Produced For
PreCollege Philomaths
1984 -by 1985
Professor Samuel L. Qreitzer 1906-1988
Volume 3
Number 1 · 5
Second Edition Original Copyright © 1984
Second Edition Copyright © 1997
The American Mathematics Competitions
Professor Samuel L. Qreltzer 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 and 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 198.3. 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 5. Klamkin accompanied the USA team to the International Mathematical Olympiad from 1974 (the first year the USA participated) to 198.3. 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 and friendship.
Dr. Walter E. Mientka Executive Director American Mathematics Competitions University of Nebraska-Lincoin
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Preface To The Second Edition
1984 - 1985 During the five year period from 1982 through 1987 Professor Greitzer authored 25 regular issues of the ARBELOS. This Volume :3 contains the issues published from 1984 - 1985. 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 solutions manuscript and they appear in Volume 6 of the ARBELOS series. The Second Edition essentially remains the same as the First. However, it does reflect the use of current computer technology versus the original edition which was completed on a manual typewriter. In addition, page headings are 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]
-i
ICONTENTS Of THE ARBELOS, VOLUME Volume 3, Chapter 1* Cover Problem Preface to The First Edition Echoes of Summer The Binomial Theorem~Etc The Gamma Function Many Cheerful Facts Undetermined Coefficients Conjugate Coordinate Geometry Olympiad Problems
31 1-27 1 2 3 6 11 13 16 19 27
Volume 3, Chapter 2· * Cover Problem Preface to The First Edition Cayley v.s. Hamilton Week-Enders A Look at Inequalities Beware of the Obvious Olympiad Problems Partitions Kurschak Korner
29 59 29 30 31 39 40 45 52 53 59
Volume 3, Chapter 3 *. * Cover Problem Preface to The First Edition Transformations in Geometry Kurschak Korner Factoring is Fun The Cubic Equation Many Cheerful Facts Olympiad Problems
61-88 61 62 63 69 70 77 85 88
•Originally published as the ArbeJas Volume 3, Issue 1, September, 1984 "Originally published as the ArbeJas Volume 3, Issue 2, November, 1984 •••Originally published as the ArbeJas Volume 3, Issue 3, Januroy, 1985 -ii
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Volume 3, Chapter 4 * Cover Problem Preface to The First Edition From Our Readers The Parabola and the Right Angle ManyO Cheerful Facts Concurrence ~ Discovery and Invention A Combinatoric Formula Problems for Beginners Olympiad Type Problems Dimensional Analysis Kurschak Korner
Volume 3, Chapter 5** Cover Problem Preface to The First Edition Trigonometry Geometric Algebra Euler's Formula Term-Enders Math is Fun Kurschak Korner Olympiad Problems
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*Originally published as the Arbelos Volume 3, Issue 4, March, 1985 * *Originally published as the Arbelos Volume 3, Issue 5, May, 1985
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VOUJMII3, CllAl'1U 1
arbelos
Produced For PreCollege Philomaths
P
I-----=.-~---I
A
Btl.
tl 2
C
A chain of circles is inscribed in the Arbelos, each tangent to the Arb~los and to the preceding circle, From the center of each circle, let a perpendicular 0iHj be constructed to line AC. Let d j be the diameter of circle
OJ' Prove that perpendicular OjH j is i x dj'
(For example, 02H2
1984-1985:
=
2 PQ).
No. 1
September, 1984
-1
ARBELOS
VOLliI'm:5, CHAI'mR 1
PREFACE It seems incredible that ARBELOS is starting its third year. We hope to continue our present policies on the basis of the numerous letters we have received from readers. First there were those who felt that no changes were needed - they were satisfied. There were some who felt that perhaps some applications could also he included. Oddly enough, there were a number who were against the inclusion of calculus. Now we have not done so yet but we do believe that if any problem can be most easily solved by calculus, then we should use it. At any rate, we plan to continue as before, if you - the readers - wish it. We will sneak in one or two applications and use a spot of calculus if that makes for a simple solution. At the June meeting of ARML, we were asked about the origin and meaning of ARBELOS. As the diagram on the cover shows, the word is Greek for a shoemaker's knife. Just leave out the diameter ABC. It was a favorite diagram for Archimedes, who worked out a number of properties of the diagram, which have appeared in previous issues. The cover illustrates another ARBELOS property.
Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903 -2
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ECHOES Of' SUMMER
I·
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Usually, beginning in June, we get problems of varying age and difficulty presented to us. For once, this year, we did not get our usual angle trisection solution. This normally takes the form of an unholy mess of lines and arcs, accompanied by the claim, "I have trisected an angle ! Show me where I'm wrong." However, we did get a few "golden oldies", which we think might be of interest to our readers. The first came to us from Professor Leon Bankoff. He is too good a mathematician to have been stuck by this one, but his question puzzled me, too. Consider the diagram at the right. In the circle, M is the midpoint of chord PQ. AB and CD) are chords that intersect PQ at X and Y. Prove that PX = QY. This is well known as the Butterfly Problem. If you have not come on it yet, try to prove it on your own. Professor Bankoff merely wanted to know when it had acquired the name, "Butterfly Problem" . I didn't know either.
D
There are many references to this problem in the literature. It is mentioned in Johnson's "Geometry" on p. 78. You will also find it in Eves', "Survey of
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ARBELOS
VOWMII
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Geometry", Vol. L p. 171. We also put it into our "Geometry Revisited", pp. 45-46. The simplest proof we have uses projective geometry. The second "oldie" came to us from Dimitrios Vathis, of Chalkis, Greece. We like best the name "Propeller Problem". In the diagram to the right, we have the hexagon ABCDEF inscribed in a circle. Chords BC, DE, FA have lengths equal to the radius of the B circle. P, Q, Rare the midpoints of the three segments AB, CD, EF. Prove that the triangle '. C with vertices P, Q, F' R is equilateral. Again, if you have not met up with this r . one, you should try to derive a proof on E your own. Of the many proofs, we again recommend Eves", "Survey of Geometry", Vol. II, p.184. Also, we have before us an article by Bankoff, Erdos, and Klamkin, entitled "The Asymmetric Propeller", which extends the theorem. Of course, Mr. Vathis has included his own proof. For the extended proof, consult "Mathematics Magazine", for November 1973, pp. 270-272. Incidentally, I suggest that Mr. Vathis consult The proof found in Eves' book.
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The third oldie came to us from Professor Meisters, of the University of Nebraska. In the diagram to the right ABC is an Isosceles triangle with vertex angle A = 20 . Cevian BP divides angle B into two angles of 60 and 20 as shown, and cevian CQ divides the angle at C into two angles of 50 and 30 as shown. Problem: Find the number of degrees in angle BPQ. We have included this in our "Geometry Revisited on p. 26. A very ingenious solution is to be found in, "Mathematical Gems II", by Honsberger, pp. 16-18. Again, try to solve this for yourself, if you have not met with it before.
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I might mention that the Butterfly Problem is at least 100 years old, a proof by Mackay being provided in the Proceedings of the Edinburgh Society in that year. As for the others, I remember them from my own high school days, at the beginning of the century.
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ARBELOS
VOUlME
3, CtIAFmR 1
TilE BINOMIAL TllEOREM . ETC. At the risk of seeming to be too simple, we are nevertheless going to examine the Binomial Theorem. Surely, our readers know the expansion of (x + y)n. However, it appears that there are many who are not aware that the expansion holds for all n -natural numbers, fractions, negative numbers, even complex numbers. We discovered this while visiting classes and students recently. When n is a natural number, we know that the coefficients of the expansion of (x + y)n can be obtained from the Pascal Triangle. Also, we can use the "binomial coefficients", which we assume known to our readers. That is, (X+
y)n ~ xn+ (~}n-~ + (~}n-2y2 +.
.
However, when n is not a natural number, we can use Newton's form for the expansion -namely: (x + y)n = xn + n xn - 1y + n(n-l) xn-2y2 + n(n-l)(n-2) . xn - 3 y3 1
2!
3!
and so on. We hope the rule is clear. We can use this form of the Binomial Theorem to
..J1O, we write
compute roots. For example, to find it as (9 + 1) 1/2 and expand it to obtain
9 1/ 2 + 0/2) 9- 1/ 2 .1+ 0/2)(-1/2) 9-3 / 2 .12 + ... 1 2! and, on calculating each term, arrive at 3.0000 + .1666 - .0046 which is quite close.
-6
=
3.162
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ARBELOS
When the exponent is a negative integer, we have
e
Ie je Ie
e !e Je e e e e e e ,e e e
(1 + x)-n = I- n
Ie
e e
+ (-n) X + (-n)(-n-I) x 2 + ... 1
2!
and the coefficient of x r in this expansion is
t
Ie
Ie
je
VOWMll3, CrtArmR 1
(I) (-n)(-n-I)(-n-2)...(-n-r+l) = (-I)rt
+ r-l) r
rl
If we elect to write ( I ) as en), then we have r en) = (-1{ (n + r - 1). r r This is a useful result for combinatorics and for probability theory. Consider, for example, (1- ax)-I. Without using the Binomial Theorem, 'but just by division, we have _1_ = 1 + ax
I-ax
+ a 2x 2 + a 3 x 3 + ....
Similarly,
je
je Ie
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up to Let us multiply these series. Then the coefficient of x r will consist of all products a, b, c, ... , t taken r at a time, repetitions allowed. If we wish to have the number of terms in each such product, we have merely to let a, b, C , ... , equal 1. However, in this case, the product becomes (l-x)-n. The coefficient in this case is (-n). r -7
ARBELOS
VOWl'm 3, CItAFI1lK 1
That is, the number of combinations of n things taken r at a time, with repetition allowed, is
t+ r - 1). r
In using the Binomial Theorem for calculating purposes, one must be careful. When the exponent n is a natural number, the number of terms in the expansion is finite, and there is no trouble. However, when n is not a natural number, the number of terms is infinite and questions of convergence arise. For example, we have (l-x)-I=I+x+x 2 +x 3 +x 4 + .... Let x=2, and we get -I = 1+2 + 4 + 8 + ... , which is obviously not so. In fact, this expansion holds only when Ixl < 1. Even for a negative integral value for n, one gets differing expansions for (1 +xrnand (x+ 1r n . In finding .flO, we would have had some difficulty had we developed (l +9) 1/2. Try it and see. However, suppose we have (x+y)n where x>y. Then we may rewrite it in the form xn(l + 1.)n . x
We can then
expand the parenthetic term and then multiply by x n since 1. < 1. x Before continuing with the subject, we should note that we can use the Binomial Theorem with fractional exponent to compute any root of a number. The reader might be amused to compute the cube root of 123. -8
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Let us now examine, say, (x+y+z)n, where n is a natural number. We can "group" Y+z and expand (x + y + z)n. However, for (x+Y+z+ ... +w)n this would obviously be quite complicated. Hence we examine (x+y+z)(x+y+zl(x+y+z) ... (x+Y+z), where we have n factors. Now to obtain a term in the expansion, we select one object from each factor. Here, we are into combinatorics. For the term xaybzc, for example, we select ax's, by's and c z's. This can be done in nl al bl cl ways, where a + b + c = n. Thus, given (x+y+z)5: if we want the coefficient of the term x 2 yz 2, this will equal
51 =30, and the 211!21
term is 30x 2 yz 2. We have not been able to find any references to expansion of powers by the Multinomial Theorem except for the little we have presented here. Yet such expansions do exist, although they must be quite complicated. We can do something with a few simpler cases. For example, given (a+bx+cx 2 )6. Let us develop in terms of a, b, c. Then the term involving, say, a 3 b 1 c 2 , would have coefficient -2L 3!l!2!
60 ·108· x 5 .
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ARBeLOS
Vowm 3. CJ1AP'lllR 1
In the development above, we have found one term involving x 5 . To find all such terms will take a bit longer. Let us try. Suppose we want the coefficient of x 5 in (l + 2x + 3x 2 )6.
Rewrite this as (a + bx + cx 2 )6. The general term of the expansion of this in terms of a.b,c is
where r+s+t=6 and s+2t=5. We may have s= 1. t=2, r=3; or
s=3. t= 1. r=2; or
s=5, t=O. r= 1. Therefore the coefficient of x 5 will be
It might have been easier to use the Binomial Theorem, but we wanted to exhibit this result. If one wanted to find the coefficient of x 5 in (1 + 2x3x2 )1/2, the use of the Binomial Theorem would be strongly recommended. In a short article, we have had no opportunity to examine either the Binomial or Multinomial Theorems for complex exponents. We apologize.
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The Gamma Function
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This article can be considered as being just a continuation of our article on binomial coefficients, but it so interesting on its own as to merit a separate treatment. It seems that it struck Euler (of whom we have written before) that it might be possible (and fun) to construct an expression that would yield values for all factorials, not merely the positive integers. He found two - one an integral which we will not examine, and one as follows:
r(x)
j.
Ie
I
=
lim
nln x - I (n x(x+l)(x+2)...(x+n+l)
~
00).
How he derived the two expressions makes a fascinating story, which can be found by referring to the English translation of his works. We will work with the limit form, where we shall assume the limit (n ~ 00) understood for ease in typing. First, let us replace x by x+ 1 in the expression above. Then r(x + 1) = lim
I
x
n.n (x+l)(x+2)...(x+n)
=
lim
I x-I n.n nx x(x+l)(x+2)...(x+n-l)(x+n)
=
lim
n.n . ....K... x(x+l)(x+2)...(x+n-l) ~+1
I x-I
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=
Now, lim
we nl
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x r(x). can
easily
find
that
= I. Hence r(2) = Ir(I) = 1, -11
r(l)
equals
ARBELOS
VOWM!l3, CtfAPTIIR 1
r(3) = 2r(2) = 2!, r(4) = 3r(3) = 3!, etc.
Certainly, Euler's expression, called the Gamma
Function, does what he set out to have it do, for
any positive integer, r(n) = (n -I)!. We now manipulate the expression a bit. Let us replace x by 1 ~x. Then we have i(1-x) = lim
I
-x
n.n (1- x)(2 - x)(3 - x) ... (n - x)
•
• •
Multiply,
and we find that r(x)f'(1- x) is equal to
~n~!.!..!.n~!n~-_I
lim
_
2 x(1- x 2 )(4 - x 2 )(9 - x 2 )... (-n_-1 - x 2 )(n - x)
If we distribute the (n!)2 in the denominator, we get lim
._1_ 1 x2 x2 x 2) ... 1+ x
x(1- -)(1-)(1n
12 22 32
At this point, we recall that, in the Arbelos of January, 1984, on page 11, we found that sin 1tX = 1tx(1- £)(1- £)(1- £) ....
2
1
22
3
2
Finally, in this last expression, let x= 1.. Then
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2
r(1.) r(1.)= 1t, or r(1.) = 2 2 2
~.
That is, we have the
value equivalent to factorial 1., and it involves 1t. 2
Note:
The interested reader can now evaluate the area under the normal probability curve. -12
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ARBELOS
Vowm 3, CItAI"'mR 1
MANY CtlEERFUL fACTS We take the liberty of presenting three theorems which may be known to some readers, which have been used in solving problems in Olympiads and in the Putnam Exam, and which therefore merit presentation
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1
The first (in alphabetic order) is known as Beatty's Theorem. Simply stated, it asserts: Let a and ~ be two irrational numbers such that
l + 1 = 1. Now consider the sets Urn a)} and Un ~)},
ex ~ where m and n are natural numbers taking on the values 1,2,3.... Then every natural number will occur in just one of the two sets, with no repetition. We first met this theorem as a problem in Niven and Zuckerman's, Introduction to the Theory of Numbers, p.84, problem 23. It also appears in the USSR Olympiad Problem Book, p.26, problem 108, and there is a full discussion in Honsberger's Ingenuity in Mathematics, Volume 23 of the "New Mathematical Library", p.93 etc. The last two have proofs of the theorem. We suggest that the reader try to construct his or her own proof. Also, we suggest that the reader refer to one or more of the references.
It may be mentioned that Problem No.3 of the IMO for 1978 was easily solvable using Beatty's Theorem.
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VOWl'm
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The second theorem is known as Nelly's Theorem. We assume that the reader knows what a convex figure on the plane is . Let us call such a figure an oval. Then the theorem states: if each three ovals of a finite (or infinite) family of ovals have a common point, then all the ovals of the family have a common point. We first met this theorem in, Combinatorial Geometry in the Plane, by Nadwiger, Debrunner, and Klee. It is also in, Euclidean Geometry and Convexity, by Benson, p. 53, and there is a very fine exposition in Mathematical Time Exposures, by Schoenberg, Chapter 5. The first reference contains a number of allied theorems, including Sylvester's conjecture, If a finite set of points on a line is such that on the line determined by two of the points, there is always a third point of the set, then all the points lie on a line. It should be noted that this last dualizes to: If a finite set of lines is such that through the intersection point of any two of the lines, there passes a third line of the set, then all the lines are concurrent.
The reader will find several instances where Nelly's theorem has been used to solve problems on the Putnam exam. One such is Problem 3 of the Ninth Putnam Competition.
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The third theorem should be well-known, since it dates from 1899. Due to George Pick, it states: The area of a simple polygon whose vertices are lattice points is equal to the number of lattice points in the interior of the polygon plus half the number on the boundary minus 1. Or, K = i + lb -I. 2
We first met the theorem in, Mathematical Snapshots, by Steinhaus, p.94. An excellent development appears in Honsberger's, Ingenuity in Mathematics, pp. 27 - 30, and a short, snappy proof followed by examples in Coxeter's, Introduction to Geometry, p. 2<;>9. We recommend the proof, on p.211, of the theorem: If cevians be drawn to the trisection points of a triangle so as to enclose a triangle, the area of this triangle is one-seventh the .area of the entire triangle.
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ARBELOS
HOW TO CATCH A WE
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•
ARBeLOS
VoWPm 3, Ct1ArmR 1
UNDETERMINED COEffICIENTS The enterprising reader has undoubtedly met with undetermined coefficients, possibly in connection with partial fractions. We met them first at the start of the century, where they were used to develop series expressions for various functions. For example, we assume that: e X =A+Bx+Cx 2 +Dx 3 +Ex 4 + .... Let x=O, and we find that A= 1. Now differentiate the series, and we have: e X =B+2Cx+3Dx 2 +4Ex 3 .... Let x=O, and we find that B= 1. Differentiate the last series and we find: e X =2C+3x2Dx+4x3Ex 2 + .... Let x=O, and we find that C=.l!. 2
Similarly, D= 1-! 3
E= 1-!, etc. This gives us the usual 4
eX
2
.3
4
2!
3!
4!
= 1+~+1L+1L+1L+ ... 1
If you want to use calculus, there is an interesting development of Arctan x that you might try. Another application is to deriving Simpson's Rule. Suppose we are given a sequence of ordinates, Yj in terms of which we wish to develop a formula for the area under the curve joining the upper ends of this set of ordinates. We assume a rule in the form K=AYO+BYI+CY2 and we try to evaluate A, B, C. If there is such a formula, it should work when the curve is Y= 1 - that is, when Yo = YI = Y2 = 1. In this case, let the distance between ordinates be h. -16-
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Then the area is 2h=A+B+C. (see figure a)
y=l 1
Again, the formula should work when y=x. In this case, (see figure b) the area is 2h 2 =O+Bh+2Ch.
h
h (0)
Finally, the formula should work when y=x 2 . In this case (see Figure c), Yo=O, YI =h 2 'Y2=4h 2 . Hence we get 8h 3
3
y=x
= 0 + Bh 2 4ch 2 . h
Note: we remind you that we have already shown that the area
h
h
(b)
under the parabola is l. 3
that of the circumscribing rectangle. Solving the three equations A+B+C=2h Bh+2Ch=2h 2 Bh 2 +4Ch 2 =
8h 3 3
-
lh h
(c)
we arrive at A= h, B= 4h , C= h. That is, our formula 333
is K = h(yo + 4YI + Y2)' 3
This is Simpson's Rule. To
use it, divide the x-axis Into an even number of equal pieces, and find the area piece by piece. -17
ARBELOS
VOLUMl\ 3, ClIAI"mR 1
For a third application, consider a chemical reaction. We wish to write a balanced equation for this reaction. We know the valences of the elements involved (we must know something I), and will use the deep postulate: Whatever you put in will come out! We start with Aluminum (AI) and hydrochloric acid (HCl). We note that hydrogen comes off (H 2 ) and that aluminum chloride is left (A 1C 13 ), Write A(Al)+B(HCl)=C(AICI 3 )+0(H 2 ). We try to find A, B, C, O. Now for each element, we find: A=A for AI: for H:
A = C} B = 20
B=3A so that
for CI: B = 3C
C=A 0= 3A 2
Let A=2, and we have A=2, B=6, C=2, 0=3.
The
balanced equation is: 2AI+6HCI=2AICI 3 +3H 2 Let us try to balance the following reaction: ACu+BHN03=CCu(N03)2+0H20+ENO. Going by elements, we have (Cu) A=C; (H) B=20 (N) B=2C+E (0) 3B=6C+0+E. Solving in terms of E, we find A= 3E 2
B=4E
C= 3E
0=2E
2
E=E.
Now let E=2, to get integer coefficients, and we have: 3Cu + 8HN0 3 = 3 Cu(N0 3 )2 + 4H 20 + 2 NO. This will balance any equation without charges. -18-
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ARBELOS
••• •• • •• •• •• •• z. • •
••• ••• •• •
z, • >. • •• •• ••
,.
Vowm 3, CHAP'TIIR 1
CONJUGATE COORDINATE GEOMETRY
Weare all acquainted with p the diagram (due to Wessel) im g showing the complex number z=x + iy graphically. It is also
usual to use the facts that
r y x=r cos 8 and y=r sin 8 to write z=r(cos 8+i sin 8). It is also usual to write z = x - iy . "-_8 ---'-_r:-:e::..;:a=1 Then z= r(cos 8-i sin 8). 0 x It struck several early mathematicians, among them Gauss, that one could express x and y in terms of z and In that case, we would have Y = z-z .
2 2i One could then express a relation in x and y in the form of another relation in terms of z and z. In a number of cases, this procedure makes it easier to derive results in coordinate geometry. We might note, for example, that zz= r 2 . Let us start with the linear equation y=mx+b. We x
= z+z
substitute and find z-z 2i
,. ,.i.i.
•
we arrive at
= m z-z + b.
On solving this for
2
z=
I-mi z - 2bi . First, the coefficient of l+mi l+mi z is not a slope. If we note that, in the diagram above, m=tan 8, we find I-mi l+mi
= l-itanS = cosS-isinS l+itanS
cosS+isinS
Now cos 8+i sin 8 = e i8 , cos 8-i sin 8 = e i8 . Hence, in the
linear equation above, the coefficient of z is e-2i8. In some texts, this is called a clinant. Let us represent it by
M. -19
;
ARBELOS
VOWl"m 3, CtlAJ"mR 1
If we consider the segment or to represent a vector, which is often done, then, given two points z 1 and zz' then z 1+zz is the diagonal of the parallelogram of which z 1 and Zz are adjoining sides, and zZ - zl is the line segment joining the other vertices of that parallelogram. Moreover, every point on the line on z 1 and Zz may be written in the form z = zl + t(zz - zl) or z number.
= (1- t)zl + zz.
Here, t is a real
However, consider the determinant
This vanishes when Z=Z 1 and when z=zz' and is linear. Therefore this determinant, equated to zero, is also the equation of the line on zl and zz. By dint of a little algebraic manipulation, we obtain -
z = (Zt-Z2 )z + (Zt-Z2 - Z2Zt ) Zt- Z2 Zt- Z2 Comparing the two forms we have for the equation of a line, we may write this last z=Mz+B. Again, M is not a slope. When we solve M = (l-mi) for m we find m = 1. I-M. . (l+mi) , i l+M Now suppose we have intersecting lines with slopes m 1 and m z . Let the angle between the lines be (J). Then tan
(J)
m = (m2- t). (1+mtm2)
We will not show the
intermediate algebra, but just -20
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ARBELOS
show the result:
Vowm 3,
tan co = i
CtfAPTER
2
1
(MI-M ). (Ml+M 2)
From this, it
follows that the two lines are parallel if M1 = M2 , and will be perpendicular if MI
= -M2 .
We can always proceed as we have thus far -start with an equation in x and y, replace with z and and thus arrive at a rule in Gaussian coordinates. Unless we can find something new and useful, it does not seem practical to do so. We will try to proceed further.
z,
r--
~b
First, we detach ourselves B from the origin. Given a line segment with endpoints a and b, suppose we select a segment with end point on the origin and at (b-a). Then 0 u======P~_ this line and the line with endpoints a and b form a parallelogram. Hence the line or is equal and parallel to the line AB in the diagram. Next, consider two triangles whose Guassian coordinates are zl' z2' z3' and u l , u 2 ' u 3 ·
If we refer back to the u J - u 1
triangles with one vertex at the origin, we get the configuration shown in the diagram at the right. Now assume that these two triangles are similar. Then
is z2 - ZI = Ile I; U2 - ul
= kqe i9 2;
z3 - zl
= qe i(8 1+a)
u3 - ul
= kqe i(9 2+
Then (U2 - ul)(z3 - Zl) = kq2ei(91+92+
= (u3 -
ul)(z2 - Zl)'
ARBf:LOS
VOWMe
3, CI1AFmR 1
Because of our previous theorem (or lemma), it follows that this last equality holds for the two original triangles - with vertices at Zl' ZZ' Zy and at u l' u z ' u 3 respectively. Finally, if we expand (uZ-uI)(z3-zl)=(u3-uI)(zZ-zl)' it turns out that we zl can write the result very simply as Zz
ul Uz
~ = O.
z3
u3
1
The necessary and sufficient condition that two triangles be similar is that the determinant above be zero. We leave it to the reader to determine why, when a triangle with vertices Zl' Zz' z3 is equilateral, then Zl z2
Z2 z.3
z.3
Zl
~ = O.
We also remind the reader that a rotation about an angle of 8 about the origin can be expressed in the form t = e i8 . There are many results that can be useful in solving problems. For example, the midpoint of the line segment with endpoints zl and zz' is (zl +zz)/2; the centroid of the triangle with vertices at Zl' Zz' z3' is (zl +zz+z3)/3, etc. The reader might take time out to show that a necessary and sufficient condition that a triangle b e equi'1 a t eraI 'IS zl z + Zzz + z3 z = zlzZ + zZ z 3 + z3zl. -22
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i.,. i.
ARBeLOS
Vowm 3, CIIAYI1IR 1
B'
A
C'
b ~_---,a,,--
~
C
A'
In the diagram above, we have directly congruent triangles CBA', ACB', and BAC' constructed externally on the sides of triangle ABC. We wish to show that triangles ABC and A ' B' C' have the same centroid.
If need be, we can refer all points so that, for example, points A and C are equivalent to a line from the origin to the point a - c, and so on. Then a
= a' + keie(C -
b)
= b' + ke ie (a - c) c = c' + keie(b - a). b
On adding (and dividing by 3) we arrive at a+b+c _ a'+b'+c' 3 3
It must be admitted that this solution is short. As another problem, a challenge to the reader, prove that, given the quadratic equation z2 + az + b = 0 with its roots in the form of Gaussian coordinates, the origin and these roots will form the vertices of an equilateral triangle if a 2 =3b. -23
ARBELOS
VoWPm 3, CHAmlR 1
Thus far, we have hardly enough material to be able to attack complicated problems. We have, in addition to the clinant and some linear equations, a rule for determining when two triangles are similar. Let us look at our linear equation once more. We can write it (z-zl) = (ZI- Z 2). When we solve for Z, we get a (z-zl) (zl-z2) very useful result:
z = M(z - zI) + zi for a line with
clinant M and on the point zi' We will use this soon. The equation for a circle with center at the origin and radius r is obviously z. Z =r z . For a circle with center at . Zo
and
radius
r.
we
therefore
have:
(z - zo)(z - zo) = ~. Solving this for z we obtain 2
Z=Zo+_r_. z-zo
For an example using what we now have, let us start with a unit circle with center at the origin. Its equation is z. z = 1. In this circle, we place a triangle with vertices z I' zz' z.3' Then the centroid of this triangle is (zl+z2+ z3). We let zi +zz+z.3=p, so the centroid is at E. .3 .3 Now M = (z.3 - zZ)(z3 - z2) = (_1 - _1 )(z.3 - z2) which Z3
Z2
simplifies to ----=.l..-. Therefore, the equation of the (z2 Z 3) line on Zz and z.3 is Z = ----=-L(z - z2) + z2' (z2 z 3)
The equation of the line on z 1 perpendicular to side (zZz.3) is therefore Z = _I_(z - zI) + zi' When we solve to Z2 Z3
find the intersection of these two lines, we get -24
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VOLUME
3,
CHAP'l'eR
1
e
•e e e e e e e e e
e e
e
e e
e e
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e e e e
•e e •e e e
Z
=-1 ( Zl + Z2 + Z.3 2
e e e e
e
=-1 ( p 2
z
z2 3 ) (--)
zl
This locates the foot of the altitude from Z 1 to side (z2 z .3).
In the same manner, we can write the equations for two altitudes. Let these be Z
=
1
+ .l;
(Z2 Z 3)(Z-ZI)
Zl
Z
-
=
+ -.L. If we solve
1 (Z3 ZI)(Z-Z3)
Z2
these equations for z, we find their intersection at z=zl +z2+ z .3=P. Let us translate this. We have: the
centroid at p/2, the orthocenter at p and the circumcenter at o. Therefore these points lie on a line with the centroid 1. of the way between the .3
circumcenter and orthocenter. The line on these three points is called the Euler line. Finally, the midpoints of the segments joining the orthocenter to the vertices are called Euler points (honest!). Now we collect all our data: midpoints of the sides are (zl+z2) 2
(z2+ z 3)
2
(z3+ z l)
2
feet of the altitudes
(p_(~:2»
e
e e
z
(z2 - -3) ) or Z zl
(p_(Z~:3
» (p_(:;l»
2
2
2
Euler points
(P+zI)
(P+z2)
(P+z3)
2
2
2
-25
ARBf:LOS
VOWMII
3, CtlAI"mR 1
Consider now the circle
z=
1
"4 +p. z-£. 2
2
Its radius is half that of the circumcircle. Its center is at E. Therefore this center lies on the Euler line,
z
halfway between the orthocenter and circum center.
the
We invited the reader to check that the nine points just listed all lie on this circle. The circle is well known as the Nine-point Circle. Note: It is obvious that we had previous knowledge of all this, so we were able to locate the center and radius and thus write its equation. We conclude with a problem: Let isosceles right triangles be constructed on the exterior of a quadrilateral so that the sides are hypotenuses of the triangles. Prove that the line segments joining the opposite vertices of the right triangles are both equal and perpendicular.
R
t
,
L
!
h
••••
Q
B PR = QS PR ..1 QS
P
-26
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VOWI'm 3, CIfAFmR 1
OLYMPIAD PROBLEMS
I) Let a be a positive integer and let (an) be defined by: ao = 0, and, for n
=
L 2, 3, ... ,
e
a n+l = (an + I)a + (a + I)a n + 2~a(a + I)an(a n + I)
e
Show that for each positive integer n, an is a positive il1teger. (Canada)
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2) Given a circle K and a line 1 tangent to it at B.
From a point A on circle K, let a perpendicular
AP be constructed to 1 with P on I. The point
M is symmetric to P with respect to AB.
Determine the locus described by M as A
moves on K.
(Bulgaria)
3) Consider the expansion
(l+x+x 2 +x 3 +x 4 )496 =aO +alx+ ... +a1984x1984; a) Determine the greatest common divisor of the coefficients a 3 , a 8 , a 13 , ... , a 1983" b) Show that one has
10 340 < a 992 < 10 347 . (Romania)
4) Three of the roots of the equation
4
x - px 3 + qx 2 - rx + s = 0 are tanA, tanB, and tanC, where A,B,C are angles of a triangle. Determine the fourth root as a function only of p, q, rand s.
(USA) -27
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VOWMll3, CttAnu 1
-28
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:5, CIfAP11!R 2
arbelos
Produced For PreCollege Philomaths
,. :.
;.I. I. •
VOWMII
A
•
C
Q
The external tangents to the two circles intersect the internal tangent at P and Q. Prove PQ=AB.
1984-1985:
November, 1984
No. 2
-29
ARBELOS
VoWPm 3, CIfAmIR 2
PREFACE Mathematics is a live and Vital discipline, and it keeps on growing. There are new areas of study that were not there ten years ago. Also, the computer has made for an increase in the study of suitable algorithms. With this in mind, we cannot help noticing that the content of the problems on the International Mathematical Olympiad is limited to what we would call "traditional" materials. Other nations also have the same opinion, and we anticipate that future Olympiads will contain additional content. That is one reason for our inclusion of an article on matrices. We hope you will find it interesting, and useful as well. In our last issue, we included an article on applications of mathematics to chemistry. We have had no reaction to this article as yet. If you believe that applications of mathematics are usefuL please let us know. We have more. Once more, we feel that the important thing is to solve a problem in the most efficient way. If calculus will do it, then we should use calculus.
Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903 -30
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ie
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VOLUMe 3, CIW"IM 2
CAYLEY vs UAMILTON We assume that the reader knows about detenninants and how to evaluate them, and also has had sufficient contact with matrices to be able to: (a) add two matrices; (b) multiply a matrix by a number; (c) multiply two matrices. Of course, this would include knowing about the matrix all of whose elements consist of zeroes, which we will write (0) and of matrices with ones in the main diagonal and zeroes everywhere else, which we will write (I). As definitions go, (0) is called the additive identity and (I) the multiplicative identity. First, we will be dealing with square matrices whose related detenninants do not equal zero. Much of what we show will apply to matrices which are not square or whose detenninants equal zero or both.
First, the transpose of a matrix is obtained. by "reflecting"
it about the main diagonal. This will enterchange rows and columns. The transpose of the matrix (A) we will write (A)'. Recall that if the row and column containing the element aU be crossed out, what is left is called the minor of aU and written AU' In evaluating the detenninant, the sign of a minor depends on the position of aU' being + when i+j is even and -when i+j is odd.
au
Now if the row and column containing is crossed out, a matrix (A)U is left which is related to the minor of the determinant. In fact, it looks like this minor, but is multiplied by (-1 )i+j. We call this object the cofactor of aV' Finally, if we construct a matrix in which the element aU IS replaced by (A)U and then write -31
ARBELOS
Vowm:5, CI1AFI1lR 2
the transpose of this matrix, we obtain a new matrix, called the adjoint of (A). We write this adj(A). We exhibit our first useful result for the case of a 3x3 matrix, but point out that it is general. all Let(A) = a2I [ a3I
al2 a22
aI3J [All a23 adj(A) = Al2
A21 A22
A3lJ A32
a32
a33
A23
A33
Al3
For the product of these two matrices, we find that: det(A) (A).adj(A) = [
0
0
0 det(A)
J
o
0
det(A)
0
= det(A).(I)
Thus, (A). adj(A) = (I). Now we know that if the product det(A)
.
of two matrices equals (I), either is the inverse of the other. Thus, we have (A)-I = adj(A) • det(A)
Theoretically, therefore, if we are given the matrix equation (A)(x) = (b) (where (x) and (b) are matrices with one column (vector matrices)), we can find (A)-I, and solve for (x), thus: (A)-I. (A)(x) = (I)(x) = (A)-I(b) = (x). We have found that, on the whole, it hardly seems worth the trouble. We did have a student once who solved the equations
x + y = 13 x-y=3 with matrices. We were not amused. -32
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VOWl'm 3,
Ctwrr.R 2
At this point, we resist the urge to wander off along the many paths opened to us involving the properties of and operations on matrices, and aim for our goal. Matrices arose naturally in the theory of transformations. Among such transformations, there is one which leads to the matrix equation (A)(x) = A(X) . This transforms the vector (x) into A(X), which is a dilation. When we manage to write such an equation, we call (x) an eigenvector of (A) Now (A) (x) = A(X) may be written in the form (A - AI)(x) = (0) . Again we illustrate for a .3 x .3 matrix, though the result is general. The determinant of (A - AI) is all -
A
al2
a21
a22 - A
a31
a32
This is a function of A, of course, and we shall write it as f( A). We call f( A)= 0 the characteristic equation of (A), and its roots are called (usually) eigenvalues of (A). Note that the degree of the characteristic equation of (A) is (usually) of degree not greater than the order of the matrix. For the case above, the degree of f( A)= 0 is three, of course. Consider now (A - AI) . adj(A-AI) = I. det(A-AI)
That is, (A - AI) . adj(A - AI) = f(A) . (I) . -33
ARBELOS
VOWKII
3, Ctw1P.R 2
Expanding adj(A - AI) in terms of powers of A, we B o + B1A + B 2 A2 +
Bn_1An- 1 = adj(A - AI).
Assume that f(A)=aO+alA+
+an_lAn-l+An. We
have
shall equate corresponding powers of A in the product (A - AI)(BO + BIA + B2A2 + .. , + Bn_IAn - l ) = (aO + alA + a2A2 + ... + an_IAn-1 + An)(I), we have (A)(B o )
=ao(I)
(A) (B 1 )-(B o) =a 1 (I)
(A)(B 2 )-(B 1) =a 2 (1)
If we multiply the equalities above (in order) by (I), (A), (A)2, (A)3, •.. , (A)n,on the left, and then add, we find that the terms on the left cancel completely, and we are left with (0) = ao(I) + al(A) + a2(A)2 + ... (A)n. That is, a matrix satisfies its characteristic equation! We find this result surprising. We are told that when Cayley first suspected this, he told this to Sir Wm. Rowan Hamilton (inventor of quaternions), who didn't believe it at first.
·34·
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VOWM~3,CIfAl'TER2
At first sight, all this may be academic, since it appears that one must derive the characteristic equatio~. However, we present, witl10ut proof, our method. We start with a matrix, (A), which, for simplicity, has four rows and columns. Let (A) be: all
al2
al3
al4
a21
a22
a23
a24
a31
a32
a33
a34
a41
a42
a43
a44
and suppose the related characteristic equation is
3 4 f(A) = A - PIA + P2~ + P3 A + P4 . First the elements in the main diagonal make up what it called the "trace" of the matrix. We let
PI = all + a22 + a33 + a44· Next, we find the sum of all the 2x2 determinants whose main diagonals consist of elements from the trace. These are: all
al2
all
al3
all
al4
a22
a23
a21
a22
a31
a33
a41
a44
a32
a33
a22
a24
a33
a34
a42
a44
a43
a44
Their sum is P2.
There are four 3 x 3 determinants whose main diagonals consist of elements from the trace. The sum of these determinants is P3. Finally, there is one determinant whose main diagonal consists of a II' a 22 , a 33 , a44· This is actually the determinant allied with the matrix. Let this be P4. Then we have the coefficients of the characteristic equation. -35
ARBELOS
VOWI'm
3, CIiAmlR 2
We can show only a few applications of the Cayley Hamilton theorem. First, consider the matrix
(B) =
1 1 °OJ . ° ( 2
123
Its characteristic equation is f(A) = A3 - 6A + IIA - 6 and the eigenvalues are A= L 2, 3. Therefore, (B)3 - 6(B?
+ II(B) - 6(1) =
(0).
When we multiply by (B)-I, we derive (B)2 - 6(B) +
11(1) = 6(B)-I.
It is easy to compute here and determine that
(B)-I = (
~
-33 ° OJ x~
-2 -1 2
In the (we hope, unlikely) case where we wished to solve: x+ Y = 7 2y = 4 x+2y+3z=I8
we could rewrite it as
(Bm
= (::
J
and multiply on the left by (B)-I. For most large matrices, with many zeros as elements, special methods have been devised that make for more efficient machine programs. -36
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VOLUME
3,
CIfAP"mR
2
It is often useful to compute a high power of a matrix. This may occur with Markov processes, for example. We illustrate with a simple case. Let
A=(~ ~J.
Then
2
f(A)=A -4A+3,
and
the
eigenvalues are A= 1, 3. Now consider A8 , and suppose it is divided by f( A). We get a quotient and a remainder, which can only be of degree at most one in A. That is, A8 = Q . f(A) + aA + b.
Then
(A)8 = a(A) + b(I), since f(A)=O. In the characteristic equation, on substitution of the eigenvalues, we get
3 8 = 3a+ b l=a+b We easily solve these equations, arriving at a=3280 b= -3279. Then (A)8 = 3280(A) - 3279(1) ,from which (A)8 = (3281
3280
3280J 3281
We could have found (A) 100 almost as easily. Finally, suppose that we have defined the function
(e)A = (I) + (A) +
(A)2
21
+
(A)3
31
+....
Let
J.
1 (A) = (2 2 1
Then f(A) = A2 - 4A + 3 as before. Again, we divide Ae by f(A) and get a quotient and a remainder, at
most of degree one. That is, eA. = Q. f(A) + aA + b. Hence e 3 =3a+b and e=a+b. -37
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When we solve these equations, we find that 3
3
a = e -e and b = 3e-e .
2
2
Substitution in the matrix equation yields us e (A) =
~ (e 22 + 1
2
IJ .
e 2 2 e -1 e + 1
The subject of matrix functions, their properties and their values, is a growing area. Mathematicians are finding more problems whose solution is easier when treated by matrices. We end this short essay by presenting two very simple problems. These are: Let (A) be the 2x2 matrix we have used in the previous problem, and suppose we define sin(A) = (A) _ (A)3
31
+ (A)5
_ ...
51
Compute sin(A) for the matrix given.
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Next, assume that two persons are matching coins. One has three coins and the other has four coins. They play until one has lost all his money. What is the expected number of games that will be played? Incidentally, for the same 2x2 matrix (A), what .!
does (A)2 equal ?
It
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ARBELOS
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WEEK-ENDERS 1) Having dealt yourself an odd number of cards from an ordinary deck of cards, how many of them are from black suits if there Is a fifty-fifty chance of selecting two cards at random and finding that they are both black?
2) The regular polygon (ABC ... ) has side AB = 1".
The sum of the diagonals from vertex A to the other vertices is just about 50". If the vertices are labeled in alphabetic order, what is the "full" name of the angle with vertex A ?
3) A pole is being transported along an endless belt. When a man walks in the direction in which the pole is moving, he goes from one end of the pole to the other end in 140 steps. Walking in the opposite direction, he goes from one end to the other in 20 steps. If each of his steps is one yard in length, how long is the pole?
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A LOOK AT INEQUALITIES Inequalities seem to have interested mathematicians almost as much as has Number Theory. Many of those who contributed to one discipline have contributed to the other. For inequalities, as for Number Theory, there are many results that are simple to state and hard to prove . We have already written on Inequalities. Reference to the Arbelos for January, 1983, contains such rules as the Cauchy Inequality, Holder's Inequality, and the inequalities of Minkowski, the Triangle inequality and those involving the Arithmetic mean (A), the Geometric mean (G) and the Harmonic mean (H). A reference to a text like Analytic Inequalities, by Mitrinovic, will convince one that there must be thousands of inequalities, developed by hundreds of mathematicians. Obviously, therefore, the problem solver must rely on a very few of these (possibly no more than those mentioned above) together with native ability (or cunning?). First, note that the same inequality can appear in many forms, and it takes some skill to recognize these. Thus, from (a - b)2 ~ 0, (a, b real)
we can get
(a + b)2
~ 4ab,
(a+b)~.j;;b, (A+Q.)~2, A~G~H~O, and so on. 2
b
a
At this point, we will represent the sum
at
+ a2 + ... + an
It makes things easier to write. Thus, from L(ax + b)2 ~ 0,
we get x2(La2)+2x(Lab)+(Lb2)~O, so the discriminant or
(La 2 )(Lb2 ) ~ (Lab)2,
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by La, albI + ... anb n by Lab, and so on, hoping that we will not be misunderstood.
(Lab)2 - (La 2 )(Lb 2 ) ~ 0, Cauchy inequality.
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This is not as pretty as Cauchy's proof, but it will serve. -40
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There are some simple concepts that will help in proving (or deriving) an inequality. For example, let La be constant, and assume that a j and aj are not equal. Then, when we replace these two terms by
(a·+a·) I
2
J,
we note that
La
remains the same, but
(a·+a·) (a·+a·) I J x 1 J > aja , . That is, we have kept the J 2 2 sum unaltered, but we have increased the product
that
ala2" .a n . To summarize, if the sum of a number of terms is constant, the product of these terms will be a maximum when all the terms are equal. We note that, if the product of a number of terms is constant, the sum will be least when the terms are equal. (We leave this for the reader to prove.) For example, consider the expression
f + (3 + x)4(3 - x)3 . This will be a maximum when
is
a
g = (3+ x) 4( 3- X) 3 4 3 maximum. However,
4 . ( 3+ X) + 3 . ( 3- X) = 6, a con s tan t. 4
3
note
that
Hen c e the
product g (and with it, f) will be a maximum when (3+X) = (.3-X) or x =~. 4 3' 7
Just for fun, one might prove that, for a rectangle with fixed perimeter, the area will be a maximum when the rectangle is a square. Sometimes, resource can be had to algebra. Let f(x)=xn-nx+n-l (n an even integer). By Descartes' rule, this, as an equation has no more than two positive roots, and has no negative roots. It has a double root at x = I. Therefore f(x) ~ o. -41
ARBELOS
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a+b
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Sometimes, notions of convexity will help with an inequality. A function is convex when the line ACB joining two points on a curve lies above (or on) the curve.
B
a
VOWME
The curve tan x (0 ~ x < 90)is convex. Suppose we have to show tan (45+x) +tan (45 - x) exceeds 2. Then
b
2
tan(45+x)+tan(45-x) > tan 45 = I, and we are done. (of course, 2
the sharp reader will have noticed that this is an example of the inequality (J! + .Q.) > 2, shown on the first page of this article.) b a Jensen's Inequality (see diagram) states that, if f(x) is convex, then f(a)+f(b) ~ f( a+b). 2 2 How to attack a given inequality depends on the form of the inequality and the skill of the solver. To show ~ -t- ~"+ ... + an-l + ~ > n, one may use the Arithmetic a2 a3 an al Geometric means, and note that the product of the terms of the (.!l+ a2 +...+.!n.) series is 1. Then a2 a3 al > n
prove that the series 1+ multiply by
,J;;
~, and we have it. However, to
-k + }; + ··· + ~ > ,J;;
I
all one does is
and note that each fraction (except the last)
exceeds unity. And to show that J... + _1_ + _1_ + ... + _1_ < 1., one 23
33
43
n3
4
could use mathematical induction. We could continue in this manner to show that there are many and varied ways of solving a problem. We will close with finding the minimum value of f
= x2 -
12x + 40 . -42
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We could use calculus, of course, but, instead note that f=x 2 -12x+36+4=(x-6)2+ 4 , which is obviously a minimum when x=6. We know we have a minimum without getting a second derivative. Finally, let us apply what we have developed to the problem of finding maximum and minimum values for f=xy+yz+zx-2xyz. (x+y+z=l). (The original problem included the limits, to which we objected.) First, let us change f to a homogeneous function by multiplication by x+y+z=l. We have f = (x + Y + z)(xy + yz + zx - 2xyz, or f = xyz + x 2 (y + z) + y2(z + x) + z2(x + y). Now we were told that x, y, Z were non~negative. Therefore each term in the above equals or exceeds 0, and the lower limit is f> O. The limit is attained when x= I, y=z=O, for example. Now let us examine (l-x)(l-y)(l-z)=g. On multiplying, we find that g = (xy + yz + zx) - xyz. Hence g = f+xyz. Now the sum of the factors of g equals 2, a constant. Therefore g is a maximum when the three factors are equaL that is, when x = y = z = 1.. 3
But then we get .J!... ~ f + _1 , or f ~.L. Equality 27 27 27 To summarize, .L ~ f ~ O. 3 27 This problem was No.1 on the 1984 IMO. It should not have been. occurs when x = y
= z < 1..
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Let us try a problem involving conjugate coordinate geometry (see September, 1984 issue.) We are given f(z) == azz + bz + bz + c, where a, care real numbers and b is complex. We wish to show that, if f(z) ~ 0, then ac ~ bb. We can work with g(z) == zz + bz + bz + c since this means division by a. We shall show c ~ bb. Consider (z + b)(z + b) = O. This is a circle with center at b and radius 0 - actually a point. This yields zz + bz + bZ + bb = 0 or zz + bz + bz = -bb . Therefore g(z) = -bb + c ~ 0, whence
c ~ bb. When
we return to f(z), we have ac ~ bb. The suggested solution has one return to complex numbers, i.e. z=x+iy, b=m+in, etc. It is much longer than the solution above. We recommend that those interested in inequalities will do well to read Introduction to Inequalities, by Beckenbach and Bellman, Volume .3 of the New Math Library, and "Geometric Inequalities" by Kazarinoff, Volume 4 of the New Math Library. I
Problem: 4 Show that (a + b + c)4 < 27(a + b 4 where a, b,c are real numbers.
-44
+ c 4)
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BEWARE Of' THE OBVIOUSI
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Imagination is most important in problem solving, but there are occasions when one must beware of the obvious. The history of mathematics has many cases where fine mathematicians jumped to hasty conclusions \:Vhich were later proved erroneous. We wish at this point to exhibit two cases from official examinations involving thousand of secondary school students that came to our attention.
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The first (in time) involved the following question: The discriminant of a quadratic equation with integer coefficients equals 23. What is the nature of the roots? We know the expected answer; the roots are real, irrational an unequal. But wait! The discriminant has the form b 2 -4ac. now 4ac is divisible by 4 with no remainder. Also, if b is even and equal to 2n, b 2 =4n 2 which is divisible by 4 with no remainder. In this case, therefore the discriminant is divisible by 4. Also, if b is odd and equal to 2m+ 1, then b 2 equals 4m 2 +4m+ 1, which leaves one when divided by 4. In every case, therefore, either the discriminant is exactly divisible by 4 or leaves a remainder of 1 when divided by 4. However when 23 is divided by 4, the remainder is 3. It follows that a quadratic equation with integer coefficients can not have 23 as ~ discriminant. We hope this surprised some readers, at least. The second example is geometrical in content: ·45·
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2
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In the diagram, A-BCDE is a square pyramid whose lateral faces are equilateral triangles. K-FGH is a regular tetrahedron each of whose faces are congruent with the lateral faces of the square pyramid. If face FGH is made to coincide with face AED, how many faces has the resulting solid? Again the answer is obvious. Both solids h~ve a total of five plus four faces, two faces disappear when merged. This leaves seven faces for the solid that is produced. Right? Wrong!
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B D
C
Here is the solid we have created. Let each edge be denoted by s. Let M be the midpoint of edge AD. Then lines CM, EM and KM, being altitudes of their triangles, are perpendicular to AD at M,
-46
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E
C
~.J3
M
2
K
~.J3 2
Now all lines perpendicular to a line at the same point lie in a plane perpendicular to that line. Thus, MC, ME, MK lie in a plane. Using the Law of Cosines in triangles MCE and MKE, 2s 2 s2
= .J.4 s 2 + .J. s 2 - .J. s 2 cos 42
= .J. s 2 + .J. s 2 -
x.
.J. s 2 cos y.
442
Subtraction yields s2 = 3s 2 (cosy-cosx). 2
Addition yields 0 = cos y + cos x. But cos(x + y) = cos x cos y -sin x sin y. From cosy +cosx = 0 cosy - cos x = 2 3
we find
cos y=.1
and
3
sin y= 2..]2
3
cos x= _.1 and sinx= 2..]2 3 3 When we SUbstitute in the formula for cos (x+y), we find cos(x + y)= _1. -.§. 9
9
= -1,
which means
that x+y=180. CMK is a straight line lying in face ACDK, which is not two faces, but one! In the same manner, we find that ABEK is not two faces, but one. Finally, the solid really has five faces, not seven. -47
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H
G
Once one has a solution to a problem, one often sees more ways to look at it. In this case, suppose we start with a cube (see diagram above) and join the centers of the six faces properly. One derives a regular octahedron, the dual of the cube. This is easy to show. For example, AH is half diagonal HAF, HD is half of equal diagonal HG, AD is half of diagonal FG, etc. So that HADE is a regular tetrahedron. Similarly, A-BCDE is a square pyramid with lateral faces equilateral triangles. Now, since A, C, D, H lie on plane FGH, face ACDH is planar, and so on. Lest the reader assume that this sort of lapse in intuition versus correctness does not occur in the case of professional mathematicians we show two additional examples where intuition has been at fault even for experts. Our first involves the Japanese mathematician Kakeya. In 1917, he proposed the following problem: Let AB be a unit segment in a plane. It is required to move AB in the plane so that A and B exchange places and so that segment AB sweeps out the minimum area. -48
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There seems to be a minimal area involved. We can rotate AB through a semicircle about A and then slide it down. The area covered by AB is then
.n.. Or we can rotate AB 2
about its midpoint through an angle of 180 and slide it to the desired position. The area covered by AB is then .K
4
A
Now Kakeya noted that the hypocycloid of three cusps,
produced when a circle with radius L rolls without :3
slipping inside a circle with radius r (see diagram) has
the property that a tangent to one part of the curve and
cut off by the other two arcs is constant. Therefore a line
one unit long, slid along this hypocycloid , will end up
with endpoints reversed after one complete pass.
Moreover, we know (calculus) that the area of the
hypocycloid is equal to .K. 8
Since the tangent always
remains constant while being moved, Kakeya conjectured that this was the minimum sought. Therefore, there was astonishment and surprise when Besicovitch proved that there was no minimum area that is, whatever area the line sweeps out, it is possible to sweep a lesser area while reversing the direction of the line. For more on this problem, consult, Mathematical Time Exposures, by I.J. Schoenberg, published by the Mathematical Association of America. There is also a film showing how this can be done, prepared by the M.A.A.
·49·
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Back in 1803, the Italian mathematician Malfatti proposed the problem which, intuitively, reduced to constructing three circles in a given triangle so that the sum of the areas of the circles would be a maximum. Malfatti thought he had proved that all one had to do was to construct three triangles as in the diagram, each circle being tangent to two sides of tl1e triangle and also tangent to the other two circles (see the figure above).
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Then Steiner, probably the best geometer of the time, gave a construction in 1826, without proof. Hall supplied elegant proofs in 1856. In 1929 (note the time spreads I), Lob and Richmond showed that, in this case, intuition was wrong ..
Even for the equilateral triangle, computation showed that the inscribed circle, with two small circles squeezed in, as shown at the right, above, yielded a greater area than three equal circles, as shown at the left, above. -50
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Moreover, it is easy to see (now that our suspicions have been aroused) that, in an isosceles triangle with a very short base and long sides, three circles placed one above the other must yield a far greater area that the Malfatti construction could yield.
Thus, so far, the problem of Malfatti has not yet been solved. The solution seems to depend on the shape of the triangle involved. A similar problem involving placing four spheres within a tetrahedron so that the cut out a maximum volume is also unsolved. Since an affine projection produces an equilateral triangle, it might be fun to compare the areas in the cases of the two situations shown in the diagram on the previous page. A very nice description of Malfatti's problem may be found in, A Survey of Geometry, Vol. II, by Howard Eves, In fact, this is an excellent text on geometry.
Moral:
Don't trust on intuition, all the time!
Shortie: We know that a right triangle cannot have integer sides, but still be isosceles. However, a 3-4-5 right triangle is nearly isosceles, and so is a 20-21-29 right triangle. Find the next largest nearly isosceles right triangle.
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OLYMPIAD PROBLEMS
e
1.
Prove that for every natural number n, the number ?n) divides the least common multiple of the numbers
n
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1.2,... ,(2n -1),2n. Note: ?n)is a binomial coefficient.
n (Bulgaria)
2.
Suppose that al,a2,a3 that
(X-al)(x....:.a2)
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(X-a2n)+(-1)n-l(nf)2 =0 has
tit
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2n
(Canada) Determine all pairs of positive real numbers (a, b) with a
* 1 such that
loga b < loga+l(b + 1) . (U.S.A.)
4.
5.
e
,a2nare distinct integers such
an integer solution r. Show that r = a l +a 2 +...+a 2n
3.
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The circles C1(i = 1,2,3) are externally tangent to each other. The centers of the circles are at the vertices of triangle ABC. Find the centers of all circles C which intersect each of the circles Ci at endpoints of a diameter of Ci . (Finland) Prove that the product of five consecutive positive integers cannot be a perfect square. (Great Britain)
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PARTITIONS A partition of a positive integer n is a separation of n into positive integers whose sum is n. Thus, 1+2+3 is a partition of 6. In some cases, we require partitions to be distinct. Thus, 2+1+3 and 1+2+3 are not considered distinct and, in our study, one would be discarded.
Incidentally, in addition, we are given a set of integers and
seek to find their sum. In a partition, we are given the sum and seek Integers whose sum equals n. Thus, we consider partition to be the inverse of addition - not subtraction. First, if we allow non-distinct solutions, the number of partitions of the number n is easy to obtain. Assume that we . have produced a line of n dots. There are (n-l) spaces between them. We can either draw a separating line in any space or not. Therefore, the number of possible partitions is 2n-1. Let p(n) be defined as the number of ways in which the positive integer n can be written as the sum of positive integers. The first few p(n) are easy to evaluate. First, assume that p(O)= 1. Now p(I)=1. Next, 2=1+1 or 2, so p(2)= 2. Then, 3=1+1+1 or 1+2 or 3, so p(3)=3. By merely listing in this way, we find p(4)=S and p(S)=7, etc. To get a more efficient and useful way of evaluating p(n), we use generating functions (see Arbelos No.2, 1983-1984.) Note that (1- x)-l = 1 + x (1- x 2 )-1
+ x 2 + x.3 + x 4 + x 5 + ...
= 1+ x 2 + x 4 + x 6 + ...
(1- x.3)-1 = 1+ x.3
+ x 6 + ...
(1-
x 4 )-1 = 1+ x 4 + .
(1-
x 5 )-1 = 1+ x 5 + .. and so on. -53
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Then the coefficient of x n in the product of all these series equals p(n). Using the expressions given on the previous page, we multiply to get the first few terms of the product and find 2
4
5
II = I + x + 2x + 3x 3 + Sx + 7x ... from which we get p(O)=I, p(I)=I, p(2)=2, p(3)=3, p(4)=S, p(S)=7. We exhibit the partitions of 5 below: l I l l I , 1112, 122, 131,23, 14,5. Leave it to Euler (whom we have mentioned before) to derive a recursion formula for p(n). It looks formidable, but appearances are deceiving. p(n)
= ~(_I)i+lp(n -
.1 (3i 2 + i)) +
2 ~(_I)i+lp(n - .1 (3i 2 - i)). 2
From this formula, for the first 20 or so values of p(n), we have p(n)=p(n-I )+p(n-2)-p(n-5)-p(n-7)+ p(n-12) + p(n-IS)-.... Some theorems involving partitions are easy to derive. For example, let us denote a partition into unequal parts by U(n), a partition such that every part is odd by O(n), and one in which every part is even by E(n). Then
= (I + n)(1 + n 2 )(1 + n 3 ) ... O(n) = (1- n)-I(I_ n 3 )-I(I_ n 5 )-1 ...
U(n)
Now U(n)(I- n)(I- n 2 )... = (1- n 2 )(I- n 4 )(I- n 6 )... On dividing by (1- n)(I- n 2 )(I_ n 3 ) ... , we note that the factors containing even powers of n in both numerator and denominator cancel leaving the denominator equal to O(n). So we have: The number of partitions with no parts equal equals the number of partitions with all parts odd.
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For example, U(7)=7, 6+ 1, 5+2, 4+3, 1+2+4. 0(7)=7, 5+1+1, 3+3+1, 3+1+1+1, 1+1+1+1+1+1+1, and U(7)=O(7)=5. A partition can be represented in the form of an array of dots called a Ferrers graph. For example, we can represent the partition 1+2+4 by
• • • • • • • By reflecting this graph about the diagonal on the upper left hand, we obtain a second graph,
• • • • • • • From a partition with three parts (4,2,1) we have arrived at a partition in which the greatest part is three (3,2,1, 1). The theorem shown is thus: The number of partitions of n into m parts is equal to the number of partitions in which the greatest part is
m. The use of the Ferrers graph appears to be an example of "backsliding" to the days when figurate numbers were fashionable. Recall that the ancients arranged dots into triangles, squares, etc., to get the triangular numbers
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• •
and square numbers
•
.. ..
•
••• • • • etc,
...
or I, 4, 9, 16. Moreover, they did quite a lot with these figurate numbers. Merely by drawing a line, for example, they proved that the sum of two successive triangular numbers is a square number. -55
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Also, by adding a row and column (at the right and at the top) of every square number, one gets the next higher square number. The part added was called a "gnomon". Now, starting from just one dot and noting that the number of dots in each gnomon added is an odd number, one derives the fact that the sum of the odd numbers from one on equals a square, or 1+3=4, 1+3+5=9, 1+3+5+ 7= 16, etc. The rythagoreans did quite a lot of mathematics with these dot figures. . To continue with partitions: We define a perfect partition as being one such that, for a given n, every partition less than or equal to n can be derived in just one way. Thus, for p(7), we find that 1,1,1, 1, 1,1, 1 1, 1, 1,4 1,2,4 and 1,2,2,2 are perfect partitions. Note that, in each case, one can derive subsets whose sums are all the values from 1 through 7. One application of this result is that one has a set of values, for each partition, which yield all the values from 1 through 7. This solves the problem of finding weights with which one can find the weight of pieces weighing from one unit to seven units. Of these, the one using the least number of weights is the set 1, 2, 4. The theory of partitions has been applied in various ways. Among these are graph theory, the study of mathematical "trees", networks, as well as applications to chemistry and to physics.
It has been proved that p(5n + 4) =O(mod 5) and that p(7n + 5) =O(mod 7). Find a similar congruence for modulus 35. -56
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VOWME
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Another application of partitions is to the determination of the number of solutions of the linear Diophantine equation alxl + a2 x 2 + a3 x 3 + ... amx m = n. This involves finding the number of partitions of n into parts al,a2' ... a m . Sylvester called this the ~~denumerant" and wrote it it the form: n
D(n;al,a2, ... a m ) = ID(n;al,a2' ... am)t . The generating function for this would then be: (1- tal)-I(I_ t a2)-I(I_ t a3)-1 .... In this, we would find the coefficient of t n for a given n. Unfortunately, using this form involves a lot of careful counting, and it is easy for mistakes to occur. Euler used a recursion process, but this is still usually too complicated and subject to error. For example, let n=20, a l =2, a 2 =5. Then D(20 ;2,5)= (1- t 2 )-I(I_ t 5 )-I. When we expand we find that the coefficient of t 2 o=3. Hence there are three solutions. However, it seems more desirable to obtain these solutions, rather than merely the number. Suppose we wish to find out the number of ways of changing a dollar into dimes and quarters. We seek solutions, therefore, to the equation IOx+25y 100, or 2x+5y=20. Again, we are at D(20:2,5). In this case, we prefer to use the methods for solving Diophantine equations described in Arbelos, No.4, March, 1983. We find that x=5k-40 and y=20-2k. Positive solutions exist when 8 ~ k ~ 10. Not only do we know that there are three solutions, for k= 8, 9 and 10, but we know what these are. -57
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VoWPm 3, CIfAPTIlH 2
These are four quarters and no dimes, two quarters and five dimes, and no quarters and ten dimes. Next, how about changing a dollar using nickels, dimes and quarters? Now we have Sx+ 1Oy+2Sz= 100 to solve. Of course, we divide by S to get x+2y+Sz=20. It is simplest to consider five cases - z=O or z= 1, or z=2, or z=3, or Z=4.
When z=O, we have x + 2y = 20. Here, we can, by inspection, find eleven solutions. These are (20,0), (18, 1), (1 6,2), (14,3), (12,4), (1 0,S), (8 ,6 ), (6,7), (~, ,8 ), (2,9) and (0,10). When z= 1, we solve x+2y= IS, we easily find that there are eight solutions. These are (1S,O), (13,1), (11,2), (9,3), (7,4), (S,S), (3,6) and (1,7). There are eight solutions. When z=2, we must solve x+2y= 1O. There are six solutions. These are (10,0), (8,1),(6,2), (4,3), (2,4) and (O,S). When z=3, we solve x+2y=S. There are three solutions, namely (1,2), (3,1) and (S,O). Finally, when z=4, there is the one solution, namely (0,0). Counting, we find that there 29 solutions, and what is more, we know what these solutions are. There are variations, which can be taken care of individually. For example, in the above problem, suppose we are constrained to use all three kinds of coin ? What if we have only eight nickels? As a problem, using either the generating function or the method given above, show that the number of ways one may change a dollar into coins, using cents, nickels, dimes, quarters and half-dollars, is 292. Note:
For an excellent reference, consult Riordan's Combinatorial Analysis. ·58·
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Vowm 3, CriAr'Iu 2
KURSCliAK KORNER This column features those problems of the famous Hungarian Kurschak 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 MAA hdqrs); the problems (with brief summaries of their solutions) covering the years from 1966-1981 were translated into English by Professor Csarmaz of Hungary in 1982. Student readers are uninterrupted 4-hour well-written solutions submit their work to critical evaluation. 1/1939
also invited to set aside an period to compose complete, to the problems below, and to the address given below for a
Let aI' a z , b I' b z , c I and C z be real numbers for which alaz > 0, alcl ~ bIz, azcz ~ b z z Prove that (al+az)(cI+cz)~(bl+bz)z.
2/1939
What is the highest power of 2 which divides 2 n ! evenly?
3/1939
Assume that MBC is a given acute triangle and that C I,A I and B I are points on the semicircles constructed outwardly on the sides AB, BC and CA, respectively. Show that these points can be chosen so that ABI=AC I , BAI=BC I AND CAI=CB I . Dr. George Berzsenyi Rose-Hulman Institute of Technology Terre Haute, IN 47803 -59
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arbelos Produced For PreCollege Philomaths
On one side of line segment XV, similar triangles XYA, XYB, XYC, XYD, XYE, and XYF are constructed.
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Prove that points, A, B, C, 0, E and F line on the same cirlce.
1984-1985:
January, 1985
No. :3
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ARBeLOS
VOWPIE
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PREfACE We begin this issue with our first article to be submitted by a reader. We hope it will be followed by more such articles. We did have problems sent in by Dimitrios Vathis, of Greece. Now we have an article by Professor Szeckeres, of Australia.
Weare happy to have the Kurschak Problems. They show what sort of problems Hungarian students are expected to be able to do. They are very good practice for the International Mathematical Olympiad. Our Olympiad Problems come from the proposed problems sent in by various nations for possible inclusion in the official Olympiad, but which have not been selected. After all, each year, there are many problems proposed, and these are quite good in their own right.
Again, because the preparation of students varies across the country, we may include subject matter that some of you find too easy, too hard, or too familiar.
Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903
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TRANSFORMATIONS IN GEOMETRY Many geometrical problems lend themselves to be solved by a transformation. We can often achieve by a transformation that a difficult problem becomes easy or an unwieldy solution is replaced by an elegant one. There are many well-known geometrical transformations: we shall discuss a few simple ones. a) Parallel translation: The following problem will illuminate the method: P
Figure 1
In a given circle two chords, AB and CD are given. Find a point P on the circle such that if PA, PB cut CD in X respectively and Y, then XY =d, a given distance. Solution: Move the distance XY until X coincides with A, and Y is in a new position. Then (Fig. 1) yy 1 is parallel to AP, so LY 1YB = LAPB, which we know, being the angle subtended by the chord AB in the given circle, So Y is a point on a circular arc which subtends the angle APB to the chord Y 1B. This arc will cut CD in 2, 1 or 0 points, giving the required solution. b) Reflection: We are given a straight line, L called the axis of reflection. We replace every point, P, by its mirror image in I. Every configuration -63
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goes into a congruent- configuration, but is described in an opposite sense: clockwise changes into counterclockwise direction. We say the figures are indirectly or oppositely congruent. Problem: Given three concurrent lines, 11' 1 2 , 1 3 and a point A on 11' Find B on 1 2 and C on 1 3 such that the three given lines should be the internal angle bisectors of the MBC. Solution: If 1 2 bisects the LABC then the line BC is the reflection on 1 2 of the line AB. That means that if we construct A', the mirror image of A on 1 2 (Fig.2), then A' is a point on BC or its extension. Similarly, we construct A", the mirror image of A in the line 1 3 , Then A' A" is the line BC, and cuts 1 2 in B,I 3 inC.
A
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£1 Figure 2
c) Rotation: A figure is rotated around a given point by a given angle, thereby remaining congruent to itself. Problem: Given three parallel lines, 11' 1 2 , 1 3 , and a point A on 11' Find points B on 1 2 and C on 1 3 such that the triangle ABC is equilateral. -64
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C
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B
A Figure 3
Solution: Let B i run along 1 2 and imagine the points C i such that BiAC i is an equilateral triangle. Then it is easily seen that C i describes a line 1, some perpendicular distance from A as 1 2 , but rotated around A by 60° (Fig. 3). Then 1 intersects 1 3 in C and AC rotated by 60° meets 1 2 in B. There are two solutions, a clockwise and a counter clockwise equilateral triangle. d) Similarity Transformation: All linear dimensions of a figure are changed by the same proportion. We distinguish between central similarity and spiral similarity. Central similarity means that the lines connecting corresponding points in two similar figures all pass through a single point, which is called the center of similarity. It is not difficult to prove that if any two similar figures are drawn so that corresponding sides are parallel to each other, then it is a case of central similarity. Problem: Let ABCD be a quadrilateral with AB parallel to CD but BC intersecting AD in P. Show that the circumcircles of ~PCD and of ~PBA are tangent to each other (Fig.4). -65
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Vowm 3, CIW'I1IR 3
P
Figure 4
Solution: Let Q be the circimcenter of ~ PBA and Q' the circumcenter of ~ PCD. These two triangles are centrally symmetric with the center P; therefore the line joining the corresponding points Q and Q' also passes through P. But P is a common point of these two circumcircles, if P is on the central line, there cannot be any other common point Le., the circles are tangent to each other. Spiral similarity. If two similar figures do not have their corresponding sides parallel, then there exists a point in the plane which can serve as the center of some rotation so as to bring one of the figures into a position of central similarity to the other.
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Problem: Given two circles, with centers 1and 02 intersecting at points A and B. Construct a line through A, cutting the circles in P, resp. Q, such that PA = AQ.
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Solution: Draw the triangles 102B and PQB (Fig. 5). Then LAPB=L0 1 0 2 B and LAQB=L0 1 0 2 B. Therefore, the two triangles are similar for every position of the line PAQ. -66
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We want the position where BA is a median. Let x
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2) Given a river with parallel banks, distance d
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4) C I and C2 are two circles, tangent externally at point T. Through T we draw two lines, meet the circles at AI' A 2 , respectively at B I , B 2 . Let the incenter ~AI TB I be II that of +A2 TB 2 be 12 , Show that the line I 112 passes through T. Prof. Esther Szekeres 94 Warragal Road Turramurra, NSW 2074 Australia 1te versus e 1t A fairly well-known problem is to compare 1t e and e 1t and determine which is greater. We offer three solutions: a)
Arithmetic: 10g1te =elog1t=2.718x.4972=1.3514. 10ge 1t = 1t loge = 3. 142x.4243 = 1.3646.
e X >I+x.
Let
x = (1t -1).
Then
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Algebra:
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Calculus: Let f = x x and take logs. Let y=ln f. Then
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In= 1, or x=e. Hence e ~ > 1t;, whence e 1t > 1t e . There are other ways of solving this inequality
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KURSCfIAK KORNER This column features those problems of the famous Hungarian Kurschak 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 hdqrs,); the problems (with brief summaries of their solutions) covering the years 1956-1981 were translated into English by Prof. Csirmaz of Hungary in 1982. 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.
1/1996
Does there exist an equisided, spatial (in generaL non-planar) pentagon such that each 0
pair of its consecutive sides forms a 90 angle?
2/1996
Prove that for each positive integer n, the first n digits, following the decimal point in the expansion of (5 +
3/1996
.J26)n, are identical.
Are there infinite subsets, A and B, of the set of non-negative integers such that every positive integer is uniquely expressible as the sum of two numbers, one from the set A and the other one from the set B? Dr. George Berzsenyi Rose-Hulman Institute of Technology Terre Haute, IN 47803
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VOWPm:5. CrtArmR:5
fACTORING IS fUN Not too long ago, we were treated to a suggestion by an "educationist" that perhaps factoring of trinomial expressions should no longer be taught. Of course, no "educator" would make any such suggestion because, properly presented, factoring is fun. Even in arithmetic, factoring numbers could be fun. One can explain why a number is divisible by 2 (the last digit is even), by 4 ( the last two digits are divisible by 4), by 8 (the last three digits are divisible by 8) and so on. One can show that a number is divisible by 3 (the sum of the digits is divisible by 3) or by 9 (the sum of the digits is divisible by 9). Divisibility by 5 is easy - by 11 more complicated (the sum of digits in the odd places minus the sum of digits in the even places is divisible by 11. As for divisibility by 7, we present for your proof the following: remove the units digit, double it and subtract from the rest of the number. If this difference is divisible by 7, so is the number. We might also mention the number 1001 (=7xllxI3) and let you see how it might be used to test for divisibility by 7, 11, or 13. At any rate, the reader can see that we have rules for finding out when a number is divisible by all numbers from 1 through 16. Now we can either present numbers as problems in factoring or let the student find numbers to factor - say license numbers. We have done this last, and have found that it works very well. When it comes to factoring quadratic forms, we can use all the above. Consider, for example, x 2 +7x+12. We seek a factorization of 12 such that the (algebraic) sum equals 7. Even a weak student will discover 4 and 3 and write the result (x+4)(x+3). There is a sense of accomplishment. -70-
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Things get a bit harder when the numbers involved get larger. Far example, given the expression x 2 -19x+78, one must test the various ways of factoring 78, 2x39, 3x26, 6x13, and selecting the correct pair of factors, We get (x-6) (x-l 3). These concepts are still quite simple, but, properly presented, can be fun and a challenge, When it comes to harder possibilities, one can borrow from the Theory of Equations. Remember that, given the quadratic (x-a)(x-b)=O, then the quadratic (x-ka)(x-kb)=O has roots k times as large. That is, x 2 -k(a+b)x+k2 ab=O has factors as shown as well as roots k times as large. Let us examine 3x 2 +13x-30. Imagining that this is an equation, multiply the roots by 3. We get x 2 +13x-90=0. Factoring as above, we have (x+18)(x-5)=0. Now, since we multiplied by 3, we divide by 3, and we get: (x + 18)(x - 5) or (x+6)(3x-5). 3
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Remember that k can have any value, integral or fractional,
and that multiplying by k' = l is the same as dividing the k
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roots of the related equation. Thus, suppose we are given: x 2 -21x + 108
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This "method" works almost all the time, Try 12x2 -30x+ 12. Operating, we have x 2 -30x+ 144 then
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x 2 -7x+12, or
(x-3)(x-4). Now multiply by 3 and we have (x-9)(x-12)=x2 -21x+ 108.
x 2 -5x+4=(x-1)(x-4), next (x-6)(x-24)=(x- l )(x-2), 2
2x 2 -5x+2.
(2x-1)(x-2) or Comparing, we see that we neglected to remove the common factor at first, and we get (2x-1)(x-2)6. -71
The "system" has corrected us here. Of course, one should also have practice with factoring a difference of squares. These may seem easy, but one can always complicate. Thus, for example, we might be faced with x 2 -256 or (752)2_(352)2 or (x+y)2_ z 2 Most students find this form easy. When it comes to a sum or difference of cubes, however, the operation of factoring takes on a bit of complication. First, if we try to factor a sum of cubes, we get x 3 +y3=(x+y)(x 2_xy+y2). Why? Here we make use of the factor theorem and the remainder theorem. Remember that, given f(x), when we divide by (x-a), we get f(x)=Q(x-a)+R. Then (a) f(a)=R, and, if R=O, (x-a) is a factor. Let f(x,y)=x 3 +y3. Let y=-x, and f(x,-x)=O. Hence (x+Y) is a factor. Now we can divide to get the other factor. Now the other factor is a quadratic, so we can solve by the usual method. However, here the cube roots of unity enter. These are usually labelled 00 and 00 2 . That is, the cube roots of unity are 1, 00 and 00 2 . It is easy to show that oo3t+r = oor and that 1+00+00 2 =0. Now we can write x 3 + y3 = (x + y)(x + ooy)(x + oo2y ) , and, when we multiply the last to factors, x 3 +y3=(x+y)(x 2_xy+y2) Also, x 3_y 3=(x_y)(x 2 +xy+y2). For the would-be expert in factoring, it is useful to know that, if n is odd, xn+yn and xn_yn factor: if n is even, xn_yn factors. -72
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We are accustomed to think of identities as belonging to the domain of trigonometry. However, there are identities in algebra as well. Handling these may require nothing more than expansion, but they may be easier when the factor theorem is used, Suppose we are asked to prove that, given a triangle with sides a, b, c and semi-perimeter s, then (s-a? +(s-b)2 +(s-c? +s2 =a2 +b2 +c2 . This is easiest proved by mere expansion. However, consider a2 (b - c) + b 2 (c - a) + c 2 (a - b). This equals zero when b=a, so (a-b) is a factor. Also this equals zero when c=b, so (b-c) is a factor. Finally, it is zero when a=c, so (c-a) is also a factor. Therefore, a 2 (b - c) + b 2 (c - a) + c 2 (a - b) = K(a - b)(b - c)(c - a). Now each side in this equality has degree three, so K must be a constant. Let a= I, b=2, c=3, and we find (1)(-1)+(4)(2) + (9)(-l)=K(-I)(-1)(2). So K=-I, and a2 (b - c) + b 2 (c - a) + c 2 (a - b) + (a - b)(b - c)(c - a) = 0 identically . Finally, assume we have been asked to graph (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = y.
(c-a)(c-b) (a-b)(a-c) (b-c)(b-a) The left side is a quadratic form, so the graph should be a conic section. However, when x=a, y= I, when x=b, y= 1 and when x=c, y=l. Now a conic section cannot have three equal values for three different values. Hence the left side is an identity, always equal to unity, and the graph would have to be a line, namely y=l. -73
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What helps make factoring fun is the many ways one can attack a given expression to be factored. Consider, for instance, x 2 +2x-323. Rather than fish around for the factors of 323, let us instead complete the square for the first two terms, thus: x 2 + 2x + 1- 1- 323 + (x + I? - 324. Now factor: (x+l+18)(x+I-18)=(x+19)(x-17), and we are done. Try to use the same process on x 2 +5x-204. Let us apply some of what has been shown here to the following problem from the International Mathematical Olympiad for 1984: Find one pair of positive integers a, b such that (1) ab(a+b) is not divisible by 7; (2) (a + b? - a 7 - b 7 is divisible by 7 7 . Justify your answer. Note first that when a=O, the expression equals zero, so a is a factor. Next, when b=O, the expression equals zero, so b is a factor. Also, when (a+b)=O, the expression is also zero, so (a+b) is a factor. We have found that: (a+b?-a7 -b 7 =ab(a+b)P where P has degree four at most. Of course, one could expand (a+b) 7 using the Binomial Theorem and get the same result, but with greater effort and more work. Once this done, one notes that every binomial coefficient (except those for a 7 and b 7 , is divisible by 7, so 7 is also a factor. If we wished to do so, we could have noted that (a+b)7 =a 7 +b 7 (mod7), so that the whole expression is divisible by 7. We now have -74
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(a + b? - a 7 - b 7 = 7ab(a + b)Q. Here most of those who solved this problem divided to find Q and then managed to factor it. Instead, we let a =
(0
b and we find that the expression is
again zero. That is, (a-rob)is a factor. Also, let a=ro 2 b and again, the value of the expression is zero. Hence (a - ro 2 b) is also a factor. Multiply these two factors and make use of the properties of co, and we find that product 2 to be (a + ab + b 2 ). Now our expression has become (a + b)7 - a 7 - b 7 = 7ab(a + b)(a2 + ab + b 2 ). R, where R is a symmetric function of degree two. Therefore, (a+b)7 _a 7 _b 7 = 7ab(a + b)(a2 + ab + b 2 )(Aa2 + Bab + Ab2 )
First, let a=b= 1, and substitute. Then we find 2A+B=3.
Next, let a=2, b= 1, and we find 5A+2B=7. Solving for A and B, we have A=B= 1, and we have factored the expression completely. We have, in fact, (a+b)7 _a 7 _b 7 =7ab(a+b)(a2 +ab+b2 )2.
From the initial conditions, we see that (a2 + ab + b 2 )
must be divisible by 7 3 . There are ways of solving this Diophantine equation, but we shall guess, Let b= 1. We seek a solution for a2 + a + 1= 343. But, a2 + a - 342 = 0 is easy to factor. In fact, we get (x+19)(x-18)=O, so one pair of answers is a=18, b=l. As for justification, we didn't just guess an answer, as quite a few students did. We derived one. By varying the value of b, we can get other solutions. -75
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It
Finally, let us attempt the following problem: For what values of the natural number n is the number n 4 +4 n a prime number? This provides us with a chance to exhibit more in the way of factoring. We note that a
4
4 2 2 +a b +b
= a 4 +2a2 b2
+ b 4 - a 2 b 2 , which is a
difference of squares, (a2 + b 2
? - a2 b 2 . Thus,
we
may write (a2 + b 2 + ab)(a2 + b 2 - ab). For example, x
4
+ 3x
2
+4=x
4
+ 4x
2
+4- x
2
=
(x 2 +2)2 _x 2 =(x2 +x+2)(x2 -x+2). Now for the problem. First, we note that, for n=O, we have 0 4 + 4 0 = 1. Since 1 is not considered to be a prime, n cannot equal zero. Next, suppose that n is an even integer (here we are introducing the concept of parity). Then both n 4 and 4 n are divisible by 16, and, in that case, our expression is not a prime number. We are left with n an odd number, which we set equal to 2t+ 1. In this case, we take the liberty of writing our expression as n 4 +4 2t + 1 . We can factor as before, now, where upon we find that (n 4 + 2 . (2)2t + 4(22t)2 - 4. (2)2t = (n 2 + 2(2)2t + 2(2 t )(n 2 + 2(2)2t - 2(2)t). If this is to be a prime, one of these factors must equal 1, and it can only be n 2 +2(2)2t -2(2 t ). This occurs only when t=O, or n= I, from which 14 +4 1 = 1+4=5, and we are done. -76
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The Cubic Equation
In discussing the cubic, it may be that we will have
occasion to introduce subject matter familiar to some of our readers. Our experience has been that much of it will be new to others. So much of the old mathematics has been replaced by new or different subjects that we believe the article will be useful. We will work with equations no higher than the cubic. We hope the reader will recognize when some result holds for equations of any degree. Consider the cubic z3+pz2+qz+r=O. Its coefficients and its roots may be real or complex. Let these roots be r I' r 2, and r 3 . Then we know (see Arbelos, Mar. 1983) that the equation may be written as (z -li)(z - r2)(z - r3) = O. Also, Ii + r2 + r3 = -p, Iir2 + r2 r3 + r3 rI = q, rI r2r3 = -r. Now let us construct a new equation, whose roots are to be k times those of our given cubic. Then (z - kli)(z - kr2)(z - kr3) = O. By using the above expressions for p, q, r, or by multiplying, we find z3+kpz+k2 qz+k3r=O. This is true for all values of k. When k> I, for example, we are, in effect enlarging the graph magnification, so to speak. When O
= (x .... h)3 + P(x -
h? + Q(x - h) + R. -77
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However, note that both equations equal zero for the same values of the roots, r l' r2' r.3" Therefore, the two forms f(x) and g(y) must be identical.
If we can find the values of P, Q and R, we can then write an equation whose roots will be h less than those of the given equation, f(x). But this is easy - all one has to do is to divide f(x) by (x-h), and the remainders will produce the desired coefficients. Thus, divide (x-h)3+p(x-h)2+Q(x-h)+R by (x-h). We have a quotient of (x-h)2+p(x-h)+Q and a remainder of R. Now divide the quotient, (x-h)2+p(x-h)+Q by (x-h). The remainder Q and the new quotient is (x-h) +P. Divide again by (x-h) and we have remainder P. The point is that we get the same remainders when we divide f(x) by (x-h). The division is most easily done by continued synthetic division, thus: Let us start with f(x) = x.3 - 9x2 + 2.3x - 15 = 0, and derive the equation whose roots are less by 2. The work looks like this:
1 1 -
9
+
23
151~
2
-
14
18
9 -10
3
7
2 1 -
5
-
1
2 1 -
3
and the reduced equation is x.3 - .3x2 - x + .3 = O. This may well be easier to solve than the original. -78-
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In fact, it factors easily to (x+I)(x-I)(x-3)=O, whose roots are -I, + I, +3, Therefore the roots of the original equation must be + I, +3 and +5. One operation tl1at will be quite useful is the removal of the second term of an equation. For the cubic x 3 +px 2 +qx+r=O, we need only reduce each root by one-third the sum of the roots - that is, by -P, from the equation we have just worked 3 on, this gives us: 1 - 9 + 23 - 15 ~
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e e
1 1
e
e
•
-
6
+
3
-
01 -
18
:1
+ 15 0
4
and the reduced equation is now x 3 -4x=O, which is simple to solve. One should notice that this operation, in effect, moves the related graph to the left, when we decrease roots, to the right, when we increase the roots; in effect, we can now raise or lower the graph, expand it or contract it, and move it left or right, as we will. Just for fun, let us remove the second term of the quadratic ax 2 +bx+c=O. The work is: a
a
e
e e e
-
3
b
c
-b
_b 2
2
4a (4ac-b 2 ) 4a
Q
2 -b a
2 0
so we have ax2 + (4ac-b 4a
2 )
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= O.
This becomes
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2 x2 = b -4ac 4a or x = ±
(~b2-4aC).
Now add back the
2a
subtracted, and x =
-b we 2a
(-b±~b2-4ac) .
2a There is a way (called a Tschirnhausen Transformation), for removing the second and third terms from a cubic, but we will not use it. The reader may wish to look it up. We add, for future use, the following: Suppose the cubic az 3 +bz+c=O has a rational root,
p
q
the
fraction being in lowest terms. Then we get ap3+bpq2+cq3=O. Try to divide this by p, and we see that each term divides by p except the last. Since p cannot divide q3 it must divide c. Similarly, when we try to divide by q, we deduce that q divides a. This may be handy since it limits the number of trials we may have to make to discover a root. For example, the equation 4x 3-11x 2 +25x-7=O can only have ±7, ± 7, ± 7 , ± L ±.l, ±.l as possible rational
2
4
2
4
roots. A quick check shows that this equation has no rational roots. Next, we state without proof, the following very useful theorem: If a cubic equation has a root of the form p
+ q,J;., where r is not a perfect square,
then it has a second root equal to p - q,J;.. In the case where r=-I, we note that if the equation has one complex root, then its conjugate is also a root. We are now almost ready to begin. We first derive -80-
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the three cube roots of unity. For this purpose, we shall use De Moivre's Theorem, which we state without proof. Let the complex number a + bi be written in the form r(cos8+isin8) ,where
r=~a2 + b 2 ,
and tan S = b.
a
Then [r(COSS + fSinS)f
= rn(cosnS + isinnS).
Here
r
is a real number and r n is positive. Consider now the cubic z3= 1. We may write this · 0° ,or z 3 = cos 0°· + I sin
e e e e e e e e
z=cosI20° +isinI20°, or z=cos240° +isin240°. Another value possible for 8 will result in repetition of one of these three values .
It
It is usual to ascribe a special symbol to
e
e
cos 120° +isinI20°, namely co. Then
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z3 = cos 360° + i sin 360° , z3 = cos720° + isin720° .
or
Here, n= 1-, so that 3
z = cos 0 ° + i sin 0,
0
r
cos240° + isin240° will equal 0)2. Several interesting conditions arise. First, 1 + (0 + (02
= O.
Next (J) 3= 1. Also (0 and (02 are squares (and square roots) of each other. If the roots are graphed (in an Argand diagram) they form the radii of an equilateral triangle. In fact, the roots of zn=k will lie, like the spokes of a wheel, at the vertices of a regular polygon. -81
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Breather: Evaluate 1+ rot + ro 2t ,t a positive integer. Of course, we can find the cube roots of unity by fa c tor in g, t h us: z3 - 1 = (z - l)(z2 + z + 1) , etc. However, DeMoivre's Theorem can be used for any equation, regardless of the power involved. Let us turn our attention to the general cubic. We agree to remove the second term, and write our cubic as z3 + 3Hz + G = 0, following custom. We return to De Moivre's Theorem. From it, we have cos2e + isin = (cose + isin e?
= cos2 e = sin 2 e + i(2sin e· cose),
from which we get two formulas at once - for the sine of double an angle and for the cosine of double an angle.
In the same manner, we write (cose + isin e)3 = (cos 3 e - 3 cose· sin 2 e) + i(3sin e· cos2 e - sin 3 e), from which we find cos3e = 4cos 3 e - 3cose. Of course, we also derive a formula for sin 3e, but we will not need it now. Now cos3e
= 4cos3 e -
3cose
= 4 cos3 (e + 120° ) - 3cos(e + 120° ) = 4 cos3 (e + 240 ° ) - 3 cos(e + 240 ° ). Therefore the roots of the cubic 4z3-3z= cos3e are
cose, cos(e+120°), and cos(e+240o). We compare this with x 3 +3Hx+G =0. Let x=kz, and -82
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we have k 3 z 3 +3Hkz+G=0. If these two equations are to have the same roots, their coefficients must be proportional. Therefore, we must have
ll. = k3 From cos38=
these,
-3 = (-3cos38) (3Hk) G
we
h.
easily
find
k =
2~-H
,
2H -H Obviously, we will have trouble when H is positive, so we postulate that H must be non-positive. But now, we can find cos38, hence cos8, then cos(8+1200) and finally cos(8+2400) Thus, the roots of our original equation equal these three cosines, each multiplied by k. Let us apply what we have to the solution of x 3 -12x-12=0. First, 3H=-12, H=-4, G=-12. We will just go through the required steps, hoping that the reader will follow them. So - let z=kx;
k 3 z 3 -12kz-12=0 also 4z 3 -3z-cos 3 cos 8=0. 3
k = -4k = -12 4 1 cos 38 Let k=4. Then cos38= 3 Then 30= 41°24'. We take
4
8 = 14° ,8 + 120° = 134°,8 + 240° = 254° .
cosl4° =.9703,cosI34° =-.6947,cos254° =-.2756. Finally, multiply each of these by k =-4, and the roots are -3.8812, +2.7788, +1.1024. -83
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This method will not always work. We have shown why H has to be negative. It also happens that the discriminant of this cubic is 0 2 + 4H3, and this discriminant must be negative if the roots are to be real. There is another purely algebraic method we can use, called Cardan's method. Where one system will not work, the other will. The history of the cubic solution and of De Moivre are interesting. First, Card an apparently stole the method from Tartaglia, who may have stolen it from Ferraro. Next, there is a story that De Moivre noticed that he appeared to need fifteen minutes more each day for sleep at one time. This increase kept on until, when he had reached the twenty-four period, he just died. We do not vouch for either story - we just pass them along. For more such stories, why not read a history of mathematics-say, the History of Mathematics, by Howard Eves?
-84
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MANY CHEERFUL fACTS We have already written about the Dirichlet Principle, a special case of Ramsey's Theorem, also called the Pigeonhole Principle. Less well known is Fubini's Principle. Simply stated, it asserts: When you count a set in two different ways,you will get the same result. For example, if you count the number of students in a classroom, first by counting the number in each row, and then by counting across, the results should be the same. A very pretty problem using Fubini can be found as Problem #2 of the IMO for 1977. We shall use it in a few theorems in Number Theory. We have already referred to the residue system of a set of integers (mod m) . Euler invented the "reduced" residue system (mod m). This consists of those integers from 1 to m which whose remainders, when divided by m, are relatively prime to m. Thus, the reduced residues for 10 would be 1, 5, 7, 9, for example, and these are four in number. Euler used the symbol and wrote (10)=4. As further examples, (1)= 1, (2)= 1, (3)=2, (4)=2, and so on. First, note that, if p is prime, (p)=p-l. Also, (p2)=p2_p , and, in general, (pi)=pi_pi-l. Now let us write
L $(d) for the sum of (d), where
dim
d divides m.
Then
L (d) = 1+ (p) + (p2) +
dlpi
(p3) + ... (pi) = 1+ (p _ 1) + (p2 _ p) + (p3 _ p2) +
... + (pi _ pi -1) = pi . We have a telescoping series. In general, L
(d) =
n.
din
Now let m and n be relatively prime numbers. We wish to show that (m) (n)= (mn). There are many ways of proving this - using Farey -85
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sequences, the Chinese Remainder Theorem, congruencies, etc. We shall try using Fubini's Principle. We write all the numbers from I through mn in a rectangular array of m x n numbers, with n rows and m columns. I
2
3
• •
m
m+1
m+2
m+3
••
2m
••
3m
2m+1 (n -I)m + I
2m+2 2m+3
• •
• •
• •
•
nm
The number of divisors in this array relatively prime to mn equals (mn). Now the number of terms relatively prime to m in the top row is (m). If this holds for any column it holds for all the terms in that column. We may discard the other columns, if we like. Also, in any column, there are (n) terms relatively prime to n. This holds for all columns. If we count, we find that there are (m) (n) terms left, and this must equal (mn). . a(l) a(2) a(i) Now, given the number N = PI P2 ... Pi where we have different primes, then the (N) will equal the product of the individual (pt). From this it is easy to derive Euler's formula (N) = N(I- .l)(I_~)... (1- .l). PI P2 Pi Incidentally, we can also show that, if a and mare relatively prime, then a(m) == I(modm). This is usually simpler to use than is Fermat's Theorem, -86
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as aP=a (mod p)
VOImm
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(p prime).
Incidentally, if N = pap b ... p t, then the sum of the 1 2 i divisors of N may be found by using the generating function
a b t a(N) = (l + PI + ... + P )(1 + P2 + '" + P )... (1 + Pi + ... + P ) 1 2 i The number of these divisors is seen to equal 't(N) = (l + a)(1 + b) ... (1 + t). Of course, we could have used our special notation to define a(n) 't(n)
= L din d
and
= L din 1.
A function like (m), where (mn)= (m) (n)when m and n are relatively prime is called a multiplicative function. Under the same restriction, it happens that (n) and 't(n) are also multiplicative functions. The reader might try to prove that, when m and n are relatively prime, then (mn)= (m) (n) and that 't(mn)= 't(m) 't(n). There are some pretty problems that are readily solved by using these functions. For example: (a) Find all solutions (if any) of (n)=IO. (b) Compute (985) (c) If 't(n) is odd, then n is a perfect square. In a future issue, we hope to introduce the Mobius function Il(m) and apply it to multiplicative functions. ·87·
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Vowm 3, CIfAYI"r.R 3
OLYMPIAD PROBLEMS I) Determine all real values of the parameter a for
which the equation
°
4 3 2
16x - ax + (2a + 17)x - ax + 16 = has exactly four distinct real roots which form a geometric progression. (Bulgaria)
2) Find the point inside the triangle ABC for which BC + CA + AB PD PE PF is a minimum, where PD, PE, PF are the perpendiculars from P to BC, CA, AB respectively. (Gt. Britain)
3) Prove that if a, b, C are integers and a+b+c=O, then 2(a 4 +b 4 +c 4 ) is the square of an integer. (USSR)
4) 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)
5) Let the set {I, 2, ... , 49} be divided into three subsets. Prove that at least one of these subsets contains three distinct integers a, b, c such that a+b=c. (Mongolia) -88
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Produced For PreCollege Philomaths
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A .,.---
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D
B Given Triangle ABC: Construct line segment DE so that BD=DE=EC.
1984-1985:
March, 1985
No.4
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Vowrom 3, CHMI'U 4
PREFACE It is our policy to include, in each issue of Arbelos, some mathematical morsels that are new to our readers. Considering the great variety in the preparation of readers, this is not an easy task. In addition to the Arbelos, we take every opportunity to visit schools to see what is being done and what might possibly make for an article. In one school we visited, we found that the students had been taught the theorem of Menelaus, but not Ceva's Theorem. The teachers had made their decision of what would be more useful. We disagreed and presented the theorem, with proof, to the class. In another class in another schooL the students had no idea that the binomial theorem was correct for any real exponent. All they knew was the Pascal Triangle. Moreover, they had no idea that there might be a multinomial theorem. We found two trends that were somewhat disturbing. One was the rush toward introducing calculus to. students unprepared for it, as a result making the whole development a matter of rote. The other was what we consider an undue dependence on the use of hand calculators and computers, at the expense of understanding. We once attended a lecture on teaching algebra by hand calculator at Columbia. The lecturer was most enthusiastic about the results. One of the stories he told about the results was that about one student who got all the right answers. When asked how he happened to be so good, he answered, "It's only buttons man - only buttons". And there was the boy on TV who was at a computer. He told the interviewer that he had studied the computer for six years, and that he was four and a half years old. So much for his understanding of numbers.
To go into more advanced study of any kind, first prepare
yourself by learning the preliminaries first.
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fROM OUR READERS The letters we receive from some of our readers are enlightening and informative. We would like to get more. Mark Kantrowitz has been a good correspondent to date. We can see the improvement in his problem solving ability. His solutions to the problems from the November issue of Arbelos were quite good . We wish him success on the Westinghouse. We have received, for the first tine, solutions to some of the problems that are worth mentioning. C. Kenneth Fan submitted two good solutions. David Stephenson did nicely with his solutions, and
Japheth Wood also did well. We hope to hear further
from them and others as well. We read (in the Mathematical Digest for October, 1984), an analysis by Dimitris Vathis of a paragraph from von Daniken's "Signs of the Gods." He clearly pointed out where von Daniken had confused names and dates, and ascribed abilities that did not exist in some of the people he named. There are probably still some people who believe in von Daniken just as there those who accept the "explanations" that apply to the Bermuda Triangle, for example. They prefer to accept what is asserted instead of testing for accuracy. Just one note - those who wish to improve their knowledge in any field should do a lot of reading, examine what they read critically, and draw their own conclusions from what they read. We ourselves like to read old texts in mathematics, for example. For instance, we have an old (1890) arithmetic book used in the eighth grade then, containing problems that would glaze most people's eyes. It's fun! -91 ..
ARBeLOS
Vowm 3, CtfAFIM 4
THE PARABOLA AND THE RIGHT ANGLE by Isaac Ji KUO, 7th grade MCKinley Middle Magnet School Baton Rouge, LA
One day I looked at a corner. Hmmm. I looked at it again. I really began to wonder: how would it look if those walls were mirrors? Pretty funny, I guessed. Well, not really. I began painting a picture in my mind. This picture is shown in Fig. 1.
Fig. 2
Fig. 1
"That reminds me about a parabola!" I said right aloud. (Good thing no one heard me). OK, so I had a pretty wild imagination. But I still had somewhat of a point. Think of it this way: a ray of light, which is perpendicular to directrix, hits a parabolic mirror and goes right through the focus. It hits the mirror again and goes away parallel to its original path. "So what ?", you may be thinking. Well, Fig. 2 shows the resultant picture in my head. Amazing, I thought in my mind, when a chord passes through the focus in a parabola. the tangents of its endpoints are -92
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perpendicular!! I was quite excited about L~ discovery. So I went to Dad. I asked him if he COUlL prove it. After almost no time, he proved it. He said he used CALCULUS. I did not know any calculus at the time. When I gave him my proof, he was quite surprised. In fact, so surprised that he asked me to write this article. And now, for anyone wondering how I prove.d it all, here it is:
e
e
e
e e e
e
e e e e e
e e
e e e
e e e
e e e e e e e e
e
1.
2x+A+2y+B= 360°
2.
A+B= 180°
3.
Hence 2x+2y= 180° and so x+y= 90° and so C= 90°
Note: The article above was sent to us by Professor Hui-Hsiung Kuo, who has every reason to be proud of it. The boy's teacher had suggested having the Arbelos publish it! This makes us proud! We have attained the status of a "prestigious" journal! If any reader has an idea of his own that he or she is willing to submit for publication, we will do so after due consultation with a referee, as do all other IJprestigious" journals. -93
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MANY CHEERFUL FACTS Educationists are always at work. Recently, one has suggested that factoring of quadratic expressions no longer be taught. We happen to believe that factoring is fun, and that it would be wrong to keep students from the enjoyment it offers. The history of mathematics has many examples of great mathematicians relaxing with factoring. Thus, Gauss relates that he used to take time out to calculate a "chiliad" of prime numbers. Someone once sent Fermat a huge number for him to factor, if possible, and, by return maiL he sent back the factors of the number. And Euler showed that 2 32 _1 was composite. Factoring offers a lot. First, let us start with numbers and their factors. We all know that a number is divisible by 2 when the units digit is divisible by 2. Most of us know that a number is divisible by 4 when the number formed by the last two digits is divisible by 4. Similarly, a number is divisible by 8 when the number formed by the last three digits is divisible by 8, (etc.). We do not have to mention when a number is divisible by 5 or 10. Also, we know that a number is divisible by .3 when the sum of its digits is divisible by .3, and by 9 when the sum of the digits is divisible by 9. By combining these results, we are able to find out when a number is divisible by 2, .3, 4, 5, 6, 8, 9 and 10. Continuing, a number is divisible by 11 when the difference between the sum of its digits in the odd places and those in the even places is divisible by 11. We offer without proof a method for determining when a number is divisible by 7. Let that number be lOt + u. Subtract 2u from t. If the result is divisible by 7, so is the original number. One can practice this sort of factoring by trying to factor license numbers on passing cars for speed and accuracy. -94
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As something less elementary, consider the number 100 1. This is (obviously) divisible by 11, and (less obviously) divisible by 7 and 13. If this number be subtracted from (or added to) a given number, and if the result is divisible by 7 or 11 or 13, so is the original number. As an example, take 1183 we find that 1183-1001=182, and that 18-4=14. Hence 1183 is divisible by 7. It happens that 182 is also
divisible by 13, so that 1183 is divisible by 13. We
can now find factors for all numbers whose factors
include any number from 2 through 16.
a) Factor 1, 111 , 111 .
b) Invent a rule for finding when 19 is a factor of a
given number. If the reader thumbs through previous issues of Arbelos, he or she will find, in articles involving number theory and congruencies, more methods for factoring numbers. We continue with quadratic forms and their factors. Suppose we use x 2 -6x-247. When we subtract 4x247=988 from 1001, we see that the remainder is 13. This gives us 247= 13x 19, so x 2 -6x-247=(x - 19)(x + 13). We hope you will agree that just guessing at the factors of 247 might be somewhat time-consuming. We might try removing the second term of the expression. The work looks like this: 1 1 -
6
-
247
3
-
9
3
-
256
3
3 1
0
x 2 -256=(x-16)(x+16).
whence we have On adding back what was subtracted, we get (x+13)(x-19). -95
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Again,
x 2 - 6x - 247
Vowm 3, CtlMmR 4
= x2 -
6x + 9 - 9 - 247
or
(x-3?-256, or (x-3+16)(x-3-16), or, finally (x+13)(x-19). On the other hand, given x 2 - 6x - 616, one can alter the expression by dividing the second term by 2 and the constant term by 4. This gives us x 2 - 3x - 154, which is easy to factor. We get (x-14)(x+ll), so that, originally, we must have had (x-28)(x+22). Once again using what has been presented in other issues of Arbelos, suppose we start with 5x 2 +26x+24. We "multiply the roots" by 5, to get x 2 +26x+ 120. This is easy to factor (or one can manipulate the expression above further) . In either case we get (x+6)(x+20). Now we "divide the roots" and get (x + 6)(x + 20),
5
5
which simplifies to (5x+6)(x+4). We hope enough has been provided to make solving a problem involving just numbers or quadratics easy. The fun occurs when you can use some original method of your own, instead of merely mechanical means. When it comes to expressions of degree higher than the second, problems may become more difficult, but opportunities for originality also increase. There are some forms whose factors it is useful to remember. For example, if one remembers that the cube roots of unity are 1, ro, ro 2, and that 1+ ro + ro 2=0, then x 3 + y3 = (x + y)(x + roy)(x + ro 2 y) as well as (x+y)(x 2-xy+y2). In fact, x n - yn has the factor (x-y) for all n, and xn+yn has the factor (x-y) for all odd values of n. -96
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xn+yn is an example of a symmetric function. If one exchanges x and y, the function is unchanged. If y=kx makes the function take on the value 0, then y-kx will be a factor. In the example above, y=-x makes the expression zero, so x+Y is a factor. Again, consider (a-b)3+(b-c)3+(c-a)3. This is a symmetric function, and, as one finds by checking, becomes zero when b=a, c=b, a=c. Therefore the expression equals M(a-b)(b-c)(c-a) where
M is a constant. One merely compares terms to find OLlt
that the constant equals -1 so that (a-b)3 +(b-C)3 +(c-a)3 =-(a-b)(b-c)(c-a). c) Factor (a3+b3)(a-b)+(b3+c3)(b-c)+(c3+a3)(c-a) d) Factor a 3 +b 3 +c3-3abc. (Remember this one.) Sometimes, one gets extra mileage from factoring. We will try to simplify and solve the quadratic (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) (c-a)(c-b) (a-b)(a-c) (b-c)(b-a)
= 1.
e e
This appears to be symmetric. Let x=a, and the expression on the left equals 1, so one root is a. Let x=b, and the expression on the left equals 1, so another root is b. Let x=c, and the expression on the left equals 1, so another root is c. However, a quadratic equation cannot have more that two roots hence we have been trying to solve an identity. This way, we have avoided much useless manipulation.
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As another example, let us attempt to factor f=(x+y)5- x 5_y 5. This equals zero when x=o and when y=O. Hence two factors are x and y. Next, let y=-x, and we find that f=O. Therefore (x+Y) is another factor. The remaining factor is quadratic, at most, so f=xy(x+y) (Ax 2 +Bxy+Ay 2)
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The coefficient A appears twice in the third factor because f is symmetric. Let x= 1, y= 1, and we have 30=2(2A+B). Let x=2, y=-l, and we have -30 = -2(5A -2B). When we solve these for A and B, we find A=B=5.
Therefore, f=5xy(x+y)(x 2 +xy+y2). e) The following problem appeared on the fMO for 1984: Find one pair of positive integers a and b such that: (i) ab(a+b) is not divisible by 7; (ii)(a+b)7- a 7_b 7 is divisible by 7 7 . Justify your answer.
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. POSSIBLE TO THAT THE
CARRY OUT SUCH LION
CAN
THREE-SPACE IN A
DEFORMATION
BE RETURNED INTO
KNOTTED
HE IS THEN
A
IT IS THEN
CONDITION.
HELPLESS.
ASPACE FilLER Suppose one can find a point P in the interior of the quadrilateral ABCD such that the four triangles PAB, PBC, PCD and PDA have the same area. Show that P is on one of the diagonals AC or BD. Let a, band c, be positive numbers. Prove that abc ~ (a + b-c)(a+c-b)(b+c-a) (These are from a Swedish Mathematical Olympiad.) -98
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CONCURRENCE There are many occasions when one is required to show that two or more lines are concurrent - that is, meet at the same point. We wish to present five ways in which this may be done. These range from the well-known to the (possibly) unknown. First is the well-known theorem of Ceva. In the diagram at the right, the necessary and A sufficient condition that the three "Cevians" AP, BQ, CR concur is that AR RB
x
BP PC
x CQ = 1
QA
With this theorem, it is very easy to show that the medians of a triangle
B
p
C
are concurrent. Remembering that, if the three Cevians are angle bisectors, so that BPjPC=cja, one can also show that the three angle bisectors are concurrent. Now, as an exercise, suppose that P is halfway around tIle perimeter of the triangle from A, Q is halfway around the perimeter from B and R is halfway around the perimeter from C, one can show that the lines AP, BQ, CR will concur. One can show in a similar manner that, if P, Q, R are the points where the incircle is tangent to the three sides of the triangle, then °AP, BQ and CR will also be concurrent. In the same manner, the three Cevians will be concurrent when the points P, Q, R are the points of tangency of the three escribed circles of the triangle. A second (and fundamental) theorem is Desargues' Theorem. As most readers know, it states that if two triangles (in two-space or three-space) are such that corresponding sides, produced if necessary, meet in three points that are collinear, then the three sets of corresponding vertices will lie in three concurrent lines. ..99..
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In this diagram, which shows two triangles in a plane, sides AB and A'B', BC and B'C', CA and C' A' meet on a line at P, Q and R. It follows that the lines on A and A', B and B' and C and C' will meet at the point O. For a diagram
C' .............
---Q .......
.......
showing Desargues' Theorem in three dimensions, see the cover of the May, 1983 issue of Arbelos. The diagram can be considered as R consisting of ten points and ten lines, with three points on every line and three lines on every point, which makes it a self-dual figure, technically called a "configuration".
If we recall that we have referred to the "line at infinity" when discussing projective geometry, and have agreed that lines that intersect on this line at infinity are parallel, one can easily show that if two triangles have their corresponding sides parallel, then their corresponding vertices lie on concurrent lines. A good exercise would be for the reader to try to show how, given two lines on a sheet of paper that would intersect somewhere off the paper, and given a point P, say, one can construct a line on P that, if extended, would pass through the intersection of these two lines. We have also presented a third method for showing that lines are concurrent by using Brianchon's Theorem. This is the dual of Pascal's Theorem and may be stated thus: If a hexagon is circumscribed about a conic, the lines joining opposite vertices are concurrent. The conic may be a circle, an ellipse, a parabola, an hyperbola, or even two lines. The diagram at the top of the next page shows some examples. In each diagram, we have AD, BE and CF concurrent at O. -100
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A
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B--==----=-~F
B F
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One can have some fun by trying to create special problems. For one example, what does Brianchon's Theorem look like when the hexagon 11as been squeezed down to form a triangle?
We come now to one theorem that is not too well
known -the dual of Pappus' Theorem. First we state Pappus' Theorem: Let A, B, C be points on line L, and A', B', C' be points on line L'. Then the A
L
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intersections of AB' and A'B, AC' and A'C, BC' and B'C are collinear. There may be some readers who do not know that we can prove Desargues" Theorem by means of Pappus " Theorem and vice versa - provided one assumes the concept of commutativity. Otherwise Desargues Theorem cannot be proved for triangles on a plane. There are non-Desarguesan geometries. When one tries to prove either theorem by using coordinates and linear algebra, one must beware of commutativity sneaking in. However, let us dualize the figure above. l
l
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f
c'
--=t-=:-bf------->,,------~
f'
a'
a
b We have the lines a, b, c On point f, and the lines a'l b', c' on point f'. Then rom the diagram, the lines (ab') (a'b), (be') (b'c), (ac') (a'c) meet on the point o. Note that, although Desargues' Theorem is self dual, (as is the dual of a triangle), Pappus' Theorem is not self-dual. The two diagrams look entirely different. (The same is really true of a complete quadrangle - four points and the six lines on them, and the complete quadrilateral - four lines and the six points on them. But that is another story). Just for fun, examine the following problem:
D
H
C
Ek:::--\"-'+---I F AL.--~-----=~
ABCD is a square and EF is parallel to AB, GH parallel to AD, as in the diagram. Prove that the lines on BE, DG and CK are concurrent. Can you recognize the upper diagram as equivalent to the lower one? Will there be concurrence when ABCD is a parallelogram? One can construct many hard-looking problems by slightly altering the figure. -} 02
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We have been trying to present theorems that are progressively less well known to the reader. We think this last one is least known. Examine the figure.
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Let PX be perpendicular to BC, Py perpendicular to CA and PZ perpendicular to AB. Then
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(BX)2 - (XC)2 + (Cy)2 - (YA)2 + (AZ)2 - (ZB)2 = 0 The proof is extremely simple - one uses the Pythagorean
e
Theorem six times. (For example, (BX)2 so on)
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= (BP)2 -
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The converse is also true, and is equally easy to prove. We leave these proofs to the reader. Now see how easy it is to prove that the altitudes of a triangle are concurrent. Also, see how easy it is to show that the perpendicular bisectors of the sides of a triangle are concurrent (at the circumcenter). And that the lines perpendicular to the sides of a triangle at the points of tangency of the inscribed circle are concurrent (at the incenter). This theorem can be used to prove otherwise difficult problems. For example, the reader might try it on the case when the point P lies on the circumcircle, in which case, X, Y, Z are collinear. This line is the well-known Simson line. Should a problem ever arise where the reader is asked to prove lines concurrent (as in a test, a contest, or the USAMO or IMO, perhaps the reader will remember one of these methods for sl10wing concurrence and be able to use it. ~103~
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DISCOVERY AND INVENTION It is our belief that, in the sciences, mathematics, and when new products are invented, new ideas arise when the world is ready for them. We cannot come up with any thing in mathematics, for example, that has not been developed by two or more people. The story of the calculus, with its question of priority between Newton and Leibnitz comes to mind, as does the development of non-Euclidean geometry by Lobachevski and Bolyai simultaneously. It is interesting to note the simultaneous invention of logarithms by Napier, Briggs and Burgi. Fermat and Descartes are given simultaneous credit for analytic geometry, etc.
Nor have we been able to locate any invention that has not been simultaneously invented by at least two persons. Edison's life was full of lawsuits by others who claimed to have invented the same thing he had but earlier. Morse had the same trouble with the telegraph and Bell with the telephone. One of the reasons for reading up on the history of mathematics is the surprise aroused by the discovery of such simultaneous events. Another is, of course, reading about the lives of famous mathematicians, some of which were just as remarkable as the lives of, say, famous generals. All this is preliminary to a look at arithmetical computations before the invention of logarithms (and calculators). It is generally assumed that one of the difficulties ancient astronomers, arithmeticians, and accountants was with ordinary arithmetic. It is true that such men as Tycho Brahe and Johannes Kepler may have done a lot of long, complicated and boring arithmetic, but we take leave to doubt this. They had methods of their own to ease the burden, if it was a burden. I have already referred to Gauss who used to calculate a "chiliad" of prime numbers in his spare time, just for fun. -104
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Of course, the problem was usually multiplication. The arithmeticians naturally tried to convert such products to sums. This is what logarithms do.
One method used before logarithms was trigonometry. It was known that the following identity held: cos(A+B)+cos(A-B)=2 cos A cos B. We are told that calculators used this formula. Let us see how this could have been done by finding the product of, say, 652x.374. let cos A=.652
cos B= ..374
Using a table of the cosines of angles, we find that
(A + B) = 117°20', (A - B) = 18°44'. cos(A+B)=-.45917 cos(A-B)=.94702. cos(A+B)+cos(A-B)=.48785, and finally, dividing by 2 and placing the decimal correctly, 652x.374=24.3920 (approximately). The result is approximate because we used four place tables of cosines. The medieval calculators had the ten-place tables of Rhaeticus and the fifteen-place tables of Pitiscus. They also had lots of practice in arithmetic.
Then
We do not know why these calculators did not use the very simple rule (A+B)2_(A-B)2=4AB. All we would need in this case is a good table of the squares of numbers. We use this formula to calculate 652x.374. Let A=652 and B=.374. Then (A+B)=l026 (A-B)=278.
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(A+B)2= 1052676 (A-B)2=77284 4AB=975392 and AB=243848. This result is exact. We checked it on our little hand calculator. We suddenly have the urge to explain a method for multiplying numbers mentally that we happen to like. For two three-digit numbers, consider 2 ax + bx + mx
2
+nx +
When these are multiplied, the product containing no x-term is cpo To find the term in x, proceed as shown below: ax
2
+
mx 2 + to derive (bp+cn)x. For the term in x 2 , proceed as shown below:
~
~
to derive (ap+bn+cm)x 2 . For the term in x 3 , proceed as shown below:
~c ~+p
to get (an+bn)x 3 Finally, the term x 4 is obviously amx 4 . Now assume that the (x) has been replaced by 10, and we have a nice fast method for calculating a product. What's more, it looks pretty.
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Applying this method to finding the product of 652 and
374, the work locks like this:
3 6 2
4
3
8
4
8
6
(The column at the right is a record of what is to "carried."
This is what is done in multiplication, anyway.)
If a product is to be found in which the two numbers
have three digits or less each, this method can, with a
little practice, be faster than even a hand calculator.
We have seen a colleague "race" a hand calculator and
win every time. Of course, when it comes to multiplying
two two-digit numbers, there is almost "no contest." Finally, a method for multiplying two numbers which are alike except that the sum of their units digits equals ten, might be in place now. The two numbers are written as (lOa+b)(I0a+d) (b+d= 10).
If we multiply the two expressions and simplify, we find that the prodUct can be written as 1OOa(a+ 1) + bd. For two numbers like 57x53, which fulfil the conditions, we get the product 100(5)(6)+21 =3021. Of course, the same method will work on a product like, say, 257x253. Here we can use the device twice. Thus the product is 100(25)(26)+21=65021. What we have done is to use algebra (and trigonometry) in the service of arithmetic. Perhaps you might be able to invent some short-cut for finding the product of, say, numbers of the form (x+y) and (x-y). We have found this to be fun as well as a challenge. Perhaps you might agree. -107
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A COMBINATORIC fORMULA It happens that we encounter problems that sound quite different but that yield to the same process for their solution. We have met with several such problems recently. First we are asked to determine the number of solutions to the linear equation
xI + X2 + X3 + ... + Xr = n. If we write n ones and then insert r-I spacers between them, this will give us the r numbers wanted. We have n+r I objects, and can choose the spacers in r-I ways. Hence, . b·momIa . I coe ff·· usmg IClen t s, we h ave
(n +r-Ir- I) soIu t·Ions.
Actually, this formula gives us the number of nonnegative solutions. Some of the xj may be zero. Incidentally, if we seek solutions in positive integers only, we can use the same method and make sure no two spacers are adjacent. In this case, we will find that the number of · . soIu t Ions IS
(nr-I-I) .
What struck us was the frequent appearance of the first binomial coefficient. The same function showed up again in finding the number of arrangements, with repetition, of r objects selected from among n objects. Here we were reading our Euler, and found his solution very clever (as usual). Take a combination of r elements selected from the full set of n objects. Written in ascending order, let these be al,aZ,ay ... ,ar · It is understood that these a j are not necessarily distinct. Note: We assume that the reader understands the meaning of
t>. r -108
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Construct the associated set of elements a 1 + 0, a2 + I,
.a3 +2, ... ,a r + r-l. These elements are all distinct. Therefore, the number of ways or selecting them will equal the nLlmber of arrangements without repetition of elements from 1 to n+r-I. This will equal t + r - I ).
r We have already mentioned the fact that the Binomial Theorem holds for any exponent - not just positive integers. Applying this to a negative integer, we have (1- x)-n = en + (-n) x + (-n)(-n-I) x 2 + ... in which the
1
coefficient of
xr
Ix2 takes the form
(-n)(-n-I)(-n-2)...(-n-r+l) = (-I/t + r-I), and, once r! r
again, the same form occurs.
Incidentally, we have shown that the generating function that
generates these forms is (I-x)-n. The reader might like to
-n
show, by easy algebra, that (
) yields the same form as above.
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It should be noted that we have a solution for the following problem: Find the number of ways of putting n like objects · · (n + r - I) . Into r d·f'& Ilerent b oxes. T h e answer IS r
If no box is to be empty, the answer is
tr-I - I). We seem to
have entered the realm of pigeonhole problems. We next examine the following problem: In n trials, some of which are successes and some failures, when will the r-th success tum up? -109
We shall assume that there are equal probabilities of success as of failure. According to elementary probability theory, we take p=q=1/2. There will be r+k-l trials, assuming k failures. At trial r+k, there is success. The probability sought will therefore be (r + k - 1)pr-1q k x p = {+ k -l)prqk Again, we have come to the k k same form as before. Before we come to our final problem, there are one or two theoretical matters to be presented. First, note that Newton and Gauss proved that: (
x r -1
x x+1 . )+ ( ) = ( ) where x may be any number and r an mteger. r
r
This is far more general than what Pascal was able to show. Next, consider an expression of the form (a+b+c+d)n. the expansion of this leads to the multinomial theorem. We state without proof the rule for expansion: A term of the form aibickct' (i+j+k+l=n) will have a coefficient of the form
(----.nL). i!j!k!l!
It is easy to show that the multinomial coefficient above is equal to the product of binomials <")(n - a)(n - a - b)(n - a - b abc
c).
d
We come now to the Banach matchbox problem: A smoker has two packs of matches in the same pocket. He selects either matchbox with probability 1/2 and uses a match. When he picks a matchbox and finds it empty, what is the probability that the other matchbox has k matches in it? Hints: Let each matchbox start with n matches. Box 1 will be empty when box 2 has k matches if n-k failures precede the n+ 1 -st. success. We find that ?n - k)2-2n+k is the answer. Can n
you derive this? -110
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PROBLEMS FOR BEGINNERS
1) Given two positive integers, we find, first, that their sum equals S, then that their difference equals 0, next that their product equals P and finally that their quotient equals Q. If S+0+P+Q=243, what are the numbers? 2) A rectangle whose dimensions are whole numbers has its perimeter equal to its area. What are its dimensions? 3) There are five successive integers which have the property that the sum of the squares of the three smaller integers equals the sum of the squares of the two larger integers. What are the five integers? 4) Given a twenty-digit positive integer, what is its sixty-fourth root? 5) The product of two two-digit numbers is not changed when the operation is written "in reverse." (Thus, 46x96=69x64). How many such pairs of numbers are there? (The products are not to be multiples of 11. This is to eliminate such pairs as IIx22=22xll). Note:
These problems are given as exercises for those who want practice in attacking "different" problems in different ways.
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OLYMPIAD TYPE PROBLEMS The following problems were among those submitted for inclusion in the International Mathematical Olympiad. We suggest that the reader try each problem and spend one hour on each.
1) The equation z2_(x2_1 )(y2_1) = 1984 is solvable in integers. Find one solution. (Federated Republic of Germany)
2) Show that for any n= I, 2, 3, ... , there is a natural number x n such that x
2 n
+ 3xn + 5 is divisible by
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(Bulgaria) 3) Find the point P inside the triangle ABC for which BC + CA + AB is a minimum, where PD, PE, PF are
PD PE PF
the perpendiculars from P to BC, CA, AB respectively. 4) The sum of all the face angles about all of the
vertices except one of a given polyhedron is equal to 5160 degrees. Find the sum of all of the face angles of the polyhedron. (USA)
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DIMENSIONAL ANALYSIS
There is no reason for mathematical methods to be confined to mathematics, and they are not so restricted. Engineers use mathematics, as do those using statistics, etc. If we can use mathematical methods in other areas, so much better. We have already shown how one can use mathematics in chemistry. Let us now try to do so in physics. We start with the observation that all pl1ysical entities can, in the long run, be represented in terms of three fundamental units. We shall agree to have these three limits be length, time, and mass.
fit
tit
It is assumed that these three units will be our fundamental units. There are other possible units, especially when we deal with electricity and heat, for example. Or we might like to take another entity from physics, say force, as a fundamental unit. Now, with just a little imagination, we can express other concepts in terms of these three units. Let us begin by using the abbreviation L for length, M for mass and T for time. Then, since an area is the product of a length by a length, we may express an area by LxL or L2, and a volume by LxLxL or L3. Again, using the rule ratextime=distance, we can use our abbreviations to find (rate) xT=L, so that a rate or velocity is expressed as Lit. Also, since an acceleration is a change in velocity with time, acceleration would be expressad as LT-2. If we recall the formula force equals massxacceleration, we have, in units, force=MLI 2 . Remember that weight is a force, so weight is also ML1 2 . One does have to know a mite of physics or have a little imagination, but with little more, one can express many physical entities in terms of L, M and T.
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Thus, recall that potential energy of a body is equal to the product of the weight multiplied by the height it is above its equilibrium level, we find that energy=ML 2 T- 2 . If we need to work with a density, we remember that density equals mass per unit volume, so that density=ML-.3. We have here determined just a few expressions for physical entities in terms of the three fundamental mental entities. There are good texts that have far more such expressions, usually in physics or chemistry. Let us now collect the few expressions we have obtained. These are: L length density ML-.3 2 L area weight MLT- 2 L.3 volume force MLi2 M mass power ML 2 i.3 T time energy ML 2 T- 2 l L i velocity work ML 2 T- 2 LT- 2 acceleration momentum MLT-l We have included a few new physical entities, and hope you will try to determine why the corresponding expressions in terms of the fundamental units are correct. Now we come to the purpose of using these units. We shall assume that, if a physical formula is expressed in terms of them, then that formula must balance as far as they are concerned. As an example, consider the formula s= 16t2 , which we have seen in some texts. If we write in the fundamental units, we find (s)L=(t 2 )T 2 . Dimensionally, we have L=T 2 , which does not balance in terms of the units. Hence this formula must be incorrect. If we compare with the correct formula, s= 1/2 gt 2 , we find we get L=(LT- 2 )(T 2 ), which does balance dimensionally. -114
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It would appear that Newton first used the idea of fundamental units or dimensions in his work. Of late, it has received attention because of its value in simplifying and even determining physical formulas. We can best explain with an example.
When we inquire into the nature of centrifugal force of an object traveling in a circular orbit whose radius (r) is given, and whose mass(m) and orbital velocity (v) are also given, we agree to write (f) varies as (m)a(r)b(v)c. Here we go beyond mere substitution and must consider what physical entities might enter into the desired formula. Usually if we select unnecessary physical quantities, that shows up in the result. At any rate. in this case, we cannot think of other quantities to include. Now we analogs. is force, mass is,
replace the quantities by their dimensional We find (MLI 2 )=(M)a(L)b(LII)C since the (f) the velocity is (LT- I ) the radius is (L) and the of course, (M).
If the dimensional equation dimensionally, we must have, for (M) 1=a for (L)
is
to
balance
l=b+c
for (T) -2=-c. When we solve these (mentally, I hope), we get a= L b=-l, c=2. Hence the corresponding formula will be f
= C (mv
2
). Here C is just a number, which we must r obtain from experiment or by other means.
Let us try another problem. The energy of an electron would seem to depend on its mass and on its velocity. That is (E)=C(m)a(v)b -115
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Dimensionally, therefore, we may write ML2 T- 2 = Ma(LT-1)b Comparing the exponents, we find that for (M) 1 = a for (L) 2 = b for (T) -2 =-b Fortunately for us, these equations have one solution. We find, in fact, that our formula should read E=Cmv 2 . Now we can do it a bit more. First the electron has a maximum velocity when v=c, the speed of light. Second, by measuring in electron volts, we can make C= I. Therefore, our final formula is E=mc 2 . Before we continue with another problem, there is one matter we must correct. Newton asserted that "any two objects in the universe attract each other with a force that is proportional to the product of their masses and that varies as the inverse square of the distance between them."
Written as a formula, we have F= k(mm') . When we 2 r replace or insert the dimensional entities, we have MLT-2 = (k)MaMbL-2. Now the right side of this equation has no term in T. Therefore (k) is not dimensionless. When we solve for (k), we find that k = M- 1L3 T-2 . The quantity (k) is usually called the "world constant." There are numerous discussions in which the world constant should be considered, but in which one tends to forget it. One does have to a little bit of physics, enough to determine what should and what should not enter into a discussion of this type. -116
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For our next example. we go to astronomy. For a satellite traveling around a central sum of mass S, we decide, after some thought that the period p of the satellite should depend on the mass of the sun, the world constant, of course, and the distance d between the satellite and the sun. That is, p=Sakbd c . We have made some assumptions to arrive at this. First, we ignore the mass of the satellite as being very small compared with S. Next we are assuming that the orbit will be nearly circular, the diff~rence being negligible. Now we substitute the dimensional units, to get T = Ma(M-IL3T-2)bLc.
When we compare exponents for each unit separately, we have for (M) 0 = a-b for (L) 0 = 3b + c for (T) 1 = -2b . These are easily solved for each variable. We find
a=-l. 2
b=-l.
c=3.
2
2
.J
Our formula takes the form p = Cd 2
1
•
(Sk)2
More simply, p2 = C £. (Sk)
Now this does not tell us much, possibly because we are not expert astronomers. However, consider two satellites traveling about the same sun. Then we have
P2
2 _ Cd.3 2
---sk
2
from which, on division, we arrive at
P)2 P2
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.3
= d).3 . d2
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In words, the squares of the periods of two satellites traveling around the same sun are proportional to the cubes of their distances from that sun. For a satellite to remain stationary above a point on the earth, we can now determine why the satellite will have to be about 23,000 miles above the surface of the earth. Also, since we can easily find the periods of the planets, we can find their distances from the sun. Note: It took Kepler twenty years to discover this formula. He examined many sets of data to get it. As another problem, we examine the motion of a pendulum. We assume that the swing depends on the length (f) of the pendulum, the mass (m) of the weight at the end, and the gravity (g). That is, t=C fagbmc. We insert the dimensional equivalents, and get T=La (Lr2 )b(M)c. Note that c=O, so the mass of the weight has no effect on the motion. Next, solving for a and b,we get -2b= 1 and a+b=O. Therefore, t =
C~.
Dimensional analysis is useful even in mathematics. using it, one cannot accept 1t r 2 as anything but an area, abc as anything but an area, etc. 4R It is also possible to use dimensional analysis to compare formulas expressed in different units. Thus, one BTU raises one pound of water one degree Fahrenheit, or 454 grams 5/9 of a degree centigrade or (454)(5/9) grams one degree Centigradge or 252 grams one degree Centigrade . Therefore, one BTU equals 252 calories. However, this takes us too far afield, so we must regretfully stop here. The applications to model building must wait. -118
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KURSCI1AK KORNER This column features those problems of the famous Hungarian Kurschak 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 MANs 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. 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. 1/1938
Prove that an integer n can be expressed as the sum of two squares of integers if and only if 2n has the same property.
2/1938
Prove that if n is a positive integer greater than 1, then 1_+ ... +_1_+_1_ > 1 .1+_ n n+l n 2 -1 n 2
3/1938
A segment XY is said to be a transversal of ~ABC if X is one of the vertices of the triangle and if Y is a point on the side (or extension thereof) opposite to X. Prove that if ~ABC is an acute triangle, then there is a point 0, not in the plane of ~ABC so that OX is perpendicular to OY for every transversal XY of ~ ABC. Dr. George Berzsenyi Rose-Hulman Institute of Technology Terre Haute, IN 47803
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1984-1985:
May, 1985
No.5
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PREFACE This issue marks the end of three- years of the Arbelos. Time has certainly flown! So has content. We hope to use the summer months to compile an index, and fear that it will take all that time. Thus far, we have limited ourselves to subject matter that has appeared or may appear in various contests - The USA Olympiad and the International Mathematical Olympiad, for example. We suspect that this content is due to change. Some nations have been suggesting that problems involving matrices, linear algebra and calculus be accepted. Our future issues will therefore have occasional articles using these disciplines. Where possible, we have had articles submitted to us that we have published, and we hope that there will be more. The letters we have received from readers have helped make the labor of preparing these issues most pleasant. We acknowledge again, as we have before, the work of Mark Kantrowitz. We also acknowledge problems proposed by our readers. One such proposer is David Morin, of South Paris, Maine. He has rediscovered two problems on his own. One is to be found in College Geometry, by Altshiller-Court (p.133). The other is also in Altshiller-Court, and appears in this issue. David is to be congratulated on discovering these problems. We have also found results that we thought original, only to find that others had already discovered them. It happens to all of us. S. Greitzer -122
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TRIGONOMETRY The advent of linear algebra and computers, it seems, has been at the expense of much of the old-fashioned sUbjects as geometry and trigonometry. The tide is turning a bit in the case of geometry, especially as Dieudonne, who previously had said, "Geometry must go !", has changed his mind. However, trigonometry is slowly fading from the secondary school curriculum. True, we may not find it useful to measure
heights and other distances, but there are aspects of
trigonometry that make it worth while to keep it in mind. These
aspects are in its nature -they are periodic and they lend
themselves to interesting identities. Of course, it would not be wise to memorize the many formulas that have been developed, but one should be able to manipulate a few fundamental rules to take care of problems. First each function can be expressed in terms of any of the other functions. Assuming one is aware of sin e, cos e. tan e and their inverses, esc e, sec e, cot e, respectively, the problem solver could, given an identity or equation, alter it so that it does not contain unfamiliar forms. Try this tan e + cot e = sec e. csc e. In previous issues, we have derived formulas for functions of sums or differences of angles, (using ptolemy's Theorem, for example). We know sin(A+B) =sinA.cosB+cosA.sinB sin(A-B) =sinA.cosB-cosA.sinB These look like simultaneous equations, and used as such, will quickly yield, for example, sin(A+B)+sin(A-B)=2sinA.cosB or, letting A+B= a. and A-B= ~, sin a. + sin ~ = 2 sin(a.+~).cos(a.-~)
2
2
Similarly, starting from
cos(A+B)=cosA.cosB-sinA.sinB
cos(A-B)=cosA.cosB+sinA.sinB -12.3
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one derives
cosu + cos ~ =
2COS(u+~).cOS(u-~). 2
2
More formulas can be derived by subtraction. Thus, given a problem involving sums of angles, one can replace by one containing products, and vice-versa. The point is that it is so easy to derive one set of formulas from the other that one does not have to memorize more than one set. Notice that, from the formulas for sin(A+B), one easily gets sin.2A=2.sin A.cos A. From the formula for cos(A+B), one gets cos2A=cos 2 A-sin 2 A and from that for cos(A-B), one derives cos 2 A+sin 2 A= 1. At this point, just to refresh the memory, one should recall that functions may differ in sign but not absolute value in different quadrants. Our way of remembering this is to use the CAST diagram: In quadrant I, All functions are positive: in quadrant II, the Sin (and cosecant) are positive: in quadrant
IV
III, the Tangent (and cotangent) are positive; in quadrant IV, the Cosine (and cosecant) are positive. Or one can remember the form of the graph of each function and find the correct sign from it. As for the values of special angles, we use:
0°
0°
30°
45°
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2
2
2
2
2
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2
2
2
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2
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cos
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sin
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We admit that we find it fun to solve a problem especially by means of a way that is different. We present the following example. Given ~ABC. Prove: a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=O. This can be done by using the formulas for the differences of the trigonometric functions, but it becomes easy if we use the Law of Sines. Try it!
Again, consider the problem: If the sum of angles A and B
45°, prove that (1 +tanA)(I +tarlB) =2. We have our short way
of doing this. Find yours!
Manipulation of known formulas gives new ones. From the two formulas on the first page of this article, we can easily find: cos2 x = cos2 x sin 2 x and cos2 x + sin 2 x = 1. A little 2 2 2 2 algebra gives the results sinK 2
= ~l-COSX
cos K
2
2
= ~1+COSX 2
Without intending to give any hint as to solution, prove: {sin a + cos a)2 = 1 + sin a. Also, in a triangle, given 2 2 cos A+cosB=4sin 2 Cj2, show that a+b=2c. It is in working with three or more angles that one is often required to use special skills and ingenuity. In this connection, there are three formulas that appear to be quite useful. These are sinA+sinB+sinC=4cos A cos B cos C 222 tanA+tanB+tanC=tanA. tan B. tan C, tan B tan C +tan C tan A +tan A tan B = 1.
2 2 2 2 2 2
These hold for A, 5, C the angles of a triangle. We have found
the second formula quite useful, and shall give a proof of it.
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But first, since we are not sure that functions involving tangents are of frequent occurrence now, we derive som-e formulas that should be remembered. First, tan(A+B)= sin(A+B) = sinA.cosB+cosA.sinB cos(A +B) cos A. cos B-sin A. sin B and, if we divide each term in the expression at the right by cosA.cosB, we derive tan(A+B)= tanA+tabB . I-tan A. tanB Now for our formula for the sum of three tangents. since A+B+C=I80, A+B=I80-C; so tan(A+B)=tan(l80-C)=-tan C, by the CAST rule. Therefore tanA+tanB =-tan C. I-tanA.tanB Now cross-multiply and rearrange, and we have tanA+tanB+tanC=tan A. tan B. tan C. In addition, since we already have formulas for sin x and cos x we may divide to get tan x . Then 2 2 2 tan x = I-cos x I+cosx 2 If we rationalize either the denominator or the numerator in this last expression, we arrive at tan x = I-cos x = sin x sin x I+cos x 2 Now you can derive the formula Number 3 on the previous page. Try it. We present an identity for the reader to try: sin A+sinB+sin(A+B) = cot A .cot B. sinA+sinB-sin(A+B) 2 2 -126
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De Moivre's Theorem, which has appeared in our earlier issues, is sometimes useful, either for finding functions of multiple angles or for deriving values of special angles. We illustrate two such applications.
From (cosA + i sin A):5, we get cos.3A+i sin .3A equal to cos:5A+.3cos2 A.sin A.i+.3 cos A.sin 2 A.i 2 +sin:5A.i:5, and, equating real and "imaginary" parts, cos .3A=4.cos:5A-.3.cosA sin.3A=.3.sin A-4.sin:5A. The first of these was used in solving the cubic in a previous issue. We will use it for evaluating sin 18, which has been found useful to remember. Let x=18. Since sin.36=cos 54, we write sin 2x=cos .3x
2 sin x cos x = 4.cos:5x -.3.cos x
2 sin x = 4.cos2 x-.3
2.sin x = 4( 1-sin2 x)-.3
4.sin 2 x + 2.sin x-I = 0
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On solving this quadratic, we finally find that sin sin x =
(.f5 -1) . 4
The other application arises in the solution of • 0 0 =, 1 cos 0 0 + 1• sm 0
0
0
0
cos .36 0 + i sin .36 0 = 1, cos 720 + i sin 720 =1,
which they are, because of the periodicity of the trigonometric
functions. Using De Moivre, we have o
x x x
0
= (cos~ + isin~) = cosO° + isinO°. :5
o
:5
0
= (cos .360 + i sin .360 ) = cos 120° + i sin 120°. :5
o
:5
0
= (cos 720 + isin 720 ) = cos240° + isin240°.
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Inserting the values of the angles (use the CAST rule and the table of values provided), we finally x= 1, x = -1+i.J3,x = -1-i.J3, or using the usual
2
2
symbols, x=I, (0, 00 2 . These values appear in unexpected places. For example, Solve x x2
+
y
+
z
=
2
+
y2
+
z2
=
0
x3
+
y3
+
z3
=
-1
Finally, we may be faced with the problem of summing a series of trigonometric terms, usually in arithmetic progression. Thus, for S = sin a + sin(a +~) + sin(a + 2~) + ... + sin(a + (n - 1~),
~e
have found it expedient to multiply by 2sin
~
2
and use the "product to sum" rules (see the first page of this article). The transformed series will be found to be "telescoping", and we end with S=
sin(a+(n -1)~. sin n~
sin~
2
2
2
It is advisable to invent your own mnemonic device to remember this rather than derive it every time. The similar formula for a sum of cosines has the same form as that for the sum of sines, but with sin(a+(n-l)Q.) replaced with cos(a+(n-l)Q.).
2
2
Now try to solve these problems from the IMO. Solve cos 2 x+cos 2 2x+cos 2 3x= 1; Show cos 1t - cos 21t + cos 31t
7
7
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GEOMETRIC ALGEBRA We know that the ancient Greeks were preeminent in geometry. Yet they appear to have been able to solve problems in algebra without our knowledge of algebraic methods. The question is how they were able to do so, We have our algebraic geometry, and use coordinates, etc., in solving our problems.
•e
First, we should recall that, to them, the product ab was an area, and abc a volume. Hence they could do little with algebra problems above the third degree. Next, they solved problems by means of constructions. Hence they became quite expert at devising constructions which they used in solving their algebra problems.
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Solving x=a was quite simple, even when a was irrational and a square root. That had been taken care of by Eudoxus. He made irrationals respectable.
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When it came to solving equations of the form ax=b, the Greek geometer resorted to proportions in some form.
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B a
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x
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Q
b
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Ib
He could draw ~ OMN (Fig. 1a) and then PQ parallel to MN, or he could draw the circle on A, B, C and in either way, find x (Fig. 1 b). By using similar devices, he could solve the equation aX+b=c, as follows: C
x 1
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We leave to the reader the construction needed to solve ax=bc. To solve aX+b=cx+d, the ancient student would first rewrite this (a-c)x=(d-b), and then either construction to find x. In short, it would appear that the geometric solution of any linear equation was relatively easy. Even when a coefficient was the square root of a number, the process was simple. The student had a method for constructing a mean proportionaL and could proceed thus:
A
a
1
H
B
Fig.2 On AB= 1+a, he would construct a semicircle, and at H construct an altitude, getting .j;; . Incidentally, he also knew how to find a third proportional - that is, to draw a diagram for a:b=b:c. Of course, solving x 2 =a or x 2 =ab was easy, as the diagrams below show.
3a 3b It is when one selects the quadratic equation x 2 +ax+b=O that one encounters difficulties and has to use ingenuity. Let us have a try at it. -130
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x xA
x B
x
E
C Fig. 4
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In Figure 4, we are given lengths AB=a and length BD=b. We construct BD perpendicular to AB at B. Let us now construct the perpendicular bisectors of AB and BD, intersecting at o. With OD=OB as radius, we draw the circle, which cuts AB extended
e
at X and X. Call BX=x. Then
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BX. BX = BC.BD, or x(x + a)=bc. This is because BX= AX. Now we have derived x 2 +ax=bc. The solutions of this quadratic are BX and BX.
D
0,
__._ _ .
A
Fig.5 Again, consider Figure 5. We are given AB=a and BD=b as before. The perpendicular bisectors of AB and BD meet at 0, and the circle with radius OD is drawn, meeting AB at M and M as shown. -131
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First, MA=MB. Next, BM.BM=BC.BD. Therefore, if BM=x, then x.(a-x)=bc. Now we have a line solution for x 2 -ax+bc=O. By suitable altering the same diagrams, one can also solve x 2 +ax+bc=O and x 2 -ax-bc=O. We have shown how geometrical methods can be used to solve any quadratic equation in which the coefficient of the first term is unity. It is a simple matter to extend these methods to more complicated equations. The point of this article is not to recommend the solution of quadratics by geometric means. It is easier to use the quadratic formula or factoring. We do recommend the use of geometric methods where their use is simpler. Consider the famous Pythagorean Theorem. In the diagram below, we have b MI---,,-<----t-------=A M
Fig.6 right triangle ABC, labeled as usual (AB=c). When the circle with center B and radius BC is drawn, then AM.AM=AC.AC, or b 2 =(c-a)(c+a)=c 2 -a 2 , from which we get a 2 + b 2 =c 2 . I can find nothing simpler! Perhaps the reader might wish to try solving x 2 -5x+6=O geometrically. Hint: let a=5 and b=BD=6. -132
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2R
Af------\-----'I--------">IB
.....x ....
x . t
:
M
R
This article is a result of a problem given me by a student in Bogota, Colombia: Construct a right triangle ABC given the lengths of the median CO and the angle bisector, t (see diagram at left). We noted that CTxTM=ATxTB, or R2 -s 2 =xt. Also, we saw that R2 +s 2 =x 2 . Elimination gave us 2R 2 =x 2 +tx. This quadratic we solved as in the article to derive x. This derivation is shown in the diagram at the right. Now it is easy to construct right triangle MOT, extend MT to C, and draw the desired right triangle. However, something (intuition) keeps telling us that there must be some neat simple way of doing the same thing. We have not been able to find any such solution. Therefore, we appeal to our readers. If anyone can derive a simpler solution, please let us know. The whole situation is making us nervous! How THERE ARE DESERT.
NO
TO
CATCH
WILD
A
LIONS
THE CAPTURE OF A BE LEFT TO THE
IN THE TAME
READER.
H. PETARD -133
LION
SAHARA LION
CAN
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EULER'S FORMULA The reader who happens to have kept the November, 198.3 issue of Arbelos will have noted that we did introduce the formula V-E+F=2 for polyhedra topologically equivalent to a sphere. This is known as Euler's Formula (although it was known to Descartes before him). The proof is quite simple, and we omit it here. In the same issue, we used the formula to show that a map or graph on a sphere (or plane) must contain regions with five or fewer edges as bounds. We also showed that a graph consisting of vertices and edges where we call the number of edges that meet at a vertex the valence of that vertex must have an even number of vertices of odd valence. We used this to draw unicursal and allied graphs. At the same time, we derived the formula for the sum of all the face angles of a polyhedron namely S=21t (V-2) where V equals the number of vertices in the polyhedron. However, we have some new results, one older and one newer, that we would like to present. First, The number of "regular" polyhedra is just five. By a "regular" polyhedron we shall mean a figure in three dimensions such that (a) every vertex has the same valence - that is, there are exactly m edges on each vertex; (b) every face has the same number of edges· that is, every face has exactly n edges. We permit the polyhedron to be as distorted as we wish. We do not require that each face be a regular polygon, nor that the angles made by the edges at a vertex be equal. In fact, we do not assume any straight lines in the solid. -134
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We assume there are V vertices, so, by our previous assumption, mV=2E. We also assume that we have nF=2E. Now we substitute in Euler's Formula: thus:
£ + 2 = 1+ 2
> 1. m n E From this inequality, we derive, in succession, 2m+2n>mn, O>mn-2m-2n add 4, and get 4>mn-2m-2n+4=(m-2)(n-2). There are just five possible solutions in integers to this inequality, namely: From this, we get
m-2= 1 n-2= 1
m-2=2 n-2= 1
m-2= 1 n-2=2
m-2=3 n-2=1
m-2=1 n-2=3.
Thus, even if the polyhedron is twisted and distorted, there are only five possible polyhedra. The result m-2= 1, n-2 =2 leads to the tetrahedron, which has four vertices and four faces. The result m-2=2, n-2= 1 leads to the octagonhedron. It has eight faces, all triangles, and four edges per vertex. The result m-2= 1, n-2=2 leads to the cube. It has eight vertices and six faces, all four-sided. The result m-2=3, n-2= 1 leads to the icosahedron. It has twenty triangular faces, and five edges at each vertex. Finally, the result m-2= 1, n-2=3 leads to the didecahedron. It has twelve faces, and three edges at each vertex. -135
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Note that cases two and three above are duals, as are also cases four and five. Just switch the numbers for vertices and faces. On the other hand, the tetrahedron is self-dual -switching edges and faces produces no change. Note that Euler's Formula is a special case of a far more general theorem due to Poincare. We will state a partial result for information only. First, consider a cube with a hole in it, from top to bottom. Now the top and bottom are no longer simple regions with a single boundary. We must add a line to each of these faces. Apply Euler's formula to this figure, and you will find that V-E+F= 16 26+ 10=0. To put it more simply, consider a mapping on a sphere with a handle to it. For this figure, V-E+F=O. In generaL let a sphere have p handles on it. Then Euler's Formula takes the form V-E+F=2-2p. But we are going too far afield. Suffice it to say that when we add non-orientable surfaces, the formula becomes more complicated. All this was brought on by a problem we received recently. This asserted that if a polyhedron had no vertices of valence three, then it must include at least eight triangular faces. Naturally, we started to play with Euler's Formula to try to prove (or disprove) the assertion. Assume that the figure has V vertices, E edges and F faces. We can write F = f3 + f4 + f5 + ... + fm where f j stands for the number of faces with i edges Similarly, we can write V=v3+ v 4+ v S+··· v n· -136
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Since every edge bounds two faces and every edge has two ends, we can now write 3v3 + 4v 4 + 5vs + ... + mV n = 2E 3f3 + 4f4 + 5fs + ... + nfn = 2E Now we multiply Euler's Formula by four, to get 4V-4E+4F=8. We will use the formulas above equally to substitute for E. We do so, and get 4( v 3 + v 4 + v S + ... + Vn) - (3 v 3 + 4 v 4 +
+ mv m)
-(3f3 + 4f4 + 5fs + ... + nfn ) - 4(f3 + f 4 + + f n ) = 8 On performing the operations and grouping terms, we arrive at (v3 + v4) - (v5 + f5) - 2(v 6 + f6) - ... = 8 If this result is to be valid, it follows that we must have (v 3 + f 3 ) > 8. Now, if there are no vertices of valence three, then, in the above, f 3 must be at least eight. Note that this occurs in the octahedron. Incidentally, suppose that no face is triangular. Then v 3 must equal at least eight. We get this effect in a cube, for example. Now we conclude with an oldie -can a solid have just seven edges? Prove or disprove. Better still, draw a figure. By working (or playing) with Euler's Formula as we have done here, you may discover surprising results. Try it!
-137
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TERM-ENDERS A
1) If OP. OQ. OR are perpendicular to sides BC, CA, AB (as shown), then (AR)2_(RB)2+ (BP) 2_(PC)2 + B L----~-=-----------=C (CQ)2_(QA)2=O and conversely.
C
2)
AL--_ _----"_-"" -----"_ T In right triangle ABC (right angle at C), the incircle is tangent to the hypotenuse at T.
Prove: the area K or the triangle is equal to AT x TB.
3) In a triangle with sides a. b, c. prove that a 2 (b+c-a)+b 2 (c+a-b)+c 2 (a+b-c)<3abc.
S 4) Line segments PQ and RS. intersecting at T, are drawn in square ABCD, as shown. Prove that line segments PB, DR and CT are concurrent.
P
Q R
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In the diagram at the left, lines have been drawn parallel to the sides of the triangle and tangent to the inscribed circle (as shown). Prove that the sum of the perimeters of the three small triangles equals the perimeter of the large triangle.
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(b) Let circles be inscribed in the three small triangles. Find the sum of the areas of all four circles. A P B
q
C 6) Perpendiculars p, q and r are drawn to a line passing through the centroid G of triangle ABC. Is it true that p+q=r?
Note: Look for the simplest method of solving these problems. Nos. 1 and 2 are easy. However, I have not been able to find No.1 in the literature. No.3 and No.4 are harder. No.Sa and No.6 are easy again. We will publish solutions or acknowledge them in our next Arbelos. s. Greitzer Editor ~139~
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MATti IS fUN Any mathematician of any stature finds his work fun as well. Fermat was a lawyer but, as his many letters show, worked number theory for fun. We have already mentioned that Gauss found pleasure in occasionally factoring a "chiliad", or thousand or so numbers when he needed recreation. All this is an introduction to an exercise donejust for fun. We had worked out the formula for the area of an inscribed quadrilateral, and, by letting one side equal zero, got the area of a triangle by means of Hero's formula, when it occurred to us to try to seejust how many ways we could invent to get the area of a triangle. Of course, we started with the well-known formula(s): K
= a. h a = b. hb = c. h c h
2
A
2
which we decided not to count among any rules we derived. Next, we recalled that, for example, h a = b.sin C so we
c
b
B <------'---:-:H,.-------------'" C
substituted in the above and got (1) K
= -l.absinC = -l.bcsinA = -l.acsinB. 222
We agreed to count these as one formula. They could be used to find an area, given SAS. Now we used the Law of Sines to get b= a sin B and substituted sin A in (1) to get (2) K =
1. .a2 . sin B. sin C = 1. .a2 . sin B. sin C
2 ~nA 2 ~n~+q
with two similar formulas for b 2 and c2 . Now we could find the area of a triangle, given ASA. Finally, we worked out Hero's Formula, just to be thorough and complete. The development follows: -140
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a First, a 2 =b 2 +c 2 -2bc cos A or b 2 +c 2 -a 2 =2.bc cos A.
Also, using our own Rule (1), we have bc sin A=2K
or 4K=2.bc sin A. Squaring and adding these to eliminate the trig functions, we get 4.b2 c 2
= (b2
+ c2 _ a 2 )2 + 16K2
or 4.b2 c2 - (b2 + c 2 - a2 )2 We factor on the left, to derive (2bc + b
2
+ c 2 - a 2 )(2bc - b 2
- c
= 16K2
2
+ a 2 ) = 16K2
"Rearranging terms in each of the parentheses, we have ((b + c)2 - a2 )(a2 - (b - c)2) = 16K2 . Factoring each bracket again, we arrive at (b + c + a) (b + c - a) (a + b - c) (a - b + c)
= 16K2
We agree to use the semi-perimeter s= (a+b+c) and, 2
16K2
after sharing the 16 in equally, get s(s-a)(s-c)(s-b)=K 2 , and finally, (3) K
= ~s(s -
a)(s - b)(s - c) .
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e
Now we can find the area of a triangle, given SSS.
e
So far, we have derived just the usual formulas.
Now let us try for some that are net so usual. We
It
recall that _a_
~
e
e e
e
e e
sin A
= _b_ = _c_ = 2R. sinB
sin C
Hence sin C = ~, 2R
substitute in Formula (1), and get (4) K = abc. 4R
This could be used to find the
circumradius when the sides and area are known. -141
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b
a Use of the circumradius R naturally makes us think of r , the inradius. From the diagram, it is obvious that we can write (5) K=rs. Incidentally, from rules (4) and (5), we can write abc=4Rr.
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It
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It We frequently find that we get different results, extras", so to speak, when we play around this way. II
Going back to the Law of Sines, we note that a=2R.sin A b=2R.sin B c=2R.sin C. We can substitute these into our Formula (4) and get, first, abc=8R3 sin A.sin B.sin C, and then (6) K=2R2 sin A.sin B.sin C.
This gives us the idea of doing the same with our
Formula (5). Here,
s
= a+b+c = 2R.sin A+2R. sinB+2R. sin C 2
2
so that our Formula (5) becomes (7) K=Rr(sin A+sin B+sin C) Actually, what we have done in these last few cases is what we do when trying to invent identities. We will try to be a little more original in our next. We go back to our first diagram and. draw all three altitudes and the orthic triangle. There may be possibilities there. ..142..
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A
B e..------'L..------~C Ha There are lots of results we can get from this diagram that have little to do with areas. For example, note that HH a BH c is inscriptible, since there are right angles at H a and H c' so that
Finally,
the orthic triangle. However, we return to areas. Now we draw the circumcircle with center at O. Then it is too obvious that K
= a(OM a ) + b(OMb ) + c(OM c ) 222
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A
C \.:.:.:
:.:.:.:.::B
B
////../ .. c
\,/'7\
for us to list this as one of our formulas. However, let us construct BC parallel to BC, CA parallel to CA and AB parallel to AB. We now have two similar triangles with every line in the larger just twice as long as the corresponding line in the smaller. But line MA in this diagram corresponds to line OM a in the previous diagram, so that MA=2.0M a. It may not look like it, but it is true. Diagrams should be an aid, and not dictators. Similarly, MB=2.0M b and MC=2.0M e Now we can use our (unaccepted) rule to get (8) K
= a(AH)+b(BH=c(CH)) 4
The orthic triangle diagram seems to attract us a lot. Let us examine it more closely. First, BCHbH c is inscriptible, since there are right angles at H b and He Hence triangles ABC and AHbH c are similar. Also, AHcHH b is inscriptible, and moreover, the circle about these has diameter AH. Therefore, by the Law of Sines, HbHc=(AH)sinA. In the same manner, we get HbHa=(CH)sinC and HaHc=(BH)sinB. The perimeter of the orthic triangle is equal to p=(AH)sinA+(BH)sinB+(CH)sinC. But, from the Law -144
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of Sines, we get; sin A
= -.1L,
sin B
2R
= J;L
, sin C
2R
= ~. 2R
When we substitute these in what we have so far, we get p = (AH)a + (BH)b + (CH)c 2R
2R
2R.
Now compare this with our Formula (8). We find p
= _1 (a(AH) + b(BH) + c(CH)) = 4K. Therefore, 2R
2R
(9) K
= (pr) . 2
If you combine this with our Formula (4), you get p = (abc). What is more, using suitable substitutions 2R 2
of what we now have, we can invent more such formulas. If we introduce new elements, we naturally have more chances to invent new formulas for the area of a triangle. We have not used the centroid, 0, for example nor the incenter, I. Enough has been presented to show what we mean by saying that this is fun. There are just two more remarks we would like to add. First, prove 2
2
K = b sin2C+c sin2B . 4
Next, what new formulas can you invent for the area of a triangle? Note:
If you would prefer another 'problem on which to try your skills, consider the special points in a triangle -namely, 0, I, H, o. Find the distance from one of these points to any or all of the others. We found this very pretty and, in some cases, quite hard. -145
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KURSCHAK KORNER This column features those problems of the famous Hungarian Kurschak 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 MM's New Mathematical Library seri~s (available from the MM 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. 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. 1/1963.
Let p and q be integers greater than 1, and assume that pq chairs are arranged in p rows and q columns. Each chair is occupied by a student of different height. A teacher chooses the shortest student in each row; among these the tallest one is of height a. Then he chooses the tallest one from each column; among these the shortest is of height b. Determine which of the three cases, ab, are possible, and whether each of these cases can be attained by an appropriate rearrangement of the seating of the students.
2/1963.
Prove that if
a
is an acute angle, then
(1+_1_)(1+_1_) > 5
sinn
3/1963.
cosn
Prove that if a triangle is not obtuse, then the sum of the lengths of its medians is less than or equal to four times the radius of the triangle's circumscribed circle.
Dr. George Berzsenyi Rose-Hulman Institute of Technology Terre Haute, IN 47803
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OLYMPIAD PROBLEMS 1) A line through vertex A divides the triangle ABC into two isosceles triangles. It is given that one of the angles of triangle ABC equals 30. Determine in all possible cases the size of the other angles of the triangle. (Netherlands) 2) Solve in non-negative integers the equation 5 x 7 Y+4=3 z .
(Bulgaria) 3) Prove that if a, b, c are integers and a+b+c=O, then 2(a 4 +b 4 +c 4 ) is the square of an integer.
(U.S.S.R.) 4) A strictly increasing function f defined on (0, 1)
satisfies f(O)=O, f(1)=1, and .1< (f(x+y)-f(x)) <2 2 (f(x)-f(x-y)) for all x and y such that O<x+y< 1. f(.1)
< 76 .
3
135
Show that
(Finland) 5) Inside tri'angle ABC a circle of radius 1 is externally tangent to the incircle and to the sides AB and AC. A circle of radius 4 is externally tangent to the incircle and to sides BA and BC. A circle of radius 9 is externally tangent to the incircle and to sides CA and CB. Determine the inradius of the triangle. (U.S.A.) -147
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