, for example.· Now Desargues' theorem t are also is self-dual. Two triangles perspective from a
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• Arbelos, November, 1982 111
perspective from a ~int . For perspectivity from a line, see the second page of this article.
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An article in the Arbelos for May, 1983, took up the question of loci and envelopes. The dual of a curve as a locus is a curve as an envelope.
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We have shown that a conic is completely detennined by five points. The dual conic will be completely determined by five lines which are projective but not perspective with each other.
~
As an example of what this means, consider the question: How many parabolas can be constructed tangent to four lines? The answer is simple - the parabola is already tangent to the ideal line. This line and the four others determine just one parabola, which answers the question.
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ARBELOS Volume 2 Chapter 5
In addition to using the ideal line and projection, therefore, it is sometimes useful to use the dual of a figure to solve a problem. Of course, not all figures are self-dual. One of these is Pappus's theorem and its dual
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B'
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Pappus
Dual
Where the dual doesn't look anything like the original. Nevertheless, if you write out a proof of Pappus and replace, in this proof, point and line, concurrent and collinear, and collinear and concurrent, you will arrive at the dual. It is always possible that, by dualizing a figure, you may arrive at a simpler figure that can be easily suited to solution. Solve the dual figure and dualize your proof.
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ARBELOS Volume 2 Chapter 5
In spite of the length of this article, we have merely touched on the subject of the projective plane and the affine plane. We cannot forebear mentioning a few curiosities just to pique the imagination. First, we have not mentioned that, in addition to the real projective plane, to which we have adhered for the most part,
there is a complex projective plane. On this plane, two curves will intersect in real or complex points. As one strange result, every circle will cross the ideal line in the same two points, called the circular points at infinity. The easiest way to show this is by the use of homogeneous coordinates. Thus, how many conics can be drawn through two given points and tangent to three given lines? Project the given points into the circular points, and ask how many circles can be drawn tangent to three lines. The answer - four; one inscribed and three escribed circles. We close by presenting some problems that may be susceptible to solution by using projections, cross-ratios, harmonic ratios, duality and/or affine methods. The opposite sides of hexagon ABCDEF are parallel and the diagonal CF is parallel to AB. If BC and AF intersect at P, CD and EF intersect at Q, and BD and AE intersect at R, show that P, Q, R are collinear. ABCD is a complete quadrilateral. AB and DC intersect at P, AC and BD intersect at Q, and AD and BC intersect at R. Segment PR is added. Now, through Q, a line is drawn parallel to AB, intersecting BR at G and PR at H. Prove that the segment QH is bisected at point G. Lines OX and OY are intersected by three concurrent lines PA'A, PBB, PC'C. Points A, B, C are on line OX and points A', B', C' are on line OY. ABO and A'C intersect at M, and AC' and A'B intersect at N. Prove that MN will pass through point P.
Finally, try your hand at the problem shown on the cover.· Interested students might refer to the texts listed below for more information. First comes, "Projective Geometry" by Veblen • "A poor thing, Sir, but mine own."
Lewis Carroll, in Pillow Problems. 113
ARBELOS Volume 2 Chapter 5
and Young, still an excellent text. Next, try "The Real Projective Plane", by H.S.M. Coxeter. Third, there is an interesting Carns Monograph, "Projective Geometry", also by Young. Only after one has read these will a student be able to read algebraic treatments critically.
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ARBELOS Volume 2 Chapter 5
KURSCHAK KORNER
This column features those problems of the famous Hungarian Ktirschak: Contest which have not yet appeared in English translation. For the problems of 1894-1928, the reader should consult Hun~arian Problem Books I & II in the MAA's New Mathematical Library series (available from the MAA headquarters); the problems (with brief summaries of their solutions) covering the years 1966-1981 were translated into English by Prof. Csirmaz of Hungary in 1982. (The author of this column will gladly send a copy of the latter to interested readers of this column.)
Student readers are also invited to set aside an uninterrupted
4-hour period to compose complete, well - written solutions to the
problems below, and to submit their work to the address given
below for critical evaluation. 1/1949
Prove that if 0 0 < a. < 1800 , then 1 1 sino. + 2"sin2o. + '3sin3o. > 0 .
2/1949 Let P be a point on the base of a given isosceles
triangle, and let Q and R be the intersections of the equal sides
with lines drawn through P parallel to these sides. Show that the
reflection of P in the line QR is on the circumcircle of the given
triangle.
3/1949 Identify those positive integers which can not be expressed as sums of two or more consecutive positive integers. Dr. George Berzsenyi
115
1) The check came to $3.61 (= 19x19). Any other amount would involve a number with three or more factors. Only this would provide a definite answer. There were 19 members who paid 19¢ each.
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2) The census taker can see the house number. The possible factors of 72 are:
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ARBELOS Volume 2 Chapter 5
ANSWERS
1 x 1 x 72
1 x 4 x 18
2 x 2 x 18
2 x 6 x 6*
1 x 2 x 36
1 x 6 x 12
2 x 3 x 12
3 x 3 x 8*
1x3x24
1x8x9
2 x,4;c. 9
3x4x6.
It must have been the case that two sets of factors had the same sum, which the two starred sets have. The oldest daughter is
eight, since in the other case, the oldest would be twins.
3) Each man ate 8/3 rolls. One contributed 3 rolls or 9/3, so he gave the third man 1/3 roll. The second contributed 5 rolls, or 15/3, so he gave 7/3 rolls to the third. The division must be in the ratio 117, so one man gets 1¢, the other 7¢. 4) Each person spent a square number of money. Also, (husband)2 -(wife)2 = 63= 63x1 or 21x3 or 9x7. These h 2 + w 2 = 21 h3 +w 3 = 9 . hI + wI = 63 yIeld h 2 -w 2 =3 h 3 -w 3 =7. h I -w I =l Now hI - W 2 = 23 and h 2 - w 3 = 11 . Hence the couples are Hendrick and Katrun, Elas and Anna, Cornelius and
Gurtrun.
5) We use modular arithmetic here. Modulo 3, 15=0 31=1 19=1 20 = 2
16 = 1
18 = 0 .
Now the total wine sold is = O(mod 3) . Hence the 20-gallon barrel has the champagne. The dealer sold 31 + 19 + 16 = 66 gallons to one customer and 15 + 18 = 33 gallons to the other.
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