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«Contents Ü Foreword Elwyn Berlekamp and Tom Rodgers ½ I Personal Magic ¿ Martin Gardner: A “Documentary” Dana Richards ½¿ Ambrose, Gardner, ...»

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H18. The planet Alpha Lyra IV is an oblate spheroid. Its axis of rotation coincides with the spheroid’s small axis, just as one observes for Sol III. It’s internal structure, however, is unique, being made of a number of coaxial right circular cylindrical layers, each of homogeneous composition. The common axis of these layers is the planet’s rotational axis. The outermost of these layers is nearly pure krypton, and the next inner layer is an anhydrous fromage. Cosmic, Inc., is contemplating mining the outer layer, and the company’s financial planners have found that the venture will be sound if there are more than one million cubic spandrals of krypton in the outer layer. (A spandral is the Lyran unit of length.) Unfortunately, the little that is known about the krypton layer was received via sub-etherial communication from a Venerian pioneer immediately prior to its demise at the hands (some would say wattles) of a frumious snatcherband. The pioneer reported with typical Venerian obscurantism that the ratio of the volume of the smallest sphere that could contain the planet to the volume of the largest sphere that could be contained within the planet is 1.331 to 1. It (the Venerians are sexless) also reported that the straight line distance between it and its copod was 120 spandrals.

(A copod corresponds roughly to something between a sibling and a rootshoot.) The reporting Venerian was mildly comforted because the distance to its copod was the minimum possible distance between the two. By nature the Venerians can only survive at the krypton/fromage boundary and by tragic mistake the two copods had landed on disconnected branches of 64 R. I. HESS the curve of intersection of the krypton/fromage boundary and the planetary surface.

Should Cosmic, Inc., undertake the mining venture?

H19. Taurus, a moon of a-Lyra IV (hieronymous), was occupied by a race of knife-makers eons ago. Before they were wiped out by a permeous accretion of Pfister-gas, they dug a series of channels in the satellite surface.

A curious feature of these channels is that each is a complete and perfect circle, lying along the intersection of a plane with the satellite’s surface. An even more curious fact is that Taurus is a torus (doughnut shape).

Five students of Taurus and its ancient culture were discussing their field

work one day when the following facts were brought to light:

¯ The first student had dug the entire length of one of the channels in search of ancient daggers. He found nothing but the fact that the length of the channel was 30 spandrals.

¯ The second student was very tired from his work. He had dug the entire length of a longer channel but never crossed the path of the first dagger digger.

¯ The third student had explored a channel 50 spandrals in length, crossing the channel of the haggard dagger digger.

¯ The fourth student, a rather lazy fellow (a laggard dagger digger?), had merely walked the 60 spandral length of another channel, swearing at the difficulties he had in crossing the channel of the haggard dagger digger.

¯ The fifth student, a rather boastful sort, was also tired because he had thoroughly dug the entire length of the largest possible channel.

How long was the channel which the braggart haggard dagger digger dug?

H20. Define

–  –  –

Solutions to Easy Problems E1. Anywhere within 10 miles of the north pole.

E2. At first glance, it appears that the rule might be subtraction, and Ü ½.

But this is not right because 18 is not the difference of 38 and 16. Instead, the rule is that two numbers combine to give a number that is the sum of the digits of the two numbers. Thus Ü ½½.

E3. In each case the numbers combine by summing the products of their digits. Thus x = 4 ¢ 6 + 2 ¢ 2 = 28.

–  –  –

E5. 15 moves: (15, 0, 0), (9, 0, 6), (9, 6, 0), (3, 6, 6), (3, 10, 2), (3, 10, 0), (3, 4, 6), (7, 0, 6), (7, 6, 0), (1, 6, 6), (1, 10, 2), (11, 0, 2), (11, 2, 0), (5, 2, 6), (5, 8, 0), (0, 8, 5).

–  –  –

E11. The figure below shows how to trap the knight after 15 moves.

E12. See the array below.

E13. Yes; A and B go across, A comes back; C and D go across, B comes back; A and B go across.

(a) 1 and 3 go across, 1 comes back; 8 and 9 go across, 3 comes back; 1 and 6 go across, 1 comes back; 1 and 4 go across, 1 comes back; 1 and 3 go across.

(b) 1 and 2 go across, 1 comes back; 8, 9, and 10 go across, 2 comes back;

1, 6,and 7 go across, 1 comes back; 1 and 2 go across.

E14. The solutions are shown in the following figures.

–  –  –

E16. If the hour hand is exactly on a second mark then the second hand will always be on the 12. For the second hand to be 18 second marks ahead of the hour hand the hour hand must be at the 42 nd second mark, and the time is 8:24.

E17. Cut the lace in half, producing pieces A and B. Burn A from both ends, noting the point that burns last. Cut B at that corresponding point, producing pieces C and D. At the same time, start burning C from both ends and D from one end. When C is consumed, light the other end of D.

In 3.75 minutes, D will finish burning. This is the shortest time interval that can be measured.

Solutions to Medium Problems

(a) 32 moves; (37, 0, 0), (31, 0, 6), (31, 6, 0), (25, 6, 6), M1.

(25, 11, 1), (36, 0, 1), (36, 0, 0), (30, 0, 6), (30, 6, 0), (24, 6, 6), (24, 11, 1), (35, 0, 1), (35, 0, 0), (29, 0, 6), (29, 6, 0), (23, 6, 6), (23, 11, 1), (23, 11, 0), (23, 5, 6), (29, 5, 0), (29, 0, 5), (18, 11, 5), (18, 10, 6), (24, 10, 0), (24, 4, 6), (30, 4, 0), (30, 0, 4), (19, 11, 4), (19, 9, 6), (25, 9, 0), (25, 3, 6), (31, 3, 0) (b) 40 moves; (37, 0, 0), (26, 11, 0), (26, 1, 10), (26, 0, 10), (36, 0, 0), (25, 11, 0), (25, 1, 10), (25, 0, 10), (35, 0, 0), (24, 11, 0), (24, 1, 10), (34, 1, 0), (34, 0, 0), (23, 11, 0), (23, 1, 10), (33, 1, 0), (33, 0, 0), (22, 11, 0), (22, 1, 10), (22, 0, 10), (11, 11, 10), (21, 11, 0), (21, 1, 10), (31, 1, 0), (31, 0, 1), (20, 11, 1), (20, 2, 10), (30, 2, 0), (30, 0, 2), (19, 11, 2), (19, 3, 10), (29, 3, 0), (29, 0, 3), (18, 11, 3), (18, 4, 10), (28, 4, 0), (28, 0, 4), (17, 11, 4), (17, 5, 10), (27, 5, 0).

–  –  –

M3. The hand below does the job if the play goes as follows. The first five tricks are alternate spade and heart ruffs by North and West, with East underruffing North each time. North wins the sixth trick with the 10 of diamonds, East playing the 9. North next leads a heart which West trumps with the queen. The next four tricks are won by South with the 5432 of clubs; North and East discard all their diamonds on these tricks. The final two tricks are won by South’s long diamonds.

–  –  –

M9. Assume that three vertices do fall on grid points. The triangle they form will always include a vertex with an angle of 36 degrees as shown below.

–  –  –

M10. (a) Weights of 1, 3, 9, and 27 do the job.

(b) Weights of 2, 6, and 18 do the job. Note that a 1-pound package is the only one that wont balance or lift a 2-pound weight.

M11. The longest musical composition without three consecutive repetitions of any sequence are AABABAABABAABAAB and its reverse.

–  –  –

M16. An eight-triangle solution is shown below.

M17. (1) The three possibilities are that you have 5, 7 or 11 on your forehead.

(2) If you have a 5, then person A with a 7 sees (5, 5) and concludes that he must have 3 or 7. But if he has 3 then he reasons that person B sees (5, 3) and would know his number is 5 or 3. B can eliminate 3 because anyone seeing (3, 3) would immediately know he had 5. Since B doesn’t know his number, A would conclude that he has a 7. Since A doesn’t draw this conclusion you know you don’t have a 5.

(3) If you have a 7 then person B with a 5 sees (7, 7) and concludes he has 3 or 5. But if he has 3 then he reasons that person A with 7 sees (7, 3) and would know his number is 7. Since A doesn’t know his number, then B would conclude he has a 5. Since B doesn’t draw this conclusion you know you don’t have a 7. Therefore you have 11 on your forehead.

–  –  –

M20. There are 7 Good Eggs. There are 5 subjects with 15 marks possible in each. The scores for the Good Eggs are (15, 14, 13, 12, 6), (15, 14, 12, 11, 7), (15, 14, 13, 10, 8), (15, 14, 12, 11, 8), (15, 14, 12, 10, 9), (15, 13, 12, 11, 9) and (14, 13, 12, 11, 10). Humpty Dumpty got a 10 in arithmetic.

Solutions to Hard Problems

–  –  –

(b) The corresponding problem of maximizing the ratio of volume to surface area within the unit cube remains unsolved.

H9. The circles need to be packed as shown below. For Ò ½ there is just enough room for 329 circles. There are 7 circles on each end with 105 sets of 3 circles in the middle. The smallest rectangle found to date containing 329 circles has 13 circles on each side of 101 sets of 3 and measures 2 ¢ 163.9973967.

H10. A view of the pyramid looking down from S perpendicular to the plane ABCD will look like the figure below.

The point S may be raised above the plane to give unit altitudes if the inscribing circle has radius ½. Now consider two such circles of different sizes as shown. Each has a radius ½. They define the four tangent lines drawn to produce ABCD as shown. Raise S and T appropriately above the


plane so that all altitudes are unit. Clearly ST is not perpendicular to the plane.

–  –  –

H12. A computer program was written to calculate the probabilities shown in the table below. My uncle is expected to take the longest time to dress on Saturday and on Friday he is least likely to get a pair from the first three socks chosen.

–  –  –

H13. We look for cases where (1) the oldest child is under 20, (2) the younger two children have different ages, and (3) there is a product and sum that give rise to ambiguity for the ages both this year and two years ago. There is only one set of ages that accomplishies this: (5, 6, 16). The product and sum could be achieved by (4, 8, 15), which must have been guessed by Smith. The product and sum two years ago could be achieved by (2, 7, 12), which must have been guessed by Jones two years ago.

H14. (a) Length ½ ¿ ¾½ ½ ¼ as shown.

(b) Length ¾ ¼¾½½¾ ¿ as shown.

(c) Unknown.

–  –  –

Six in a row is impossible because three consecutive even special years cannot occur.

H18. The venture should be undertaken since the volume in cubic spandrals can be determined using calculus to be

–  –  –

H19. The figure below shows a cross-section of the torus.

Let Ö and Ê. There are three classes of channels that can be dug on its surface.

(a) It is clear that many channels with radius Ö are possible.

(b) Channels with radii between Ê   ¾Ö and Ê are possible.

(c) A less well-known third type of circular channel with radius Ê  Ö is possible, which is the intersection of the plane shown by the slanting line and the torus.

From the descriptions of the first two students they must have dug channels of type (b). The third and fourth students must have explored channels of type (a) and (c) in some order. One case gives Ê   Ö ¾ and Ö ¿¼;

the other case gives Ê   Ö ¿¼ and Ö ¾. The first case is not possible because Ê   ¾Ö ¼. Thus the second case applies and ¾ Ê ½½¼.

H20. (1) Define

–  –  –

is the counterclockwise path about the Ø pole of Õ´Þµ, not inwhere cluding its pole at Þ ¼.

(5) On Þ Ê for sufficiently large Ê, Õ´Þµ is bounded and the integral on the path tends to zero as Ê tends to infinity. Thus ´Òµ is the sum of integrals on small circle paths about the poles of Õ´Þµ. A pole of Õ´Þµ occurs at ½ · Þ Þ, where Þ Ü · Ý. The resulting equations in Ü and Ý are ´½ Ý ÓØݵ and Ü ÐÒ´Ý × Ò Ýµ. Since the equation in Ý is even, × ÒÝ Ý we can look for poles in the upper half-plane only and reflect each one into the lower half-plane. Call the Ø such pole in the upper half-plane Þ Ü · Ý. The table below gives numerical values of the first few. For larger ´¾ ·½ ¾µ. Then Ý approximately equals   ½·ÐÒ´ µ.


–  –  –

(b) For Ò ½¼, ´Òµ is dominated by the first pole and its reflected pole.

Thus ´½¼¼¼µ is very nearly ½  ½¼¼¼Ü½ × Ò´½¼¼¼Ý½   ½ µ, giving ´½¼¼¼µ ¢ ½¼  ¼  ½ ½ (c) From observing the behavior of × Ò´Òݽ   ½ µ one can determine the smallest Ñ where the magnitude of ´Ñµ is less than the magnitude of ´Ñ · ½µ. It is at Ñ ¼¼.

–  –  –

Sources All problems written by the author unless otherwise indicated.

E1. From David Singmaster (private communication).

E2. From Nobuyuki Yoshigahara (private communication).

E3. From Nobuyuki Yoshigahara (private communication).

E4. From David Singmaster (private communication).

E5. Problem 3, “Puzzle Corner,” in MIT’s Technology Review, edited by Alan Gottlieb, ­ May–June 1990, p. MIT 55, 1990. All rights reserved. Reprinted with permission.

E7. Unknown.

E8. From Nobuyuki Yoshigahara (private communication).

E10. From Nobuyuki Yoshigahara (private communication).

E11. From Yoshiyuki Kotani (private communication).

E12. Unknown.

E13. Unknown; (a) and (b) original problems.

E14. From Nobuyuki Yoshigahara (private communication).

E15. From Dieter Gebhardt (private communication).

E16. Unknown.

M2. Unknown.

M3. Modification of Problem 1, “Puzzle Corner,” in MIT’s Technology Review, edited ­ by Allan Gottlieb, October 1987, p. MIT 59, 1987. Submitted by Lawrence Kells.

All rights reserved. Reprinted with permission.

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