A Mathematical Mosaic 1895997046

Table of contents :
Acknowledgments
About the Author
Foreword
Table of Contents: Part I
Table of Contents: Part II
Number Theory
Combinatorics
The Fibonacci Sequence
Game Theory
Geometry
Combinatorics Revisited
Chessboard Coloring
Number Theory Revisited
Fibonacci & the Golden Mean
Geomety Revisited
Infinity
Game Theory Revisited
Concepts in Calculus
Complex Numbers
Infinity Revisited
Afterword
Annotated References
Index

Citation preview

A MATHEMATICAL MOSAIC

A MATHEMATICAL MOSAIC //

Patterns & Problem Solving

by

R avi V akil

Brendan Kelly Publishing Inc. 2122 Highview Drive Burlington, Ontario L7R 3X4

SEP 81998

Cover design: Pronk&Associates Toronto, Ontario Illustrations:

Taisa Kelly

Brendan Kelly Publishing Inc. 2122 Highview Drive Burlington, Ontario L7R 3X4

Copyright © 1996 by Brendan Kelly Publishing Inc. All rights reserved. No part o f this work may be reproduced or transm itted in any form or by any means, electronic or mechanical, including photocopying and recording, or by any inform ation storage or retrieval system, without permission in writing from the publisher. ISBN 1-895997-04-6

ATTENTION EDUCATIONAL ORGANIZATIONS Quantity discounts are available on bulk purchases of this book for educational purposes or fund raising. For inform ation, please contact: Brendan Kelly Publishing Inc. 2122 Highview Drive Burlington, Ontario L7R 3X4 Telephone: (905)335-5954

4

Fax: (905) 335-5104

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m z m o z y

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5

athem atics is an intensely social d iscip lin e. M y first d eb t is to those who have shared their love o f m a th e m a tic s w ith m e: m y te a c h e rs , professors, colleagues, students, and friends. I am especially grateful to the young men and women who agreed to share their unique perspectives on the joys o f mathematics for the seven profiles included in this book. M ost of the ideas contained within A M athem atical M osaic have been passed on to me over the years by word o f mouth. Although they are presented here in new and, I hope, engaging ways, the origins o f many o f these gems have been lost in m athem atical folklore. I have cited references where I am aware o f them, but there are many other sources that I m ust gratefully, if silently, acknowledge here.

M

M ichael Roth, a graduate student in mathematics at Harvard University, and Naoki Sato, a talented undergraduate at the U niversity o f Toronto, both gave essential advice on the final draft. M ark W underlich, form erly a student o f m ine at H arvard College and now a graduate student in philosophy at the University o f Arizona, provided feedback during the developm ent of much o f this material. The book is far m ore readable thanks to their efforts; any flaws that rem ain are solely the responsibility of the author. From helping to develop the original concept to suggesting final revisions, Brendan Kelly transcended his role as publisher. This book has evolved during long conversations with him, and his influence is visible on every page. I greatly appreciate his creative vision and com m itm ent to this rather unorthodox project. Finally, I must thank Alice Staveley o f Oxford University for her meticulous editorial advice and endless patience.

6

Publisher’s Introduction A Celebration o f the Human Intellect well-known story from the folklore o f mathematical history tells how Archimedes bounded from his bathtub and ran naked through the streets of ancient Greece yelling, “Eureka!” [I have found it]. Apparently a period of intense meditation in the bathtub had provided the inspiration for A rchim edes’ discovery of the law of hydrostatics which now bears his name. W hether apocryphal or authentic, this legend captures the ecstasy which accompanies a flash o f insight into a deep problem. It is to those who relish the joy of mathematical discovery that this book is dedicated. It is for those who chase relentlessly the thrill o f eureka that this book is written.

A

Through the past three millennia, the mental giants of each generation have pitted their problem solving skills against the m ost intractable problems to challenge the human mind. W hen a problem yielded like a dragon to a deft sword, the successful strategy was refined, polished, and added to the growing body o f intellectual technique known as m athematics. One of the greatest problem solvers of all time, Isaac Newton, characterized his favorite discoveries as “smoother pebbles or prettier shells than ordinary ” . 1 From an assortm ent of the most beautiful pebbles and the prettiest shells o f mathem atical technique, Ravi Vakil, author of this book, has created an intriguing mosaic. It is a mosaic which links various branches of mathematics through powerful overarching ideas. Ravi him self is a preem inent problem solver, having won top honors on virtually every mathematics competition and olym piad. His rem arkable achievem ents are outlined on page 9 and on the back cover. T hro u g h o u t this book, R avi Vakil pro files several other recent w inners o f m athem atical competitions and olympiads: young people who have honed their intellectual gifts to w orld class levels. As you read these human interest features, you will observe that these fine young minds all share a passion for the eureka sensation. It is som ething they all understand and something that bonds them together in friendly collegial competitions. We are pleased to publish this celebration of the human intellect. It is offered as a salute to the problem solvers o f the present and future and as a tribute to the intellectual giants o f the past. To the reader, we issue a challenge. We dare you to flip through the pages o f this book w ithout finding any problems to pique your curiosity! T o r the complete quote, see page 8.

7

/ do not know what I may appear to the world; but to myself I seem to have been only like a boy playing on the seashore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean o f truth lay all undiscovered before me. — Isaac Newton

Shout tfje Suttjor PHOTO BY JEWEL RANDOLPH

avi Vakil's resume reads like a dictionary o f superlatives. D uring his high school years he won every m ajor mathematics com petition. These include first place standing in the Canadian M athem atical Olympiad for two successive years, first place standing in North A m e ric a on th e 1988 USA M a th e m a tic a l Olympiad, and two gold medals and one silver m e d a l in th e I n te r n a tio n a l M a th e m a tic a l O lym piad. He also w on firs t p rize in the Canadian Association o f Physicists Competition and led his high school com puter team to three p ro v in c ia l c h a m p io n s h ip s . R a v i w as valedictorian o f his 1988 graduating class at M artingrove Collegiate Institute in M etropolitan Toronto and was aw arded the Academ ic Gold M edal and the G overnor General's Aw ard for excellence.

R

During his undergraduate years at the University o f Toronto, Ravi raised his brilliant achievem ents to new heights. In all four years he placed among the top five com petitors in the prestigious North American Putnam M athematical Competition, qualifying him as Putnam Fellow in each year. He graduated in 1992 with B. Sc. and M. Sc. degrees; for his B. Sc., he was awarded the G overnor General's M edal for the highest graduating marks at the University o f Toronto. Ravi's extraordinary achievem ents have not prevented him from pursuing his many other interests. These include squash, debating, student government, journalism , and A m nesty International. F or this breadth o f involvem ent he has received num erous scholarships including the John H. M oss M em orial Fund Scholarship for the best all-round graduating student at the University o f Toronto. In spite o f these personal successes, Ravi Vakil is a com passionate person with an eager w illingness to help others. He has worked extensively as a coach o f the Canadian Team to the International M athem atical O lym piad from 1989 through 1995. H e is also co-founder o f M athem atical Mayhem, a mathematical problem ­ solving journal for high school and undergraduate students — the only student-run journal o f its kind in the English-speaking world! Ravi Vakil is currently a Ph.D. candidate in pure mathematics at Harvard University, working in Algebraic G eom etry under the supervision o f Dr. Joseph Harris.

9

Jforetoorti he public face o f mathematics is sometimes grey and utilitarian: math is a useful tool, a haphazard collection o f recipes and algorithm s, a necessary prerequisite to understanding science. The private face is much more beautiful: M athematics as Queen, not servant, of Science. M ath is a uniquely aesthetic discipline; m athematicians use words like beauty, depth, elegance, and pow er to describe excellent ideas. In order to truly enjoy m athem atics, one m ust learn to appreciate the beauty o f elegant arguments, and then learn to construct them.

T

In A M athem atica l M osaic, I hope to get across a little o f w hat it is that mathematicians actually do. M ost “big ideas” and recurring themes in mathematics com e up in surprisingly simple problem s or puzzles that are accessible w ith relatively little background. M any o f them have been collected here, along with an inkling o f how they relate to the frontiers o f m odem research. These themes will also be traced backwards; often ideas centuries old take on new m eaning and relevance as mathematical understanding advances. And if in the process o f looking at these important ideas we get a little playful and irreverent, well, that is also in keeping with the nature o f mathematics. This book does not purport to teach problem solving, although it might communicate what is interesting and exciting about grappling with a problem. Instead, the reader should be left with a large chunk o f mathematics to digest slowly. This is not a book intended to ever be “finished w ith” or completed. Instead, it should be read bit by bit. Like all mathem atics, it should be read with pencil in hand, with more tim e spent in deep thought or frantic scribbling than in actual reading. Mathematics is not a spectator sport! You will also notice that, like any mosaic, this book conies in many small pieces o f different sorts. It is loosely arranged in order o f difficulty, and the headers often indicate the general them e o f the article. You m ight enjoy just starting with one section which especially interests you; from there, you can skip to other sections which link with your earlier choices. You will soon discover that ideas from one field carry over into many others. And if you d o n ’t end up at a section that amazes and perplexes you, I will be sorely disappointed.

10

M any sections end with some Food fo r Thought. M ore than just exercises, these excursions are intended to be starting points for independent thinking. The material beforehand will help, but often only tangentially. I have deliberately included very few solutions. Indeed, often there is no single “correct” solution, and som etim es the problem s are so open-ended that there is no way to completely “finish” them. Don't try to solve every one; choose one or two that catch your eye and think about them on and off for a few days. Success should not be defined by how many problem s you solve, but by how many new ideas you have. And remember: patience is a virtue. Every so often throughout the book, references are mentioned. This is not to intim idate you into reading many weighty tomes. Instead, they are there to provide some ideas for further reading in case you find some field particularly interesting. I have also included short profiles of interesting and talented young people who have been involved in mathematics for many years. I have chosen them fairly random ly from the large number o f fascinating people I have had the good fortune to m eet through my involvem ent in mathematics education. Some o f them are going on in mathematics (and indeed one has already made his name in the field), while others are pursuing their interests in related subjects. Through these profiles, I hope to share their unique perspectives on the joys of mathematics. You w ill notice th a t th ese young peo p le have all done extrem ely w ell in m athem atical com petitions; collectively they have won eleven Gold M edals at International M athem atical Olympiads. In noting their achievements, I certainly do not mean to im ply that com petitions are the only gateway to mathematics, or that they are a necessary prerequisite to success in the field. M y selection merely reflects the fact that my own involvem ent with young mathematicians has grown out o f my association with mathematics competitions. There is no background required to begin enjoying this book. You should be able to pick up many o f the ideas as you go along. W hen you truly get stuck, ask a friend or a teacher for help. You m ight be surprised at how much you learn! For som e o f the later sections, a fam iliarity w ith m ore advanced ideas such as mathem atical induction, indirect arguments, set theory, complex numbers, and even calculus will be useful or even necessary, but the vast m ajority can be read with nothing m ore than early high school mathem atics and a little chutzpah. But enough o f this talk. On to the mathematics! Cam bridge, Mass. January, 1996

11

TABLE OF CONTENTS: Part I Number Theory Calculating Tricks o f the Trade M ission Im possible Calculating Prodigies and Idiot Savants The M agic Birthday Predictor Cards D ivisibility Rules The Factor Theorem M ore Com plicated Divisibility Tests The Secret Behind the M agic Birthday Predictor Cards M agic Squares Two Theorem s about M agic Squares Personal Profile: D o What You Enjoy!

18 19 20 21 22 26 27 30 31 35 40

Combinatorics A M athem atical Card Trick Counting the Faces o f Hypercubes How the M athem atical Card Trick Works The Water and W ine Puzzle Questions A bout Soccer Balls Personal Profile: Cracking Intractable Problems — That's a Big Buzz

44 45 51 52 53 54

The Fibonacci Sequence Elvis Numbers H istorical Digression: Fibonacci: The Greatest European M athem atician o f the M iddle Ages The Solution to Elvis the E lf s Eccentric Exercise A Form ula for the nlh Term o f the Fibonacci Sequence

56 57 59 60

Game Theory The Theory o f Games The Game o f Nim Game Theory and Politics: Arrow's Theorem H istorical Digression: Social Choice Theory Elem entary My Dear Watson!

64 66 68

72 74

TABLE OF CONTENTS: Part I Geometry Historical Digression: The M athematician Who Anticipated Calculus by Alm ost 2000 Years Lengths, Areas, and Volumes: Arguments Both True & False Part 1: The A rea o f a Circle Part 2: The A rea o f a Segment o f a Parabola Part 3: The Surface A rea o f a Sphere Part 4: The Volume o f a Sphere Archimedes Strikes Again The Falling Ladder Problem An Easy Proof that arctan 1/3 + arctan 1/2 = 45° The Ailles Rectangle Solution to the Ailles Rectangle Problem Personal Profile: One o f the Purest Forms o f M ental Exercise

78 79 79 79 80 81 82 85 86

87 88

89

Combinatorics Revisited Parity Problems: The World Series The "Brute Force" Solution The Elegant Solution More Parity Problems The Triangles of Pascal, Chu Shih-Chieh, and Sierpinski Solutions to the Parity Problems

92 93 95

96 98 103

Chessboard Coloring The Invention of Chess A Chessboard Tiling Problem A Solution to the Chessboard Tiling Problem A Tetromino Tiling Problem Another Chessboard Tiling Problem Solution to the Tetromino Tiling Problem Personal Profile: A Sudden Flash o f Insight

106 107 108 110 11\ 113 114

TABLE OF CONTENTS: Part II Number Theory Revisited Numbers, Numbers, and M ore Numbers! First Steps in N um ber Theory: Rational and Irrational Numbers H istorical Digression: The M ost Famous Conjecture in M athematics Prime Numbers in N um ber Theory The Painted Lockers G eom etry M eets N um ber Theory: Constructing Pythagorean Triples A Strange Result in Base 2 and Base 5 A Rem arkable Coincidence? O r An Even M ore Rem arkable N on-Coincidence? H istorical Digression: The Poor Clerk who K new Num bers on a First-Name Basis Solution to the Painted Lockers Problem Sum o f kP Powers H istorical Digression: A M athem atical Genius o f the H ighest Order Bernoulli Numbers H istorical D igression: The Incredible B em ouilli Family Personal Profile: Nothing M akes Me Less Aware o f the Passage o f Time

118 120 123 124 127 128 131 132 133 134 135 137 140 141 142

Fibonacci & The Golden Mean Funny Fibonacci Facts The G olden M ean Some Interesting Properties o f the Golden M ean The Pythagorean Pentagram, The Golden M ean, & Strange Trigonometry

146 150 151 154

Geometry Revisited A D o-It-Yourself P roof o f Heron's Formula A Short Route to the Cosine Law Solution to the Falling Ladder Problem Locus Hokus Pokus The Triangle Inequality: From "Common Sense" to Einstein M ore Locus Hokus Pokus A Short Question on Sym m etry

160 163 164 165 166 173 174

Infinity The M athem atics o f the Birds and the Bee An Early Encounter with the Infinite The Flaw in Zeno's Paradox The H arm onic Series H istorical Digression: The H uman Com puter A nsw er to the M athem atics o f the Birds and the Bee

176 177 178 179 184 184

TABLE OF CONTENTS: Part II Game Theory Revisited Sherlock Holmes' Secret Strategy A Three-W ay Duel: W hy M ulti-Player Games are M uch H arder to A nalyze Why Holmes' Strategy Works How to Win at Nim

186 187 189 191

Concepts in Calculus Historical Digression: The M an Who Solved the System o f the World A Question of Continuity An Integration by Parts Paradox A nsw er to A Question o f Continuity Dual Numbers Personal Profile: A Big 'Yes!!!' Goes Through my H ead

196 198 199 200 201 205

Complex Numbers i‘ and Other Im probabilities Historical Digression: The Oddly Polyglot Statem ent ein + 1 = 0 Addition Formulas for Sine and Cosine C o m p u tin g /' Roots o f Unity Three Impossible Problem s o f Antiquity Galois Theory Historical Digression: Pistols at Dawn — The Teenager who Launched A bstract Algebra Using Complex Numbers in Geometry

208 209 211 212 213 215 218 219 220

Infinity Revisited Paradox! Two Different Infinities An Infinitude of Infinities H istorical Digression: The M an Who Shook the Foundations o f M athem atics The Existence o f Transcendental Numbers Personal Profile: The Youngest Tenured Professor in H arvard History

Afterword Annotated References Index

226 229 234 236 237 241 246 248 251

-

Mi/ mbcr Thcory

Mathematics is the Queen of the Sciences, and Number Theory the Queen o f Mathematics. — Karl Gauss

I

Number Theory Calculating Tricks o f the Trade T h ere are m any “c u te ” little n u m e rical trick s for perform ing calculations quickly. On the surface, they may not seem to be serious m athem atics, but they are good practice for elementary algebra. For example, here’s a little trick to multiply two positive integers with the same tens digits and with the units digits summing to ten. Study these exam ples to see if you can find a pattern in the answers and discover the trick. 38

13

79

25

118

x 32

X 17 221

X 71

x 25

5609

625

X 112 13216

1216

The Trick •M ultiply the tens digit by the next largest number. Call your result “A ” . •Then multiply the units digits together, and call your result “B ”. •Write A im mediately followed by B (where B is considered a tw o-digit number), and read it as a single number. •You have your product! For example, to multiply 32 and 38 in your head, you get A = 12 and B = 16, so 32 x 38 = 1216. Use the trick to verify the other answers above. (The last exam ple may not appear to follow the rule, but you can easily adapt it.)

The Proof The proof isn’t too difficult. (You m ight w ant to derive it yourself.) Let x be the tens digit o f both numbers, and y be the units digit o f the first number. Then 10-y is the units digit o f the second number. The two numbers are therefore 10 x + y and lOx + 10-y. M ultiplying them together, we get: Some steps have (10x+y)(10x + 10 -y)= IOOjc2 + lOOx + y(10-y) been omitted. = 1 0 0 jc(jc+1 ) + y ( 1 0 -y) = 100 A + B 34

8

11

10

5

—> 34

12

7

6

9

—> 34

13 i

2

3

16 —> 34

1

34 34

1

1

34 34

34

This constant sum is called the magic sum of the magic square. These mathematical objects w ere sacred to the ancient Chinese, and were thought to have mystical powers. M agic squares o f order n exist for all positive integers n except n = 2. (It isn't hard to see why there are no m agic squares o f order 2.) In general, there are many magic squares o f any given order. In the next section, Two Theorems A bout M agic Squares, we will look at a little theory behind them. M eanw hile, here is a m ethod to make magic squares o f all possible sizes.

Constructing Magic Squares of Order n (n odd) W e’ll construct a magic square of order 5. All other odd squares follow the same pattern. O Place tw o other 5 x 5 arrays around the central 5 x 5 array, one directly above and one to the right. Start off by placing a 1 in the middle square in the top row o f the central 5 x 5 array. © Then place the integers in order from 2 to 25, m oving diagonally up and to the right at each step. If you run off the central magic square onto an abutting array, go to the same position in the central magic square and record the number. 17 24 1 15 © Every so often (at the multiples o f n) you’11 23 5 7 14 16 be prevented from continuing. In that case, 4 6 13 2 0 2 2 7 go instead to the square immediately below 1 0 1 2 19 2 1 3 w here you were. (Notice where the 6 was 9 11 18 25 2 placed in this sample magic square.)

0 I8

1 1 1

1

6

3 33 7 9 2

111

30 L 34 1 2 14 16 36 29 13 18 11 'y l i I

111

1 1 1 1 1 1 1 1 1 111 1 1 1 1 1 1

111

8

1

6

3 5. 7 V) 9 2

26 19 24 21

23 25

22

27

20

§f>) 28 33 17 1 0 15 30 © 34 1 2 14 16 31 36 29 13 18 11 x

6

example becomes:

111

O ur square is now magic!

33

Number Theory I

Constructing Magic Squares of Order n ( n = 4m) First o f all, write the numbers from 1 to n 2 in the square, filling in the rows from left to right, beginning with the top row and working down. (In the following exam ple we construct a m agic square o f order n = 8 , i.e. m = 2 .)

1

2

3

4

5

9

10

11

12

13 14 15

17

18 19

20

21

25 26 27 28

6

22

7

8

16

23 24

29 30 31

32

33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60

61 62 63 64

Divide the magic squares into m 2 subsquares of order 4. In each 4 x 4 subsquare, circle the diagonal entries as shown in the figure above right. Finally, in the large square, replace all o f the circled numbers by switching each one with the “opposite” entry in the magic square (with respect to the center o f the square). That is, interchange all the circled numbers with their images under a rotation o f the magic square 180° about its center. For example, the entry in the third row and second colum n (i.e. 18) w ould be switched with the entry in the thirdlast row and second-last column (i.e. 47). (If this is too confusing, you can replace any circled num ber x with the num ber n2 + I - x.) O ur com pleted magic square is shown in the figure on the right.

There are many places to read more about magic squares. Mathematical Recreations and Essays by Coxeter and Rouse Ball is an excellent source. See the Annotated References for bibliographic information.

34

I

Number Theory Two Theorems about Magic Squares

In M agic Squares, we saw how to construct magic squares of any order greater than 2. In this section, we will prove a couple o f theorems about them. Theorem I

t

• r- . , n(n2 +1) In a magic square of order n, the magic sum is ----------- .

In M agic Squares, we had three different methods for creating magic squares. You can check that in each o f the exam ples we created ( w = 4, 5, 6 , and 8 ) that the theorem holds. (This makes us happy, because otherwise the theorem w ouldn’t be very good!) Proof. Let s denote the magic sum for a magic square o f order n. Since this is a magic square, the sum o f the elements in each o f its n rows is s. Therefore the sum o f the elem ents in all n rows is ns. But, the elements in the magic square are the numbers 1 ,2 ,3 ,..., n2, so the sum of the elem ents in the magic square is the sum of the first n2 natural numbers; that is, 2

n ^n

2

This formula will be proved in Sums o f k'h Powers (p. 135).

. n2(n2+1) Com bining the two preceding statements, we have ns = ----from which we

deduce that

5

n(n2 +1) = ----------- .

So our first theorem tells us, for example, that the magic sum o f a 10 X 10 magic square is 505, before we even look at a magic square of order 1 0 .

35

Number Theory 0

Our second theorem says that there is essentially only one m agic square o f order 3. Before we discuss it, w e’ll have to elaborate on what we mean by “essentially” . For example, at first glance, the following magic squares all look different. I

II

III

CO

1

6

6

1

8

4

3

8

3

5

7

7

5

3

9

5

1

4

9

2

2

9

4

2

7

6

But square II is obtained from square I by switching the first and third colum ns — in effect, by reflecting the square in the m iddle column. Square III is obtained from square I by rotating the square clockwise 90 degrees. (It can also be obtained from square II by reflection in the diagonal passing from the top-right o f the square to the bottom-left.) We say that these squares are identical up to rotations and reflections. (This is one instance o f the mathematical concept o f symmetry.) Theorem 2 There is only magic square o f order 3 (up to symmetry). One way o f proving this is to try all possible ways o f placing the integers 1 through 9 into a 3 X 3 matrix. There are 9! = 362,880 possibilities to try, so this would take some time. With a bit o f cleverness, we can avoid this hassle. Proof. The magic sum must be 15.

(This is ju st Theorem 1 for the special case n = 3.)

The number in the middle square must be 5. The proof o f this is very similar to the proof o f Theorem 1. Fill up the magic square with 9 unknowns a, b, c, d, e , f g, h, i as follows:

36

■f• J Number Theory a

b

d

e f h i

g

c

Now, for exam ple, a+e+i = 15, because the sum o f the elements on the main diagonal must equal the magic sum. Therefore, 3e = (a+e+i) + (b+ e+ h) + (c+ e+ g) - (a+ b+c)- (g+h+i) = 15 + 1 5 + 1 5 - 1 5 - 1 5 = 15 So e = 5. From here, there are several directions to go. If you are feeling intrepid, you may want to explore on your own before continuing. The numbers 9 and 8 can’t be in the same row, column, or diagonal, because their sum is already greater than 15, the magic sum. Similarly, the 9 and 7 can ’t be in the same row, column, or diagonal, and neither can the 8 and 7. So w here can these three “big” numbers fit? A little thought shows that they must lie in one o f the following configurations (where the big numbers are represented by a “B”). B

B 5 B B

B

5 B

B

B

5 B

B 5 B

B

Now, the 4 and 6 must lie on opposite sides o f the 5; 4 -5- 6 must be either i row, column or diagonal passing through the center. But, looking at where the big numbers have to be placed (see step 3), we see that the 4-5-6 must be a diagonal. As w e’re considering magic squares to be the same up to rotations, we can assum e that the 4 is in the top right comer. So our magic square looks like:

37

D

Number Theory

W here can the 9 be now? The 9 can ’t be in the same row or colum n as heir sum is already 15, and the addition o f the third term in that row or colum n would knock the sum over 15. So the 9 must be in one o f the circled positions in the following diagram. 4 5

;)

6

But w e’re counting magic squares to be the same up to reflection, so we may as well assume the 9 is at the middle position o f the first row. O ur magic square now looks like:

ji^ n From here, using the fact that magic sum is 15, it’s easy to fill out the remaining elements. So the only m agic square, up to rotations and reflections, is:

7

9 5

4 3

6

1

OO

2

We observe that this is the same as the 3 X 3 m agic square in M agic Squares (p. 32), rotated by 180° about the center.

38

I

Number Theory Food fo r Thought

O.

If we take away the “sym m etry” condition, how many magic squares o f order 3 are there? (Answer: 8 )

& In how many ways can 15 be expressed as the sum o f three distinct positive integers between 1 and 9? Do all such triplets appear as rows, columns, or diagonals o f the 3 x 3 magic square? (This fact can be used to provide another proof of Theorem 2.)

© . The Magic Fifteen Game1 N ine cards, the ace through nine o f spades, are rem oved from the deck and placed face up on a table. Two players alternate in picking up cards. The first player to have three distinct numbers summing to fifteen wins. (Some games will end in a draw.) Can you see a trick that will enable you to play really well? (Hint #1: It is identical to another w ell-know n gam e m entioned elsew here in this book. H int #2: W hy has this problem been included in this particular section?)

'In Berlekamp, Conway, and Guy's Winning Ways, The Magic Fifteen Game is attributed to E. Pericoloso Sporgersi.

39

D

^ W ( L a ,t

^ h fo u .

£ n jo y !

‘ .P. G rossm an’s playful approach to life w as ev id en t from an early age. In kindergarten, he developed an enduring interest in taking things apart: clocks, c a lc u la to rs , w a tc h e s, ra d io s — ev en com puters and a mechanical record player. His ingenuity d id n ’t alw ays m eet with popular approval, however; he was once sent to the principal’s office for working too far ahead o f his class.

J

“Taking things apart” to understand them typifies J. P.’s method of learning. In Grade 9, he and a friend read an article in D iscover m agazine about fractals, at the start o f the fractal craze. They were curious to find out how fractals worked. The first hurdle was to u n d erstan d this m y sterio u s “ i” th at appeared in the equations as the alleged square root of -1. Rather than ask for advice and risk being told that their questions were “too hard”, they played around with the concept o f “i” and figured out how it worked. They then wrote com puter program s to generate fractals on w hatever machines they could find, starting with a Vic 20 and working their way up, beginning with BASIC and learning other languages as they needed them. In Grade 10, J. P.’s competition career took off, thanks to his intuitive approach to mathematical challenges com bined with his ability to pick problem s apart. In Grade 11, he won the Canadian M athem atical O lym piad for the first o f three consecutive years, easily qualifying for the Canadian Team to the International M athem atical Olympiad. During his three years com peting at the IM O, he won two Silver M edals and a Gold M edal. Moreover, in his second-last year o f high school he placed first in the USA M athem atical Olympiad, the last year Canadians were officially allowed to compete. W hile studying m athematics, physics, and electrical engineering at the U niversity o f Toronto, J.P. took the N orth American Putnam M athem atical Competition three times, twice achieving the highest honor o f Putnam Fellow. He has one more year o f eligibility before he graduates from Toronto.

40

Z

P

z r s o

n

a

f

Along with his talents in mathematics, J. P. pursues many other interests, including chess, debating, com petitive soccer, and skiing. A jazz enthusiast, he is also an accom plished tenor saxophone player. A fter graduation, J. P. plans to work in VLSI (Very Large Scale Integration) chip design. For J. P., chip design is problem solving of a different sort, requiring (like m athem atics) tinkering, fiddling around, and taking things apart. W hat J. P. loves most about mathematics are the ingenious leaps required, “putting together two facts that seem unrelated.” He finds that he has become experienced in “m aking connections betw een different things ju st by recognizing certain patterns,” both within and outside the field o f mathematics. One of his favorite exam ples o f this principle is The M agic Birthday Predictor Cards (p. 21). W hen he saw this trick at a young age and realized how it worked, he had a sudden insight into binary numbers. J. P.’s curiosity is typical o f the young m athem aticians profiled in this book. Also typical is the vigor with which he pursues his interests. Rather than seeing his undiscrim inating intellectual acquisitiveness as a distraction that holds him back and wastes his time, he finds it constantly driving him forward. His philosophy of life? “Do w hat you enjoy!”

41

CoMewAToms

Music is the pleasure the human soul experiences from counting without being aware that it is counting. — Gottfried Leibniz

Combinatorics I

A Mathematical Card Trick This card trick is incredible when you first see it. But be prepared for a let-down when you figure out how it works!

The Trick I ask you to shuffle a deck o f cards thoroughly. Then I ask for them back (face down). Carefully examining the backs o f the cards, I separate them into two piles. I then claim that, through the power o f magic, I’ve made sure that the number o f black cards in the first pile is the number of red cards in the second pile! How did I do it?! arzivjex £1 in c d f o w t lz . 3

T h e n th E lv is n u m b e r is a c tu a lly th e nth Fibonacci number! H ow does Elvis see this? Rather than skipping up the stairs as is his wont, c Elvis steps carefully up one at a time. On the n* step, he writes down E . E lvis begins by lab o rio u sly co u nting the num ber o f ways of climbing to the nth step in his green suede shoes:

1 i

1 or $

p 1 E =1

iKi

'

1 F3 =

2

or

1 or

f

c i

Cl *4 = 3

To avoid intense boredom, Elvis thinks o f a way to w ork out E s without doing lots o f counting. So he thinks, “H ow can I get to the fifth step? Well, I can get to the third step, and then jum p up two steps. There are F 3 ways of doing that. Then again, I m ight get to the fourth step, and then jum p up one step. There are E4 ways o f doing that. But I have to do exactly one o f those two things, so the num ber of ways I can get to step 5 is E 3 + E4. Thus Es = E4 + E 3 ” Elvis quickly generalizes this procedure to show that E n = En + En 2 for n > 2. A long with E = 1 and E2 = 1, this is enough to determ ine E n in general. But the F ib o n acci num bers are given by the sam e recip e (F ; = 1 and F 2 = 1, and p - Fn] + F 2), so these two sequences must be the same! Elvis finishes his proof with the letters Q.E.D., w hich is short for Quod Erat Demonstrandum, a Latin phrase m eaning “which was to be proved” . It’s ju st a slightly pompous way of showing when a proof is complete.

59

V

T

The Fibonacci Sequence I

/ > i

A Formula for the nth Term o f the Fibonacci Sequence A few pages earlier, we observed that the Fibonacci numbers, Fn, are defined according to the following rules:

Then, remarkably, it can be shown that Fn is given by the following equation, often called B in ets Form ula: Teachers! Give your students B inet's form ula. Invite them to guess the sequence — it’s quite a surprise to find nice Fibonacci numbers coming out of that hideous formula! Then they can try to prove it. This is quite a mouthful! We can express Fn m ore simply using some constants that w e’ll pull out of our hat:

x=

1+V5

1—s/5 G

=

x and G are Greek letters, called tau and sigm a respectively. Tau (which rhym es with “O w !”) is a constant known to the ancient Greeks as the “golden m ean” . It has a habit o f turning up in the oddest places in mathem atics; some o f them are described in the chapter Fibonacci & The Golden M ean. In terms o f x and g: n

F =

T

n

-CT

S There are several other w ays o f expressing Fn that you m ight prefer.

Since o = — — (Check this!),

F = n

V5

G can be expressed in term s o f x in a second way: o = 1 - x. (Check this too!) Thus F = n

V5

60 N

The Fibonacci Sequence We know from the definition o f the Fibonacci sequence (although not from the form ula for its general term) that Fn is a positive integer for all n. It follows from n n T a the equation above that F = — is an integer for all n. Since < - for n S Vs & 2 n T all n (verify this for yourself), — is closer to Fn than it is to any other integer. n

V5 t

n

a

.

n

That is, Fn is the integer closest to — . Since — becom es very small as n gets V5 T

V5

n

large, — gets very very close to F . (Try it on your calculator with n = 20.') V5

"

You can prove that the crazy form ula for Fn actually works by employing the sam e m ethod we used in the Solution to Elvis the E lfs Eccentric Exercise (p. 59) — the m ethod of m athem atical induction (although induction w asn’t actually m entioned by name in that section). Basically, induction is a great way o f proving things w hen you already know (or have guessed) what the answ er is. U sing the m agical formula, or induction, or a lot o f ingenuity, it is possible to prove some rem arkable results about Fibonacci numbers, which appear in Funny Fibonacci Facts (p. 146).

Foodfor Thought

'

H ow many ways can you spell “E L V IS”? Start at the E and move either dow n or to the right at each stage.

a n s v js z

Li.

cjLv e h

on

tlis

ftto v a L n cj

jia c jc .

*It is indeed remarkable that for any large n, f " / \!5 is very very close to an integer. Even more remarkable is that for any large n, xn is also very very close to an integer — try it on a calculator and see! These oddities, and others, come up in Some Interesting Properties o f the Golden Mean (p. 151).

61

I

The Fibonacci Sequence

Answer to Foodfo r Thought on the Previous Page There are 16 ways to spell Elvis. Is it a coincidence that the answ er turned out to be a pow er o f 2? As a hint, find out how many ways there are to spell “N O ” in the following diagram, starting with the “N ” in the upper left-hand comer, and moving either down or to the right. N -> O I O If you’re not convinced (and you’re probably not), then count the num ber o f ways o f spelling “NEGATIVE” :

N- >E- >G I 1 E^G A i G A T

A

T

I

T

I

V

I

V

E

A

T

I

V

E

T

I

V

E

If you are still not convinced, read the upside-dow n hint below.

[ B u o § B ip o q i

I

V

E

s q jB d 821

= LZ

• y u b o i jsj o q j

uo

g ub

oi

t\[ u i o j j

3 -ratp p u i j 3 M ‘ u J o q B d s u p g u m u p u o j u io jj

s q jB d g o jb o j o q i o s ‘ s A b m o m j u i y

UB qOBOJ UBO 3 M ‘ 9 Ip B O U IO J J -Q B 01 JvJ o q i u i o j j s q jB d

V

E

p o jb o j o q i s n q x -sA b m a s o q j jo qoBO u i o j j 's A b m 2

ju o jo jjip

2 u i 9 b qOBOJ u b o o m

s ,j

ui j

ju a o B fp B u b qOBOJ

u b o o m ‘j q o q i u i o j j j u M o p u o j i f S u d i p 01 o a o i u t i j u o fo u i am f i ] d u o S Dip dip uo 3 uo 01 f t tuouf sipDd duDiu m ° h ‘J S B o j s i u io [ q o jd s i q i o j X [d d B u b o o m X S o j b j j s o u q

Here is another question: there are eight E's in the diagonal above. How many paths are there ending at each particular E? Answer: 1-7-21-35-35-21-7-1. H ow and w hy does this relate to Pascal's Triangle (discussed in The Triangles o f Pascal, Chu Shih-Chieh, and Sierpinski, p. 98)?

62

Gam c THcory

The game isn't over till it's over. — Yogi Berra

Game Theory Ik *

—-^..'d ...» r"

The Theory o f Games Very loosely, game theory is the field o f mathem atics that analyzes strategies in a wide variety o f situations, such as the com petition for scarce resources. It is very im portant in economics, and also o f interest in political science, psychology and philosophy. One o f the 1994 Nobel Prizes in Econom ics was awarded to John Nash, a mathematician who laid much o f the groundw ork for game theory in his 1950 Princeton Ph.D. thesis, com pleted w hen he was 21. (This w asn’t the first N obel Prize in Econom ics given for brilliant m athem atical work; for another exam ple, see Game Theory and Politics: A rro w ’s Theorem, p. 6 8 .)

One important exam ple o f gam e theory in action is “The Prisoner’s D ilem m a” . There are many variations o f this problem , but the basic structure is as follows. Two suspects are taken into custody and separated. The District Attorney is certain that they are guilty o f a specific crime, but she Hmmm...is he does not have enough evidence to convict them gonna rat on me? at a trial. She explains to the suspects that they each have two alternatives: they can confess to the crime, or not confess. If neither confesses, then she will charge them on some trivial offense, and their punishment will be relatively minor— one year in jail. If one o f them confesses, then he will get off scot-free, while the other will serve ten years in prison. If both confess, then they will both be prosecuted, but the D istrict A ttorney w ill recom m end a less severe sentence— eight years each.

C

What should the suspects do ? Put yourself in the shoes of one o f the suspects. No m atter what your partner in crim e does, you are better off by confessing. (If your partner is silent, you will get off completely, and if your partner confesses, you will receive a sentence o f eight instead o f ten years.) However, if both suspects act in their individual self-interest and follow this logic, then both will confess and both will receive eight year sentences. This is much worse for them than if they had both rem ained silent. Ironically, the strategy which would serve the best interests o f each individual fails to serve the best interests o f either when applied by both individuals.

64

Game Theory A nother w ell-know n idea with a gam e-theoretic flavor is the following: two distrustful people wish to share a pie equally. One way of doing this is to have one person cut the pie and let the second choose the piece. That way the cutter will try as much as possible to cut the pie fairly. (How could we generalize this procedure if there were three people rather than two?) Game theory is also used to analyze things that we usually think of as “real” games. For exam ple, the popular gam e Connect-Four™ was com pletely solved in 1984 — there is a perfect strategy for the first player. In other words, if you are the first player and you know the strategy, you can always win, no m atter how brilliant your opponent is. One way o f analyzing certain games is to look at all possible ways in which the gam e can be played out. For example, with Tic-Tac-Toe one would realize that there are essentially only three opening moves:

X

X X

In term s of strategy, a move to the lower right com er is really just the same as a m ove to the upper left; just consider the entire board rotated 180° about the center square. You will notice that the same principle o f symmetry was invoked in Two Theorems about M agic Squares (p. 35). Each o f these first moves can be followed by only so many second moves. All possibilities can be explored and the best possibilities can be found. By this sort of analysis one can confirm the well-known fact that if both players play reasonably well, a gam e o f Tic-Tac-Toe will end in a draw. (A lesser-known fact: in some sense the best opening m ove is the comer. If the second player does not play in the center, then the first player can force a win.) The best way to get a feel for gam e theory is to play around with some games and see what makes them tick. Sometimes games with very simple rules can have some subtle and beautiful mathematics lurking in the background. The most famous exam ple o f this is the gam e o f Nim which we investigate in the next few pages.

65

Game Theory The Game ofNim N im is perhaps the best-know n gam e for w hich w inning strategies have been com pletely analyzed. The rules are simple, but the w inning strategy is far from evident. In fact, the strategy is simple, unexpected, and has connections to other parts o f mathematics.

*Nim is played by two players with several rows o f toothpicks. •Each player alternates rem oving some toothpicks from a single row. •The person who removes the last toothpick is the winner.

For example, a common starting configuration consists o f five rows, with 1, 2, 3, 4, and 5 toothpicks respectively as shown below.

To whet your appetite The winning strategy requires that the first player must remove exactly one toothpick from one of the odd rows, or the second player will be able to win.

If a gam e began with 3 rows as shown below, it m ight proceed as follows:

nitia y

66

^*FSt player t^ en removes two toothpicks from the third row, leaving:

Player 2 then takes the lone toothpick from the first row, leaving:

Game Theory Player 1, sensing victory, takes a single toothpick from l_ the middle row leaving:

m u you

Player 2 now has no choice but to take one of the remaining toothpicks, but then Player 1 can win by taking the one which remains.

w in n in g s tra te g y is. e x p la in e d in c d {o w t o (W in a t c d V im ff i. IQ lJ .

B e fo re

r e a d th a t s e c tio n , y o u s h o u ld f i n d a n o jijio n e n t a n d fi^ay a fe w g a m e s o f

< e N im , in o r d e r to g e t a f e e l f o r w h a t m a le s a

g o o d s tra te g y , e s p e c ia lly in

th e e n d g a m e w h en th e re a r e n ’t m an y t o o t h p ia ls le ft .

D f you

haue

enough

s t ra t e g y y o u r s e lf , f i r s t

s e lf - c o n t r o l, y o u in

s h o u ld try

the. s m a l l cases,

to f ig u r e

a n d la t e z

in

out

th e

w in n in g

c o m p le te g e n e ra lit y .

O h i s m a y m e a n p o n d e r in g th e g a m e o n a n d o f f fo z w e e lis o z m o n th s.

B u t th e

s a t is fa c t io n in c o m in g u p w ith th e w in n in g stza te g y y o u z s e lf is fa n t a s t ic - it s n o m e a n fe a t o f in g e n u it y I

U f y o u ’re t h in h in g o f t r y in g

t h is ro u te ,

sn ea h a

p e a l a t th e d r y p t ic x e K im c d f in t .

C p y p tie

N im

H in t

SBUIFS UIBO XSJU JOH EPXO UIF TJAF PG U IF QJM FT JO CBTF 10, X SJU F UIFN EPXO JO CBTF 2.

To decode the Cryptic Nim Hint, write the above statement replacing each letter with the letter which precedes it in the alphabet.

67

Game Theory Game Theory and Politics: Arrow's Theorem Philosophers and laypeople alike have debated the question, “W hat is the best form o f governm ent?” for millennia. In this day and age, our society has com e to the consensus that dem ocracy is in some sense the best. (O f course, this begs the question, “W hose version o f dem ocracy is the best?”) The quest for a “perfect” system o f government can be viewed from a mathematical perspective. There are certain “irrationalities” in some o f the candidates for the best system. For example, under the “majority rules” system, a decision between two choices is made simply by taking a vote. But som ething odd can happen when m ore than two choices are possible. For example, there m ight be four candidates for the presidency of a country. One o f them leans (politically) to the right, and is popular with 28% o f the voting public. Three o f them lean to the left, and have similar platforms — they are each supported by 24% of the voters. Thus, a leader will be elected from the right, although the vast majority (72%) of the population would have preferred one from the left. Some people would ju st consider this “tough luck” ; others would suggest that a slightly different system would be fairer. Kenneth Arrow, an econom ist at Harvard University, addressed this issue in 1951. (He won a Nobel Prize for this and other w ork in 1972.) He started by dem anding that a system satisfy the barest m inim um standards of rationality. We want some system to turn our individual desires into a collective choice. Imagine that we have a country (or state, or com mittee, or school) with three or more individuals, and we want to make a collective choice am ong several (more than two) options. Everyone writes down their preferences (an ordering o f the options) on their ballot. These are then sent to a computer, and the com puter produces a single result, listing the options in order o f collective preference. The com puter can use any crazy algorithm, subject only to the follow ing minimal reasonable restrictions: 1. First o f all, a non-restriction. We w on ’t dem and that all votes be counted equally. This allows the possibility o f a system in which the value of a vote depends on the voter’s age, education, profession, income, or eye color. For exam ple, the com puter may decide that college professors are out o f touch with reality, and that their votes should be ignored.

68

2. If everyone prefers option X to option Y, then society as a whole will prefer option X to option Y. This is certainly a reasonable m inim um condition! 3. (Independence of Irrelevant A lternatives) Society’s decision between two choices should depend only on how individuals feel about those two choices. In terms o f the 1992 U.S. Presidential election, this means that society’s preference between Clinton and Bush should not be affected by whether Perot is on the ballot. 4. W e’ll assum e that individuals aren’t fools — they w on’t say that they prefer A to B, B to C, and C to A. So we will dem and the same of society. Note that the “majority rules” system fails this condition. For example, consider a society with three people, with the following preferences: Voter 1 prefers Voter 2 prefers Voter 3 prefers

A to B to C B to C to A C to A to B

Then if society operates by “m ajority rules” , then society would prefer A to B, B to C, and C to A, each by a 2-1 vote! The four conditions above seem like reasonable bare m inim um conditions for an election. We have seen that “majority rules” fails this test, but one would think that surely some dem ocratic system would pass. N ot so, says Arrow: A rrow ’s Rem arkable Theorem The only system satisfying these minimal is a dictatorship by one person. This theorem has numerous disturbing implications, many o f w hich should be im m ediately evident to the reader. One possible response is that m any other factors need to be taken into account when judging a decision-making system, and although the varied dem ocratic systems all fail one or another o f the rationality conditions, they still w ork very w ell fo r o th e r reaso n s. R ead ers m ig h t s y m p a th iz e w ith W in sto n C h u rc h ill’s w ell-k n o w n comment: D em ocracy is the w orst system devised by the w it o f man, except fo r all others.

69

Game Theory Food fo r Thought Partial answers and discussion are given on page 72.

O.

There are certain paradoxes with a flavor sim ilar to what w e’ve seen here. Consider three (fair) dice, num bered unusually:

Two people play a game. They choose different dice. They then roll their respective dice, and the player with the higher number wins. If one player has A, and the other has B, it is fairly straightforward to see that A will win 2/3 of the time. B will always roll a 3, and A will beat that 2/3 of the time (by rolling a 5), and lose 1/3 o f the time (by rolling a 1).

W ho will most likely win in a com petition between B and C? W ho will m ost likely win in a competition between C and A? W hy is that surprising? If you are unsure how to proceed, take three normal dice, change their numbering, and experiment!

70

Game Theory Food fo r Thought © . In 1956, the U.S. H ouse of Representatives voted on a bill calling for federal aid for school construction. An amendment was proposed stipulating that federal aid would be provided only to those states whose schools were integrated. The voters were loosely divided into three equal-sized interest groups: Republicans, northern Democrats, and southern Democrats. O f the three options available, the three groups had differing opinions. The Republicans favored no bill at all, but if one were to pass they preferred one restricting aid to integrated states. The southern Democrats, being from states with segregated schools, preferred the original bill first, no bill second, and the am ended bill third. The northern Democrats preferred the am ended bill, followed by the original bill and no bill. These preferences are sum m arized in a table. Preference First Second Third

Republicans no bill amended bill original bill

1

Southern Democrats Northern Democrats I original bill amended bill , no bill original bill am ended bill no bill

Clearly, the original bill would have passed, as the Democrats together preferred the original bill to no bill at all. However, in keeping with House procedure, the first vote was on whether to accept the amendment; the Republicans and northern Democrats together ensured that the amendment passed, because they both preferred the amended bill to the original bill. The second vote was whether to approve the amended bill, or to have no bill at all. This time, the Republicans and the southern Democrats, both preferring to have no bill rather than the am ended bill, conspired to ensure the bill’s defeat. Paradoxically, the proposal o f a popular am endment to a popular bill ensured the bill’s eventual defeat! Analyze how this happened.

The author was first introduced to this topic by Paras Mehta of Oxford University. Arrow's Theorem is discussed further in Hoffman's Archimedes' Revenge. (See the Annotated References for bibliographic information.) The example of the federal bill for school construction is taken largely from that source.

71

Game Theory Answers to Foodfor Thought (pp. 70-71) 1. Com petition between A and B: B will always roll a 3. A will therefore win two-thirds o f the time by rolling a 5 and will lose one-third o f the tim e by rolling a 1. Thus A will usually beat B. Com petition between B and C: By a similar argument to the one above, B will usually beat C. Competition between A and C: One method for determining who will win the com petition between A and C is to list the possible outcomes in a table like the one below.

C rolls a 2 ( 2/3 of the time)

C rolls a 6 (1/3 of the time)

A rolls a 1 (1/3 of the time)

A rolls a 5 (2/3 of the time)

A rolls a 1 and C rolls a 2

A rolls a 5 and C rolls a 2

2 l 2 ~ x - = — 3 3 9

of the time.

A rolls a 1 and C rolls a 1

1

_ x _

1

“ 9

3

3

o f the time.

6

2 2 4 — x — = — o f the time. 3 3 9

A rolls a 5 and C rolls a 1

2

2

3

3

9

- x - =-

6

o f the time. |

From this table, it is clear that the probability that C will beat A is: 2

1 2 I

9

9

5

1-------------—

—

9

9

Incredible as it seems, we have just shown that A will usually beat B, B will usually beat C, and C will usually beat A! 2. This paradox is very sim ilar to the paradox we encounter in the problem of the dice given above. It also relates to the failure o f the “majority rules” system to satisfy condition 4 on page 69.

72

^

\

H is to r ic a l D ig r e s s io n

\

Social Choice Theory

THE BETTMANN ARCHIVE

Social choice theory is a part o f gam e theory that is studied by people in a num ber o f different fields, including econom ics, political science, and philosopy. Its ties to political science relate to theories o f voting, elections, and the state and its links to philosophy pertain to theories o f equity, ju s tic e , and ra tio n a l actio n . A lthough it is a young field, its roots lie as far b ack as the eig h teen th cen tu ry w ith Jerem y Bentham (philosophical foundations o f utilitarianism and welfare analysis), John C h arles de B o rd a, and th e M arq u is de C o n d o rc e t (th e o ry o f e le c tio n s and com m ittee decisions). C harles Lutw idge D odgson, better know n as Lew is Carroll, author o f Alice in Wonderland and Through the Looking Glass, anticipated developments in social choice theory when, as an Oxford mathem atics tutor, he m ade the first com plete analysis o f voting paradoxes.

K enneth Arrow established the current com binatorial framework o f social choice theory. In Archim edes' Revenge, Paul H offm an states: The N obel Prize-w inning work o f Am erican economist Kenneth Arrow shows that achieving the ideals o f a perfect democracy is a m athem atical impossibility. Indeed, undesirable paradoxes can arise not only in voting but even before voting takes place, in deciding how many representatives are allocated to each district in a system o f indirect representation, such as the H ouse o f Representatives (p. 213). (See the A nnotated References for bibliographic inform ation.)

73

Game Theory Elementary My Dear h erlock H olm es, m aster o f lo g ical deduction, chases a hardened criminal through Europe. One player plays the dashing and daring H olm es, and the other player plays the evil, nefarious, dastardly, re p re h e n s ib le , u n s c ru p u lo u s , n o to rio u s crim inal— hereafter denoted by the acronym, ENDRUN. T h e G am e This battle o f wits can be played in the airports o f Europe. A less expensive alternative is to use the map on the opposite page. Each of the dots represents a city where the chase can take plaee, and the players can indicate their position by pointing at a city with a peneil or a pen. The straight lines indicate air routes where the players may travel. Sherlock Holmes begins in Stockholm, w here he begins his famous chase. The criminal, ENDRUN, starts off in Paris, w here he has ju st stolen the M ona Lisa. Each player, in his or her turn, may jum p from one city to another connected by an air link. Holmes, being true and noble, goes first. If Holmes lands in the same city as ENDRUN, he captures the thief after an exciting and excessively destructive car chase and wins the game. If Holmes does not eatch ENDRUN in fifteen or few er moves, EN D RUN escapes and wins the game. T h e P ro b lem This game is fixed. If you pick the right side, and play perfectly, you can always win, no matter how well your opponent plays. The big question is: who has the winning strategy? (And w hat is it?) A Hint This is less a hint than plain com mon sense. D on’t give up too soon! Try playing a few games with som eone else, just to get a feel for how it works. Play both sides. M ake a conjecture. Then try to figure out why it’s right (or wrong!). a n u v E t Li cjLuzn o n jia c jz

74

1S6.

£~a

(S

C7£Z.

rD rU

Game Theory

I

Sherlock Holmes' Continental Chase

GeOMCTRV

There is no royal road to Geometry. Menaechmus (to Alexander the Great)

THE BETTMANN ARCHIVE

The Mathematician Who Anticipated Calculus by Almost 2000 Years When historians o f mathematics attempt to identify the g reatest m a th em a tic ian o f all tim e, they in e v ita b ly p ro d u c e a lis t o f th re e g ia n ts : A rc h im e d e s, N ew to n , and G au ss. A lth o u g h A rc h im e d e s liv e d in a n c ie n t G re e c e , h is mathematical techniques were comparable in rigor to those o f the greatest m athem aticians o f the 17th and 18th centuries. A m ong his many treatises was one titled On Spirals in which he presented twentyeight propositions about spirals — including one which now bears his name. H is m eth o d o f “e x h a u s tio n ” an tic ip a te d the developm ent of integral calculus by alm ost 2 0 0 0 years! In his treatise Q uadrature o f the P arabola, he applied the m ethod of exhaustion to derive the theorem presented on page 79, expressing the area o f a segm ent o f a parabola in term s o f the area o f a particular inscribed triangle. Archim edes was unable to find the area o f a general segm ent o f the other conics, the ellipse and the hyperbola, because these involve transcendental functions not known in his time. A rchim edes also m ade significant contributions to applied m athem atics and engineering, including his famous law o f hydrostatics. He developed a variety o f practical devices such as the lever which he used to catapult huge quarter-ton rocks against an attacking Roman fleet during the seige o f Syracuse in 212 B.C. O f his countless contributions to many branches o f m athematics, Archimedes was most proud of his discovery that the volume o f a sphere is two2R thirds the volume o f the cylinder in which it is inscribed. (His method is described in Archimedes Strikes Again on page 82.) Archimedes requested that this discovery and the corresponding diagram be carved on his tomb. This request must have been granted, for we are told by the great The volume of a sphere is two-thirds Rom an orator Cicero that he had restored the volum e of the right circular Archimedes' tomb and the engraving when cylinder in which it is inscribed. he was quaestor in Sicily.

78

Geometry Lengths, Areas, and Volumes Arguments both True & False Part 1: The Area o f a Circle The circumference C and area A o f a circle with radius r can be calculated using these fam iliar formulas: C = 2 7t r and A = Kt2 From these formulas, we can readily deduce the formula Here is another intuitive method to see why this is true. We divide the circle into n equal “pie slices” (no pun intended). The full circle divided into n equal parts

One slice of the full circle is one of n equal parts

A = y

.

This entire argument can be better expressed in the language of limits. In fact, if you work out what is really going on, you’ll rediscover the famous result: lim x sin X —>00

\x j

-■ 1 or lim *^0

sin x

=

1

*

This example is a great motivation for why we might expect the above limit to be true. Each pie slice looks a lot like a triangle with base ~ and height r (especially when n is very large). Then each triangle has area ^ . As there are n such pie slices, the total area o f the circle is ) = rC. 2n 2 Part 2: The Area o f a Segment o f a Parabola Som e readers may think that the foregoing argument is bogus. However, this is the method by which the ancient Greeks computed the area. Calculus o f some sort is necessary to m ake this parabolic segment argum ent rigorous, but even without calculus = colored region the Greeks were able to prove some remarkable results. For example, Archimedes wished to find the area of a segment o f a parabola. (A parabola is a plane curve sim ilar to the curve y = ax2, for some real num ber a. A segm ent o f a parabola is the closed region bounded by that parabola and any straight line.) Archimedes gave a rigorous proof that the area of the segment o f the parabola is four thirds the area o f a triangle having the same base and equal height.

79

Geometry Part 3: The Surface Area o f a Sphere You m ight still be feeling queasy about the lack o f rigor in this style o f argument — and rightly so. Here is a sim ilar argum ent that gives a w rong answer. Can you find the flaw ? (This is trickier than the usual “find-w hat’s-wrong-w ith-this-proof” puzzles, as this argument doesn’t claim to be rigorous in the first place.) We divide the sphere of radius r into two hem ispheres at its equator, and partition 2nr the equator into n segments, each o f length ~ . We draw great circles through these partition points and the north and south poles of the sphere. Now im agine the sphere sliced along these great circles, peeled back and flattened out like a M ercator projection (for readers with a background in geography).

• 2 TTr Each o f these little triangles has base — and height one-quarter the circum ference o f the sphere, ^ 4

, and hence has area

1 ( 2tcr^\f 27tr^

2n

Since there are 2n triangles, the total area 2 2 . is n r , which is the correct form ula for the surface area of the sphere — N O T !!! Oh n o ! Some of you know that the surface area of a sphere o f radius r is 4 7 t r \ so we are off by a factor of ^ . W hat has gone wrong? How can the m ethod that w orked so well in Part 1 have failed us in Part 3? Philosophical pseudom athem atics tells us that two wrongs do not make a right; but what do a right and a wrong make?

80

"Yes, yes, I know that, Sidney .. everybody knows th a t)... Bui look; Four wrongs squored, minus two wrongs la the laurth power, divided by this formula, da make a right.”

THE FAR SIDE COPYRIGHT © FAR W O RK S, IN C . distributed by U niversal Press S yndicate. Reprinted with perm ission. All rights reserved.

2V n A 4

2 2

n r

Geometry Part 4: The Volume o f a Sphere We can add to this confusing stew by working out the volume of the sphere. The reader should be able to show (by the same argum ent as in Part 1, this time by d isse c tin g the sp h ere into n alm o stpyramids with vertices at the center o f the A sphere) that if a sphere has surface area A n and radius r, then the volum e V o f the sphere is given by V =

.

4tcf^ As the volume o f the sphere is V = and the surface area is A = 4 7 tr2, the form ula is correct. The m ethod appears to be working again.

Food fo r Thought

O.

a) Define the “radius” r o f a square to be the length of the altitude from the center of the square to one o f its sides. Show that the form ula from Part 1 (A = rC) still holds. b) How should the “radius” o f a regular rc-gon be defined so the form ula still holds? c) W hat should the “radius” o f a cube be for the formula in Part 4 to hold? How about a regular tetrahedron? W hy are all o f these answers similar? © . The “hypervolum e” o f a “four-dim ensional sphere” o f radius r is 2 4

H =

n r

. Can you use a m ethod sim ilar to that o f Part 1 and Part 4 to find

the “surface volum e” ?

Some of this material appeared in Mathematical Mayhem (Vol. 4, Issue 4, (March/April 1992)). Part of the erroneous argument is taken with permission from the Mathematical Digest (No. 86, (Jan. 1992)), 15. For more information on either of these publications, see the Annotated References.

You’ve probably m em orized the form ula for the volum e V of a sphere (in term s of 4

3

the radius r : V = —7rr ), but you probably haven’t seen w hy it is true. Here is an intuitive proof, originally by Archimedes.

First we recall one fact: the volume o f a right circular cone with base o f radius r and height h is V — —Ttr2h . 3

This is a special case of something more general.

If you have a strange plane

shape S with area B, and a point P at a height h above the plane, then the cone with base S and vertex P consists of everything that lies on a line segm ent with one endpoint in S and one endpoint at P. Then the volum e o f this cone is V = —B h . 3 This is a generalization o f the fact that the area o f a triangle w ith base b and height h is

Can you propose the further generalization to four dim ensions?

82

Geometry K

f

Archimedes Strikes Again (co n fd ) Archimedes com puted the volum e o f a hem isphere follows. Place your hemisphere (of radius r) on a flat table A. Beside it, place a circular cone with base o f radius r, and height r, balanced perfectly on its vertex.

Table A

On another table (call it B), place a cylinder with base o f radius r, and height r.

Table B Now consider a cross-section over table A at height As show n in the d iag ram , th e crosssection of the sphere is a circle of radius “ '

h.

The cross-section o f the cone is a circle o f radius h. (The proof uses the fact that the the shorter sides o f a 90°-45°45° triangle are equal.)

Table A We are now in a position to execute Archimedes' beautiful heuristic proof.

83

Archimedes Strikes Again (cont'd) Thus, the total cross-sectional area at height h above table A is 2

+ t hem isphere

7th2

t inverted cone

Ttr

2

T total cross-section

The cross-section of the cylinder is always a circle o f radius r, so the total crosssectional area above table B is also 71r2. H ere’s the key step: for each height, the cross-sectional areas are the same, so the total volum e must be the same. (You m ight w ant to ponder this for a w hile and convince yourself that it is reasonable.) Thus: volum e o f hem isphere + volum e of cone = volum e o f cylinder so,

volum e o f hem isphere +

3

(ndy

= ( n r 2 )r

From this, we conclude that the volum e of a hem isphere of radius r is 2 Therefore the volume V of a sphere of radius r is given by:

A nd the proof is com plete! This clever argum ent surely deserves a shouted “Eureka” , but I’m not sure if I ’d run naked through the streets over it. If you are com fortable with integral calculus, you will recognize how A rchim edes’ ideas anticipated its fundamental concepts. And if you haven’t seen integral calculus before, keep this exam ple in mind for when you do!

84

Geometry I

Late at night, a thief with a flashlight has c lim b e d e x a c tly halfway up a ladder in o rd e r to break into a house.

© Taisa Kelly.

Falling Ladder Problem U n lo rlu n a te ly (o r fo rtu n a te ly . depending on your point o f view), he hasn't secured the base of the ladder. (D on’t try this at home!)

0

T h e la d d er slid e s o u t from under him, leaving him ignominiously on the ground.

Reprinted with permission

A police officer running to the scene observes the attempted burglary. Through the darkness, all she can see is the path of the flashlight. However, she is able to deduce from this the nature o f the misadventure and later provide surprisingly accurate testimony at the burglar's trial.

W hat path did the flashlight take? Did it travel in a straight line between point A and point B? Before looking at the answer (in the Solution to the Falling Ladder Problem, p. 164), you may want to think about it, and com e up with a guess of your own. W hat would be the path o f a point on the ladder but not at the half-way point? •

...

85

Geometry An Easy Proof that arctan 113 + arctan 1/2 = 45° Here is a slick non-algebraic proof o f the identity arctan 1/3 + arctan 1/2 = 45° Consider the triangle in the Cartesian plane with vertices at A (0,0), B (1,3), and C (2 ,l), and the point D(1,0).

B ( 1 ,3) There are several ways o f seeing that ABC is a 4 5 °-4 5 °-9 0 ° tria n g le . (F o r e x a m p le , 1

AC = BC = ^ j A B . )

Hence Z A B C = 45°.

But Z A B C = Z A B D + ZD BC. C an you see w hy ta n (Z A B D ) = 1/3 and tan (ZC B D ) = 1/2? If you can, then you’ve basically completed the proof!

The especially intrepid reader m ight w ant to try to find sim ilar proofs of the following: a) arctan 1 + arctan 2 +arctan 3 = 180°. b)

1

1

1

arctan —= arctan - + arctan 2 3 7 c) M ore generally, if a and b are positive integers with a > b, arctan

1

1 b = arctan — I- arctan -----------a~b a

anot(z£.r E x a m f-h o f a [ n o o f by dLagtcun,

see

on

tflE O^ljlOiLtE fza-CjE. '

‘Many of these “proofs by diagram” have been collected in Nelsen's Proofs Without Words. See page 39 of Nelsen's book for a proof (due to Edward M. Harris) of the identity: arctan 1 + arctan 2 + arctan 3 = 180°. (For bibliographic information, see the Annotated References.)

86

Geometry The Ailles Rectangle Everyone who has learned trigonom etry has mem orized the sines, cosines, and tangents o f “easy” angles such as 0°, 30°, 45°, 60°, and 90°. Doug Ailles, a high school teacher at Etobicoke Collegiate Institute in Etobicoke, Ontario, came up with an incredibly sim ple m ethod o f com puting the trigonometric functions of 15° and 75°. In the long history o f m athematical thought, it would not be surprising if som eone had already invented a sim ilar construction. But, in the absence of such inform ation, I hereby christen it the Ailles Rectangle. H ere’s how it works.

ABC is a 30°-60°-90° triangle with sides AB = 2, BC = 1, CA = V3. ABC is inscribed in rectangle A D EF such that ZFA C = 45°. Fill in the missing angles in the problem , and then (using the 45°-45°-90° triangles and the fact that A F = DE and AD = FE), find all missing lengths necessary to calculate the trigonom etric functions of 15° and 75°.

The solution to this problem is given on the next page. A nother way to compute these values is to substitute x = 45° and y = 30° into the sine, cosine, and tangent subtraction formulas: sin(x - y) = sin x cos y - cos x sin y cos(x - y) = cos x cos y + sin x sin y tan .v- tan y ta n (x -y ) : 1 + tan * tan y However, it is often m ore com forting to do things concretely and geometrically.

87

Geometry Solution to the Ailles Rectangle Problem (p. 87) Since the sum of the angle measures in a triangle is 180°, we can easily deduce that A C F and BC E are 45°-45°-90° trian g les. T h erefo re A F = FC = ^2. and BE = CE = 4- . V2 U sing the above results with the equations A F = D E and AD = FE, we obtain the lengths shown in the figure below. V3

A

V2

V2

F

vr2

From triangle ABD in the figure, we can im m ediately w rite the fo llow ing expressions for the trigonom etric functions of 15° and 7 5 °.

sin 15 —

t— 2'12

1r-o a / 3 + I cos 15 = — /=2a/2

tan 15° = 2 - a/3

88

• 7Co a/3+ 1 sin 7 5 ------2 a/ 2

a/3 —l cos 75 = — 7— 2 a/2

tan 7 5 °= 2 + V3

(Dns o f t/iE ^ P w test Q o im i o f 0, B

= -

B

r j

m+ 1

where f m+1l \ j ;

7=0X J )

So, for example, we have: B=-~, l 2

B= 7 , 2 6

B = 0, 3

B=~ — , 5 = 0 , 4 30 5

B = — 6 42

Bernouilli numbers display the following rem arkable properties: a) For all positive integers m, B 2m+I = 0. B2m alternates in sign for successive values o f m. b) The zeta function is defined as follows: 1

1

1

—+ — + — + C(m) = |m 2m

1

Try This!

If m is a positive integer, then

Use property b) to prove this identity. i

( - \ ) m+\

2

m)lg( 2 rn)

„2m-l

2

l

l

+ ^2

3

Can you find 1h— ^ 2

x c) If we expand

as a pow er series in x, then e

-1

n

+--- + ^ 2 +--i

2

= —

6

2m

K (proved by Euler)

2

l +^2

3

•• +^r+- •• /

H is t o r ic a l D ig r e s sio n

The Incredible Bernouilli Family he Bernouilli family of Basel is certainly the most celebrated family in the history of mathematics. Between Nicolaus Bernouilli (1623-1708) and Jean G u stav e B ern o u illi (1 8 1 1 -1 8 6 3 ), the fam ily p ro d u ced tw elve outstanding m athem aticians and physicists.

T

The Bernouilli numbers were discovered by Jacques Bernouilli (1654 -1705), who developed them to solve the problem of summing the k'hpowers of integers (see Sum s o f the k?h Powers, p. 135). Bernoulli numbers are surprisingly linked to Fermat's Last Theorem (described in The M ost Famous Conjecture in Mathematics, p. 123). The German mathem atician Ernst K um m er (1810-1893) proved that if p is an odd prim e that does not divide the numerators o f the Bernoulli numbers Br B4, ..., Bp 3, then Fermat's Last Theorem holds for n = p (and for any n divisible by p). Such a p i s called a regular prim e. Along with the case when n is divisible by 4 (the only case that Ferm at proved himself), this theorem proves Fermat's Last Theorem for all n less than 100 other than 37, 59, 67, and 74. Jacques was also falsely credited as being the first person to notice that the following series diverges. (This series is called the harmonic series and is discussed more fully on pp. 179-183.) 1

1

1

+ - + - '+ —+ ... 2 3 4 One of the more bizarre anecdotes about this impressive family concerns l’H opital’s rule, which appears in virtually every elementary calculus class. 1

L’H opital’s Rule:

I f /0 0 an(l § 0 0 are functions differentiable at x = a such t h a t / / ) = g(a) = 0 , and if / '( * ) / ~TTT exists, then ’im , , g'(x) X->CI g(x)

_ -

/' ,, , X^a g (x)

Jean Bernouilli (1667-1748), the younger brother of Jacques, had instructed a French marquis, G.F.A. de L’Hopital (1661-1704), in the art of calculus. In return for a regular salary, Jean agreed to send L’Hopital all of his mathematical discoveries. In 1694, Jean Bernouilli discovered “L’H opital’s Rule” , and the rest is history.

I

d V otliL n ej ^J\J[o£iES < cA/\je.

P fak cPj-vjaxe. o f tb . ^Pa&scuje. o f ^Jirne.

h e first m a th em a tic al ex p erien ce Jordan Ellenberg rem em bers dates from when he was about five. After staring at a rectan g u lar grid o f holes, he su d d en ly realized w hy m u ltip licatio n is com m utative (that is, why it is that when you multiply tw o num bers together, the answ er doesn’t depend on the order in w hich you multiply them, so m X n = n X m).

T

W hen he was seven, Jordan was discovered by Eric W alstein, a teacher at th e nearby M ontgom ery Blair High School. Jordan had Jordan Ellenberg (USA) already learned how to multiply three-digit Born Oct. 3 0 ,1971 numbers in his head (som ething he has long since forgotten), and had shown other signs of incipient talent. Walstein took Jordan under his wing and oversaw his m athematical development. In G rade 4, Jordan began participating in the Am erican Regions M athem atics League, where most o f the com petitors were already in their final year of high school. At about this time, Jordan discovered the following theorem: If a is an integer and p is a prim e number, then ap - a is divisible by p.

He proved it by looking at Pascal’s triangle modulo p. He soon heard, however, that his theorem had been proven before, and that it was called F erm a t’s Little Theorem. H e was sorely disappointed that it was only a “little” theorem. At W inston Churchill High School in Potom ac, M aryland, Jordan really began to shine. In 1989, he placed second in the Westinghouse Science Talent Search, a national science fair. In his last three years o f high school he did spectacularly well on the USA M athem atical Olympiad, rising from ninth place, to second, and finally to first. He com peted for the United States at the International M athem atical O lym piad in each of those three years, winning tw o Gold M edals and one Silver M edal.

142

W hile studying mathematics as an undergraduate at Harvard, Jordan competed in the North American Putnam M athematical Competition, twice achieving the highest honor o f Putnam Fellow. In three of hi s four years, he was a member of the Harvard team that w as in the process o f w inning the Putnam com petition for eight consecutive years. Despite his mathematical successes, Jordan spent much of his undergraduate career avidly pursuing his interest in literature. He regularly wrote for the Harvard Crimson and was the Fiction Editor for the Harvard Advocate. After graduation, he completed a M asters Degree in Creative W riting at Johns Hopkins University in Baltimore, where he wrote a novel, The G rasshopper King. W hen asked about this unusual detour, Jordan responds, “It is always good to have some element of resistance to w hat’s easy and w hat’s obvious.” H e always intended to return to mathematics, and this was his last opportunity to write seriously. After completing his Masters, he returned to Harvard to begin a Ph.D. in N um ber Theory. Jordan has strong views on the role of mathematics in society: “Math occupies a strange cultural place. If you have an aptitude for math, that often makes you the ‘brain’ o f your class, more so than for other fields such as history or writing.” He believes that mathematics is too often seen as the province o f a special class of people. There is even a disturbing cheerfulness to people’s admission that they don’t like math, and aren’t good at it. As Jordan reflects, “You d o n ’t hear people say, ‘Long words make my head spin!” ’ But everyone needs basic math skills to be a functioning m em ber of society. As a result, Jordan feels that the strength o f a mathematical curriculum should not be judged on the basis of abstruse theorems in calculus. People should learn to use mathematical thinking to make them better critics of the world around them. Jordan cites new spaper polls as one exam ple: “In order to tell when you are being manipulated, you need to have some feel for probabilities and quantities. W hat does it mean when you hear that ‘50% o f Bud drinkers would prefer M iller’? W hat does it m ean w hen you hear that the ‘rate o f grow th o f the deficit is decreasing’?” For Jordan, the most rem arkable feature o f mathematics is its ability to completely engage his m ind and soul: “Hours can go by. Nothing makes me less aware o f the passage o f tim e.” His parting advice for young people interested in the subject? “Do math for m ath’s sake, not because your parents will be proud o f you, or because people will think you are sm art.”

143

**

f/eomcct THE GOLDEN MEAN

The merit o f painting lies in the exactness o f reproduction. Painting is a science and all sciences are based on mathematics. No human inquiry can be a science unless it pursues its path through mathematical exposition and demonstration.

Leonardo da Vinci

Fibonacci & The Golden Mean Funny Fibonacci Facts ‘G c z tfiE dzjbzLtLon o j tizz S d fo n a cziL n u n zljzzi i z z

3 3 ifo n a a a i: F 7 /ie ^ te a te s t £ u x o ft£ c u z

1. f o r" -2 2.

- F j = F„, for

> 2.

d)7 If m is a factor o f n,7 then F m is a factor o f F n . e) Let

2 ti ,, ~ ~ . ( It follows from part d) that Gn is an integer.)

n

Then Gn = Gn_{ + G„_2. oo * > s n= 2

1 , 1 = - a rc ta n --

arctan V

F2n J

3. Bees have unusual reproductive habits. Each m ale bee has only one parent (a mother, the queen bee), whereas each fem ale bee has two parents (both a m other and a father). So each male bee has one parent, and you can quickly check that he has two grandparents (the two parents o f his mother). How many great-grandparents does he have? How m any great-great-grandparents? Can you guess the pattern? Can you prove it? (Drawing a fam ily tree m ight help.)

146

2

2

2

4. a) Can you prove this identity, Fj + F, + ... + Fn = FnFn+l' This diagram may suggest a proof to you.

b) Can you find a simple formula for Fl + F, + ... + Fn in terms o f Fn+11 It m ight remind you o f the form ula for 2° + 2 1 + ... + 2" in terms o f 2n+'.

5. Use a calculator to evaluate 10000/9899 to eight or ten decimal places. W hat pattern do you notice? Can you guess the next few digits? (The remaining questions will m ake no sense unless you have already guessed the pattern.) The digits must eventually repeat, as 10000/9899 is a rational number. Do you find this surprising? Can you find more numbers with a similar property? (Hint: Try 100/89. Then try to find more.) W hat if you w anted powers o f 2 to appear instead?

6

\+ S . . „ . Consider a circle whose circumference is the golden mean, t = —-— ~ 1.61803.

Start at any point, on the circle and take some num ber o f consecutive steps o f arc length one in the clockw ise direction. N um ber the points you step on in the order you encounter them, labelling your first step P p your second step P2, and so on. Prove that when you stop, the difference in the su b sc rip ts o f any tw o ad jacen t n u m bers is a Fibonacci number.

147

Fibonacci & The Golden M ea

7. The rem arkable interplay between Fibonacci num bers and Pascal’s triangle is best shown in a diagram w ithout any words o f explanation: There are several ways o f proving why this is true. W hat may be the easiest method is to let P denote the n* diagonal P a s c a l sum , an d th e n o b se rv e th a t P 3 = P 2 = 1 and show that P n = P n - 1, + P n -2„ for n > 2 (which is not obvious). Another method is to use the “Elvis” way o f looking at Fibonacci numbers. (See any o f the sections dealing with Elvis Num bers.) In how m any ways can Elvis get to the n* step? Well, we know (from The Solution to Elvis the E l f s Eccentric Exercise, p. 59) that he can m anage this in Fn ways. But let’s try an alternate way of looking at it. He can take n single steps and 0 double steps (which he can do in f n\

=

1

way); or he can take n - 2 single steps and

1

double step (which he can do

vOy ' n - 1' in

=n V

-1

ways, because out o f the n - 1 steps he takes, he must choose one to

1 J

be the double step); or he can take n-4 single steps and 2 double steps (which he n—2 can do in

ways, because out o f the n - 2 steps, he m ust choose two o f them to \ ^ j be double steps); and so on. The total num ber o f ways in which he can get to the nth step is thus fn \ ,0 )

+

fn -1 ] ,

1

,

+

f ” “ 2l S ^ j

+ ..

which is ju st P . (Just look at the diagram again if you don’t believe it!) Thus P is the nth Elvis number, which is F !

148

Fibonacci & The Golden Mea it

. Elvis the E lf decided to create a landing o f marble tiles at the foot o f his front hall staircase. He m easured off a rectangular area o f length 13 feet and width 5 feet and determ ined that he would need 65 marble tiles measuring one foot square to cover this landing. He purchased 65 tiles but upon arriving hom e discovered that one tile was dam aged beyond repair and only 64 o f the tiles could be used. U ndaunted by the challenge, he arranged the 64 tiles in an 8 x 8 square array and then cut the large square into four regions, A, B, C and D as shown below left. He then rearranged these four pieces into the 1 3 x 5 rectangle shown below right. H ow did Elvis the E lf create a rectangle o f area 65 square fe e t by rearranging the pieces o f a square o f area 64 square feet? 8

'A _

_____

5

-

-----------------------------------------------------------

Take a piece o f graph paper, and cut out an 8 x 8 square. Then cut the square into four pieces as shown above. Rearrange these four pieces to form a 5 x 13 rectangle. W hat do you discover?

Notice that the numbers {5, 8 , 13} are consecutive Fibonacci numbers. Try to create a paradox like the one above by replacing them with another triplet o f larger consecutive Fibonacci numbers, such as { 8, 13, 2 1 }. You m ight notice a similarity to fact # 2 a).

I learned of some of the more unusual results in fact # 2 from Geoig Gunther of Memorial University of Newfoundland. The bee problem is from H. R. Jacobs, Mathematics, A Human Endeavor, p. 98-99. Fact # 6 was related to me by Greg Kuperbeig of the University of Chicago. Fact # 8 is taken from H. S. M. Coxeter, “The Golden Section, Phyllotaxis, and Wythoffs Game”, Scripta Mathematica. (Vol. 19, No. 2-3, (June-Sept. 1953).), 135-43.

149

Fibonacci & The Golden Meam The Golden Mean \+ S

The golden mean is the num ber T = —- — (= 1.618033989), often represented by the greek letter T (spelled “tau”, and pronounced so it rhym es with “Yow!”). The ancient Greeks felt that the 1 XT rectangle was the m ost aesthetically pleasing, and much classical architecture is based on that proportion. (Any rectangle sim ilar to this one is called a “golden rectangle” .)

T h e P a rth e n o n o f A n c ie n t Greece, com pleted in 438 B.C., was constructed so that its front e le v a tio n fo rm s a g o ld e n rectangle. But nature also uses th e g o ld e n m ean in its o w n architecture . 1 4— The aesthetically pleasing l x r rectangle THE BETTMANN ARCHIVE

T , sometimes called the golden section or golden ratio, is the first letter o f the ancient greek word TO|iT| m eaning “the section” , t has already appeared in our discussions about the Fibonacci sequence, w here we saw that:

'In Coxeter's “The Golden Section, Phyllotaxis, and Wythoffs Game”, further unusual occurrences of the golden mean are discussed, including a surprising appearance of Fibonacci numbers in pineapples. (For bibliographic information, see the Annotated References.)

Fibonacci & The Golden Mea y> Some Interesting Properties o f the Golden Mean The form ula for Fibonacci numbers (p. 60) is just one o f a series of surprising situations in which ^m ysteriously appears. Here are some o f them, many of which you can try to prove yourself. 1 . a) r is an irrational number; in other words, there are no positive integers m and n such that T = ni/n. But T can be approximated by rational numbers. For exam ple, 3/2 = 1.5 is pretty close; 55/34 (= 1.61764...) is closer still. In fact, the “best” rational approxim ations to the golden mean are given by ratios o f consecutive Fibonacci numbers!

b) A related fact is that the fraction

Fn+l p? converges to r a s n gets large.

c) In fact, if you take any two positive real numbers a and b, and define G,n + 1 the sequence Gn by Gj= a, G 2= b, and G n= G n_1+Gn_2, then the fraction

q

gets closer and closer to r as n gets large.

2. a )

t

2-

t

-1 = 0-

b) Using this equation, you can show that if you take a golden rectangle ABCD (where AB is the short side), and subtract a square, ABEF, then the rem aining rectangle FECD is also golden. --------------------------------

B

1

----------------------------------------------

E

r

-------- T - 1 --------

A

F

BC

DC

AB

CE

D

This fact is related to fact # 4 a) in Funny Fibonacci Facts (p. 146).

151

Fibonacci & The Golden Mean Some Interesting Properties o f the Golden Mean (cont'd) 3. Here are a couple o f calculator experiments (that can also be done on a computer). a) Choose any positive number. W rite it down on a piece o f paper, and enter it on your calculator. Then press the following keys: iw i Q

Q

B

Record the new number. R epeat by pressing the same sequence o f keys, and then once again write down the num ber you get. Keep doing this. W hat do you notice? b) R epeat the process in part a), but press this sequence o f keys : 1

BO 1 1 1 l -------- b - + — 2 3 4J ,6 u

1

7

1

1 1— 8

9

|+

1

,11

12

13

T—1

b)

sin 18° =

c)

cos 36° = — 2

5. a) A regular decagon (ten-sided polygon) o f side 1 can be inscribed in a circle o f radius T.

b) The icosahedron is a platonic solid with twenty faces. It will be fam iliar to some readers as a tw e n ty -sid e d die. I f A and B are tw o “neig h b o rin g ” vertices, and A and C are two vertices “once rem oved” , then AC AB

= T.

'This sequence is not universal to all calculators.

152

I

Some Interesting Properties o f the Golden Mean (cont'd) . By pure coincidence (or some evil conspiracy), 1 mile is approximately equal to T kilometres. This means that you can use Fibonacci numbers to quickly translate between miles and kilometres. For example, 13 km is about 8 miles; 89 km /h is approxim ately 55 mph.

6

7. (M ore experimentation.) a) The greatest-integer function [ ] is defined as follows: [x] is the integral part of x. For example, [9.7] = 9, and [ K ] = 3. Make two lists o f integers: [T],

[ 2 t ] , [ 3 t ],

...

[ 1 * ] , [ 2 t ?] , [ 3 t ?], . . .

W hat do you notice? b) (You will only need about four decimal places of accuracy for this one.) M ake another list o f real numbers: T, Y, T3, t , ... W hat do you notice? c) Do the same thing as in part b) with the sequence: T/V5,

TVV5, T W 5 ,

T * / f5 ,

...

A strange fact about these strange properties: a surprising number of them have in some way to do with the num ber 5. 5 appears in the definition o f T. It also appears in properties # 4 (notice that 18° = 9075), #5, and #7 c). M any o f these properties are discussed elsewhere (see Funny Fibonacci Facts, p. 146, and The P yth a g o re a n Pe n t a g r a m, The G o ld en Me a n , a n d S tra n g e Trigonometry, p. 154). If you want to learn more about the golden mean, you should visit these pages. But if you haven't already done so, you should first check out the very first section, Elvis Num bers (p. 56).

153

Fibonacci & The Golden Mean The Pythagorean Pentagram, The Golden Mean, & Strange Trigonometry

The golden mean r

(r =

1 +^

« 1.618) is one o f those m athem atical constants

(such as 7t or e) that pops up in the most unusual places, as we have seen in the previous section. For example, if you use your calculator to com pute 2cos 36° or 2sin 18° + 1, you’ll get t. If you com pute the reciprocal o f 2sin 18°, you’ll get T again. L et’s see why this is true. Consider the regular pentagon with all o f the diagonals drawn in, and labelled as follows:

D

C

This figure, called a pentagram , was the symbol o f the Pythagorean brotherhood. The Pythagoreans were a secret society dedicated to the pursuit of mathem atics and philosophy, established by Pythagoras o f Samos (c. 580-500 B.C.) on the southeastern coast o f what is now Italy.

154

Fibonacci & The Golden Mean

With a little thought, you can fill in all the angles. (Hint: start by showing that the angle betw een consecutive sides o f a pentagon is 108°.)

We choose our unit o f length so that the pentagon has side o f length 1.Therefore, AB = 1. Let t denote the length of AC. AI

AB

By checking angles, you’ll see that AAIB is sim ilar to AABC, so Since AB=1 and AC= t , AI = - . t AIBC is isosceles, so IC = BC. BC is 1, so IC = 1. t = AC = AI + IC = - + 1, so t 2 - t - 1 = 0 . t The roots o f this quadratic equation in t are — -— and —-— . As t i s a length, it is positive, so we can throw out the second solution. Therefore, t = 1 + — , the golden m ean !

155

Fibonacci & The Golden Meat Equipped with this inform ation, we have two basic triangles; AACD and AABC.

A

A

A

You should be able to see lots of triangles sim ilar to both o f these in the pentagram. For example, AEAB and AABC are equiangular and therefore similar. From here, we can w ork out lots o f trigonometric values. For exam ple, if we drop an altitude from A to EB in AEAB, as shown below,

B

we see that

cos 36° =

EK

T

EA

2

You can derive a lot m ore yourself. The next page provides some ideas to get you started.

156

Fibonacci & The Golden Mea Food fo r Thought

O.

Apply the cosine law to one of the basic triangles to compute cos 108°.

sin 12° Apply the sine law to com pute sjn36° , and (using sin 72° = 2 sin 36°cos 36°)

com pute cos 36° in another way.

_

o

19. Prove that sin 18° = —— . Compute cos 72° and sin 54°

_

© . For the original pentagram, show that

AC n D

Com pute the ratio

AB

Al

= —— = — = T ■ /A l u

Area ABCDE Area FGHIJ '

0 . A regular decagon (10-gon) of side 1 is inscribed in a circle. W hat is the radius o f the circle? (Hint: See Some Interesting Properties o f the Golden Mean, p. 151.) © . Using the relationships in exercise 3, and the fact th a t th e ra tio s o f consecutive Fibonacci num bers (especially large ones) are approximately T, we can fill in lengths of the pentagram that are approximately integers, and Fibonacci numbers to boot. Can you determ ine the missing length? (Warning: these are only approximations!)

a/r oady w ay to coayietto otAe/' orAA t/'iyoaom et/'ic oaAted, sin 15° =

to

ad

V3-1 2V2

’

//i. S /J .

157

Ge o m e tr y

rey / s / ted

That vast book which stands forever open before our eyes; I mean the universe, cannot be read until we have learned the language. It is written in mathematical language, and its characteristics are triangles, circles and other geometric figures, without which it is humanly impossible to comprehend a single word; without these one is wandering about a dark labyrinth. Galileo Galilei

Geometry Revisited A “Do-It-Yourself” Proof o f Heron's Formula H eron’s form ula provides a quick way o f com puting the area of a triangle given the lengths o f the sides. Let the sides o f the triangle be denoted by a, b, and c, and let s denote the semi-perimeter, w here s is defined by: a+ b+ c

S

Heron's Form ula

=

A

---------------

2

If the area o f the triangle is K, then B K = yj s(s —a )(s—b )(s—c )

H eron (c. 75 A.D.) was a m athem atician from A lexandria (in m odern-day Egypt). Although the form ula bearing H eron’s nam e was known to Archimedes several centuries earlier, the proof by H eron is the earliest know n to us. In many parts of N orth America, this form ula is still taught in high school, but very few people know why it is true. H ere is a do-it-yourself proof; the key steps are indicated, but you’ll have to fill in the blanks yourself. Step 1 We’ll need a little notation first. Let ABC be the triangle in question, with sides a, b, c. The values s-a, s-b, and s-c will com e up repeatedly, so w e ’ll define them as new variables. L e t* = s-a, y = s-b, z = s-c. Show that a = y + z, b = x + z, c = x + y, and 5 = x + y + z. Let D, E, and F be the points o f tangency of the inscribed circle with sides BC, CA, and A B respectively. Show that A E = A F = x, B F = BD = y, and CD = CE = z.

160

Geometry Revisited Step 2 Next, w e’ll get two form ulas for the area K. Let r be the radius o f the inscribed circle. Show that K = rs as follows. Let I be the incenter. Work out the areas of IAB, IBC, and ICA in terms o f r, a, b, c, and add the results together to get K.

Let ra be the radius of the escribed circle opposite A. (The escribed circle away from A is defined to be the circle tangent to all three sides of ABC, lying outside the triangle, away from A, as shown below.) Let Ia be its center. Show that K = r x by working out the areas of I A B , I BC, and I C A and showing that if K = Area(ABC) then: b K = A rea(/ BA) + A rea(/ CA) - A rea(/ BC)

c Sim ilar arguments show that K = rb y - rc z w here r.b ande r are the radii o f the escribed circles opposite B and C respectively.

Geometry Revisited 0 Step 3 Let G be the point of tangency o f the escribed circle opposite A to BC, as in the figure below. Show that BG = z. (This m ight be tricky, depending on how you go about it.)

Show that triangle BG Ia is sim ilar to triangle IDB. Conclude that ^ 2 and therefore that rra = yyz. Sim ilar arguments show that rrb = zx and rr. = xy.

= BG ^ a

Step 4 Using Step 2, show that

K 6 = (rs) (rs) (rs) (ra x) (rj ) (r z)

By using the identities from Step 3 (rra = yz, rrb = zx, r r= xy) and a little algebra, show that: K 6 = (sxyz) 3 From there, recalling that x = s - a, y = s - b, z - s - c , prove H eron’s formula: K Congratulations — you’ve done it!

s(s-a )(s-b )(s-c )

Geometry Revisited A Short Route to the Cosine Law If you know (and understand) the dot product, you can prove the Cosine Law really quickly. (This is a handy way o f rem em bering the Cosine Law!)

Theorem (Cosine Law) In any AABC:

a 2 = b 2 + c 2 - 2 be cosA b 2 = a 2 + c2 - 2 ac cosB c 2 = a 2 + b 2 - 2 ab cosC

where a, b, and c are the lengths of the sides opposite angles A, B, and C respectively. Proof. We prove the last equation; the proofs o f the others are the same. _2 —> —» c =c•c — (a — b ) • (a - b) —> —> —> —> —> —> = a • a + b • b —2 a • b = a » a + b » b - 2 a b cos C = a 2 + b 2 —2

cos C

If you have only seen the properties o f the dot product proved using the Cosine Law, this isn ’t technically a proof, o f course. But it shows how integrally the two concepts are related.

163

Geometry Revisited "

C

j

I

Solution to the Falling Ladder Problem (p. 85) Surprisingly enough, the flashlight describes a quarter circle. H ere’s why. Picture the ladder at some point in mid-fall.

Label the points in the diagram W, X, Y, Z as shown, w here W is w here the wall meets the ground, X is w here the ladder meets the wall, Y is w here the ladder meets the ground, and Z is the location o f the flashlight. X Z = YZ, since the flashlight is is at the midpoint of the ladder. Let this distance be 1. Using simple Euclidean arguments, it’s not difficult to prove that WZ = 1 as well. So, as the ladder falls to the ground, the flashlight is always a distance 1 from point W. This seems very much like the definition o f a circle! M ore generally, if the burglar weren't exactly halfw ay up the ladder, the flashlight would travel through a quarter ellipse. This problem is an exam ple of a “locus problem ” . The locus of a point is the path o f a point as something happens. In this problem, w e’re w ondering about the locus o f the m idpoint of the ladder as the ladder falls to the ground. You can read about other (seemingly unrelated) locus problem s in the next section. I am thankful to Ed Barbeau of the University of Toronto for introducing me to this classic.

164

Geometry Revisited Locus Hokus Pokus On

!Z7fie.

of

to a ui .

JlaAAex O^xoHrHsm f fi.

85

j , w s H iisfC ij in t r o d u c e d tils, c o n c s fit

c d is rs . a-is. tw o t o a u i- r s t a t s d j i r o U rtam w itfi i.uzjizii.incj a n tw s z t to

ts.±t t jo u i m s t tts .

© In a circle C o f radius r (r > 10), a chord o f length 10 travels around the perim eter o f the circle. The midpoint o f the chord traces out another circle D. W hat is the area lying between C and D?

© . Fix a circle C in the plane. Take another circle B with half the diameter o f C, and “roll it around the inside" o f circle C. Fix a point £P on B. How does £P m ove as B rolls around C?

!Z7’( is

1u.1j 11ii.Lncj a n iw s r i to t f is is jir o H t s n ii ----- w it h o u t jix o o f i

BC' = BE + EC' = BE + EC = s BE' + E'C > BE + EC The left side is the distance that Farm er Brown would have to travel if he aimed for E'. The right side is the distance he would have to travel if he aimed for E. This proves that the route by way o f E is at least as good as any other route. Q.E.D.

Geometry Revisited You m ight have noticed in the previous problem that ABAE is sim ilar to ACDE, so ZAEB = ZDEC. Thus if there were a ball at B and a wall at AD, and if you pushed the ball to bounce off the wall at E, then it would follow the path from E to C. This reflection principle works in general. That is, when a ball bounces off a wall, the angle o f reflection (ZDEC) is equal to the angle o f incidence (ZAEB). Furtherm ore, the path BEC is the shortest o f all paths from B to C by way o f the wall.

D

Som ething similar happens even when the wall is curved, which leads to some unusual results. For example, im agine that you had a pool table in the shape o f an ellipse. The equation of an ellipse can be given by:

where a and b are non-negative real numbers and a > b.

\-----------------------2a ------------------ 1

There are two points in the ellipse called fo c i (the plural o f fo c u s ) situated symmetrically about the center, along the longer axis. If the ball is at one focus and there is a hole at the other, then no m atter which direction you hit the ball, it will bounce off a wall and go in the hole. You can ’t lose! (W hat if the ball isn ’t at the focus? In w hat direction should you hit it so that it banks off a wall and goes in the hole? W hat happens if the table is shaped like a parabola?)

S ound w aves in an e llip tic a lly shaped room d isp lay the sam e reflectio n characteristics as billiard balls on an elliptical pool table. Any sound emitted at one focus of the elliptical room can be heard with rem arkable clarity at the other focus. The whispering gallery in the Tower o f London has two foci a considerable distance apart and yet a whisper at either focus can easily be heard at the other focus. John Quincy Adams (1767-1848), the sixth president of the United States, understood this principle and placed his desk at one focus o f the elliptical Old House Chamber in the Capitol building in Washington D. C. He placed the meeting table at the other focus. In this way he was able to eavesdrop on the conversations of his colleagues and m onitor their loyalties. All this talk o f reflection may also remind you of how light behaves. As we will soon see, this analogy can be pursued much farther. Imagine now that Farm er Brown is standing in a field o f grass, and he sees Einstein standing in a field of arom atic m anure in an adjacent field. Once again, Farm er Brown wants to get to Einstein as quickly as possible. Naturally, he can m ove much faster over grass than he can through manure.

field o f manure 1

field o f grass

Farm er Brown

He is best off making a beeline to some point on the boundary between the grass and the manure. Because he m oves faster over grass, he will choose to spend a greater proportion o f time in the grass than he would have had he headed straight for Einstein.

169

Geometry Revisited l

«

3

If y ou’ve seen S n ell’s Law in optics, you will recognize Farm er Brown's situation. If light moves at different speeds in two different media, it will “bend” (or refract) at the boundary according to a simple rule. In fact, in this situation there is a similar rule (“Sm ell’s Law ?”) relating the angles a and /3 to Farm er B row n’s speeds over grass and manure. S nell's L aw W hen a beam o f light passes from m edium 1 at an angle a to a perpendicular to the interface, it enters medium 2 at an angle /j, where a and ft are related by the equation, sin a

Vj

sin P

v2

and vy and v2 denote the velocities o f light in m edium 1 and medium 2 respectively.

W hat if, instead, there were no distinct boundary between the grass and the manure? W hat if the manure ju st got deeper and thicker? In this case, we can well im agine that F anner B row n’s optimal path would be curved, as in the figure.

Geometry Revisited How does all this discussion tie into our original question, "What is the shortest distance between two points?" W hat happened to our com mon sense proposition that “the shortest path between two points is a straight line” ? M ust we discard it? It may surprise you to learn that the answ er is no— we ju st change the definition of straig h t lin e to m ean q u ickest path. (The technical name for this is geodesic.) Then we can transfer our notions of straight lines from ordinary flat two-dim ensional space to odder spaces such as the surface of a sphere (where geodesics are great circles) and Farmer Brow n’s field o f manure. T he theory o f G eneral R elativity developed by A lbert Einstein (not the cow!) implies that the space we live in isn’t flat. Light is attracted by gravity and follows a path that curves toward gravitational sources. Before Einstein, this fact seemed to contradict the belief that light takes the shortest path and still travels in straight lines. Einstein’s theory asserts that light indeed travels in geodesics, but th a t sp ace b eco m es cu rv ed (m uch less “fla t” ) n ear gravitational sources. That is, the shape of the path traveled by a ray o f light is determined by the geometry o f the space in which it is traveling.

This field of mathematics is very active today. It falls into many different categories — differential geom etry and mathem atical physics among others — and research on these issues will continue to shape the way we view our universe.

171

O.

©

T he U.S. C hess F ed eratio n is p lan n in g a to u rn am en t fo r all the Grandmasters in the country. It wants to locate the tournam ent so the total distance traveled by all the players is a minimum . M ore than half o f the G randmasters live in New York City. Can you prove that the best site to hold the tournam ent is in New York regardless o f the locations o f the other players? (Assume that New York City is a point, that the U.S. is flat, and that the players always take direct routes and never get caught in New York traffic.)

a) Choose some general point P on side AB of an equilateral triangle ABC. How would you construct the shortest path P —> Q —» R —> P so that Q lies on side BC and R lies on side AC? b) (This is quite tricky!) Given an acute triangle ABC, we construct a triangle PQR with P on AB, Q on BC, and R on CA so that APQR has minimal perimeter. Show that P is the foot o f the altitude from C to AB, Q is the foot o f the altitude from A to BC, and R is the foot of the altitude from B to CA.

172

Geometry Revisited More Locus Hokus Pokus In Locus H okus Pokus (p. 165), two problem s were posed. The answers might surprise you.

O.

In a circle C of radius r (r > 10), a chord of length 10 travels around the perim eter of the circle. The m idpoint of the chord traces out another circle D. W hat is the area lying between C and D? \ r 2 - 25

2

Area o f C is nr Area of D is

n | V r2 -2 5

Area of C minus D is n r 1 - n y \ l r ^ - 2 5 J The answ er is 2571. Surprisingly, the answer is independent o f r\ Here is a fam ous problem that is similar. A length o f string is wrapped tightly about the earth’s equator. Then the string is cut, a length o f 1 m eter is added, and the string is retied. It is then suspended a fixed distance above the earth’s surface. W hat is that distance? (Assume the earth is a perfect sphere.)

© . Fix a circle C in the plane. Take another circle B with half the diam eter o f C, and roll it around C. Fix a point on B. How does < P ' move as B rolls around C? The point LP travels along a fixed diam eter of C. All o f the circu lar m otion is som ehow transform ed into linear motion! (This unusual result m ight not com e as a surprise to those readers who have played with Spirograph™ or worked with gears.)

173

There is an even m ore surprising generalization to the first problem. Take a large random convex shape C, large enough so that a chord of length 10 can travel around the perimeter. The m idpoint o f the chord will sweep out another odd shape D. (The convex condition, w hich essentially m eans that C doesn’t have any indentations, is there to ensure that D actually sits inside C.) It turns out that the area between C and D is always 2571, regardless of the shape of C! One proof o f this general result uses a clever application o f G reen’s Theorem, often taught in second-year university calculus. But with much less machinery, it is possible to prove special cases of this result. The first problem from Locus Hokus Pokus proves the theorem when C is a circle. If you try to prove it when C is a rectangle, you will discover The Falling Ladder Problem (p. 85) in a new guise. H ow about other shapes? An equilateral triangle?

A Short Question on Symmetry This is an old classic: W hy is it, when you look into a mirror, that left and right are reversed but top and bottom are not? (If you think you know, try explaining it to a friend.)

174

The infinite! No other question has ever moved so profoundly the spirit of man. David Hilbert

The Mathematics o f the Birds and the Bee Two birds are racing towards each other in the heat o f passion. They are initially 10 km from each other, and they are flying at the speed of 0.5 km/min. A voyeuristic bee, which can fly at a speed o f 1 km/min, starts with one o f the birds and flies towards the other. W hen it reaches the second bird, it turns around and flies back towards the first bird again. The bee keeps this up until the two birds meet. How far does the bee travel before this happens? (The answ er is on the bottom o f page 184.)

I'll bee there beefore the bird.

lothing sungs like "'N. double rejection! x

C 176

©

c c s , ;

An Early Encounter with the Infinite The paradoxes o f Zeno o f Elea, a philosopher in ancient Greece, reveal some early encounters with the notion o f infinity and its strange properties. About 450 B.C., he made the following assertion known as Z e n o ’s racecourse paradox. Z en o’s Assertion: A runner can never reach the end o f a racecourse in a finite time.

16

8

4

2

Statem ent

Reason

1. The runner must first pass the point -?> located halfway between herself and the finish line before she can finish the race.

^ is between the runner and the finish line.

2. It will take a finite time to reach the point ± .

It is a finite distance from 1 t0 2 -

3. Once reached, there is another halfway point 1 T he rem aining in terv al is divided in half. which the runner must reach before she can finish. 4. There are an infinite number o f such halfway points which the runner m ust reach and each will take a finite time.

S ta te m e n ts 1, 2, an d 3 repeated an infinite number o f times.

5. The total time for the race is infinite.

T h e sum o f an in fin ite n u m b er o f fin ite tim es is infinite.

But we know from experience that the runner can reach the finish line. W here is the flaw in Z eno’s argument?

177

The Flaw in Zeno's Paradox The following story may help you discover where Zeno’s argument breaks down. A fam ous pirate, L ong John G litter, found a cylindrical bar o f gold one m eter in length. To divide the bar into m ore manageable units he cut it in half (perpendicular to its axis o f symmetry). One piece he placed in his treasure chest. The remaining piece was cut in half and one o f those pieces was then placed in the chest. Again the remaining piece was cut in half and one of the pieces was placed in the chest. This process was repeated n times. WR/TE THE LEA/GTH O f THE IIth P ffC f PLACED /A/ THE CHEST AA/D THE LEA/GTH O f THE REMA/A//A/G PfECE, WR/TE AS A SUM O f fRACT/OA/S THE TOTAL LEA/GTH /A/METERS O f THE f/RST H P/ECES PLACED /A/ THE CHEST,

WR/TE AA/ EXPRESS/OA/ fOR THE TOTAL LEA/GTH O f THE f/RST 11 P/ECES /A/ THE CHEST

S u ppo se th/s pro cess could b e coa/ t/a/ ued a a/ /A/E/Ai/TE a/ u m b er o e r/MEs. W ould th e total lea/ gth o e th e P / eces o e th e gold BAR EVER EXCEED

f

METER?

This exam ple shows that the sum of an infinite num ber o f finite quantities may indeed be finite. This calls into question the reason supporting the fifth statem ent in Zeno’s argument above. Paradoxes such as this one prom pted m athem aticians to put mathematics on firm er foundations by specifying a basic set of “self-evident” truths or axioms from which all m athem atical theorems could be logically and unambiguously derived. This w ould (hopefully) expose all hidden assumptions and remove paradoxes. O ver two millenia later, deeper paradoxes arose which shook the axiom atic foundations of m athem atics m ore profoundly than the challenges made by Zeno. One such paradox appears on page 226.

178

The Harmonic Series An infinite series isa sum o f a sequence with an infinite number of terms. When the sum is finite, the sequence is said to converge. Otherwise the series is said to diverge. For example, the series generated by the lengths o f the pieces thrown into the basket (see the previous page) is given by: 1 1 1 - +- +- + 2 4 8

1 + — + ... 2 ”

It converges (to 1), butthe series 1 + 1 + 1+ 1 + ... clearly does not converge. It is important to recognize that there is a limit implicitly involved here. For example, 1 1 1 1 the value o f T + T + ^ + - + —7 + --is defined to be the limit of the sum o f the 2 4 5 2 first n terms as n approaches infinity. That is, the lim it o f this sequence is the limit o f the sequence of partial sums, 11 1 2 ’

1 1 1 2

4’

2

4

8

The study of convergence is an im portant one, and often much o f an undergraduate course is devoted to understanding how one can tell when a series will converge.

The harm onic series is the series: 1 H

I 2

1

l

l 1 h — + ...H----- 1-... 3 4 n

For many reasons, it com es up repeatedly in many different fields o f higher m athematics. There is a beautiful proof that this famous series diverges. If you have never thought about it before, you m ight want to ponder it yourself (or even sleep on it) before reading this rem arkable argument.

179

The Classical Proof that the Harmonic Series Diverges This proof is very slick. We ignore the 1, and instead sum 1/2 + 1/3 + 1/4 + ... . By the time we get to 1/2, the partial sum is equal to 1/2. By the tim e we get to 1/4, the partial sum is greater than 2/2. By the time we get to 1/8, the partial sum is greater than 3/2.

By the time we get to l/2 n, the partial sum is greater than n/2. This means that if we are given some positive number, we can go far enough to make the partial sum bigger than this number. The proof works by considering intervals between adjacent powers of two.

Sum between 1/2"1 and 1/2" 1/2 1/3 + 1/4 1/5 + 1/6 + 1/7 + 1/8

Lower Bound

Value

= 1/2 > 1/4 + 1/4 > 1/8 + 1/8 + 1/8 + 1/8

= 1/2 = 1/2 = 1/2

•

*

•

•

•

•

•

•

•

1/(2"-'+ 1) + . . . + 1 /(2 "-1 )+ 1 /2 "

> 1/2" + ... + 1/2" + 1/2" = 1/2 J

V Adding all of these together, we get: , 1 /2 + 1 /3 + 1 /4 + . . . + 1 /2 "

n terms_______ (

> 1/2 + 1/2 + ... + 1 /2

That is, the partial sums increase w ithout lim it as n increases.

= n/2

©

©

s

3

§

j

There are many variations on this result. Here are a few. V ariation

1

Instead o f evaluating

lH

2

1----- 1----- h ... 3 4 1

we can try to evaluate l +

1 +

1 +

Each term in the second series is only a tiny tiny bit sm aller than the corresponding term in the harm onic series. B ut remarkably, the second sum converges! It converges to something very large, though. If you want to see how large and know a little about integration, you can w ork it out yourself, using the information that the sum is approxim ately the same as the following integral: oo J 1.00001 1

J X

Variation 2 Instead o f evaluating 1 + ^

^

^ + ... , we can try to sum the reciprocals only

of those num bers with no 9 ’s in their decim al representation. So the sum would begin:

1H

1 1 1 1 1 1 1 h ... H 1 h ... H 1 —+ ... 2 3 8 10 18 20

It doesn’t seem as though w e’ve thrown out many numbers, but strangely, this sum also converges.

181

Variation 3 This time we will ju st sum the reciprocals of the prim es, so our series will begin: 1 2

1 1 1 1 1----- 1----- 1----- 1-------h ... 3

5

7

11

Now w e’ve thrown out a lot o f numbers. There aren’t that many prim es — there are only 25 less than 100, 168 less than 1000, and 78,498 less than 1,000,000 — and they keep on getting rarer. But amazingly, especially in light of the previous variation, this sum diverges! Rarer than primes are the so-called twin primes, which are any pair of prime numbers that differ by 2, such as 5 and 7 or 41 and 43. The series o f the reciprocals o f all twin primes converges. This tells us that twin prim es are very rare indeed.

Variation 4 In Prime Num bers in N um ber Theory (p. 124), we showed using a simple m ethod that there are an infinite num ber o f primes. Leonhard Euler (1707-1783), the great Sw'iss m athematician, had another proof based on the fact that the harm onic series diverges. The proof is subtle, but you should be able to figure it out. We use a m ethod known as the indirect method — we assume the opposite o f what we want to prove, and try to find a contradiction. Im agine that there are only a finite num ber o f prim es,

, P2, P3, • ••, Pr ■

Then

every positive integer n can be expressed (uniquely) as n=P

r

™1 P

P 3™3 •• • P™r w here the m, are integers.

Consider the product P given by: \ 1 1 1 1 1 + — + — + — + ... 1 + — + — + — + ... P1 Pi 2 PV,i33 V p2 P22 P i J

. 1 1 1 1H 1-------- 1-------- h V

Pr

P r2

Since the num ber o f prim es is assum ed to be finite, P is finite . (W hy? )

182

P r3

J

L

o

g

' A

N

Infinity

,

N ow im agine that you expanded out all o f the brackets (in the same way that when you expand (a + b){c + d) you get ac + be + ad + bd — you get every possible product o f something in the first bracket w ith something in the second bracket). By analogy, we would expect P to be some huge sum 1+

1

P\

h

1

1 h

Pi

1

K. . +

Pi

1 h

Pn

+ ...

P\Pl

In this sum should appear every num ber o f the form: n m. n m n m „m Px 'P 2 *P3 3— Pr ' We said earlier that every positive integer is o f the form, p f p f 2 p f so P is in fact the harmonic series. Thus P is infinite.

3

•• •p f

But we said that P is finite, so we have found a contradiction. There must be an error somewhere, so our original assumption (that there are only a finite number of prim es) must be false. Hence there are an infinite num ber o f primes. V ariation 5 We can use calculus to get more evidence that the harmonic series diverges. (As if we w eren’t already convinced!) Define f i x) to be the following infinite series: 2

,

fix) = J

X

H

X

2

X

3

X

4

1---------1---------h . .

3'

4

Notice that/(0) = 0, and th a t/f 1) is the harmonic series. Differentiating with respect to x, we get: f (jc)= { + x + x i + x i + x * + _ This is an infinite geom etric series, so we evaluate this sum to obtain: f (■*) = Thus

' 1 / ( l ) = / ( l ) - / ( 0 ) = j f i x ) dx = \ o

1

0 1 -

dx X

J

= —In (1 —jc)|

'o

= “ C-00) (To be fair, substituting x =1 into -In (1-x) is not quite "cricket"; rather we should look at the limit of -In (1-x) as x approaches 1 from below.)

183

v

H isto ric a l D ig r e s s io n

27k? Human Computer

Stories about John von Neumann's rem arkable capacity for mental com putation abound. On one occasion he was at a cocktail party when a guest challenged him with the problem given on page 176 about the Birds and the Bee. Johnny, as he was fondly called, contem plated the problem for a second or two and then replied, “ten kilom eters” . The surprised interrogator retorted, “Oh, you've heard that trick before?” In typically innocent fashion, Johnny responded, “W hat trick? I merely summed the infinite series.”

Answer to The Mathematics o f the Birds and the Bee (p. 176) One could take von Neumann's approach and calculate how far the bee travels in each direction. This requires that you find the sum o f an infinite geom etric series. However, there is a faster solution. The birds each take ten m inutes to reach the halfw ay point; in this time, the bee w ill travel 10 km because it travels 1 km/min. T hat’s it! (Here is a follow-up question: How many times does the bee turn around during this journey?)

184

THE BETTMANN ARCHIVE

John von Neumann (1903-1957) was one o f the preem inent mathematicians o f the 20th century. A t the age o f thirty, he was appointed (along with A lbert Einstein) as one of the first professors of the Institute for Advanced Study in Princeton. His work in mathematics and physics spanned a wide spectrum o f fields and his contributions in any one o f these would have w on him recognition. Von N eu m an n 's e a rlie s t c o n trib u tio n s lay in h is a x io m a tiz a tio n o f s e t th e o ry ; th a t is, h e reform ulated the axioms o f set theory to deal with some o f the problems such as Russell's Paradox, discu ssed in P aradox (p. 226). F rom this he p ro c e e d e d to a re f o r m u la tio n o f Q u a n tu m M echanics. He also pioneered the developm ent o f G am e Theory, the digital computer, and the study of cellular automata. N ear the end o f his career he worked on the atomic bomb, perform ing detailed calculations (often in his head) related to pressures created during implosions.

Gami Meow Remtreo

“Come, Watson, come! The game is afoot. ” — The Return of Sherlock Holmes Sir Arthur Conan Doyle

Sherlock Holmes' Secret Strategy: The Reykjavik Gambit TJo leazn How to jlajj eSds.xloalL cTfoCm&s.’ (2o n tin sn ta £ Oa(Lus, zecuI SCsm ejtkaxy c=Mij U s eat ^W ation! j . 2. 74)-

If you’ve played around with the game awhile, you’ve probably convinced yourself that EN D RU N can always stay one step ahead o f Holmes. But in fact, Holmes can always win. Holmes' secret trick is to start off the gam e by m aking a beeline to Reykjavik (he can get there in two moves), and then head to Edinburgh. After this, he should have no trouble chasing down ENDRUN. Try this out and check that it works. Loosely speaking, Holmes just has to try to back ENDRUN towards a comer, usually by m oving to the opposite com er o f a quadrilateral from EN D RU N . F or exam ple, if H olm es w ere in B russels and EN DRUN were in Venice, then Holmes should move to M unich. M ira c u lo u s ly , b y g o in g to Reykjavik Holmes has removed the potential difficulty in chasing down E N D R U N ! T he fifteen m oves allow ed H olm es are m ore than enough — usually, six will suffice. This is a great game, because you can play the part of EN D RUN against an unsuspecting friend, and then (after winning repeatedly), offer to switch sides. You then appear to walk about randomly for the first few moves, and ju st happen to w ander through Reykjavik. A round the eighth move, you suddenly seem to rem em ber the object o f the game, and quickly hunt down your bew ildered opponent. H ere’s a rem arkable fact: Reykjavik is essential to w inning the game. If you erase Reykjavik from the map (so neither Holmes nor EN D R U N may go here), then Holmes can never catch EN D RUN — even if there is no time limit and EN D R U N tries to help! (Rem em ber — Holmes has to m ove to the position o f ENDRUN; EN D RUN is not perm itted to m ove to Holmes' position and turn him self in.) If you can figure out why Reykjavik is essential, then y o u ’ll have an excellent understanding of how this gam e works. Perhaps m ore im portant, you’ll have a vital insight into other im portant themes addressed in this book, including parity, coloring, and graph theory. (You can take that last sentence as a hint!) W hen you think you've figured it out, turn to Why H o lm es’ Strategy Works (p. 189).

186

A Three- Way Duel Why Multi-Player Games are Much Harder to Analyze This section has three different elements mixed together: game theory, a good probability problem, and a paradox. The Set-Up Three players, A, B, and C, are involved in a gunfight. They take turns shooting one bullet at a time until only one player remains. C is a perfect shooter; he kills his target every time. B has an accuracy o f two kills in three attempts. Finally, A only hits his target one out of every three times. To be fair, it is decided that A should begin the gunfight, to be followed by B and then C (if they are alive), and then back to A, and so on. (Warning: Do not try this at home! These characters are trained mathematicians and use only theoretical bullets!) The Strategy W hen it is C ’s turn to shoot (assuming he is still alive), it is clearly in his best interest to shoot his strongest opponent. So if B is still alive, C will shoot him. B ’s strategy is also clear: C is the target he m ust shoot first. The Paradox This pattern would seem to suggest that A should shoot at C if possible, and otherwise aim at B. But it turns out that A’s best strategy is to deliberately miss his first shot. If A aims at C with his first shot, and B and C follow their best strategies, A will win (approxim ately) 31% o f the time, B will win 54% o f the time, and C will win 15% of the time. But if A deliberately misses on his first shot, he will win 40% o f the time, B will win 38% o f the time, and C will win 22% of the time. So A’s odds o f winning becom e much better by following this counter-intuitive strategy — and he becomes the one m ost likely to win!

187

Game Theory Revisited

The Probability Problem Try to work out all o f the probabilities m entioned above. You can check your answers against the given percentages. The G am e-Theoretic M oral Two-player games are much easier to analyze than m ulti-player games for a number of reasons. As in this example, a w eak third party may initially choose not to take advantage o f its position, preferring instead to let the stronger powers w ear each other down. W eaker parties can hold the balance o f pow er between tw o evenly matched strong opponents. Also, it can sometimes be dangerous to grow too strong too quickly, as the other players may team up if they feel threatened. O f course, all of these reasons often make m ulti-player games more fun to play! Napoleon III hopes to pick up the spoils o f the A ustro-Prussian War.

no c jf o

S' 5o X ) CD

3. CD

O

c«

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K 3 S-

BRITANNIA: “W ELL! I’V E D O N E M Y BEST. IF THEY W ILL SM ASH EAC H OTHER, TH EY M UST.” NAPOLEON: [Aside]

188

“A N D SO M EO NE M AY PICK U P TH E PIECES!”

Game Theory Revisited

Why Holmes' Strategy Works Why Iceland is Essential Imagine that continental drift causes Reykjavik to disappear from the playing board. Im agine also that each o f the cities were colored black or white as in the map shown here. W henever a player moves, he will move from a city of one color to a city of another color. (If you play chess, this pattern will remind you o f the knight, w ho always moves from a square o f one color to a square of the other color.) When the game begins, Holmes is in Stockholm, which is a black city. EN D RUN is in Paris, which is also a black city. L et’s change the gam e a little. Instead o f Stockholm, H olm es has to start in some arbitrary black city. Instead o f Paris, EN D RU N has to start in som e other black city. W e’ll see that even in t. Petersburg th is m ore g en era l iStockholi situation, H olm es O can never win. M oscow

•■Edinburgh

W arsai ?don A A m e rs te rd a m •gssels

Ovienria Munich •Budapest [enice

Lisbon

189

Holmes will start by jum ping to a white city. (ENDRUN isn ’t there; h e’s in a black city.) ENDRUN will jum p from a black city to a white city. Holmes will then jum p from his white city to a black city, but he can’t catch EN D RUN in this move, because EN DRUN is in a white city. EN D RUN will then jum p from the w hite city to a black city. So Holmes can’t catch EN D RUN in either o f the first two moves o f the game. But the two players are once again in a starting position, as they are both in black cities! So, by the same argument, Holmes can’t catch EN D R U N in the next two moves. This can go on forever. So Holmes can’t win, no m atter how well he plays and how badly EN D RUN plays! (The argument above is short, but tricky. You m ight w ant to think about it and play around on the map a little before m oving on. You should think about how the argum ent fails if the players are allowed to go to Reykjavik.) This gam e is related to many other issues such as parity discussed earlier (see p. 92). The gam e was played on a graph (where graph is m eant in a technical sense), so graph theory was involved. We solved the problem by introducing a coloring, so this is also an exam ple o f a coloring problem (discussed in the chapter on Chessboard Coloring). In particular, the graph we looked at had a very special property. The cities (called vertices in the lingo of graph theory) could be colored black or white in such a way that two connected vertices are always different colours. Such a graph is called a bipartite graph. These graphs appear rem arkably often in a w ide variety o f situations.

Food fo r Thought O If you look at the altered map (with Reykjavik removed), you’ll notice that if you take a round trip starting anywhere and returning to the same place, the round trip must have an even num ber o f stops (counting the beginning and end point only once). For example, one possible round trip is Madrid-M arseilles-Zurich-Venice-Vienna-M unich-Brussels-ParisM adrid. Why is this true? © . (This one is tougher!) If you have a graph such that any round trip has an even num ber o f stops, must the graph be bipartite?

How to Win at Nim

The winning strategy for Nim is simple, but hard to figure out. Some calculations are required, but with a little practice you should be able to do them in your head, at least for small games. If it is your turn to play and you are confronted with this configuration:

then you are guaranteed to lose. Less obviously, if you are confronted with the configuration: ■

and your opponent plays perfectly, then you are also guaranteed to lose. (Can you figure out why this is so?) W e’ll call configurations like this losing configurations. T here is a quick m ethod to check w hether a given configuration is a losing configuration. Armed with this, we can play perfectly: if we are faced with a non­ losing co n fig u ra tio n , w e ’ll see how to leave our o p p o n en t w ith a losing configuration. (If we are confronted with a losing configuration, then we are doomed, unless our opponent slips up and makes a non-optimal move.)

• First o f all, write the number appearing in each row in base 2 in a colum n (as though you were about to add them). • Next, draw a horizontal line. U nder each column, put a 1 if there are an odd number o f 1’s in the column, and a 0 if there are an even number in the column. This is called the Nim-sum of the numbers. (Com puter aficionados may recognize this as XOR.) It isn’t the same as adding, because we never have any “carrying” .

191

Game Theory Revisited As an example, we evaluate the configuration shown below.

1 1

100 1 10 If there are only zeroes in the bottom row, then we have a losing configuration. (So we see that the configuration shown above isn ’t a losing configuration.) This rule helps us decide how to move. If confronted with the configuration shown above, we want to leave our hapless opponent w ith a zero Nim-sum, so w e’ll need to change the number o f 1’s in the first and second colum n by altering only one number. By looking closely at the table, you will soon see that the only w ay to do this by reducing one o f the numbers is to change 100 to 010 (i.e. 4 to 2), rem oving tw o toothpicks from the b o tto m row. Then o u r op p o n en t is le ft w ith the configuration shown, which has N im -sum zero (check it!), and therefore is a losing configuration. ■ 1 Try finishing this gam e by playing according to the strategy w hen it is y o u r tu rn and ran d o m ly w hen it is y o u r opponent’s. Then you can tackle the practice questions in the F ood f o r Thought section below . A fter you have m astered them, you w ill be ready to start playing (and beating) other people!

Practice Questions

O. a)

Food fo r Thought

W hat moves would you make in the following situations? W hy? | b) | | c) || d) '

1 1 10

00

71! Lin,

.-far

V I Z 7 £ li

JL

Game Theory Revisited ttrm .

[

Food fo r Thought If there are two equal rows left in the game, who will win, the first or the second player? W hat course will the game take? Further Research (You certainly don’t need to solve these to play Nim perfectly!) © Consider the three-pile configuration shown in the figure (0 < a < 6). For which a is this a losing position? (M ake a table.) For which a is

odd? Is this a co­

incidence? Replace 6 with other numbers and find out! (For related discussion, see The Triangles o f Pascal, Chu Shih-Chieh, and Sierpinski, p. 98.)

6-a

i i . . .

i

0 . How would one prove that the strategy described in this section will always work? © . If you are ready to tackle the winning strategy to another game, try these. a) Change the rules of N im slightly, so each player is allowed to remove as many matches as he wants from up to two piles (although at least one match must be removed). W hat is the winning strategy? (As in Nim, think in base 2.) b) There are n toothpicks in a row. The first player can take up to n- 1 toothpicks. The second player can take at most the num ber o f toothpicks that the first has just taken. The first can then take at most the number o f toothpicks that the second has just taken. This continues until all the toothpicks are gone. The player who takes the last toothpick wins. For which n does the first player have a winning strategy? The answ er m ay surprise you. (As with Nim, write the num ber in base 2. Try to find out if the first player can force a win with n = 1, 2, 3, 4, 5. If the first player can force a win, what is her winning first move? Can you think of a winning first move that will work for all odd n? If so, how about numbers that are even but not divisible by 4?)

C oa/ c b p ts / a/ Calcuu / s

I f I have seen a little further than others, it is because I have stood on the shoulders o f giants. — Sir Isaac Newton

t

H isto ric a l D ig r e s s io n

\ \

The Man Who Solved the System o f the World Prior to the 17th century, people did not understand how the earth m oved in relation to the sun. Although Copernicus had asserted in 1543 that the earth was not stationary, his theory was not generally accepted until scientists like G alileo and Tycho Brahe gathered more evidence. Furtherm ore, the shape o f the earth's orbit and the orbits o f the other planets around the sun „ , , was still unknow n until Johannes K epler enunciated Johannes Kepler 1571 - 1630 , , „ , his three laws o f planetary m otion in publications dated 1609 and 1619. Kepler's laws were em pirical and it rem ained for scientists to explain in terms o f some theory why these laws were true.

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In 1663-64, Cam bridge University was closed on account o f the bubonic plague. Isaac Newton, who was a tw enty-one year old undergraduate, was sent hom e to continue his studies. During that highly productive period, Newton formulated his law o f universal gravitation which states that the force of attraction between any two masses is proportional to the product o f their masses and inversely proportional to the square o f the distance between them. At this time, Newton also developed a new branch o f mathIsaac Newton 1642 ' 1727 em atics called calculus. Using the methods o f calculus, he was able to deduce the three laws of Kepler from his own law of universal gravitation. He waited until 1687 to publish this work in a treatise titled Philosopliiae Naturalis Principia M athematica. The scientist Pierre-Sim on Laplace exclaimed, N ewton was surely the man o f genius p a r excellence, but we m ust agree that he was also the luckiest: one fin d s only once the system o f the world to be solved! H ow d/d HewroA/ OiDVce H c p u r 's SecouD La w ? Newton reasoned as follows: 1. If there were no force, a planet would move from P to P 2 in a small time, At. In the next time interval At it would m ove from P, to P 3 so that P ,P 2= P 2Py The areas swept out in successive time intervals would be equal; that is, A rea AS P ,P , = Area AS P 2Py (Why?) 2. Suppose the sun at S exerted a force on the planet in its direction. Suppose also that the “average” direction o f this force over the infinitesimal time interval 2At w ere in the direction of P,S. Then the net effect of the sun's force on the planet would be to m ove it parallel to P 2S away from its destination P} to a point PA. The area swept out by the planet in moving from P 2 to P 4 would be Area A SP 2 P f. Furtherm ore A reaA S P ,P 4 = Area A SP,Pr (Why?) As observed earlier, A rea A SP,P3 = Area A S P ^ . Q. E. D.

i

C o n ce p ts

in Calculus

A Question o f Continuity

Can you prove that right now, as you read this, there are two diametrically opposed points on the earth’s equator that are exactly the same tem perature? The answ er highlights an im portant idea. Think about it, and then turn to page 200 for the answer.

198

I

An Integration by Parts Paradox Recall the “Integration by Parts” formula:

J fix) J

M ore simply:

d r = f ( x ) g ( x ) - J g( x) dx

J

dx dx

\fdg =fg -\g d f

L et’s substitute / ( x) = —,

g (x ) = x in this formula.

x

[!< & = [ - ] * - [ * < / Jx \xj J 1 xv (

= ‘ -

-1

J

\ dx

V

= 1 + J—d r Thus f —d r = 1 + f —d r , so 0 = 1 . W hat has gone wrong? 1 x Jx

199

I

Concepts in Calculus Answer to A Question o f Continuity (p. 198)

It is true that, right now, there are two diam etrically-opposed points on the earth’s equator that are exactly the same temperature. H ere’s why. Im agine that a bizarre new therm om eter has been invented. It doesn’t display the temperature; it displays the difference between the tem perature where you are and the tem perature at the point on the globe diam etrically opposite to you. For exam ple, if you are standing in Quito, Ecuador and the tem perature is exactly 20°C, and the tem perature in the mountains o f Sumatra in Indonesia diam etrically opposite Quito is 25°C, then your strange therm om eter w ill read -5. Now, walk to some point on the equator, and look at your strange thermometer. If it reads 0, then you’ve found the point you’re looking for. If it doesn’t, then it is either positive or negative. Say it is positive. (The negative case will turn out to be exactly the same.) Then, quickly run (or swim) around the equator to the opposite side o f the globe. (You’ll have to do this infinitely quickly to make sure that the tem perature doesn’t change.) Now your strange therm om eter should have the negative of the reading it had before. So if you watched it carefully while you were globe-trotting, you would have seen it change continuously from being positive to being negative. At that point, it would be zero, so that would be the spot you were looking for. The simple idea we used was that a function (in this case, the reading on your odd therm om eter) that varies continuously from one value to another must take on every value in between. This principle has the slightly pompous name o f the Interm ediate Value Theorem. Although it sounds quite trivial, it is in fact very useful. Unfortunately, in order to prove it rigorously, you will need to define exactly what you mean when you say that a function is continuous. (Waving your hands in the air doesn’t count.) If you have seen one o f the definitions o f continuity that m athematicians use, you can try to use it to prove the Interm ediate Value Theorem. But watch out: it is far harder than it looks!

Food fo r Thought Can you prove that, right now, there are two points on the earth’s equator separated by 120° that are exactly the same tem perature?

200

I

Concepts in Calculus Dual Numbers izit le c t io n iv i H u ts to m s z l ’z m sn ta z ij c a h u f u i.

(J o m jt ls x n u in b s u a z s u t s d a t

latocjtj, and toms fanufiazity iv itfi tfis m iv o u t d (js . a b iy fzslfi. Part 1 : Complex Numbers Com plex numbers are numbers o f the form a + bi, where a and b are real. They are added together in the natural way: (a + bi) + (c + di) = (a + c) + (b + d)i imaginary axis Ui + c, b + d ) (c, d) = c + cU— - ^ ^ 7 = (a + c) +(b + d)i (a, b) = a + bi real axis The sum of two complex numbers can be located by drawing the diagonal of the parallelogram defined by the two numbers. They are multiplied using the fact that i2 = -1: (ia + bi)(c + di) = a(c + di) + bi(c + di) = ac + (ad + bc)i + bd i2 = ■(ac - bd) + (ad + bc)i Division can be worked out with a little cleverness. The top and bottom of a fraction are m ultiplied by the conjugate o f the denom inator, and everything magically works out: a + bi

(a + b i )( c -d i )

c + di

(c + d i )( c -d i ) _ ( a c + bd) + ( b c - ad)i

, j2 c 2 +d ac + bd

be-ad .

c2+ d 2

c2+ d 2

(How are the line segments corresponding to two complex numbers related in m agnitude and direction to the line segments corresponding to their product and quotient?)

I

D ual Numbers Part 2: From Complex to D u a l Numbers The dual numbers are sim ilar to com plex numbers, they are numbers o f the form a + bE, where a and b are real. (The symbol £ is an epsilon. Epsilon is the fifth letter in the Greek alphabet.) Dual numbers are added in the natural way, so (3+4e) + (5-2e) = (8+2e), and they are multiplied using the fact that e2 = 0. (But e isn’t zero! Yes, this is strange!) For exam ple,

(3 + 4e)(5 - 2 e) = 3(5 - 2 e) + 4e(5 - 2e) = 15 - 6e + 20e - 8 e 2 = 15 + 14e

You should try an easy exam ple to make sure you understand how m ultiplication w orks. For exam ple, you can check that (2 + e)(6 - e) = 12 + 4 e, or that ( 3).

Let A = jc", B = y 1, and C = zn, so A + B = C. Then the product of the primes dividing A B C is at most xyz < Ci/n, which is much less than C. If we could show that this is impossible (when phrased more rigorously) then we could prove Fermat’s Last Theorem. The A BC Conjecture is basically ju st that: if A, B, and C are relatively prime, and A + B = C, then the product o f primes dividing A B C should be much greater than C [~£- Put very loosely, if a little carelessly, in exam ples when C is very large, the product must be at least o f the same order of m agnitude as C. The ABC Conjecture, altered slightly, is know n to be true if A, B, and C are polynomials instead of integers. The proof o f this result is very sim ilar to the solution to the following problem from the Sixteenth William Lowell Putnam M athem atical Competition in 1956: The polynom ials P{z) and Q(z) with com plex coefficients have the same set of numbers for their zeros but possibly different multiplicities. T he sam e is tru e o f th e p o ly n o m ials P(z) + 1 and Q(z) + 1. Prove that P{z) = Q(z)-

N oam has also suggested another unsolved problem that you m ight w ish to investigate. This problem is a m ore general form o f the following question first form ulated by Sam Loyd (1841-1911), a renow ned A m erican puzzlist. What is the greatest num ber o f line segments which can be drawn through 1 6 points so that each line segm ent contains 4 points? In the diagram below, there are sixteen dots and ten line segments, with four dots on each line segment. We will w rite this as “ 16= 10 ® 4” . (Here, the symbols = and ® do not have their usual m eanings.) Number of line segments Number of dots

*^ „ 16^10® 4 ^ can be arranged

Number of dots per segment

in such a way that

However, in the following diagram, there are sixteen dots and fifteen lines, with four dots on each line. (We write this as “ 16 = 15 ® 4” .) This is the best-know n result for 16 dots and 4 dots per line segment.

This case can be generalized to get n 2 = (3n+3) ® n, where n is even (and at least 4). Can you see how ? Try it with n = 6. Is this the best that can be done? The following diagram shows the best result for n = 3:

9 = 10® 3

In 1981, N oam discovered the following exam ple that is the best known for n = 5:

25 = 18® 5

It isn’t even know n if n 2 = (3n + 3) ® n is possible for n odd and greater than 5. Can you find an exam ple that shows that 72 = 24 ® 7? How close can you get? Also, can you prove that one can never get n 2 = m ® n for m > 3n+3?

245

S tte r to o r t I hope you have enjoyed these forays into mathematics as m uch as I have, and that you have been left w ith m uch to think about. The m oral to this story is that beautiful, surprising results are always worthw hile for their own sake, but they can often lead into deeper and even m ore exciting waters. W hile a student at Northern Secondary School in Toronto, Kevin Purbhoo cam e up with the following beautiful problem:

On a rem ote Norwegian mountain top, there is a huge checkerboard, 1000 squares wide and 1000 squares long, surrounded by steep cliffs to the north, south, east, and west. Each square is m arked with an arrow pointing in one o f the eight compass directions, so (with the possible exception of some squares on the edges) each square has an arrow pointing to one o f its eight nearest neighbors. The arrows on squares sharing an edge differ by at most 45°. A lem tm ng is placed random ly on one of the squares, and it jum ps from square to square following the arrows. Prove that the poor creature will eventually plunge from a cliff to its death.

—>

T -4 T

7> —>

—>

A lthough Kevin did not know it at the time, this problem anticipates several subtle and im portant results in topology, a type o f geom etric thinking useful in most areas of m athematics. This is only one exam ple w here a sim ple idea or problem opens up new m athem atical horizons. If you are a high school student and want to learn m ore interesting m athem at­ ics, there are many avenues you can pursue. First o f all, becom e com fortable writing proofs. Putting ideas to paper in a form convincing to others sharpens the mind; like any new skill, it requires practice. Euclidean geom etry pro­ vides a perfect opportunity for this, as the tools required are m inim al and the rewards great. (Sadly, geom etry seems to be quietly disappearing from the high school curriculum.)

The beauty o f mathematics has inspired many authors to write expository books. As a result, there are numerous excellent mathematics books aimed at many levels. The A nnotated References is a good starting point, but there are m any other w orthwhile books not listed. Browse around, and follow up on authors and subjects you really enjoy. If your school library has only a few titles, convince your librarian to buy more. A lthough private reading and independent study are useful, it is im portant to share your ideas with teachers and other students. A M ath Club is a one way to form such a com m unity; arguing about ideas with friends is another. There are also m any sum m er and w eekend program s aim ed at students; your local university is a good source. Finally, you m ight enjoy taking part in mathematical competitions. In Canada, the U niversity o f W aterloo runs an excellent series o f competitions at many grade levels. (For m ore inform ation, contact the Canadian M athem atics C om petition, Faculty o f M athem atics, U niversity o f Waterloo, Waterloo, O ntario, Canada N 2L 3G1, tel. (519) 885-1211 ext. 2248, fax (519) 7466592.) Success here opens the door to many other competitions, including the Canadian M athem atical O lym piad and the International M athem atical Olympiad. In both Canada and the U.S., the M athem atical Association of America runs a cycle o f competitions beginning with the American High School M athem atical Exam ination (contact: Dr. W alter E. M ientka, Executive Director, A m erican M athem atics Competition, 1740 Vine Street, University o f Nebraska, Lincoln NE, 68588-0658 USA). Teams from parts o f the U.S. and Canada com pete every M ay at the fun and inform al American Regions M athematics League, (contact: Mr. M ark E. Saul, ARM L President, Bronxville School, B ronxville NY, 10708 USA, (914) 337-5600, e-mail 73047.3156 @ compuserve.com). Finally, the International M athem atics Talent Search (contact: Dr. George Berzsenyi, D epartm ent o f M athem atics / Box 121, RoseHulm an Institute o f Technology, Terre H aute IN 47803-3999 USA) is a non­ c o m p e titiv e y e a r-lo n g p ro b le m -so lv in g p ro g ram . W h ile m o st oth er com petitions have stringent tim e limits, the IMTS allows more reflection on the part o f the participants, fostering not only ingenuity, fast thinking, and creativity, but also com m itm ent, reliability, and perseverance. Once again, I hope you have been challenged and excited by this book. If you have any com m ents, please write. Best o f luck in future explorations! Ravi Vakil Departm ent o f M athematics Harvard University One Oxford St. Cam bridge, M A USA 02138-2901

247

Annotated References Berlekamp, Elwyn R., Conway, John H., and Guy, R ichard K. Winning Ways fo r your M athem atical Plays. New York: A cadem ic Press, 1982. These two volumes explain mathematical game theory with many examples of fun games to play and win. W ell-written but intellectually dem anding. Boyer, Carl B. and Merzbach, Uta C. A H istory o f M athem atics. 2nd ed. New York: John W iley & Sons, 1991. Ball, W.W. Rouse and Coxeter, H.S.M. M athem atical Recreations and Essays. 13th ed. New York: D over Publications Inc., 1987. Beckm ann, Petr. 1971.

A H istory o f Pi.

Boulder, Colorado: The Golem Press,

Castellanos, Dario. “The U biquitous 7t” . M athem atics M agazine. Vol. 61, No. 22, Apr. 1988, 6 7 - 9 8 . Coxeter, H.S.M . “The G olden Section, Phyllotaxis, and W ythoff’s G am e” , Scripta M athem atica. Vol. 19, No. 2-3, June-Sept. 1953, 135 - 43. *Coxeter, H.S.M . and Greitzer, Sam uel L. Geometry Revisited. W ashington D.C.: M athem atical A ssociation o f A m erica (M AA), 1967. This is Volume 19 in the M AA’s excellent N ew M athem atical Library. F o r m ore in fo rm atio n ab o u t any o f th e M A A ’s p u b lic a tio n s, call 1- 800-331-1622 or (202) 387-5200. Crux M athematicorum. Crux is a high-level problem -solving journal published by the Canadian M athem atical Society; it appears ten times per year, and includes a regular “O lym piad C om er” . For more inform ation, contact: CM S/SM C, 577 King Edward, Suite 109, POB/CB 450, Station A, Ottawa, Ontario, Canada K IN 6N5, tel. (613) 564-2223, fax (613) 565-1539. Curl, Robert F. and Smalley, R ichard E. “Fullerenes” , Scientific American. Vol. 264, No. 4, Oct. 1991, 54-63. Gardner, M artin. “Six sensational discoveries that som ehow or another have escaped public attention". Scientific Am erican Vol. 232, No. 4, Apr. 1975, 126-133. Keep an eye out for anything G ardner has written!

248

Annotated References

(cont'd)

Gleick, James. Genius. New York: Vintage Books, 1992. *Greitzer, Samuel L., Arbelos. W ashington D.C.: M AA, 1982-1988. The Arbelos was a first-rate journal o f problem -solving mathematics for high school students that was essentially the work o f one man. Hardy, G. H. A M athem atician’s Apology. Cambridge: Cambridge University Press, 1993. Hoffman, Paul. A rchim edes’Revenge: The Joys and Perils o f M athematics. New York: W.W. Norton & Co., 1988. Hofstadter, Douglas R. Godel, Escher, Bach: A n Eternal Golden Braid. New York: Random House, 1979. Brilliant; w inner o f the Pulitzer Prize. Holton, Derek. L e t’s Solve Some M ath Problems. Waterloo, Ontario: Waterloo M athem atics Foundation, 1993. For more inform ation, contact the Canadian M athem atics Competition, Faculty o f M athematics, University of Waterloo, Waterloo, Ontario, Canada N 2L 3G 1, tel. (519) 885-1211 ext. 2248, fax (519) 746-6592. *Honsberger, Ross. Ingenuity in Mathematics. Washington D.C.: MAA, 1970. Ingenuity is another book from the M AA’s N ew M athem atical Library. The M AA also has other titles by Honsberger, all o f them excellent. Jacobs, H arold R. M athematics, A Human Endeavor. San Francisco: W.H. Freeman, 1982. A teacher’s guide is also available. *Larson, Loren C. Problem-Solving Through Problems. New York: SpringerVerlag, 1983. T his is an ex cellen t book in g eneral. A lthough it is in tended for undergraduates, many parts are accessible (and interesting) to gifted high school students. Chapters include: Induction and Pigeonhole, Arithmetic, Algebra, Complex Numbers, Inequalities, and Geometry. Call SpringerVerlag at 1-800-777-4643 for more information.

249

Annotated References (cont’d ) M athem atical Digest. M athem atical D igest is an excellent jo u rn al for high school students appearing four tim es per year, edited by Dr. John Webb at the U niversity o f C ape Tow n in S o u th A fric a . F o r m o re in fo rm a tio n , w rite to M athem atical Digest, D epartm ent of M athem atics, U niversity o f Cape Town, 7700 Rondebosch, South Africa. *M athem atical M ayhem. M ayhem is a non-profit problem -solving journal appearing five tim es per year, written by and for high school and university students. There are inform ative and irreverent articles as well as problem s at different levels. For more inform ation, contact the Departm ent o f M athem atics, University of Toronto, 100 St. George St., Toronto, Ontario, M5S 1A1, or e-m ail mayhem @ math.toronto.edu. N elsen, R oger B. Proofs W ithout Words: Exercises in Visual Thinking. Washington D.C.: M AA, 1993. *Polya, G. How to Solve It. Princeton: Princeton U niversity Press, 1973. Rucker, Rudy. Infinity and the M ind: The Science and Philosophy o f the Infinite. Boston: Birkhauser, 1982. Swetz, Frank J. ed. From Five Fingers to Infinity: A Journey through the History o f Mathematics. Chicago: Open Court Publishing Company, 1994.

*1 strongly believe that every high school should have these books on hand.

250

Index ABC Conjecture, 243 Ailles, D., 87, 88 Archimedes, 78-84, 160 Argand plane, 220 arithmetic series, 35, 135, 137 Arrow, K., 68-72 base five, 131 base ten, 18, 19, 20, 23-29, 131,181 base three, 102 base twelve, 29 base two, 19, 21, 30, 41, 66-67, 100, 131, 191-193 Bernouilli family, 141 Bernouilli numbers, 139-141 binary numbers, see base two B inet’s formula, 60 bipartite graph, 190 Brown, F„ 167-171 Cantor, G„ 226, 232, 236 cardinality, 114, 226, 229-240 Carroll, L., 73 Cartesian plane, 86, 216 Cauchy, A.-L., 219, 236 Chain Rule, 204 com plex numbers, 40, 118-119, 201-202, 207-223, 237-240 com posite numbers, 25 constructible numbers, 118-119, 215-217 correspondence, 231 -240 Cosine Law, 163 cryptography, 20, 67, 126 de M oivre’s Theorem, 203, 208-214 Descartes, R., 216 Dodgson, C. L., 73 dot product, 163 Einstein, A., 132, 171, 184 Einstein the Cow, 167-171 ellipse, 78, 164, 168-169, 196 escribed circle, 161-162 Euclid, 120, 124, 241 Euler, L„ 182, 242

Index

(cont'd)

Factor Theorem, 26 Fermat, P. de, 123 Ferm at’s Last Theorem, 120, 122, 123, 141, 243 Ferm at’s Little Theorem, 142 Feynman, Richard, 209 Fibonacci, 57 four dim ensions, 46-50, 81 fractals, 40, 89, 101-102, 228 Galois, E„ 216, 218, 219 Galois Theory, 216, 218, 219 Gardner, M., 132, 241 Gauss, K.F., 17, 22, 78, 137, 236 G elfond-Schneider Theorem, 122 General Relativity, 171 geodesic, 171 geom etric series, 106, 131, 147, 153 G reen’s Theorem, 174 Hardy, G.H., 117, 120, 121, 124, 133 Heron, 160 Hilbert, D., 122, 175 L’H opital’s Rule, 141 L’Hopital, G.F.A. de, 141 indirect proof, 108-109, 120-121, 124-125, 182-183, 235, 236 induction, mathem atical, 47-48, 59, 61 Interm ediate Value Theorem, 200 Kepler, J., 196-197 Kepler's Laws, 196-197 Leonardo of Pisa (Fibonacci), 57 limits, 58, 79, 146, 151, 152, 175-184 locus, 164, 165, 173-174 Loyd, S., 244 M achin, J., 211 m odular arithmetic, 22-29, 102, 213

252

Index (cont'd) Napoleon, 219, 223 Nash, J., 64 Newton, I., 8, 78, 195, 197 numbers, algebraic, 118-119, 236, 237-240 com plex, 40, 118-119, 201, 202, 207-223, 233, 237-240 constructible, 118-119, 215-217 irrational, 118-119, 120-122,218, 237 rational, 118-119, 120-122, 147, 214, 218, 229, 236, 237, 240 real, 118-119, 120, 232-233, 240 transcendental, 118-119, 122, 237-240 parabola, 78, 79, 168 Pascal, B., 98 Pascal’s Triangle, 45, 62, 93, 95, 98-102, 142, 148, 193 platonic solids, 50, 53, 152 pow er series, 140, 210 Prisoner’s Dilemma, 64 probability, 70, 73, 92-97, 103-104, 123, 143, 187-188 proofs by diagram, 86, 87, 88, 135 Pythagoras, 154 Pythagoreans, 121, 128-129, 154 radians, 208, 210, 211 Ram anujan, S., 133 reflection, 168-169, 174 Rem ainder Theorem, 26 Riemann Hypothesis, 122, 243 Russell, B., 228, 233 R ussell’s Paradox, 184, 228, 233 Snell’s Law, 170 symmetry, 36, 3 9 ,4 4 , 51, 52, 65, 92-97, 100, 103-104, 112, 174 Tic-Tac-Toe, 39, 65 trigonometry, 86, 87, 88, 130, 146, 152-153, 154-157, 208, 210-214, 217, 222 twin primes, 182 Von N eum ann, J., 184 Wiles, A., 123 Zeno, 177-178

253

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254

MATHEMATICS

Some of the favorite problem solving strategies of one of the world’s most accomplished iad Winners! About the Author Ravi Vakil’s pre-eminence in the world of mathematical competitions is evident in the long list of his many triumphs. To highlight just a few: 1988 USA Mathematical Olympiad

• placed first in North America International Mathematical Olympiad

• won two gold medals and one silver medal by Jewel Randolph

Putnam Mathematical Competition

Photo

• placed among the top five competitors in North America in each of his four undergraduate years. In the pages of this book, Ravi shares some of his most powerful problem solving ideas in mathematics. He also profiles seven other Olympiad winners including Noam Elides, the youngest professor to receive tenure at Harvard.

This book is a must for teachers seeking to challenge their best students, and for students

58879 ISBN

L - f l B S B B T - DM- b

9781895997040

preparing for mathematics competitions. 9 781895 997040