a gardner's workout (gnv64)

A Gql'"cinel'"is WOl'"kout Trinin9 the Mind nd Entertinin9 the Spirit

A K Peters Natick, Massachusetts

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A K Peters, Ltd. 63 South Avenue Natick, MA 01760 www.akpeters.com

Copyright 2001 by A K Peters, Ltd.

All rights reserved. No part of the material protected by this copyright notice may be reproduced or utilized in any form, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the copyright owner.

Library of Congress Cataloging-in-Publication Data

Gardner, Martin A gardner's workout: training the mind and entertaining the spirit /

Martin Gardner. p. cm.

ISBN 1-56881-120-9 1. Mathematical recreations. 1. Title.

QA95 .G 158 2001 793.7' 4--dc 21

Printed in Canada 05 04 03 02 01 10 9 8 7 6 5 4 3 2 1

2001021427

To all the underpaid teachers of mathematics, everywhere, who love their subject

and are able to communicate that love to their students.

Tqble of Contents

Pef-lce xi

Pqrl: I 1

1 The Op'lque Cube 3

2 The Squ'le Root of 2 = 1.414 213 562 373 095 ... 9

3 Flip, the Psychic Robot 21

4 The Poposition'll c'lieulus with Diede G'lphs 25

5 M'lthem'ltics 'ln Worpl'ly 35

6 Steiner Trees on 'l Checkebo'lr 39

7 Tiling the Bent Tromino 61

8 Covering 'l Cube with Conguent Polygons 73

9 M'lgic Tricks on 'l Computer 77

10 V'li'ltions on the 12345679 Trick 83

11 Moe c'lieul'ltor Whimsies 85

12 !(qsP'lrov's DeFe'lt by Deep Blue 91

13 Computes Ne'l the TheshoW 97

14 Comehng the King 109

15 T oroi'll Cunency 117

VII

VIII A Gardner's Workout

16 Six ChClllenging Dissection TClSks 121

17 Lewis CClrroll's Pillow-Problems 129

18 Lewis CClrroll's Wor LClers 133

19 The Ant on 1 x 1 x 2 139

20 Three-Point Tiling 143

21 Lucky Numbers Cln 2187 149

22 3 x 3 MClgic SCJuClres 157

23 Some New Discoveries Clbout 3 x 3 MClgic SCJuClres 167

24 Primes in Arithmetic Progression 175

25 Prime MClgic SquClres 181

26 The Dominono GClme 187

27 The Growth or RecreCltionClI MClthemCltics 191

28 MClximum Inscribe SquClres, RectClngles, Cln TriClngles 203

29

30

31

32

33

34.

SeriClI Isogons or 90 Degrees

Aroun the SolClr System

Ten AmClzing MClthemClticClI Tricks

Moeling MClthemCltics with PIClying CClrs

The Asymmetric Propeller Theorem

Chess Queens Cln MClximum VnClttClcke Cells

Pq rl: I I

35 Lion Hunting

36 Two Books on InAnity

215

231

235

241

249

257

269

271

275

Table or Contents IX

37 The Universe cmd the Te'lcup 281

38 A New Result on PerFect M'lgic SCJu'lres 285

39 The Number Devil 291

40 Prob'lbility 1 295

41 Fuzzy New New M'lth 301

Prefqce

For 25 years I had the honor and pleasure of writing the "Mathematical Games" column in Scien tific American. All those columns have now been reprinted, with updating , in fifteen volumes, starting with The Scientific American Book of Mathematical Puzzles and Diversions and ending with Last Recreations.

Since I stopped writing the column I have from time to time contributed articles and book reviews about mathematics to both academic journals and popular magazines. Forty-one of these pieces are gathered here. The most controversial is the final review in which I criticize a current teaching fad known as the "new new Math."

By the time this book is published I would guess and hope that new new math is being abandoned almost as rapidly as the old new math faded. I could be wrong . In any case, it may be decades before our public education is able to attract competent teachers who have learned how to teach math to pre-college students without putting them to sleep . There are, of course, many teachers who deserve nothing but praise. It is to them I have dedicated this book.

XI

Martin Gardner

Hendersonville, NC

C h'1pte r 1 The 0p'1q ue Cu be

The Opqq ue C u be

I want to propose the following unsolved problem. As far as I know, I am the first to ask it.

* * *

What is the minimal area of surfaces inside a transparent cube that will render it opaque?

* * *

By opaque I mean if the surfaces are opaque, no ray of light, entering the cube from any direction, will pass through it.

The answer may or may not be the minimal surface spanning the twelve edges of the cube. This question also is unanswered. See the discussion of it by Courant and Robbins ( [ 1 ] , Ch. 7) .

That the minimal spanning surface may not be the answer to the opaque cube problem is suggested by the fact that the minimal Steiner tree spanning the four corners of a square is not the answer to the opaque square problem. The best known solution for the square (also not proved minimal) is shown in Figure 1 . The square problem is discussed by Ross Honsberger ( [2] , p. 22) .

My best solution for the cube is to join the center to all the corners. These lines outline 12 triangles with a combined area of 3V2.

I believe the opaque cube problem to be extremely difficult. It is keeping me awake at night !l

IThe note appeared in the Dutch periodical Cubism for Fun (No. 23 , March 1990 , p. 15). I followed this with a second note, "The Opaque Cube Again," in the same periodical (No. 25, December 1990, Part 1, p. 14) .

3

1. The 0P.'l'iue (ur,e

Opqque Cubes by Kenneth Btqkke

Figure 2. Twelve triangles from the edges to the centre. Area = 3v'2 ';':; 4.2426.

Figure 4. Generalization of opaque square solution to three dimensions.

Topologically, this is the opaque square solution stretched vertically with the

top and bottom faces of the cube added.

Area,;,:; 4.2343.

Figure 3. Soap film formed by dipping a cubical frame in a soap solution. Area

';':; 4.2398.

Figure 5. My best solution. This is l ike Figure 4, except i t has three-fold

symmetry in place of two-fold. Area ';':; 4.2324.

5

1 . The 0p'lque Cube

, "

" ... ,'

'\\

C hqpte r 2 The Sq uq re Root of 2

1.414 213 562 373 095

Roses are red, Violets are blue. One point 4 14 . . .

Is the square root of two.

I confess that I wrote the above jingle only to have some light verse top this article . The dots at the end of the third line indicate that the decimal fraction is endless and nonrepeating . In other words, v'2 is irrational. Although its decimal digits, like those of other famous irrationals such as 7r and e, look like a sequence of random digits, they are far from random because if you know what the number is you can always calculate the next digit after any break in the sequence. Such irrationals also should not be called "patternless" because they have a pattern provided by any formula that calculates them. The square root of two, for example, is the limit of the following continued (endless) fraction:

v'2 = __ 1---:1;---2+ 1

2+ --2 + 1

From this continued fraction one can derive rational fractions (fractions with integers above and below the line) that give V'2 to any desired accuracy. The sequence 111 , 3/2 , 7/5 , 1 71 12 , 4 1129, 99170, 239/1 69, 577/408 , 1 303/985 . . . is sometimes called "Eudoxus' ladder" after an ancient Greek astronomer and geometrician. The fractions are alternately higher and lower than their limit, which is V'2. Each fraction is closer to v'2 than its predecessor. The best approximation with numerator and denominator not exceeding three digits is 577/408. It gives

This article first appeared in Math Horizons (April 1997) .

9

2. The Square Root at 2 = 1 .414 213 562 373 095 ... 13

b

al2

a

Figure 2. Another infinite descent proof.

shaded rectangle. Its sides are b and (a - b). Because a and b are integers, (a - b) must also be an integer. Therefore the shaded rectangle must have sides that are multiples of k.

We can repeat the procedure by cutting two squares from the shaded rectangle to create a still smaller rectangle, similar to the shaded one, with sides that also must be multiples of k. As in the previous proof, if this process is continued we soon produce a rectangle with sides smaller than k. We have reached a contradiction. The procedure can be carried to infinity, but one cannot have an infinite sequence of integers that keep getting smaller and smaller. Therefore a and b are incommensurable, and .J2 is irrational. Infinite descent proofs can be given algebraic forms, many of which generalize to proving that any nth root not an nth power is irrational .

b }a-b a-b b a b b

b b .

a

a+b

Figure 3. Steinhaus' infinite descent proof.

Ch'1pte r 3 Fl ip, the Psych ic Robot

Cm You Outwit q M ihd l ess Automqtoh? Here's a chance to take on an opponent who has no control over the moves he makes . Still , winning may be tougher than you think.

The game is matching pennies. You flip a coin and then Flip, the robot, will do the same. If Flip's flip matches yours (heads after you have thrown heads, or tails after your tails), Flip wins. If not (tails after your heads, or heads after your tails),