Wrap the whole group in an imaginary film. The film takes the shape of parts of cylinders, parts of planes, and parts of planets. The parts of the planets that contact the film are precisely those that are invisible to the other planets; a planet floating inside the group is entirely visible.

The invisible regions on the planets are bordered by spherical arcs, parallel two by two. If we translate all these arcs onto a single planet, they’ll partition it into several parts, each corresponding to the invisible region of some planet. Together these parts cover the planet completely.

A Soviet problem from the 22nd International Mathematical Olympiad in 1981. Here’s the same idea in two dimensions.

I create a gap between us by slowing down, then break the speed limit in catching up.

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.)