Place triangle ABC into a square and draw the grid shown here, in which each line is parallel to EB or AC and contains a midpoint or quarter point on the side of the square. With CE and BE having lengths 3 and 4, we can determine that hypotenuse CB is of length 5.

“We have therefore shown the nature of this right triangle using a cleverly constructed diagram, and we have avoided lots of tedious computation,” write Alfred S. Posamentier and Ingmar Lehmann in Mathematical Curiosities. “This is an example where we can see how an elegant solution underscores the beauty of mathematics.”

1. Qa5! leads to four different mates:

1. … Kxb3 2. Bd1#
1. … Kb1 2. Bg6#
1. … b1=Q/R/B 2. Qc3#
1. … b1=N 2. Qa2#

From American Chess Bulletin, September-October 1940. In How to Solve Chess Problems (1945), Kenneth Howard writes, “The black pawn promotion to queen or to knight, with two different mates, is the most interesting feature of the problem. … While the solution presents little difficulty, a position like this delights the solver because of its daintiness.”

Any two of the tetrahedron’s skew perpendicular opposite edges define a series of parallel planes. The plane that passes midway between these edges produces a square cross section.

Say a player is in position H if H is the number she’s named.

Any player in position 90-99 is lost, because her opponent can advance immediately to 100.

That means that a player in position 89 will win, since her opponent will be forced to enter this losing territory.

But then position 78 is also winning, since the opponent must open a road to 89.

And so on backward: 67, 56, 45, 34, 23, 12, and 1 are all winning positions.

With correct play, then, the first player always wins: She chooses 1 and then aims for each of these milestones.

The key is to notice that the 1 × 1 × 1 blocks must fall along a diagonal of the finished cube, to ensure that each 3 × 3 × 1 “slice” of the cube (in every orientation) can contain an odd number of blocks:

White’s knights began the game on squares of opposite color, but now they’re both on white squares. That means that, between them, they’ve made an odd number of moves in this game. By a similar argument White’s rooks and king have made an even number of moves in total, and we can see that White has made a single pawn move. White’s bishops haven’t moved, and his queen must have been captured on her home square. So White has made an even number of moves.

The same considerations apply to Black’s position, but he’s made an extra pawn move, so altogether he’s made an odd number of moves. Since White moved first, this means it’s Black turn to move, and he can mate with 1. Nxc2#.

(From Schach, 1957, via Noam D. Elkies and Richard P. Stanley, “The Mathematical Knight,” Mathematical Intelligencer 25:1 [December 2003], 22-34.)

Henry Dudeney presented this solution in his 1917 Amusements in Mathematics; it’s now known as the “method of buttons and strings.” No knight will ever visit the center square, so we can ignore that; regard each of the remaining squares as a button and connect it by a string to each square that’s a knight’s move away. That produces diagram B. Now if we untangle the strings we get diagram C, a unique cycle that any knight must follow in making its way around this board.

The four knights must follow this cycle, and no two knights may ever occupy the same square. The white knights start on squares 1 and 3, the black on 6 and 8. So to exchange the white and black knights, we must advance all four knights four steps along the circle, either clockwise or counterclockwise. Since the four knights make four moves each, the task is done in 16 moves. That’s the minimum. (Somewhat paradoxically, if three white knights oppose three black ones in the starting position, the minimum number of moves drops to 8.)

Yes. Put the coin in one pan, and put all 100 remaining coins in the other. Now suppose a true coin weighs t grams and a fake coin weighs f grams. If the coin is fake, then the difference between the weights will be

51t + 49f – f = 51t + 48f = 51t + 48(t ± 1) = 99t ± 48,

which is always divisible by 3. If the coin is true, then the difference will be

50t + 50f – t = 49t + 50f = 49t + 50(t ± 1) = 99t ± 50,

which is not divisible by 3.

Another solution is to put aside the chosen coin and divide the remaining 100 coins into two piles. If the difference in weight between these piles is even, then the chosen coin is genuine; if it’s odd then it’s false.

(From Sergei Aleksandrovich Genkin, Dmitrii Vladimirovich Fomin, and Ilia Itenberg, Mathematical Circles: (Russian Experience), 1996.)