AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS VOL 19 In Pólya’s Footsteps

Miscellaneous Problems and Essays

Ross Honsberger In P6lya's Footsteps Miscellaneous Problems and Essays The inspiration for mfulY of these problems cLune from the Olympiad Comer of Crux MathenulticorlUl1, now Crux Mathel1UlticorU111 }vith Mathenultical MaYhe111, published by the Canadian Mathematical Society with support from the University of Calgary, Memorial University of Newfoundland, and the lJniversity of Ottawa. Full attribution ell1 be found with each problem.

@1997 by The Mathematical Association ofAmerica (lncorporatedj Library (?f Congress Catalog Card Number 97-70506

Complete Set ISBN 0-88385-300-0 Vol. 19 ISBN 0-88385-326-4

Printed in the United States ofAmerica

Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 10.1090/dol/019

The Do/ciani Mathenlatical EX/Jositions

NUMBER NINETEEN

In P6lya's Footsteps Miscellaneous Problems and Essays

Ross Honsberger University) o.f Waterloo

Puhlished and Distrihuted hy THE yLt\THEMATICAL ASSOCL\TIO~ OF AMERICA THE DOLCIANI MATHEMATICAL EXPOSITIONS

Published by THE MATHEMATICAL ASSOCIATION OF AMERICA

Committee on Publications JAMES W. DANIEL, Chair

Dolciani Mathematical Expositions Editorial Board BRUCE P. PALKA, Editor CHRISTINE W. AYOUB EDWARD J. BARBEAU IRL C. BIVENS The DOLCJANI MATHEMATICAL EXPOSITIONS series of the Mathematical Association ofAlnerica was established through a generous gift to the Association from Mary P. Dolciani, Professor of Mathematics at Hunter College of the City University of New York. In making the gift, Professor Dolciani, herself an exceptionally talented and successful expositor of mathematics, had the purpose of furthering the ideal of excellence in mathematical exposition. The Association, for its part, was delighted to accept the gracious gesture initi­ ating the revolving fund for this series from one who has served the Association with distinction, both as a member of the Comlnittee on Publications and as a member of the Board of Governors. It was with genuine pleasure that the Board chose to name the series in her honor. The books in the series are selected for their lucid expository style and stimulating mathematical content. Typically, they contain an alnple supply of exercises, many with accompanying solutions. They are intended to be sufficiently elementary for the undergraduate and even the mathematically inclined high-school student to understand and enjoy, but also to be interesting and sometimes challenging to the more advanced mathematician.

1. Mathematical Gelns, Ross Honsberger 2. Mathematical Gelns 11, Ross Honsberger 3. Mathematical Morsels, Ross Honsberger 4. Mathematical Plums, Ross Honsberger (ed.) 5. Great Moments in Mathematics (Before 1650), Howard Eves 6. Maxima and Minima without Calculus, Ivan Niven 7. Great Moments in Mathematics (A.fter 1650), Howard Eves 8. Map Coloring, Polyhedra, and the Four-Color Prohlem, David Barnette 9. Mathematical Gems Ill, Ross Honsberger 10. More Mathematical Morsels, Ross Honsberger 11. Old and New Unsolved Problems in Plane Geolnetry and Numher Theory, Victor Klee and Stan Wagon 12. Problems .lor Mathematicians, Young and Old, Paul R. Halmos 13. Excursions in Calculus: An Intefplay oj'the Continuous and the Discrete, Robert M. Young 14. The Wohascum County Prohlenl Book, George T. Gilbert, Mark Krusemeyer, and Loren C. Larson 15. Lion Hunting and Other Mathenlatical Pursuits: A Collection 0.( Mathematics, Verse, and Stories by Ralph P Boas, if:, edited by Gerald L. Alexanderson and Dale H. Mugler 16. Linear Algebra Prohlem Book, Paul R. Halmos 17. From ErdtJs to Kiev: Proble/ns 0.( Olympiad Caliber, Ross Honsberger 18. Which Way Did the Bicycle Go? ... and Other Intriguing Mathenlatical A{vster­ ies, Joseph D. E. Konhauser, Dan Velleman, and Stan Wagon 19. In P6~ya 's Footsteps: Miscellaneous Problems and Essays, Ross Honsberger

Preface

Just as a recording of a Mozart concerto makes no pretense of teaching one to compose music~ these mathematical perfonnances are not motivated by a desire to teach mathematics; they are offered solely .for your enjoyment. There is no denying that a certain degree of concentration is required for the appreciation of their beautiful ideas~ but it is hoped that a leisurely pace and generous explanations \vill make them a pleasure to read. The technical demands are very modest; a high school graduate should be well equipped to handle many of the topics and a university undergraduate in mathematics ought to be perfectly comfortable throughout. I hope you will find something exciting in each of these topics-a sur­ prising result~ an intriguing approach~ a stroke of ingenuity-and that you will approach them as entertainment. You are certainly not required to at­ tempt these problems before going through the solutions~ but if you are able to give them a little thought first~ I'm sure you will find them all the more exciting. It is a pleasure to acknowledge the great debt this volume owes the un­ dergraduate problems journal Crux Mathematicarum, now Crux Mathemati­ carunl vvith Mathematical Mayhem, published by the Canadian Mathematical Society with support from the University of Calgary~ Memorial University of Newfoundland~ and the University of Ottawa. For interesting elementary problems~ this publication is in a class by itself. I came across the great major­ ity of the problems discussed in the present volume in the Olympiad Comer columns of Crux Mathenlaticorum. Sometimes a solution given in the present collection has been based on a solution published in this column~ but unless otherwise acknowledged~ the solutions here are based on my own work (with one exception) and in all cases the responsibility for shortcomings in presen­ tation is mine alone. The exception is the essay The Infinite Checkerboard~

vii viii In P61ya's Footsteps which was written probably twenty years ago, and with my deepest apologies~ I cannot recall where I encountered the topic. The sections may be read in any order. No attempt has been made to bring together all the problems on a specific subject; in fact, this has been avoided in the hope of encouraging a feeling of spontaneity throughout the work. I would like to extend my wannest thanks to chainnan Bruce Palka ofthe Dolciani Subcommittee and to the members Christine Ayoub and Irl Bivens for their generous reception of this manuscript and their perceptive revie\vs. Finally, it has again been my great good fortune to have had Beverly Ruedi and Elaine Pedreira see this work through publication; their unfailing geniality and excellence in all phases of this trying process are deeply appreciated. CONTENTS

Preface vii Four Engaging Problems 1 A Problem from the 1991 Asian Pacific Olympiad 6 Four Problems from the First Round of the 1988 Spanish Olympiad 9 Problem K797 from Kvant 14 An Unused Problem from the 1990 International Olympiad 16 A Problem from the 1990 Nordic Olympiad 19 Three Problems from the 1991 AIME 21 An Elementary Inequality 26 Six Geometry Problems 29 Two Problems from the 1989 Swedish Olympiad 36 Two Problems from the 1989 Austrian-Polish Mathematics Competition .42 Two Problems from the 1990 Australian Olympiad 45 Problem 1367 from Crux Mathematicorum 48 Three Problems from Japan 51 Two Problems from the 1990 Canadian Olympiad 59 A Problem from the 1989 U. S. A. Olympiad 65 A Problem on Seating Rearrangements 67 Three Problems from the 1980 and 1981 Chinese New Year's Contests .73 A Problem in Arithmetic 78 A Checkerboard Problem 81 Two Problems from the 1990 Asian Pacific Olympiad 85 Four Problems from the 1989 AIME 89 Five Unused Problems from the 1989 International Olympiad 98 Four Geometry Problems 111 Five Problems from the 1980 All-Union Russian Olympiad 121 The Fundamental Theorem of 3-Bar Motion 129

ix x In Polya's Footsteps

Three Problems from the 1989 Austrian Olympiad 132 Three Problems from the Tournament of the Towns Competitions 138 Problem 1506 from Crux Mathematicorum 143 Three Unused Problems from the 1987 International Olympiad 145 Two Problems from the 1981 Leningrad High School Olympiad 151 Four Problems from the Pi Mu Epsilon Journal-Fall 1992 156 An Elegant Solution to Morsel 26 165 Two Euclidean Problems from The Netherlands 168 Two Problems from the 1989 Singapore Mathematical Society Interschool Competitions 174 Problem M1046 from Kvant (1987) 177 Two Theorems on Convex Figures 179 The Infinite Checkerboard 184 Two Problems from the 1986 Swedish Mathematical Competition 188 A Brilliant 1-1 Correspondence 192 The Steiner-Lehmus Problem Revisited 194 Two Problems from the 1987 Bulgarian Olympiad 197 A Problem from the 1987 Hungarian National Olympiad 202 A Problem from the 1987 Canadian Olympiad 206 Problem 1123 from Crux Mathematicorum 208 A Problem from the 1987 AIME 211 A Generalization of Old Morsel 3 213 Two Problems from the 1991 Canadian Olympiad 216 An Old Chestnut 221 A Combinatorial Discontinuity 223 A Surprising Theorem of Kummer 229 A Combinatorial Problem in Solid Geometry 234 Two Problems from the 1989 Indian Olympiad 239 A Gem from Combinatorics 243 Two Problems from the 1989 Asian Pacific Olympiad 247 A Selection of Joseph Liouville's Amazing Identities Concerning the Arithmetic Functions a(n), T( n), ¢(n), Jl( n), A(n) 251 A Problem from the 1988 Austrian-Polish Mathematics Competition 267 An Excursion into the Complex Plane 271 Two Problems from the 1990 International Olympiad 280 Exercises 289 Solutions to the Exercises 293 Praise God from Whom all blessings flow.

Exercises

These exercises are generally much easier than most of the problems we have been considering and they are included just for your amusement. They are questions, some slightly revised, that occurred on various Grade 13 Problems Papers ofOntario, a defunct examination which used to be taken by candidates writing for university scholarships involving mathematics.

1. (1960). If the lengths of the sides of a triangle are in the ratios 3 : 7 : 8, show that its angles are in arithmetic progression.

2. (1947). A candidate at an examination writes four papers. Ifthe maximum number of marks obtainable on each paper is m, show that the number of ways of obtaining a total of 2m marks is !(m + 1) (2m2 + 4m + 3).

3. (1945). Find the coefficient of x 6 yzt5 in the expansion of

4. (1950). Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola is drawn through it, then 1 1 PK2 + QK2 is constant for all positions of the chord.

5. (1949). Find the roots of

4 3 2 17x + 36x - 14x - 4x + 1 == 0, given that they are in harmonic progression.

289 290 In P61ya's Footsteps

6. (1936). A and B are arbitrary points inside a given circle K. Construct a circle through A and B to touch K.

7. (1938). Reduce

to its simplest surd fonn.

8. (1937). A quadratic expression in x is posItIve except for the range - 2 :S x :S 1, and a second quadratic expression is negative except for the range -1 :S x :S 4. Each expression has the value 60 when :r == 3. For what other value of x are these expressions equal?

9. (1940). Find the equation of the locus of a point P which moves so that 2 2 the tangents from P to the circle x + y2 == a cut off a segment of length 2a on the line x == a.

10. (1937). Show that for any positive integer p and any integer s > 1 there are p consecutive odd numbers whose sum is pS, and find the first of these numbers.

11. (1945). Find a polynomial f(x) of degree five such that f(x) - 1 IS 3 3 divisible by (x - 1) and f (x) is itself divisible by x .

12. (1958). A unifonn cylindrical tub can be filled with water from the cold water tap in 14 minutes, and drained completely through a hole in the bottom in 21 minutes. When both hot and cold water taps are turned on and the plug removed from the hole, the tub fills in 12.6 minutes. How long would it take the hot water tap to fill the tub by itself?

13. (1934). A, B, C, and D are arbitrary fixed points in 3-space and P is a variable point. Show that PA 2 + PB 2 + PC2 + PD 2 is a minimum for P at the midpoint of the segment which joins the midpoints of .<4C and BD.

14. (1951). Show that the equation 1 1 1 1 --+--==-+-­ x+2 y+2 2 z+2

is not satisfied by any set of positive integers x, y, Z, in which :r ~ 4. Hence find all solutions (x. y ~ z) in positive integers. Exercises 291

15. (a) (1956). If an ellipse JI rolls without slipping on an identical fixed ellipse E, starting with them touching at a pair of vertices, prove that the locus of a focus of l\I is a circle with center at a focus of E.

(b) (1956). If a parabola 1.\1 rolls without slipping on an identical fixed parabola P, starting with them touching at their vertices, prove that the locus of the focus of ~~1 is the directrix of P. (More interestingly, the locus of the vertex is a cissoid of Diocles, a curve famous for providing a solution to the ancient problem of doubling the cube.)

16. (1934). Prove that

2 0 • • 0 • 0 • 0 sin 50 SIn 10 + SIn 3 + SIn 5 + ... + SIn 99 == . sin 10 17. (1942). Of the 9! numbers formed by pennuting the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, how many are between 678000000 and 859000000?

18. (a) (1959). Show that the segment of a tangent to a hyperbola which is intercepted between the asymptotes is bisected by the point of contact and that, with the asymptotes, such a segment fonns a triangle of constant area.

(b) (1954). A straight line cuts a hyperbola at the points Q and Rand its asymptotes at P and S. Prove the midpoint of QR is also the midpoint of PS.

19. (1961). Given that one side of a triangle is twice as long as another side, show that the angle opposite the longer of these sides is greater than twice the angle opposite the lesser side.

20. (1935). If II, 12 , 13 are the centers of the escribed circles of 6ABC, prove that the area of abc 61 1 1 == -. 1 2 3 2r 21. (1943). A triangle ABC is such that 3AB == 2AC. Also, a point D on BC is such that BD == 2DC and AD == BC. Show that tan~LADB= f5. 2 V19 22. (1950). Let x be a number between -1 and 1. Noting that n 2 n(n + 1) - n, find a fonnula for the sum of the infinite series 292 In P6lya's Footsteps

23. (1942). If m == esc () - sin () and n == sec () - cos (), show that

m 2/3 + n 2/3 == (mn)-2/3.

24. (1961). If 8 n denotes the sum of the first n positive integers, find the sum of the infinite series s== -+-+-+81 82 83 ... +--+8 n .... 1 2 4 2n - 1

25. (1941). Find the smallest positive value of

4tan-1 --1 tan-1 -1 + tan-1 -1 5 70 99 Solutions to the Exercises

1. (1960). The key to this problem is the recognition that the angles of a triangle are in arithmetic progression if and only if one of them is 60°: if an angle is 60°, the amount a second angle is less than 60° is precisely the amount the third angle must exceed 60° in order to make the sum 180° (in general, three numbers are in arithmetic progression if and only if one of them is equal to their average).

In a scalene triangle, then, the 60° angle must be the mid-sized angle and therefore be opposite the side of intermediate length. Thus if the sides are of lengths 3k, 7k and 8k, we need only show that the angle x opposite the side of length 7k is 60°. But this is immediate from the law of cosines:

2 2 2 49k == 9k + 64k - 2 . 3k . 8k cos x~ 2 2 73k - 49k 24 1 cos x == 48k2 48 2

2. (1947). The exponents in the generating function

f (x) == 1 + x + x 2 + x 3 + ... + x'm

are the possible scores on a paper. We may consider scores of a, b, C, and d on a b c d the four papers to be registered in the product f (;r)4 by the term x + + + :

f(X)4 == (1 + x + x2 + + x m)(l + x + x 2 + + xTn)x

2 r7l 2 17l (1 + x + x + + x )(l +:T + x + + x ) == ... + (xa)(xb)(xc)(.rd) + ...

293 294 In P6lya's Footsteps

From this viewpoint, the number of ways of accumulating a total score of 2m is the number of ways of generating the tenn x 2m in this product, makin~ the required number simply the coefficient of x 2m in f(x)4. Letting [xn]g(x ~ denote the coefficient of x n in the function g(x), we want to detennine tht number [x 2m]f(x)4. Now

f(X)4 == (1 + x + x 2 + x 3 + ... + Xm )4 rn 1 == (1_X + )4 I-x == (1 - x m +1 )4(1 - x)-4, and so the required number is == [x2m](1- Xm +1)4(1_ x)-4,

2m 2 == [x 2m](1 - 4xm +1 + 6x + - ...)(1 - x)-4 == [x 2m](1 - 4xm +1 )(1 - x)-4

m 1 2m (higher powers of x + do not contribute to the coefficient of x )

2m m 1 i == [x ](1 _ 4x + ) L: (4 +: -l)x i~O

2m m 1 ~ ~ = [x ](1 - 4x + ) C 3) Xi m 3 = C3+ )_4(m;2) == (2m + 3)(2m + 2)(2m + 1) _ 4. (m + 2)(m + l)m 6 6 2(m+1) 2 == (4m2 + 8m + 3 - 2m - 4m) 6 m+1 == -3-(2m2 + 4m + 3),

as required.

3. (1945). Since the term x 6 y zt5 contains only a single y, in multiplying 2 2 out the product (x + y + z+ t) lO(X + z) (x + y2), from the factor (x + y2) it 2 6 5 must be the x that is taken toward the formation of a term in x yzt . Thus 2 [X6yzt5](X + Y + z + t)10(X + z)(x + y2)

==[X4yzt5](x + Y + z + t)10(X + z) Solutions to the Exercises 295

==[x4yzf5J(x + Y + z + t)lO. 1-~ + [X 4yzt5](X + Y + z + t)lO. Z ==[X3yzt5](X + Y + z + t)10 + [x4yf5](~r + y + z + t)lO

= C30) G) G) G) + C40) G) G) 10 ·9·8 10 . 9 ·8· 7 3.2 ·7·6·1+ 4.3.2 ·6·1 ==5040 + 1260 ==6300.

2 4. (1950). Let the parabola be y2 == x, let P(a ~ a) be any point on it and K(k,O) be a point on its axis (figure 189). Then

1 1 PK2 (a 2 -k)2+ a2'

2 2 The slopem of PK ism == a/(a - k), implying a - k == aim, and therefore

1 1 PK2

y

2 Y =x

--0+------+-----.;.----- X

FIGURE 189 296 In P61ya's Footsteps

2 Let PK extend into chord PQ and that Q is (b . - b). Then .. b slope QK == 'm == k _ b2 . implying k - b2 == ~, giving

,) 1 1 1 m- QK2

Then

Now a+b 1 slope PQ == m == 2 b2 == --b· a - a- implying

a 2 + b2 N== (a-b)2 2 2 1 a b [ + (a ~ b)2 ]

2 2 Also m == al(a - k) implies a - k == aim, giving

2 a 2 k == a -- == a - a(a - b) == abo m

Thus we want to show that there is a value of ab which makes j\;'" a constant. One way this could happen is if

a 2 + b2 ----== 1 (a - b)2 + 1 ' making 1 1 N == a 2 b2 == k 2 ~ a constant for all chords through K. In this case, we would have

giving ab == ~ and N == -&- == 4. Thus N is constant for all chords through the point K ( ~ , 0), and, after observing that the focus ofour parabola is (±. 0), the required point for the parabola y2 == 4p:r: is K(2p, 0). Solutions to the Exercises 297

5. (1949). Since the roots are in hannonic progression, by definition their reciprocals are in arithmetic progression. Thus, let y == ~ and let the four values of y be a - 3d, a - d, a + d, and a. + 3d. Then x == 1.y and these values of yare the roots of the equation 17 36 14 4 - + - - - - - + 1 == O. y4 y3 y2 Y that is,

y4 _ 4y3 - 14y2 + 36y + 17 == O. Thus

the sum of the roots == 4a == 4, and a == 1, and the roots are 1 - 3d, 1 - d, 1+d, and 1+3d. Also the sum ofthe products of the roots taken two at a time is

2 2 2 2 (1 - 4d + 3d ) + (1 - 2d - 3d ) + (1 - 9d ) + (1 - d )

2 2 +(1 + 2d - 3d ) + (1 + 4d + 3d ) == -14, giving

-10d2 + 6 == -14,

d2 == 2,

d == ±J2. Both these values of d give the same set of roots,

y == 1 ± 3J2 and 1 ± J2.

Hence the required roots x are the reciprocals

1 1. _-_1_±_3_J2_2 and _ 1 ± J2. 1 ± 3J2' 1 ± J2' I.e., 17

6. (1936). Let the desired circle touch K at T and let the common tangent at T meet the chord XY, through A and B, at the exterior point P (figure 190). Then

PT2 == PX . PY == PA . P B,

implying PX PB PA PY 298 In P61ya's Footsteps

I----~----~~r__-~p A B

K T

FIGURE 190

Subtracting 1 from each side yields PX-PA PB-PY PA PY that is, XA BY AP yp' and we have XA AP YB pya That is to say, P can be found on XY extended as the point which divides AY externally in the known ratio XA: YB. Thus, after first locating P, draw the tangent to K from P to get T and then draw the circle through A, B, and T.

7. (1938). Observe that 6- 2J5 == (J5 -1)2 and 9 - 4J5 == (J5 - 2)2. We have V3-J5 J2-V7-3J5 J2 + V7 - 3J5 J2 - V7 - 3J5 V6 - 2J5 - V36 - 16J5 3J5 - 5 J5 - 1- 2( J5 - 2) 3J5 -5 Solutions to the Exercises 299

3- V5 :3)b -.5 3 - V5 3)b + .5 3V5 - ·5 . 3)b + .5 -:l)b vb -- -- 20 5

8. (1937). For some value a, the first expression must be a(x - l)(x + 2), and for some value b, the second expression must be b(x - 4)(x + 1). Since, for x == 3, we get

a(2)(5) == 60 == b(-1)(4)~

it follows that a == 6 and b == -1.5. Hence the expressions are equal when 6(x - l)(x + 2) == -15(x - 4)(x + 1), 2(x - l)(:r + 2) == -5(x - 4)(x~ + 1),

2 7x - 13x - 24 == 0 (x - 3)(7x + 8) == O.

giving x == 3 and -~.

9. (1940). Let Q(a. k) and R(a. k + 2a) be any two points on x == a which are a distance 2a apart and let the tangents from Q and R to the given circle meet at P(x,y) (figure 191). We want the locus of P. A non-vertical straight line through Q has equation

y - k == m(x - a)~ i.e., mx - y + k - am == 0, 'm finite.

To be a tangent to the given circle, we must have k-am --;=::::;:======±a. J'm2 + 1 k - a:rn == ±ay/m2 + 1.

2 2 2 2 2 2 k - 2am.k + a m == a m + a , 2 2 k - a --- == m.. 2ak implying the equation of the tangent is

2 k -a2 k 2 -02 ---.T - Y + k - == 0 2ak' 2k 300 In P61ya's Footsteps

x=a y

R (a, k + 2a)

2a

Q (a, k)

---+-----I-----+----x o

2 2 2 X + Y = a

FIGURE 191

2 2 3 (k2 - a2)x - 2aky + 2ak - ak + a == 0

2 2 2 3 (k - a )x - 2aky + ak + a == o. (1)

Similarly, the tangent through R has equation

[(k + 2a)2 - a2]x - 2a(k + 2a)y + a(k + 2a)2 + a3 == O. (2)

Thus the equation of the locus in question is obtained by eliminating k from equations (1) and (2). Subtracting (1) from (2) gives

which, divided by 4a, yields

2 (k + a)x - ay + ak + a == 0, giving a(y - x - a) k(x+a)==a(y-x-a) and k== . x+a Solutions to the Exercises 301

Substituting in (1), we get

a2(y-x-a)2 _ 2] _. a(y-x-a) a2 (y-x-a)2 3_ 2 a x 2ay + a ( )2 + a - 0, [ ()x+a x+a x+a which, after considerable but straightforward simplification, reduces to

Thus the locus consists of all points on the parabola having vertex at (a, 0) and the x-axis for its axis except for the points given by x == ±a, which require special interpretation to be admitted and are better rejected altogether.

10. (1937). (a) Ifp is odd, say 2n + 1, then we wish to solve the following equation for a:

(a- 2n) +... + (a -4) + (a- 2) +a+ (a+ 2) + (a+4) +... + (a+2n) == pS that is, (2n + l)a == pS, and since (2n + 1) == p, we may take a == ps-l, which is odd, and begin at the number

a - 2n == ps-l - (p - 1) == ps-1 - P + 1.

(b) If p is even, say 2n, we similarly wish to solve

(a - 2n) + ... + (a - 4) + (a - 2) + a + (a + 2)+(a + 4) + ... + (a + 2n - 2) == pS 2na - 2n == pS 2n(a - 1) == p(a - 1) == pS a==ps-l+1, which is odd, and again having the series stat1 at the same number pS -1 - p+1:

a - 2n == a - p == ps-l - P + 1.

11. (1945). Let

f(x) == (x - 1)3(ax2 + bx + c) + 1 == (x3 - 3x2 + 3x - 1)(ax2 + bx + c) + 1. 302 In P6lya's Footsteps

3 2 Since f (x) is divisible by x , the terms in x and :r, as well as the absolute term, must vanish. Thus

the absolute term == -c + 1 == 0, giving c == 1. the coefficient of x == 3c - b == 3 - b == 0, giving b == 3. the coefficient of x 2 == -3 + 3b - a == 0, giving a == 6. and making

3 2 2 f(x) == (x - 3x + 3x - 1)(6x + 3x + 1) + 1

5 4 3 == 6x - 15x + 10x . 12. (1958). Let V denote the volume of the tub and let the hot water tap by itself take x minutes to fill the tub. Thus the hot water tap supplies water at the rate of ¥- units per minute, the cold water tap supplies water at the rate of ~ units per minute, while the hole in the bottom drains water at the rate of ~ units per minute. Thus VVV \tl -; + 14 - 21 == 21.6' 1 1 1 1 1 7 ;; == 21.6 + 21 - 14 == 126 (10 + 6 - 9) == 126' giving 126 . x == - == 18 mInutes. 7

13. (1934). Let the midpoints of AC and BD be X and Y and the midpoint of XY be Z (figure 192). Our solution is based on the theorem that the sum of the squares of two sides of a triangle is equal to twice the square of the

c

x z

A

B "------~------...;;;:- D Y

FIGURE 192 Solutions to the Exercises 303

p

x

y

FIGURE 193

median to the third side increased by one-half the square of the third side. When the point P is at Z, we have both

and therefore

ZA2 + ZB2 + ZC2 + ZD2 1 == 2ZX2 + 2Zy2 + -(AC2 + BD2) 2 =4ZX2 + ~(AC2 + BD2) (recall Z bisects XY) 2 = Xy2 + ~(AC2 + BD2). 2 For P in general position, we have similarly that

PA2 + PB2 + PC2 + PD2 == (PA2 + PC2) + (PB2 + PD2) == 2PX2 + ~AC2 + 2py2 + ~BD2 2 2 = 2(PX2 + py2) + ~(AC2 + BD2). 2 In order to show a minimum occurs when P is at Z, then, we need to show that

(1)

But, in 6.PXY (figure 193), PZ is a median and we have

PX2 + py2 == 2PZ2 + ~Xy2 > ~XY2 22' from which (1) follows immediately, completing the solution. 304 In P6lya's Footsteps

14. (1951). If X 2: 4, then 1 1 1 1 1 1 - + -- == -- + -- < - + --. 2 z+2 x+2 y+2-6 y+2 giving 1 1 1 1 1 ---==-<----- 26 3-y+2 z+2·

But y 2: 1, implying y + 2 2: 3, and Y~2 ~ ~. Hence

1 1 1 -- - -- <-~ y+2 z+2 3' a contradiction, and it follows that x < 4. Now, if both x and yare greater than 1, then each of x~2 and Y~2 is less than or equal to ~, making it impossible for x~2 + Y~2 to exceed ~, which it must do in order to be equal to ~ + Z~2. Hence either one or both of x and y must equal 1. Also, the equation is symmetrical in x and y, implying a limit of 3 on the value of y as well as on x, and that the values of x and y can be interchanged for the same value of z. For x == 1 and y == 1, we get z == 4, and the solution (x, y, z) == (1~ 1,4); also

x == 1, Y == 2, gives (1,2,10); x == 1, Y == 3, gives (1,3,28); and interchanging x and y we also have the additional two solutions

(2, 1, 4) and (3, 1, 28) .

15. (1956). (a) In all positions, ]vI is the reflection of E in the common tangent at their point of contact P. Therefore the focal radii to P in AI are respectively the same lengths as those to P in E. However, in figure 194, the reflector property of an ellipse gives Lx == Ly, and the reflection of E in the tangent gives Lx == Lz. Thus y == z and APA' is straight. Hence at every position AA' == rl + r2 == 2a, the length of the major axis, implying the locus of A' is the circle A (2a). Similarly B' traces the circle B (2a). (b) This is similar to part (a). The solution is based on the following property: the perpendicular from the focus of a parabola to a tangent meets the tangent at the point where it crosses the tangent at the vertex. The details are left as an additional exercise. Solutions to the Exercises 305

B M start

FIGURE 194

16. (1934). Clearing of fractions, we need to show that 2 sin 1° sin 1° + sin 1° sin 3 C + sin 1° sin 5° + ... + sin 1° sin 99° == sin 50°. But this is immediate from sin A sin B == - ~ [cos(A + B) - cos(A - B)J:

1 0 the left side == - - [(cos 2° - cos 0°) + (cos 4 - cos 2°) 2 + (cos 6° - cos4°) + ... + (cos 100° - cos 98°)J

= - ~ (cos 1000 - cos 00 ) 2

2 0 = -~ [(1 - 2sin 50 )- 1] 2 == sin2 50°.

17. (1942). The initial digits of the integers in question can be classified nicely into 16 types in each of which the remaining digits may be appended in any order: 678 , giving 6! ways of putting in the remaining 6 digits, 679 , giving 6! ways of putting in the remaining 6 digits, 68 , giving 7! ways of putting in the remaining 7 digits, 69 , giving 7! ways of putting in the remaining 7 digits, 7 , giving 8! ways of putting in the remaining 8 digits, 81 , giving 7! ways of putting in the remaining 7 digits, 82 , giving 7! ways of putting in the remaining 7 digits, 83 , giving 7! ways of putting in the remaining 7 digits, 84 , giving 7! ways of putting in the remaining 7 digits, 306 . In P6lya's Footsteps

851 , giving 6! ways of putting in the remaining 6 digits, 852 , giving 6! ways of putting in the remaining 6 digits, 853 , giving 6! ways of putting in the remaining 6 digits, 854 , giving 6! ways of putting in the remaining 6 digits, (no integer can begin 855 ... since there is only one 5) 856 , giving 6! ways of putting in the remaining 6 digits, 857 , giving 6! ways of putting in the remaining 6 digits.

Hence the required total is

8 . 6! + 6 . 7! + 8! == 6!(8 + 42 + 56) == 720 . 106 == 76320.

2 2 2 2 2 18. (a) (1959). If P(s, t) is a point on the hyperbola b x - a y2 == a b , the equation of the tangent QPR is

2 2 2 Solving with the equation of the asymptotes, b x - a y2 == 0, we have

y

Q

------tf-----w----+-+------x

~ bx±ay=O /

FIGURE 195 Solutions to the Exercises 307 therefore 2 y b2 (a2b2 + a t )2 _ 2,2 = 0 b2 s a y .

2 2 2 a b + a ty )_ b ( b2s - ±ay~

ab2 + aty = ±bsy~ ab2 = y(±bs - at), giving ab2 ab2 y - or - bs - at -bs - at' Hence the ordinate of the midpoint of QR is 2 2 1 (ab ab ) y = 2 bs - at - bs + at 2 = ab (bS + at - bs + at) 2 b2 s 2 - a2 t 2 a 2 b2 t b2s2 - a2t 2 .

2 2 2 2 2 2 But (s, t) satisfies the equation of the hyperbola, giving b s - a t = a b , and implying Y = t. Substituting the values of y in the equation of the asymptotes, similar calculations give Q and R to have the coordinates

2 2 2 2 Q ( a b , ab ) and R ( a b , -ab ). bs - at bs - at bs + at bs + at Accordingly, it is easily checked that the abscissa of the midpoint of Q R is t, making the midpoint P, as required. Also, the area of 6QOR is easily found to be the constant ab for all values of sand t.

2 2 2 2 2 (b) (1954). Solving the equations b x - a y2 = a b and y = mx+ k for the coordinates of Q and R (figure 196), we have

2 2 2 2 2 2 2 2 b x - a (m x + 2mkx + k ) = a b , 2 2 2 2 2 2 2 2 (b - a m )x - 2a mkx - a (k + b ) = 0, (1) giving the x-coordinate of the midpoint At of QR to be 2 2 1 ( 2a mk) a mk 2 b2 - a 2 m 2 = b2 - a 2 m 2 . 308 In P6lya's Footsteps

y

-----t----~---_+___+----x o

s

FIGURE 196

2 2 2 Solving the equations b x - a y 2 == 0 and y == mx + k for the midpoint of PS clearly does not alter the coefficients of x 2 and x in the equation corresponding to (1) (only the absolute term is changed), and so the same x-coordinate is obtained for the midpoint. Similarly for the y-coordinates.

19. (1961). In figure 197, we would like to show that ¢ > O. By the law of sines, we have x 2x ( x) sin () == sin 2¢ == sin ¢ cos ¢ , implying sin () == sin ¢ cos ¢.

x

FIGURE 197 Solutions to the Exercises 309

Now 0 > O~ implying cos 0 < 1, and therefore

sin e < sin o.

0 But because 20 < 180 (in the triangle)~ then (J) < 90° ~ and since the sine 0 function is increasing in the range (0°. 90 ) ~ sin e< sin dJ implies () < ¢.

20. (1935). Recall the standard relations 1. AL == s~ the semiperimeter of triangle ABC~ 2. the area ~ oftriangle ABC is given by TS and by J s(s - a)(s - b)(s - c)~ 3. (from similar triangles~ see the figure) r s-a giving rs ~ rl == -- == --~ s-a s-a and similarly for T2 and r3. From figure 198, we have that 1 ~ABII == -CTI 2

FIGURE 198 310 In P61ya's Footsteps

and hence, adding, we get ABI1C == ~rl (b + c); similarly for .4BC12 and ACB13 , and we have altogether that

~111213 == ABI1C + ABC12 + ACB13 - 2~ABC

= Dr1(b+C)] + [~r2(a+c)] + Dr3(a+b)] -2~ 1 = 2h(2s - a) + r2(2s - b) + r3(2s - c)] - 2~

= ~ {rds + (s - a)] + r2 [s + (s - b)] + r3 [s + (s - c)J} - 2~ 1 [s(rl r2 + r3) + 3rsl - 2rs (recall rl(s - a) rs ==~, etc.) == -2 + - == 1 == 2s(r1 + r2 + r3 - r).

Thus we want to show that 1 abc -s(r1 +r2 +r3 -r)==-2' 2 r that is,

or equivalently, ~ ~ ~ ~ abc --+--+----==- 8-a s-b s-c s ~' or finally, ~2 ~2 ~2 ~2 -- + -- + -- -- == abc. s-a s-b s-c s Recalling that ~ == vis(s - a)(s - b)(s - c), the left side is s(s - b)(s - e) + s(s - a)(s - e) + s(s - a)(s - b) - (s-a)(s-b)(s-c)

2 2 2 2 == (s3 - bs - es + bcs) + (s3 - as - cs + acs)

2 2 + (s3 - as - bs + abs) - (s3 - bs2 - cs2 + bcs - as2 + abs + acs - abc)

== 2s3 - s2(2a + 2b + 2c - a - b - c) + s(be + ae + ab - be - ae - ab) + abe == 2s3 - s2(2s) + abe

== abc. Solutions to the Exercises 311

A

B 2y D y c

FIGURE 199

21. (1943). In LABD (figure 199), the semiperimeter s = x + ~y, and therefore

1 (s - a)(s - b) (x+~)(x-~) tan 2LADB = s(s - d) (x + ~y)(~y - x)

25 2 - x 2 4 y Now in LABD, the law of cosines gives

implying 2 /ADB = 13y2 - 4x COSL 2 2 1 y similarly, from LADC we get

and 9x2 - 10y2 cosLADC = 2. 6y But cos LADE = - cos LADC (supplementary angles), and so 13y2 - 4x2 9x2 - 10y2 12y 2 6y 2 13y2 - 4x2 = 18x2 - 20y2, 312 In P6lya's Footsteps

2 33y 2 == 22x ~

2 3y 2 == 2x . giving

Hence

1 6y2 - y2 (5 tan -LADB == 2 25 y 2 - 6y 2 == V19'

22. (1950).

== (1· 2 - 1) + (2·3 - 2)x + (3·4 - 3)x2

n 1 + ... + [n(n + 1) - n]x - + ....

Now, the order of these tenns may be rearranged provided the series is abso­ lutely convergent for -1 < x < 1. This is the case, but at this point let us simply assume it. Then

2 n 1 S == [1 ·2 + 2 . 3x + 3 . 4x + ... + n(n + l)x - + ...] 2 n 1 - (1 + 2x + 3x + ... + nx - + .. '),

which is the derivative of (1 + 2x +3x2 +... +nxn - 1 +...) minus the series 2 n 1 itself. Since 1 + 2x + 3x + ... + nx - + ... == (1- x)-2, we have

s = d(l - x)-2 _ (1 _ x)-2 dx 2 1 (l-x)3 (l-x)2 2-(I-x) (1 - x)3 l+x (1 - x)3 . Solutions to the Exercises 313

To justify our assumption concerning convergence, we need to check this result. We have

[xn-1](1 + x)(l - ~r)-3 == [xn-1](1 + x) L (i ~ 2)Xi i~O (n + l)n + n(n - 1) 2

as desired.

23. (1942). We have

1 m == esc () - sin () == -:--() - sin () SIn 2 1 - sin () sin () and 1 n == sec () - cos () == --() - cos () cos 2 1 - cos2 () sin () cos () cos () . Therefore mn == sin () cos () . We want to show that

2 C~S2 ()) 2/3 + (sin ()) 2/3 1 ) 2/3 ( sIn () cos () ( sin () cos () that is,

4 3 COS4 / 3 () sin / () 1 --- + == ------sin2/ 3 () cos2/ 3 () sin2/ 3 () cos2 / 3 ()' or 4 3 2 3 COS4 / 3 () cos2 / 3 () + sin / () sin / () 1 2 3 2 3 sin / () cos2 / 3 () sin / () coS2 / 3 () . But, since

4 3 2 3 4 3 2 3 2 2 COS / () COS / () + sin / () sin / () == cos () + sin () == 1, this is clearly so, and because all the steps in our derivation are reversible, the desired conclusion follows. 314 In Polya's Footsteps

n(n + 1) 24. (1961). Sn == , and therefore 2 S == " n(n + 1) ~ 2n n2::1

evaluated at x == ~,

d 2 3 4 1 == - (x + 2x + 3x + ...) dx at x == 2'

== -d [x 2(1 + 2x + 3x2]+ ...) 1 dx at x == 2' 2 d[x (1 - X)-2] 1 dx at x == 2' 2 - 1 == [2x(1 - X)-2 + x (-2)(1 - x)-3(-1)] at x - 2'

1) -2 1 ( 1) -3 ==1· ( 2 +4. 2 2 ==4+4

== 8.

1 1 1 25. (1941). Let tan- .g. == x, tan- /0 == y, and tan- gIg == z. Then we want the smallest value of 4x - y + z == 4x - (y - z). Now,

2 tan 2x 2 ( 1 ~t::n~ x ) tan4x == 2 1- tan 2x 1 (2 tan x ) 2 - 1 - tan2 x

4 tanx(l - tan2 x) == (1-tan2 x)2-4tan2 x 4(.g.)(1 - <15) (1 - is)2 - 4(25) 4(5)(25 - 1) (25 - 1)2 - 4(25) 480 120 -- 476 119 Solutions to the Exercises

Also, tan y - tanz _ 99 -70 tan(y - z) = 1 + tanytanz 70 . 99 + 1 29 1 == 6931 == 239 Therefore tan4x - tan(y - z) tan(4x - y + z) = tan[4x - (y - z)] = 1 + tan 4x tan(y _ z) ~ - 2k 120 . 239 - 119 == 120 1 119.239+120 1 + 119 . 239 == 28561 == 1. 28561 Hence the desired

1r 4x - y + z == -. · 4

AMS / MAA DOLCIANI MATHEMATICAL EXPOSITIONS

In Pólya’s Footsteps Ross Honsberger was born in Toronto, Canada, in 1929 and attended the University of Toronto. After more than a decade of teaching mathe matics in Toronto, he took advantage of a sab batical leave to continue his studies at the University of Waterloo, Canada. He joined its faculty in 1964 in the Department of Combinatorics and Optimization, and has been there ever since. He has published a number of best selling books with the Mathematical Association of America, including Episodes in Nineteenth and Twentieth Century Euclidean Geometry, Hendricks Photography and From Erdo˝s to Kiev. In Pólya’s Footsteps is his eighth book published in the Dolciani Mathematical Exposition Series. The study of mathematics is often undertaken with an air of such seriousness that it doesn’t always seem to be much fun at the time. However, it is quite amazing how many surprising results and brilliant arguments one is in a posi- tion to enjoy with just a high school background. This is a book of miscellaneous delights, presented not in an attempt to instruct but as a harvest of rewards that are due good high school students and, of course, those more advanced—their teachers, and everyone in the university mathematics community. Admittedly, they take a little concentration, but the price is a bargain for such gems. A half dozen essays are sprinkled among some hundred problems, most of which are the easier problems that have appeared on various national and inter- national olympiads. Many subjects are represented—combinatorics, geometry, number theory, algebra, probability. The sections may be read in any order. The book concludes with twenty-fi ve exercises and their detailed solutions. It is hoped that something to delight will be found in every section—a surprising result, an intriguing approach, a stroke of ingenuity—and that the leisurely pace and generous explanations will make them a pleasure to read. The inspiration for many of the problems came from the Olympiad Corner of Crux Mathematicorum, published by the Canadian Mathematical Society.