1. The weight is sent down; the empty basket comes up.
2. The son goes down; the weight comes up.
3. The weight is taken out; the daughter goes down; the son comes up.
4. The son gets out; the weight goes down; the empty basket comes up.
5. The Queen goes down; the daughter and weight come up together; the daughter gets out.
6. The weight goes down; empty basket comes up.
7. The son goes down; the weight comes up.
8. The daughter removes the weight and goes down; the son comes up.
9. The son sends down the weight; empty basket comes up.
10. The son goes down; the weight comes up.
11. The son gets out; the weight falls to the ground.

1. Qh1! and Black will lose no matter which knight he captures.

From Schweizerische Schachzeitung, 1907.

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.

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.)

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).

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!”

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.)

1. Rb8 Qf1 2. Rb1 Qf8 3. Ka1 Qa3#.

The distance from f8 to a3 is slightly greater than that from f8 to f1 (7 units).

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.