With the waiting move 1. Kh2! White forces Black to abandon some vital defense:

If the Nb3 moves, then 2. b4#.
If 1. … Nb5, then 2. Qa6#.
If the Nd6 moves elsewhere, then 2. Nb7#.
If 1. … Bb5, then 2. Qc7#.
If the black bishop moves elsewhere along the a6-f1 diagonal, then 2. Nxb3#.
If the black bishop moves along the a2-g8 diagonal, then 2. Qa6#.

Note that the white king’s move must be to h2 — any other move leaves him exposed to a check by Black.

Yes. The top of this trapezoid represents the unstretched snake, and the bottom represents the stretched snake. Every point on the top corresponds to one on the bottom, as denoted by the green lines. Since at least one of these lines must be perpendicular to both top and bottom, there must be some point on the snake’s body that does not move.

Imagine that each car carries a flag, and that when two cars meet they exchange flags. Now each flag travels around the course with a constant direction and speed, and so after 1 hour each flag has returned to its starting position. Because each car reverses direction when it meets its neighbor, the order in which the cars are arranged around the track does not change. This means that the starting position has now been rotated, perhaps onto itself, and that after an integral number of hours the original position must be reached again.

See Measured Steps.

1. Bd3! blocks Black’s d-pawn, forcing him to move a knight or the c-pawn — and accept a fatal check from the queen.

Yesterday was fair — no rain.

No. Color the squares as shown. Now each tile must cover a blue, a yellow, and a red square. If we’re to cover the figure completely, then it must contain an equal number of squares of each color. But it contains 17 blue squares, 18 yellow squares, and 19 red squares. So the task is impossible.

1. Qg2! fxg2 2. f4#

From the Illustrated London News, 1849.

Let x be the number of blocks in a full row. Then

n = x³ – x

= x(x + 1)(x – 1).

These factors are three consecutive integers, which means that at least one of them is even and one is divisible by 3.

Since its factors include numbers that are divisible by both 2 and 3, n is divisible by 6.

(Thanks, Allan.)

1. Ra2! and the knight will mate.

From the Australian Chess Annual, 1896.

Denote the number of trips from the first floor to the second floor by n₁₂, and so on. Now we can divide all the day’s elevator trips into three groups: trips to the third floor (n₁₃ + n₂₃), trips from the third floor (n₃₁ + n₃₂), and trips between the first and second floors (n₁₂ + n₂₁). By assumption, the first group makes up less than 1/3 of all trips. Now, if the store is empty at night, and if the floors communicate only by elevator, then the total number of customers who enter the third floor on a given day is equal to the number who leave it. This means that the first group above is the same size as the second. And it follows that the last group is the largest:

(n₃₁ + n₃₂) = (n₁₃ + n₂₃) < (n₁₂ + n₂₁)

If we substitute n₂₁ = n₂₃, we get n₁₃ < n₁₂: More customers go from the first floor to the second than from the first to the third.