Black and White

billiards chess puzzle

Billiards chess is a variant of traditional chess in which the pieces carom off the sides of the board at right angles. In the diagram above, the white bishop at a2 controls the diagonal a2-g8 as in normal chess, but its attack also “bounces” from g8 to h7 and then back along the h7-b1 diagonal. Both bishops attack and move along these “bouncing” lines. How can White mate the black king in two moves?

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The king is already nearly corralled; White just has to arrange a final attack along the long dark diagonal. 1. Ka1 e2 2. Bb2 does the trick; if 1. … Ke5 then 2. Bg7#. By F.F.L. Alexander and E.P. White, from The Problemist, June 1932.

October 21, 2016October 30, 2016 | Puzzles

Speaking Volumes

A puzzle by A. Kozlov from the Soviet popular science magazine Kvant:

Watching a solar eclipse, a girl asks her father how much farther away is the sun than the moon.

He says, “As far as I remember, 387 times farther.”

She says, “Then I can figure out how much greater is the sun’s volume than the moon’s.”

He thinks about this and says, “I think maybe you can.” How can she do this?

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During a solar eclipse the moon covers the sun almost perfectly. This means that the ratio of the sun’s radius to the moon’s radius is the same as the ratio of these two bodies’ distance from Earth. So the sun’s volume is larger than the moon’s by a factor of about 387³.

October 20, 2016October 30, 2016 | Puzzles

Royal Descent

A puzzle from Stuart Collingwood’s Lewis Carroll Picture Book, 1899:

A captive Queen and her son and daughter were shut up in the top room of a very high tower. Outside their window was a pulley with a rope round it, and a basket fastened at each end of the rope of equal weight. They managed to escape with the help of this and a weight they found in the room, quite safely. It would have been dangerous for any of them to come down if they weighed more than 15 lbs. more than the contents of the lower basket, for they would do so too quick, and they also managed not to weigh less either.

The one basket coming down would naturally of course draw the other up.

The Queen weighed 195 lbs., daughter 105, son 90, and the weight 75.

How did they do it?

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The weight is sent down; the empty basket comes up. The son goes down; the weight comes up. The weight is taken out; the daughter goes down; the son comes up. The son gets out; the weight goes down; the empty basket comes up. The Queen goes down; the daughter and weight come up together; the daughter gets out. The weight goes down; empty basket comes up. The son goes down; the weight comes up. The daughter removes the weight and goes down; the son comes up. The son sends down the weight; empty basket comes up. The son goes down; the weight comes up. The son gets out; the weight falls to the ground.

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October 20, 2016July 4, 2024 | Puzzles

Black and White

blumenthal chess problem

By Oskar Blumenthal. White to mate in two moves.

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1. Qh1! and Black will lose no matter which knight he captures. From Schweizerische Schachzeitung, 1907.

September 28, 2016October 9, 2016 | Puzzles

Black and White

dawson half-move chess puzzle

One more chess curiosity by T.R. Dawson: How can White mate in two half moves?

The answer is to play the first half of Bg1-f2, and the second half of Bf1-g2, thus getting the white bishop from g1 to g2 and giving mate.

A fair-minded reader might ask why Black can’t pull the same trick, transferring his bishop from b8 to b7 to block the check. The answer, Dawson argues, is that some of the constituent moves are illegal: Black can’t combine Bb8-c7 and Bc8-b7 because a bishop on c8 would put the white king in an unreal check on h3; and he can’t combine Bb8-a7 and Ba8-b7 because a8 is occupied.

From Caissa’s Fairy Tales (1947).

September 7, 2016September 6, 2016 | Puzzles

Crossing Guard

Suppose some 2n points in the plane are chosen so that no three are collinear, and then half are colored red and half blue. Will it always be possible to connect each red point with a blue one, in pairs, so that none of the connecting lines intersect?

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Yes. Since n is finite, there are a limited number of ways to make the pairings. And for each way of making them, the lengths of the n segments add up to some total amount. Now, if an arrangement contains an intersection, this total length can be reduced by an appeal to the triangle inequality, which says that the sum of the lengths of any two sides of a triangle must be greater than the third side. In the example above, if we reassign the connections as shown, the total length of the connecting segments is reduced, because c < a + b and f < d + e. Hence the arrangement with the minimum total length must necessarily be free of intersections. (A Putnam paper problem from 1979, via Ross Honsberger’s Mathematical Gems, 1985.) 09/04/2016 Reader Jacob Bandes-Storch shows one way to establish the pairings.

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September 3, 2016September 11, 2016 | Puzzles

Match Point

https://commons.wikimedia.org/wiki/File:Wedding_rings.jpg — Image: Wikimedia Commons

In all 50 states, it’s legal for second cousins to marry. So can third cousins, fourth cousins, and cousins of any higher degree.

Alex and Zelda are third cousins but cannot marry. Why? If they were not related, they would be perfectly eligible.

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Because they’re brother and sister. Their parents are second cousins who married. That makes Alex and Zelda both siblings (because they have the same parents) and third cousins (because they have the same great-great-grandparents). (Offered by Mark Astolfi in MIT Technology Review, September/October 2008.)

August 30, 2016September 11, 2016 | Puzzles

Black and White

dawson chess puzzle

A curious chess puzzle by T.R. Dawson. White is to mate in four moves, with the stipulation that white men that are guarded may not move.

This seems immediately impossible. The two rooks guard one another, and one of them guards the king. And moving the pawn will place it under the king’s protection, leaving White completely frozen. How can he proceed?

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That assessment isn’t quite accurate. Moving the pawn to c4 shields the king from the rook, which permits it to move and, in turn, break the connection between the rooks: 1. c4 Ka2 2. Kc5 (K moves) 3. Kb5 Ka2 4. Ra7#. There’s a philosophical quibble here: It might be argued that the black king can still escape via b1, since the rook on b4, guarded by the king, doesn’t really menace that square. “The argument is the same as that in normal chess where a pinned man may check the adverse K,” Dawson contends. “Guarded-man immobility does not apply when captures of the Black K are in question.” From Caissa’s Fairy Tales (1947).

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August 24, 2016September 4, 2016 | Puzzles

Cyclic Billiards

A puzzle from Colin White’s Projectile Dynamics in Sport (2010): Suppose a billiard table has a length twice its width and that a rolling ball loses no energy to bounces or friction but simply caroms around the table forever. Call the angle between the launch direction and the long side of the table α. At what angle(s) should the ball be hit so that it will arrive back at the same point on the table and traveling in the same direction, so that its motion is cyclic, following the same path repeatedly?

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When the ball hits a cushion, it bounces back, with the angle of incidence equal to the angle of reflection. White writes, “Why not reflect the table instead?” Conceive that the ball travels in a straight line, and that each time it reaches a cushion it travels across a boundary and onto a new table. “So, we are now in a position to severely limit the allowable values of the angle, α, even if we cannot completely define those angles. The condition for cyclic motion is that the ball travels a certain number of tables up (not including the table the ball starts on), say, p tables, and so many tables to the right, say q tables. Two degenerate cases would be α = 0, in which case p = 0, and α = π/2 when q = 0. For the cases in between, 0 < α < π/2 and p > 0 and q > 0 but they must both be whole integers. In summary, we can now state that the ball will cycle on the billiard table if, and only if: i. α = 0° ii. α = π/2 = 90° iii. tan α = p/q (i.e. is a rational number) “As I stated, this is not a complete solution, but an interesting insight into a heuristic method. In justification, even if the perfect solution could be provided, it would be of no practical use to players of the game!”

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August 11, 2016August 21, 2016 | Puzzles

Parallelogram Puzzle

parallelogram puzzle

Point E lies on segment AB, and point C lies on segment FG. The area of parallelogram ABCD is 20 square units. What’s the area of parallelogram EFGD?

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Draw EC. Now parallelogram ABCD and triangle EDC share a common base (DC), and they have the same altitude (a perpendicular from E to DC). So triangle EDC has half the area of parallelogram ABCD. But likewise, parallelogram EFGD and triangle EDC share a common base (ED), and they have the same altitude (a perpendicular from C to ED). So triangle EDC has half the area of parallelogram EFGD. Since both parallelograms have twice the area of the same triangle, their own areas must be equal. So the area of EFGD is 20 square units. (From Alfred S. Posamentier, Math Wonders to Inspire Teachers and Students, 2003.)

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August 10, 2016August 21, 2016 | Puzzles