Ingenuity in Mathematics Ross Honsberger 10.1090/Nml/023

AMS / MAA ANNELI LAX NEW MATHEMATICAL LIBRARY VOL 23 Ingenuity in Mathematics Ross Honsberger 10.1090/nml/023

INGENUITY IN MATHEMAmCS NEW MATHEMATICAL LIBRARY PUBLISHED BY THEMATHEMATICAL ASSOCIATION OF AMERICA

Editorial Committee Basil Gordon, Chairman (1975-76) Anneli Lax, Editor University of California, L.A. New York University Ivan Niven (1 975-77) University of Oregon M. M. Schiffer (1975-77) Stanford University

The New Mathematical Library (NML) was begun in 1961 by the School Mathematics Study Group to make available to high school students short expository books on various topics not usually covered in the high school syllabus. In a decade the NML matured into a steadily growing series of some twenty titles of interest not only to the originally intended audience, but to college students and teachers at all levels. Previously published by Random House and L. W. Singer, the NML became a publication series of the Mathematical Association of America (MAA) in 1975. Under the auspices of the MAA the NML will continue to grow and will remain dedicated to its original and expanded purposes. INGENUITY IN MATHEMATICS

by Ross Honsberger

University of Waterloo, Canada

MATHEMATICAL ASSOCIATION OF AMERICA Illustrated by George H. Buehler

Sixth Printing

A11 rights reserved under International and Pan-American Copyright Conventions. Published in Washington by The Mathematical Association of America

Library of Congress Catalog Card Number: 77-134351 Print ISBN 978-0-88385-623-9 Electronic ISBN 978-0-88385-938-4

Manufactured in the United States of America Note to the Reader

his book is one of a series written by professional mathematicians Tin order to make some important mathematical ideas interesting and understandable to a large audience of high school students and laymen. Most of the volumes in the New Mathematical Library cover topics not usually included in the high school curriculum; they vary in difficulty, and, even within a single book, some parts require a greater degree of concentration than others. Thus, while the reader needs little technical knowledge to understand most of these books, he will have to make an intellectual effort. If the reader has so far encountered mathematics only in classroom work, he should keep in mind that a book on mathematics cannot be read quickly. Nor must he expect to understand all parts of the book on first reading. He should feel free to skip complicated parts and return to them later; often an argument will be clarified by a subse- quent remark. On the other hand, sections containing thoroughly familiar material may be read very quickly. The best way to learn mathematics is to do mathematics, and each book includes problems, some of Yhich may require considerable thought. The reader is urged to acquire the habit of reading with paper and pencil in hand; in this way mathematics will become in- creasingly meaningful to him. The authors and editorial committee are interested in reactions to the books in this series and hope that readers will write to: Anneli Lax, Editor, New Mathematical Library, NEW YORKUNIVERSITY, THE COURANTINSTITUTE OF MATHEMATICAL SCIENCES, 251 Mercer Street, New York, N. Y. 10012.

The Editors NEW MATHEMATICAL LIBRARY

1. Numbers: Rational and Irrational by Ivan Niven 2. What is Calculus About? by W. W. Sawyer 3. An introduction to Inequalities by E. F. Beckenbach and R. Bellmun 4. Geometric Inequalities by N. D.Kuzarinoff 5. The Contest Problem Book I Annual High School Mathematics Examinations 195&1960. Compiled and with solutions by Charles T. Salkind 6. The Lore of Large Numbers by P. J. Davis 7. Uses of Infinity by Leo Zippin 8. Geometric Transformations I by 1. M. Yaglom, translated by A. Shields 9. Continued Fractions by Carl D.Olds 10. Replaced by NML-34 Hungarian Problem Books I and 11, Based on the Eotviis Competitions 189&1905 and 1906-1928, translated by E. Rapuport 13. Episodes from the Early History of Mathematics by A. Auboe 14. Groups and Their Graphs by E. Grossman and W. Magnus 15. The Mathematics of Choice by Ivan Niven 16. From Pythagoras to Einstein by K. 0. Friedrichs 17. The Contest Problem Book I1 Annual High School Mathematics Examinations 1961-1965. Compiled and with solutions by Charles T. Salkind 18. First Concepts of Topology by W. G. Chinn and N. E. Steenrod 19. Geometry Revisited by H. S. M. Coxeter and S. L. Greifzer 20. Invitation to Number Theory by Oystein Ore 21. Geometric Transformations I1 by I. M. Yaglom, framluted by A. Shields 22. Elementary Cryptanalysis-A Mathematical Approach by A. Sinkov 23. Ingenuity in Mathematics by Ross Honsberger 24. Geometric Transformations I11 by I. M. Yaglom, framlated by A. Shenitzer 25. The Contest Problem Book III Annual High School Mathematics Examinations 19661972. Compiled and with solutions by C. T. Salkind and J. M. Earl 26. Mathematical Methods in Science by George P61p 27. International Mathematical Olympiads-1959-1977. Compiled and with solutions by S. L. Greitzer 28. The Mathematics of Games and Gambling by Edward W. Packel 29. The Contest Problem Book IV Annual High School Mathmatics Examinations 1973-1982. Compiled and with solutions by R. A. Arfino, A. M. Gaglione, and N. Shell 30. The Role of Mathematics in Science by M. M. Schiffer and L.Bowden 31. International Mathematical Olympiads 1978-1985 and forty supplementary problems. Compiled and with solutions by Murray S. Klamkin 32. Riddles of the Sphinx by Marlin Gardner 33. U.S.A. Mathematical Olympiads 1972-1986. Compiled and with solutions by Murray S. Klamkin 34. Graphs and Their Uses by Oystein Ore. Revised and updated by Robin J. Wilson 35. Exploring Mathematics with Your Computer by Arthur Engel 36. Game Theory and Strategy by Philip D. Sfruffin,Jr. 37. Episodes in Nineteenth and Twenthieth Century Euclidean Geometry by Ross Honrberger

Other titles in preparation. Contents

Preface 1

Essay 1 Probability and 1 3 Essay 2 Odd and Even Numbers 7 Essay 3 Sylvester’s Problem of Collinear Triads 13 Essay 4 The Algebra of Statements 17 Essay 5 The Farey Series 24 Ewy 6 A Property of an 38 Essay 7 Squaring the Square 46 Essay 8 Writing a Number as a Sum of Two Squares 61 Essay 9 The Isoperimetric Problem 67 Essay 10 Five Curiosities from Arithmetic 72 Essay 11 A Problem of Regiomontanus 89 Essay 12 Complementary Sequences 93 Essay 13 Pythagorean Arithmetic 111 Essay 14 Abundant Numbers 125 Essay 15 Mascheroni and Steiner 130 Essay 16 A Property of Some Repeating Decimals 147 Essay 17 The Theorem of Barbier 157 Essay 18 The Series of Reciprocals of Primes 165 Essay 19 Van Schooten’s Problem 173

Solutions to Exercises 178

Bibliography 203 vii

INGENUITY IN MATHEMAmCS

Preface

This book contains short, self-contained essays on elementarq mathematics. Many describe work of outstanding mathematicians, Gauss among them, and show the incisiveness and ingenuity so characteristic of kst rate mathematical work, no matter what the level. I have chosen the topics treated here because I remember how exciting and stirring I found them on first encounter. I want to share this sense of wonder with a larger audience and believe that many people, though unaware of it, have the minimal mathematical back- ground needed to understand and enjoy the ideas described here; the basic material of a standard first high school course in algebra and geometry sdices. It is my earnest hope that some readers will come under the spell of these fascinating topics. The book is modelled after The Enjoywent of Mathmatus by Rademacher and Toeplitz, in my opinion one of the best popular mathematical expositions ever written. The references given in each essay usually include my sources as well as appropriate material for those readers who wish to pursue the topic treated a little farther. The references in the bibliography at the end of this monograph are outstanding books in the general field of elementary mathematics and will nurture a growing interest. In this connection I want to mention specifically the periodical Scripta Mathmmtica which, specially during the years 1935-1957, was a veritable treasure trove of mathematical riches. In some essays I have treated rather specific concrete problems, often neglecting or merely hinting at generalizations, while in others I have tried to generalize and abstract a bit more. The reader should 1 2 INGENUITY IN MATREMATICS be aware that the treatment of any particular topic here is just one of several possible approaches. At the end of most essays I have added related exercises whose solutions appear at the end of the book. Since the essays are independent of one another, they may be read in any order. In conclusion I wish to thank Dr. Harold Shapiro for his excellent appendix to Essay 12 on Complementary Sequences and Dr. Anneli Lax, the Editor of this series of books, for her insight, judgment, patience and kind manner. Solutions to Exercises

Essay 1

1. Let the times of amval be x o’clock and y o’clock. Then x and y occur at random in the interval [0, 11. Let P be the point in the unit square with coordinates (x, y) . In order to have a meeting, I x - y I < 4, i.e., -4 5 x - y < t. All points with coordinates x, y in the unit interval and satisfying the above inequalities x - y 2 -2 and x - y 5 2 are located in the shaded strip of Figure S1. Its area is 1 - 2(AABC) = 1 - (9/16) = 7/16, which is equal to the required probability.

Figure S1 178 SOLUTIONS 179

2. Let the length of the rod be taken as the unit length; let one end of the rod be the origin and let x and y be the coordinates of the breaks. Then x, y occur at random in [0, 11, corresponding to a point P(x,y) which has an equal chance of occurring anywhere in the unit square. Y

Rod X ox Y1 "1 Figure S2

We require the probability that the triangle inequality is satisfied, i.e., that the sum of each two of the pieces exceeds the third. This is satisfied if each piece is less than one-half the length of the rod, and otherwise not. For definiteness, suppose x Iy. Then we must have x < 3, y - x < 3, and 1 - y < 3. Each of these inequalities defines a region in the unit square (see Figure S2) and triangle ABC is the intersection of these regions. So P must be in AABC. Similarly, if x 2 y, P must lie in ABDE. The required probability, then, is given by the sum of the areas of these triangles, which is easily seen to be 2.

3. Let the circumference be taken as the unit of length. Since the circle can always be rotated about its centre to bring the vertex A into a preassigned position on the circumference (say the top), we may assume the point A to be fixed in this position. Cutting the circle at A and stretching it ht we obtain a unit segment ABCA with points B and C chosen at random in it. The requirement that AABC be acute-angled means that each arc of the circle determined by the points A,B, and C must be less than one-half the circumference. This identifies Exercises 2 and 3, giving the answer 2. 180 INGENUITY IN MATHEMATICS

4. Let the equilateral triangle be AABC with side s, and denote the lengths of PX, PY, PZ by x, y, z, respectively (see Figure S3).

A A

Figure S3 Figure S4

(i) Then

AABC = APAB + APBC + APCA = gsz + 4sx + 3sy = 3s(x + y + 2). But AABC = gsh,

where h is an altitude of AABC. This implies that for all positions of P, ~+y+~=h.

(ii) x,y, z represent the sides of a triangle if and only if each is less than h/2. This restricts P to the “medial triangle”, whose vertices are the midpoints of the sides. The required probability, then, is the ratio AA’B’C’IAABC, which is easily seen to be 1/4. (See Figure S4.) Note that Exercise 4 could also have been solved by the method used in solving Exercises 2 and 3. In those cases, we required that three numbers x,y and z satisfy the conditions 1) they are equally likely to lie anywhere between 0 and 1, 2) each is less than 1/2, and these conditions determined a region occupying precisely 1/4 of the unit square. SOLUTIONS 181

The only modification in the present case is that our numbers x, y, z 1) lie at random in the interval [0, h], 2) are each less than h/2. The modified conditions determine a region occupying precisely 1/4 of a square of side h, and the required probability is the ratio

Indeed, the special choice of unit length for the rod in Exercise 2 and the circumferenceof Exercise 3 was made for the sake of simplicity; any other constant would have led to the same answer.

Essay 4

1. Let A,, Ad, Am, B,, Bd, * denote “Arnold is the shipper”, “Arnold is the driver”, and so on. We are given C. =+ Ba, which we interpret

+ Bd = 1. Similarly we have

Hence

This gives 182 INGENUITY IN MATHEMATICS

This means

Now, clearly Ca + 4d= 1. Then

i.e., 4d(B8+ Ad + Bad) = 1, which @VeS 4d= 1 and B8 +Ad +BJd = 1. From 4d= 4,= 1, it follows that C, = 1. Consequently B, = 0, and then B, + Ad = 1. But the shipper is either Brown or Arnold (since Clark is the manager), i.e., B, + A, = 1. Thus (B, + A,) (Bu + Ad) = B,(BLI+Ad) = 1-

Hence B, = 1, and Ad = 1 follows. Thus Clarkisthemanager, Brown the shipper, and Arnold the driver.

2. Let A, B, C respectively denote “A is guilty”, etc. Then we have -A * BC, i.e., A + BC = 1, and also -A + B = 1, and -B + 4= 1. Thus (A + BC) (-A + B)(-B + 4)= 1, which reduces to AB - C = 1, implying that C is the only innocent one.

3. If X makes the statement Y, then we have that (i) X is telling the truth, and Y is true, or (ii) X is lying, and Y is false. Thatis,wehaveboth X*Y and -XJ-Y; i.e., -X+ Y = 1 and X + -Y = 1. Thus we have A *-B + -C, giving -A+-B+4=1 (1) -A*-(-B+4) =BC, A+BC=l (2) B=+-A, -B+-A=l (3) -B+A, B+A =1 (4) C--A-B, 4+-A-B=1 (5) -C*-(-A-B) =A +B, C+A+B=l (6)- Now (4) times (5) gives

4B+ 4A= 4(B+A) = 1, giving 4 = 1. SOLUTIONS 183

This means C = 0, giving A = 1 in (2). From (3), then, -B = 1. Hence only A is telling the truth.

4. Wehave

A-BC, giving -A + BC = 1 (1) -A*-(BC)=-B+--C, A+-B+--C=I (2) BaA, -B+A=l (3) -B=;s-A, B+-A =1 (4) Ce-A-B, --C+-A-B=l (5) -C*-(-A-B) =A + B, C+A+B =1 (6)*

Now (3) times (4) gives AB + -A-B = 1, i.e., either A and B are both truth-tellers or they are both liars. If both are truth- tellers, then (1) gives C = 1, while (5) gives 4 = 1, a contradiction. Hence A = 0 and B = 0. Now (6) gives C = 1. These latter values satisfy all six equations. Thus only C tells the truth.

5. Let H, J, D, denote “Harry did it”, “James did it”, etc’ The statements of the boys are, respectively, CG, DT, TC, HC, DJ. In four of the five compound statements, one statement is true and the other false; in the fifth both are false. Thus in all cases, the truth-value is 0; i.e.,

For the same reason, four of the compound statements

have truth-value 1, while the fifth has value 0. Consequently,

Using (1) ,this reduces simply to

(3) CDCCD = CD = 0. 184 INGENUITY IN MATHEMATICS

Since only one of C + G, D + T, T + C, H + C, D + J is 0, then we have

where the left side contains the sum of the products of the compound statements taken four at a time. Using (1) and (3), this gives simply CJ = 1. Hence Charlie and James stole the apples.

6. Let A,, A,, Ab, B,, ... denote “A is red”, etc. The given statements are A,, -Br, and -Cb. Now, some counter is red; hence

A, + B, + C, = 1. Similarly,

Also, counter A has one of the three colours; hence

Also, only one counter is of a given colour, so

Finally, A can have only one colour, hence SOLUTIONS 185

Now then, we know that only one of the given statements is true. Hence

But A& = 0, giving

Multiplying by A, + B, + C, = 1, we get just

Hence B is red, and C is not blue; but it is not red either (since B is). Hence C is white, and A blue.

7. Let A,, A,, A,, Ad, C,, -0. denote “Albert’s last name is Albert”, “Albert’s last name is Charles”, etc. Then we have

since no one has the same first and last names. Also, we have in particular that

Let the Christian name of Mr. Frederick be X, and let the surname of Charles be y. Then

Xf = 1, c, = 1,

and from the long complicated paragraph, we have that Y, = 1. Then for some choiceof Xand Yamong A, C, F and D (Xand Y are different) we have

XfYX, = 1. Substituting all possible values and summing gives an expression whose truth-value must be 1 (some choice is valid) ;i.e., 186 INGENUITY IN MATHEMATICS

Working this out, and using (1) and (2) ,we get just D&Ca = 1. Hence Mr. Dick's first name is Albert.

Essay 6

Let n be chosen and let the multiples a, 2a, .*., (n - 1)a be marked on a number-line. Also, let the same or another number-line have each interval between consecutive integers divided into n equal parts. (See Figure SS.)

Now wrap both number-lines around a circle C of unit circumference, in the clockwise direction, beginning with the origin at the same point 0 in each case (see Figure S6). The divisions of the second line divide the circumference into n equal arcs. The multiples determine n - 1 points on C. Now if one of these points occurs in either arc adjacent to the origin 0, the required conclusion follows.

Figure S6

Otherwise, the n - 1 points lie in the other n - 2 arcs. It follows, by Dirichlet's Pigeon-hole Principle that at least two of them occur in the same arc. Suppose they are the multiples ra and sa (r < s). SOLUTIONS 187

(Consider the arcs to be half-closed; this prevents both points from being endpoints.) This means that the difference sa - ra differs from a whole number of circumferences, i.e., a whole number, by an amount less than l/n. That is, (s - r)a differs from an integer by less than l/n. But (s - r)a is, itself, one of the given multiples a, 2a, ..., (n - 2)a, because r and s are two different numbers from the set fl,2, -0 0, n - 11.

Essay 7

1. See Figure S7.

Figure S7

2. The 9-square tiling of the 32 X 33 rectangle given at the beginning of the essay can be extended to contain any number of squares by repeatedly constructing squares outwardly on the larger side (see Figure SS) . No two squares are equal since no new square can duplicate an existing square. The same is true of the 9-square tiling of the 61 X 69 rectangle which is a solution of Exercise 1. 188 INGENUITY IN MATHEMATICS

Figure S8

3. (a) If S were to have a side BC on the base, then not all triangles touching B or C can be larger than S without overlapping. [See Figure S9 (a) .] Thus S is “point down”.

BC (a) (bl (C) Figure S9

(b) The upper side of S cannot be continued beyond S lest S not be the smallest triangle touching the base. Hence the possi- bility of a single triangle in contact with the upper side of S cannot occur. Part (b) now follows as did part (a). [See Figure S9 (0), (c) .] (c) No. The diagram shows that the single point of contact does not necessarily persist through the next stage (here T touches S at a vertex, which S could not do to the base). SOLUTIONS 189 Essay 9

1. Reflect P in each side of the angle a to obtain P' and P". (See Figure SlO.) We claim that the segment P'P'' cuts the arms at the required points Q and R. The perimeter of any other triangle PQ'R' is

PQ' + Q'R' + R'P = P'Q' +- QR' +- R'P'' 2 the straight segment P'P':

which is clearly the perimeter of APQR.

Figure S10

Figure S11

2. For base BC = 5 and area 15, the altitude to BC must be 6. Hence the third vertex A lies on a line m parallel to BC and 6 units away (see Figure Sll). To locate A on m so that AABC 190 INGENUITY IN MATHEMATICS

has minimum perimeter, reflect B in m to give B'. Let B'C cut m at A. Then for any other point A' on m, the perimeter of AA'BC is

BC + A'B + A% = BC + A'B' + A% -> BC + the straight segment B'C = the perimeter of AABC.

It is not difficult to show that AB = AC. Hence each side is

1/(5/2)2 + 62 = 13/2.

Essay 10

1. (i) Let S = 1 +2 +3 + *** +k.

We rewrite S, reversing the order of the terms:

s = k + k - 1 + k - 2 + -0- + 1.

Adding gives

2S= k+l +k+ 1 +k+l+ 0.. +k+ 1 = k(k + 1).

Hence s = K(k + 1)/2. (ii) By (i) we have that

k+1 k cc i12 - cz i12 1 1

= (qy[k2 + 4k + 4 - k2-J = (k + 1)yk + 1)

= (k + 1)' SOLUTIONS 191 (iii) If

then

&I k k c P = ce + (K + 1)8 = [Xi12 + (K + 1)8 1 1 1

k+l = [C i7, by (ii). 1

(iv) For K = 1, each side of identity (1) has the value 1.

2. There are C(10,4) = 210 numbers with 4 different digits, 10C(9,2) = 360 numbers with digits 2 alike and 2 different, C(10,2) = 45 numbers with digits 2 alike and 2 alike, 9C(10,1) = 90 numbers with digits 3 alike, giving a total of 705.

3. (a) Dl = 2088; El = 8082. Direct calculation and inspection establishes parts (b) and (c) .

7. (a) The n-th number in the middle row has nz at the bottom of its column. The number of integers in this column is nz - (n - 1)2 = 2n - 1, which is always odd. The middle one in the column, then, is (3[(2n- 1) + I])-th from the top (i.e., n-th from the top). Since it is in the n-th place past the bottom number of the previous column, which is (n - l)z, its value is (12 - 1)2 + n = n2 - n + 1. (b) The number 3 and every third number thereafter is in position

2 + 3K for K = 0, 1, 2, -0.. By part (a), its value therefore is

(2 + 3K)Z - (2 + 3K) + 1 = 9KZ + 9K + 3; and every number of this form is clearly divisible by 3. The solutions of parts (c) and (d) are similar. 192 INGENUITY IN MATHEMATICS

(e) Consecutive numbers in the n-th and (n + 1)-th positions of the middle row have the forms n2 - n + 1 and (n + 1)2 - (n + 1) + 1 = nz + n + 1. Their product is (n2 + 1)2 - nz = (n2 + 1)2 - (nz+ 1) + 1, implying that it occurs in the (n2 + 1)-th place in the row, as required. The square at the bottom of the column containing the smaller factor, nz - n + 1, is nz.

8. The entry in position (n, n) is the sum 1 + 5 + 9 + --* of rt terms and is equal to n[2 + (n - 1)4]/2 = 2nz - n. The common difference for column n is the n-th odd number, 2n - 1. Hence the number in position (m,n) (m2 n) is [entryin (n,n)]+ [(m- n) X commondifferenceofn-thcolumn] = 2n2 - n + (m- n) (2n - 1) = (2n - 1)m. The sum of the numbers in row K, then, is the sum of the ?umbers inpositions (K, l),(K, 2), -.*, (k, K), whichis

(2.1 - l)K + (2.2 - l)K + (2.3 - l)K + *** + (2.K - 1)R = K[2(1 + 2 + + K) - K] = K[K(K + 1) - K] = Ka.

Essay 11

1. The angle in question is the sum of an angle x of elevation and an angle y of depression, each of which increases continuously to a right angle as the observer approaches the statue. (See Figure S12.) C

Figure S12 Figure S13 SOLUTIONS 193

2. Use the hint and the fact that an exterior angle of a triangle exceeds either non-adjacent interior angle. (See Figure S13.)

3. (i) The line y = $(b + c), the perpendicular bisector of the chord BC, must pass through the centre Z whose ordinate therefore is $ (b + G) . The abscissa of the centre is equal to OM’, and is given by (see Figure 11.4)

(OM’)2 = 0B.W (OM‘ is a tangent and OBC a secant) ;

i.e., the abscissa = OM’ = dG. Since OM’ is a tangent, the radius ZM‘ is vertical, so its length is &(b+ c) . (ii) Let the midpoint of BC be T. We saw in (i) that 4%= OM’ = ZT, while &(b+ c) = radius ZM’ = ZB. Thus dGis a leg and 4 (b + c) is the hypotenuse of right- triangle ZTB, implying the required in’equality.

(iii) Since (b - G)~2 0, (b - c)2 + 4bc 2 4bc; hence (b + c)~2 4bc, b + c 2 246 or $(b + c) 2 4%.Here equality holds only if b = c.

4. From the given data b = p - e = 8’, G = p + h - e = 18’, and the required distance is

& = dE.=12feet.

Essay 12

1. Let f(x0) = n. Now f*(n) = maxx such that f(x) I n, i.e., such that f(x) sj(~0).Clearly XO is one value of x satisfying this inequality. Since f is strictly increasing, xo is the largest such 2. Thus j*(n) =f*(f(xo)) = xo for arbitrary XO.

2. Let F(n), G(n) denote the sequences of triangular numbers and non-triangular numbers, respectively. Then

n(n + 1) n(n - 1) f(n)= F(n) - n = -n= 2 2’ 194 INGENUITY IN MATHEMATICS

and

f*(n)= the number of positive m such thatj(m) < n m(m - 1) = the number of positive m such that 2 < 12,

ie., such that m(m - 1) < 2n.

Now just as in our treatment of the non-square numbers (see pp. 97-8) the inequality m(m - 1) < n led to the formula G(n) = rt + (*), so our present inequality m(m - 1) < 2n

leads to the formula G(n) = n + (6).

3. (a) Suppose k I < k + 1/2. Then [*]= k, and also (&) = k. NOW k2 5 n < k2 + k + 1/4

and, adding k to each member, we obtain

k2 + k I n + [*I < k2 + 2k + 1/4,

whence k2 < n + [&I < k2 + 2k + 1,

and

14% +[*I1 = k = (*>.

Part (b) is similar. SOLUTIONS 195

4. Let T be the n-th positive integer not of the form [ek], K a positive integer, and suppose that it occurs between [em] and [e"+l]. (See Figure 34.)

Then

T-m=n,or T=n+m.

Since e is a transcendental number (i.e., not a root of a polynomial equation with integer coefficients), ek is not an integer for any positive integer K; for, if ek = an integer a, then ek - a = 0 so that e would be a zero of the polynomial xk - a = 0. It follows that [ek] < ek for all K. Also, since T is not of the form [ek] and lies between [em] and [eel], we see that T I[em+'] - 1. (See Figure SlS.) Therefore

em < [em] + 1 IT I[em+'] - 1 and

em < T + 1 I[em+'] < e*1.

Taking logarithms of the first, second and last members, we obtain

and conclude

Dog (T+ 1)1 = Dog (n + 1 + m)l = m.

Figure S15 1% INGENUITY IN MATHEMATICS

Using this repeatedly we get

T=n+m = n + Clog (n + 1 + m)] = n + Dog (n + 1 + Dog (n + 1 + m)])] = ...

as far as one pleases. We need to show that

Z = n + Dog (n + 1 + Dog (n + 1)1)1 is the constant value of all these expressions. First we show that 1z exceeds [e-']. For, if not,

n 5 [e-']< e-l

would yield T =n+ml [e*']+rn,

and

em < T [em-'] + m < e*'+ m; that is, em < em-' + m. This leads to em-l(e - 1) < m and, since e - 1 > 1, to ern-' < m. But

(m- 1)2 e-1 = 1 + (m- 1) + +.*.21+(m-l) =m, 2! a contradiction. SOLUTIONS 197

Then n occurs between [e-l] and T. This interval also contains [em]. We consider the subintervals [[em], T) and ([e-ll, [PI> separately. (a) If [em] 5 n < T, then

em c [em] + 1 I n + 1 5 T < [e*lI < @I,

m < log (n + 1) < m + 1, and Dog (n + l)] = m. In this case

Z = n+ Dog (n + 1 +m)] = T as required. (b) If [e-l] < n < [PI, then

so that

m - 1 < log (n + 1) < m and Dog (n + l)] = m - 1. Thus Z = n + Dog (n + 1 + m - l)] = n + Dog (n + m)l = n + nog TI.

Since em < T < em+', we have Dog T] = my so that

Z = n +m = T, as required.

5. Ob~ously (0, O), (1, 01, (0, I), (2, 01, (0, 2), (1, 1) are all either winning positions or losing ones. Suppose the assertion holds for (x, y) if x + y < n. Consider (X, U),where X + Y = n. 198 INGENUITY IN MATHEMATICS

Now, in general, there are many positions (x, y) which can be moved into from (X,U). Imagine, then, that you receive the position (X, U) from your opponent and that you move to (x, y). For every possible (x, y) that you might choose, x + y < n, implying that each is either a winner or a loser (for you, that is). Now if any of these is a winner for you, then (X, U) is a loser for your opponent, because you will, of course, choose to move into that position (x, y) that wins for you. In this case (X, U) is a loser because it loses for the player who moves into it (i.e. passes it along to his opponent). On the other hand, if none of the (x, y) are winners for you, then (X,Y) is a winner for your opponent, because you can’t win no matter what you do. Thus from your opponent’s point of view, and this is the point of view from which to appraise (X, U),we see that (X, U) is also either a winner or a loser.

Essay 13

1. We use the formula for k-gonal numbers derived on pp. 117- 119 and find

n 12-1 pAk-1) + = - [n(k - 3) - k + 51 + 7[(n - 1) + 11 2 n =-[n(k-3) -k+S+n-l] 2 n = - [n(k - 2) - k + 41 2 = pp.

2. x = 2mn, where one of m, n is even. This implies x is divisible by 4.

3. If neither m nor n is divisible by 3, we have m and n of the form 3k + 1 or 3k + 2. In either case, m2 and n2 are of the form 3k + 1, so that y = m2 - rt2 is divisible by 3. SOLUTIONS 199

4. Suppose neither m nor n is divisible by 5. Then m,n are congruent to 1, - 1,2, or -2 (mod 5), giving m2,n2 congruent to 1 or 4. If m2 and n2 are both congruent to the same number, then y = m2- 122 is congruent to 0; otherwise z = m2 + n2 is congruent to 0.

5. Suppose q is not a prime, but equal to ab, where a, b are integers greater than 1. Then 29 - 1 = (2a)*- 1, which has the divisor 25 - 1. And 2a - 1 is not 1, lest a = 1; nor is it 29 - 1, lest b = 1. Hence 29 - 1 is composite; contradiction.

6. An even perfect number m is of the form 2*-*(2* - l), where 2%- 1 is a prime (see the theorem on page 115). By the previouS problem, n is a prime. Consequently n = 2, or n is of the form 4k + 1 or 4k + 3. (It is either even, or it is an odd prime). If n = 2, we get m = 6, which is acceptable. If n=4k+1,

m = 24k(24k+1 - 1) = 16k(2.16k- 1).

16k ends in 6 for all k; thus 2016~- 1 ends in 1, and thus m ends in 6. If n =4k+3,

m = 24k+2(8.16k - 1) = 4.16k(8.16k - 1).

Since 16k ends in 6, 4-16&ends in 4, 8.16k - 1 ends in 7, and their product m ends in 8.

Essay 16

1. (a) If k = 1, d is a power of 10 so that n/d terminates. If k > 1, then 200 INGENUITY IN MATHEMATICS

giving a terminating decimal. If k < 1, multiply numerator and denominator by 21-k to obtain an equivalent fraction with 10’ in the denominator. (b) Let the prime decomposition of b be b = p“;p;* .--p”;.

Then, if d = && 0.. p!#, where the exponents are non- negative integers, the “b-mal” expansion of n/d terminates.

Prouj: Letqbeapositiveintegersuchthateachofalq, azp, -**, alq exceeds the maximum of &, &, - , Bt.

so that the equivalent fraction is

and evidently terminates.

2. We are concerned only with the part of the division process which provides digits in the quotient after the decimal point. After the decimal point, only zeros are “brought down” to usher in successive

steps. The only possible remainders are 0, 1, 2, 0, d - 1, and if the process does not end, the remainder 0 cannot occur. In at most d steps, then, one of the remainders must repeat (Dirichlet’s principle again). After the second appearance of this remainder, the process continues exactly as it did after its first occurrence (again a zero is brought down, etc.). Thus the process repeats over and over with a period of length not exceeding d - 1.

3. (a) This is simply a matter of verifying the three properties required in the definition.

(b) Employing the hint we get

and

jij2 - jig2 = fikzp SOLUTIONS 201

whose sum is

This implies the desired conclusion. (c) We use the method of induction. Suppose the assertion holds for m = k. Now consider k + 1 pairs

By assumption,

and

fk+i = gk+l (mod p). These congruences, together with the result of (b) ,yield

Thus the assertion holds for m = K + 1. Since it holds for m = 2, bypart (b),itholdsforall m = 2, 3, 4, ---.

4. The statement and proof parallel the details given in the essay with 10 replaced by b, 9 by b - 1, and the condition p is not 2 or 5 by p does not divide b. The working out of the substitutions is left to the reader.

5. This problem may be attacked either by generalizing the dis- cussion on pp. 150-153 from a prime p to a number q relatively prime to 10, or by examining the division algorithm. We present the latter approach. It suffices to consider the case r < q (see page 150). For its decimal expansion to terminate, r/q must be expressible in the form s/lO" for some integers s and 12; i.e., there must be an integer m such that mq = 10". But this equation cannot hold if q is relatively prime to 10. Hence the expansion of r/q does not terminate; it is periodic (see Exercise 2). 202 INGENUITY IN MATHEMATICS

In the long division process repeating remainders cause repeating digits in the quotient. Suppose equal remainders, r, and rn, occur in the mth and wth steps (places after the decimal point). Since every remainder except the first results from subtracting a multiple of the divisor from 10 times the previous remainder, we have

lor,-1 = aq + m and

The difference of these equations is

and since q is relatively prime to 10, p divides r,-~ - rn-1. How- ever, each remainder is a positive integer less than q, so the difference r,-~ - r,l is less than q and hence can be divisible by q only if it is zero, i.e., if r,-1 = m-1. In other words, if r, = rn, then r-1 = m-1, and so on back to the very fmt decimal place where r1, the first remainder, starts the period.

6. This well known result is called Euler’s generalization of Fermat’s theorem, and its proof can be found in many books on number theory, e.g., Hardy and Wright, Introdzlction to Number Theory, or H. N. Wright, Theory of Numbers (First Course). Bibliography

W. W. R. Ball,Mathematbl RecreatiomandEssays (llthed.), MacMillan, 1939, London.

E. T. Bell, Men of Mathematics, Simon and Schuster, 1961, New York.

R. Courant and H. Robbins, Wha4 is Mathematics?, Oxford University Press, 1941, New York.

N. A. Court, CoZkge My(2nd ed. ), Barnes and Noble, 1952, New York.

H. S. M. Coxeter, Introdzrctiott to Geometry, Wiley, 1961, New York.

H. Dome, 100 Great Probhm of Elemmtary Mathematics, Dover, 1958, New York.

H. Eves, Int~oducthto the History of Mathematics, Holt, Rinehart, and Winston, 1969, New York.

G. Hardy and E. Wright, Introduction to the Theory of Numbers, Clarendon, 1960, Oxford.

D. Hilbert and S. Cob-Vossen, Gem@ and the Imagination, Chelsea, 1952, New York.

L. Lyusternik, Convex Figures and Polyhedra, Dover, 1963, New York.

E. A. Maxwell, Geometry for Advanced Pupils, Oxford Press, 1949, Oxford.

NCTM Yearbook 28: E&hW Mathenodics for High Schools, 1963, Washington.

D. Pedoe, The Gcntlc Art of Matktics, Penguin reprint, 1969, New York. 203 204 INGENUITY IN MATHEMATICS

G. P6lya, Mathematics and Plausible Reasoning, Vol. 1, Princeton, 1954, Princeton.

H. Rademacher, Lectures in Elementary Number Theory, Blaisdell, 1964, New York.

H. Rademacher and 0. Toeplitz, The Enjoyment of Mathematics, Princeton University Press, 1957, Princeton.

H. Steinhaus, 100 Problems in Elementary Mathematics, Basic Books, 1964, New York.

H. Tietze, Famous Problems of Mathematics, Graylock, 1965, New York.

H. N. Wright, First Course in Theory of Numbers, Wiley, 1939, New York.

I. M. Yaglom and V. G. Boltyanskii, Convex Figures, Holt, Rinehart and Winston, 1961, New York.

A.. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problem with Elementary Sohtimrs, Holden Day, 1964, San Francisco. Index abundant number, 125 electrical networks in graph Aitken, A.C., 94 theory, 51ff amicable pair, I 11 equivalence (logic), 18 Arabian rule, 112 Euclid, 113,130 amicable chain, 112 Euler, L., 112,113,119 semi-amicable pair, 112 experimental determination arithmetic (Greek), 111 ofr, 1 Barbier’s theorem, 158 Fermat, F? de, 112 Beatty’s theorem, 93 little theorem, 152 Brooks, 46 figurate numbers, 117 frequency distribution function, Chartres, R., 1 96 collapsing compasses, 130,131 Fryer, K. D., 112 complementary sequences, 95 compound tiling, 56 Gallai, T,14 conjunction (logic), 17 Gauss, 63 constructions 4thcompasses, generalized pentagonal numbers, 132ff 119 with straightedge, 138ff golden section, 102 convergent series, 166 greatest integer function, 94 convex curve, 67 Coxeter, H.S.M.,16 hidden point, 26 curve of constant breadth, 157ff implication (logic), 21 decimal, periodic, 147 inverse sequences, 96 deficient number, 125 degree of a vertex, 7 Kelly, L. M.,14 Descartes, R., 112 Kirchhoffs Laws, 54 Dickson, L. E., 113 disjunction (logic), 18 Lambek, J., 95 divsrgent series, 166 lattice point, 26 divider-type compasses, 130 Liouville, J., 72 Dud, E., 73 logistic (Greek), 111 205 206 INGENUITY IN MATHEMATICS

Mascheroni, L., 130 Shapiro, H., 96 theorem of, 132R simple polygon, 27 Moser, L., 15,95 simple tiling, 56 Smith, C.A.B., 46 non-squares, 97 Sperner’s lemma, 9ff Sprague, R., 55 Ohm’s law, 53 Steiner, J., 67,130 Olds, C. D.,124 Stone, 46 Ostrowski, A., 94 sum of divisors, (n), 114 Sundaram’s sieve, 75 Paganini, N., 112 supporting line, 157 perfect number, 113 Sylvester, J.J.,13 perfect tiling (or square), 55 Pick’s theorem, 27 ’Jbtte, W.T., 46 Poncelet, J.V,130 primitive triangle, 27 Pythagoras, 111 Van Schooten, E, 173 triples, 120 visible point, 26 primitive triples, 121 Willcocks, T. H., 55 Reuleaux triangle, 157 Wilson, J., 56 rh), 61 Wythoffs game, 101 AMS / MAA ANNELI LAX NEW MATHEMATICAL LIBRARY

Ingenuity in Mathematics

Ross Honsberger

Readers who sit with this book for half to three-quarters of an hour will have mastered an elegant and pleasing bit of mathematics. Nineteen independent essays on topics, including number theory, geometry, combinatorics, logic, and probability. The nineteen essays contained in this volume are independent of one another; they represent many different aspects of mathematical thinking. Standard high school courses in algebra and geometry furnish a suffi cient basis for understanding each essay.

The topics treated in this volume include number theory, geometry, combinatorics, logic, probability. The methods used often involve an interaction of these various disciplines. Some essays are easy to read, and others are more challenging. Some exercises are routine; others will lead the reader deeper into the subject.