Black and White

blumenthal chess problem

By Oskar Blumenthal. White to mate in two moves.

September 28, 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

Black and White

dawson chess puzzle

A “maximummer-selfmate” by T.R. Dawson, from 1934. White wants to force Black to checkmate him, and Black always makes the geometrically longest move available to him. How can White accomplish his goal in three moves?

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1. Rb8 Qf1 2. Rb1 Qf8 3. Ka1 Qa3#. The distance from f8 to a3 $\left ( 5\sqrt{2} \approx 7.071 \right )$ is slightly greater than that from f8 to f1 (7 units).

August 5, 2016August 14, 2016 | Puzzles

All Roads

Another puzzle from Kendall and Thomas’ Mathematical Puzzles for the Connoisseur (1971):

Take three consecutive positive integers and cube them. Add up the digits in each of the three results, and add again until you’ve reached a single digit for each of the three numbers. For example:

46³ = 97336; 9 + 7 + 3 + 3 + 6 = 28; 2 + 8 = 10; 1 + 0 = 1
47³ = 103823; 1 + 0 + 3 + 8 + 2 + 3 = 17; 1 + 7 = 8
48³ = 110592; 1 + 1 + 0 + 5 + 9 + 2 = 18; 1 + 8 = 9

Putting the three digits in ascending order will always give the result 189. Why?

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This repeated process of summing a number’s digits produces its digital root. One of the properties of digital roots is that when two numbers are multiplied, the digital root of the product is the same as the product of the digital roots of the two numbers themselves. Every positive integer has a digital root of from 1 to 9, and in the case of cubes: 1³ gives a digital root of 1 2³ gives a digital root of 8 3³ gives a digital root of 9 4³ gives a digital root of 1 5³ gives a digital root of 8 6³ gives a digital root of 9 And so on. As a result, the cubes of any three positive integers will always produce the digital roots 1, 8, 9 (in some order). UPDATE: Reader Hector Pefo offers a fuller explanation: Any sequence of three positive integers will include numbers that are 0, 1, and 2 mod 3 (that is, a multiple of 3, a number that is one more than a multiple of 3, and a number that is two more than a multiple of 3). The cube of a number that is 0 mod 3 is itself 0 mod 9, that of a number that is 1 mod 3 is 1 mod 9 (as can be seen by expanding the cube of 3n+1), and that of a number that is 2 mod 3 is 8 mod 9. Digital summation preserves a number’s value mod 9. For example, 3996 is 0 mod 9 and 4004 is 8 mod 9. The digital sum of 4004 comes from that of 3996 by subtracting 2 (final digit), adding 1 (first digit) and subtracting 18 (middle digits) for an overall effect of subtracting 19, which, mod 9, is the same as adding 8. And so the digital roots of numbers that are 0, 1, and 2 mod 3 will always be 9, 1, and 8.

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August 4, 2016August 14, 2016 | Puzzles

Perimeter Check

perimeter puzzle

A puzzle by Matt Parker of standupmaths:

The standard paper size A4 has dimensions in the ratio $1:\sqrt{2}$ . Hold a piece of A4 paper horizontally, as shown, and fold down the top left corner to meet the other side, creating fold AB, as if you were going to make a paper square. Then fold down the top right corner to meet the edge of this 1 × 1 square (making fold BC).

The perimeter of the original sheet was $\left ( 2 \times \sqrt{2} \right ) + \left ( 2 \times 1 \right )$ . What is the perimeter of the folded shape (the quadrilateral ABCD above)?

I’ll honor Matt’s request not to reveal the answer, but here’s a clue: The shape ABCD is a kite. See Matt’s video for a more visual explanation and a non-spoilery way to tell whether you have the right answer.

(Thanks to Dave Lawrence for the tip and the diagram.)

August 1, 2016July 31, 2016 | Puzzles