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'
ND MISCELLANY OF PUZZLES
STEPHEN BARR
ND MISCELLANY OF PUZZLES MATHEMATICAL AND OTHERWISE
THE MACMILLAN COMPANY
COLLIERMACMILLAN LIMITED, LONDON
Copyright (
1969 by STEPHEN BARR
All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any information storage and retrieval system, without permission in writing from the Publisher. Library of Congress Catalog Card Number: 7075899 First Printing
The Macmillan Company CollierMacmillan Canada Ltd., Toronto, Ontario Printed in the United States of America
Add another hue unto the rainbow ... Shakespeare We subtract faith and fallacy from fact ... Samuel Hoflenstein Be fruitful and multiply ... Genesis Round numbers are always false ... Samuel Johnson Vicious circle ... George Du Maurier Straight down the crooked lane and all around the square ... Thomas Hood Every cubic inch of space is a miracle ... Walt Whitman The god delights in an odd number ... Virgil Euclid alone has looked on beauty bare ... Edna St. Vincent Millay Measure your mind's height by the shade it casts . .. Browning
CONTENTS PL IZ Z L ES 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Rollers Owls' Eggs Conical Helix The Owl Island Flag PseudoMoebius Strip The Butler and the Crumbs The Three Clocks Slit Strips The Pot on the Crosspiece For Scrabble Players Area of Roof PaperFolding The Two Pyramids The Man Who Gave UD Srjoking Tetrahedron Angles Hypocycloids Squares on a Circle Three Coins Two Coins The Coin Collector's Nightmare The Hiq4 Set Cryptarithmetic
19
20 21 22 23 24 25
26 28 29 31
31 32 33 37
38 39 40 40
41 42 43 9
23 24 25 26 27 28
10
Origametry The Hauberk More Origametry Unique Parts of Letters For >"' A
6>g IOB
t
21f
Xd
X' B'
I'
K'
A W= 1
½21 1
(A)
A6 /s 1
\w
/1
\,1 ½21 F' FIGURE 1
Proof:
Figure 2 is drawn to the proporti ons given, but we ignore the fact that the lengthi is 2, and call it 1. 0 is the center of the unfoldled rectangle, and the folded shape is symmetirical about AO, perpendicular to FF', the fold. D, the center of the required circle, which is tang ent to FF' at 0, and to the ends of the paper A'E (taking Continued on next page 175
Continued
LAl
z
E
FIGURE 2
only one because the shape is symmetrical) at C', and passing through A, the intersection of the long edges. DC' is perpendicular to A'E; draw OB' perpendicular to A'E, and thus parallel to DC' and FE. 0 is the center of FF', therefore B' is the center of A'E. Draw a line from A to B, the intersection of OB' with the required circle; then zABO is a right angle because it is in a semicircle; therefore AB is parallel to A'B'. Draw BD; since LACD is a right angle, AC = CB; therefore AC = 1/4. Draw DC perpendicular to OB; then OG = BG = CD. B'O =1/2; let x = radius AD, and y =1/22BG. BG=CD=xy; .. y=1/22x+2y .. y=2x1/2 or 2xy=1/2 (Equation 1) In the right triangle ACD, the sides are x, x y, and 1/4; 176
ERRATA Second Miscellany of Puzzles (#48887) by Stephen Barr Page 23, Puzzle 5, line 1: For I in. x 10 in. read 1 in. x 11 in. Page 23, Puzzle 5, line 2: For 1 in. x 11 in. read 1 in. x 10 in. Page 44, Puzzle 23, line 6: For (horizontal in read (vertical in Page 44, Puzzle 23, line 9: For vertical creases read horizontal creases Page 64, Puzzle 44: Martin Gardner should be credited for permission to use his versionmuch improvedof my puzzle The Poisoned Glass, which appeared in Scientific American. Page 79, Puzzle 57, line 5: For What are its most likely proportions and shape? read The reason is that there is an empty pot of inconsiderable thickness inside, attached to the bottom so it doesn't float. What are its most likely proportions and shape? Page 81, Puzzle 60, line 17: For turns? read turns? Answer on next page, Figure 1. Page 86, Puzzle 61, line 12: For to Figs. 5 and 6.) read to Figs. 5 and 6 on page 205.) Page 96, Figure 1: Add another dot to this figure. Page 103, Puzzle 44, line 4: oology should be read as if printed at an angle. Page 111, Puzzle 4, line 11: For amount n/2, read amount 2n, Page 115, Puzzle 8, Figure 2: Note that both sets of lines cross incorrectly at the top of this figure. Page 137, Puzzle 19, line 3: For an arc read a chord Page 137, Puzzle 19, line 6: For and CD read and CB Page 141, Puzzle 20, line 11: For F read = CF _V~ V2 Page 203. Puzzle 61, line 3: For (Fig. 1). read (Fig. 1, page 82). Page 218, Puzzle 11: For Tokyo. read Shanghai.
Continued
A.
L~Lj z .. x
2
=x
2
.. 2 xy 
2xy+y
y
2
=
2
+ 1/16
1/16 (Equation 2)
Multiply both sides of Equation 1 by y; 2 xy  y2= yl/2; then with Equation 2 we get yl/2 = 1/16 y = ½8l This was the value Suzi calculated forAA', with 1 = 2; AA'= 1/16. The circle covered has radius = 1/4 + y/2, in this case 1/32 more than the circle covered by the unfolded paper.
Answer 2 1
and
12
The lower limit, I1, is when the required circle has a diameter of 1, the width of the paper, giving no advantage. The upper limit, 12 which gives the greatest advantage, is when the circle touches the long edges of the paper. Any further increase of 12 means that the circle calculated by the above method will be too big to be covered. Continued on next page 177
45
Continued
m
In the first case (Fig. 1), x = w/2, or 1/2, so that in the AADC, DA = 2/4 and AC = 1/4; .. DC =V /4 = 1/2y; y = 1/2 VJ/4= 4
2
3
/8l;
j
1
4 4 2V Q.E.D.
The upper limit (Fig. half only) gives x = 3/4 so in AADC, DA = 3/4 and V8 3 1 .34
62
2; showing the left because CE = 3/4; AC = 1/4; .. DC = Vg 1. =4 81' *
aF
Q.E.D.
Asymmetrical folding, where FE' does not go through 0, within rather narrow limits can give slightly higher values for x, but is too unwieldy to discuss here, and comes under the heading of problems rather than puzzles.
FIGURE 1
178
FIGURE 2
Continued
An m
Answer 3 It is diophantine; that is, the ratio of its sides is rational, and can be reduced to finite whole numbers. In the case given, with w = 1, 1= 2, the AADC has sides measuring 1/4, x y, and x. y = 1/16; x = 1/4  y/2 = 1/2  1/32; so the three sides are 1/4, 17/32  1/16, and 17/32. Reduced to the same denominator, 8/32, 15/32, and 17/32, and 82  152 = 172. If 1 = w, the sides are 3, 4, and 5. Since we can always express a rational w/L by reducing w to 1, the general formula for the sides of the triangle in question is: where I is a rational number, the sides are as 1, 12  1/4, and 12 + 1/4. This is another version of the more usual expression: If n is an odd number, the sides are n, 1/2(n2

1),
1/2(n 2 + 1),
If n is a whole number, the sides are n, (n/2)
2

1,
(n/2)
2
+
1.
In this way any proportion of rectangle that can be established by folding (i.e., that is obtainable by the feasible subdivisions; all the powers of 2, and a wide variety of others) allows one to mark off y on the edge, and thus fold the optimum shape within the limits given above for symmetrical folding.
179
B z
46
PURE ORIGAMI SOLUTION In Fig. 1 the creases are numbered in order, circled. 1, 2, and 3 are successive halving of the width, 1, giving AC = 1/8. Crease 4 runs from C to the corner, B (must be very exact). 5 brings the edge BD up to coincide with the top edge AB, enabling one to mark the length y with crease 6 on the bottom edge next to D. With some finaglement (Fig. 2), the y at D is transferred to the top edge next to A, with crease 7. The final fold, 8, dotted, is made by bringing D up and over next to A, and adjusting until creases 6 and 7 just meetand flattening carefully.
FIGURE 1
180
Continued
4
r
Proof:
In Fig. 1, AB is to AC(1/8) as 1 is to y, being in similar triangles. Thus 1/8 = 1; Y= 1/8 =
which is the required value, as proved
in Puzzle 45. Suzi explained to Lillian that it also works for rectangles shorter than the low limiteven a squarebut that no advantage is gained in size of circle covered. "I shouldn't have brought up all this math," she said. "I know you're not keen on it." "Mathematics, no," replied Lillian, "mathematicians, yes. Anyway I like the shape it makes: I expect it to fold its wings like the scarab, and as busily buzz away."
FIGURE 2
181
Bzl47 THE FLAT PAN He fills the pan on the table more than halffull, and then carefully tilts up one end, pouring out the water, until the level reaches E, the bottom edge of the raised end (Fig. 1). This leaves exactly 1/2 pt. in the pan, since the empty part is the same shape and size as the filled part. Being an ordinary kitchen type, the table has a straight edge; the man slides the pan over the edge so that the opposite corners, C and C', coincide with the edge (Fig. 2), and starts tilting again, with the bowl held to catch the water. He tilts until the surface of the water coincides with the corners C and C'. The bowl now contains 1/3 pt. Proof: Proportions of the pan are ignored. Figure 3 shows the pan at the end of pouring. The remaining water is in the form of a pyramid, the volume of which equals the area of the base times onethird its height. Here the area of base 2 .. volume = 2 x
= w3 h6
Since the volume of the pan (1 pt.) = Iwh, the remaining water = 1/6 pt. Thus he poured into the bowl 1/2  1/6 pt.; or 1/3 pt. TransparentView
FIGURE 2
FIGURE 3
483g z
THE SIAMESE MOEBIUS STRIP One piece, two sides, three edges (1, 2, and 3), no twists, topologically speaking, for the twists in the upper part are canceled out by those in the lower. There are also two holes. The answer was foreseeable (see Fig. 2), since EB, the lower edge of L, joins B'A, the upper edge of R and L, which joins A'E', the lower edge of R; then EBB'AA'E' together make the upper half of edge one (Fig. 1). The lower half of edge one is formed analogously by FCC'DD'F'. The inner edge made by cut X (dotted), and the outer edge made by cut Y, together make edge three (Fig. 1), and analogously the outer edge X and the inner edge Y make edge two (Fig. 1). This tells us there are three edges. By following the connections made by the joints, we see that the strip remains in one piece. Finally, in each case a twisted element is joined to one twisted oppositely; therefore all twists cancel; therefore there are (topologically) no twists, and from that we see there must be two sides, as there were before joining and cutting. 7e 3
Disc
s
FIGURE 1
D
y FIGURE 2
'
183
V)
THE
NINE DIGITS The first line is produced by multiplying the nine digits in their natural order, by three, and writing down the final digits of the products; the second is produced by multiplying the digits by seven, and writing the terminal digits of the products. When one and nine are used as the multipliers we again get all nine digits, but in natural and reversed order, respectively. The other digits give less than all nine digits. This leads to zeros and repeats, which throws everything off (see figure). Multipliers 1 2 3 4 5 6 7 8 9
Terminal Digits* 123456789 246802468 369258147 482604826 505050505 628406284 741852963 864208642 987654321
*With operation X the first and last of these produce themselves. It will be noticed that the pattern formed by this figure is like a word square, in that it reads the same down as across.
184
Vc LI)
z MINIMUM
AREA
Figure 1 shows one way to start; the sides add up to 4s, and the area is reduced to 5/9s 2 . But if we try to improve it by reducing the length of the sides, and increasing their number (Fig. 2), getting an area of 41/81s2 , and continue to increase the number of sides, we can see that the limit is area =
FIGURE 1
2
s21
FIGU RE 2
One solutionthere are others, but more complicatedis Fig. 3. Figure 4 shows the method carried further, and Fig. 5 shows the limit, with zero area. The latter would, of course, have to be reduced the appropriate amount, like the others, to give the right length
of sides, in this case by onehalf, because they are still twice as long as they look, due to the (now) infinitely small and infinitely numerous Continued on next page
185
Continued
W
z steps. Topologically the solutions just described form an open set, for the limit case is excluded since by definition the sides would be touching one another, the distance between them being zero. This does not affect the answer, however.
FIGURE 3
186
FIGURE 4
FIGURE 5
51 COCYCLIC POINTS The four points of intersection of circle E, which needs no proof, and the points A, B, C, and D. Proof: A and C are right angles; draw DB, which we consider to be the diameter of an imaginary circle. Any point, such as A or C, subtended at a right angle by D and B, will lie on the circle.
L52 LUJ z
THE TILTED
CARTON Enough milk to fill the carton to a depth of half its width, or less; a = 45°. This is true for any proportion of carton if h > w. If h = w, the amount of milk has no effect on a. Proof: Figure 1 shows the carton tilted to a = 450. C is the midpoint of the dotted line parallel to the base AB, and at a distance of ½/2w from it; when the carton is tilted, the level of the milk will still pass through C, and at 45° will also touch B. Symmetry shows the carton will balance at this angle (ignoring the weight of the carton). If any more milk is added, NN', the extra milk will be offcenter to the left, so that the new center of gravity will be to the left of the vertical axis CA, and the carton will tip. If on the other hand the amount of milk is less, the center of gravity will still be vertically above A when a = 450. When h = w the carton may be full, partially full or empty, and a will always be 450. milk less than full re
53 1 Cn
z
TOPOLOGY PUZZLE As shown in Fig. 1. The edges can be followed by eye for counting. The rule is that the number of holes is unlimited provided it is odd. Figure 2 shows an uncompleted model with three holes and the possibility of a fourth if we join the loose end to x. If the joint is made without a twist, we shall have four holes and two edges but only one side; with a twist we shall have two sides and four holes, but three edges. This applies to all even numbers of holes.
Shaded part is the second side
Twists may be in either direction FIGURE 1
FIGURE 2
189
Cd5)
z