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Puzzles

## Cube Problems

### Puzzle

Imagine a cube on a flat table, tantalizingly balanced on one of its vertices such that the vertex most distant from it is vertically above it. (a) What is the length of the shortest path an ant could take to go from the topmost vertex to the bottommost vertex? (b) What will be the projection on the table if there is a light source right above the cube? (c) What would be the cross-section obtained if we slice the cube along a plane parallel to the table, passing through the midpoint of the topmost and the bottommost points of the cube? (d) Split a large 3×3×3 cube into 27 small 1×1×1 cubes. An ant can burrow through one small cube to an adjacent small cube if these two cubes share a face. Can the ant burrow through all of the 27 small cubes, visiting each small cube exactly once? Can such a sequence have the additional property that the first and the last small cube share a face?

### Source

Heard a long time ago, in 1990s.

### Solution

(a) The length is √5 times the length of the edge. The solution is explained here.

(b) A hexagon (shown on the right).

(c) A hexagon (shown below). In fact, the hexagonal cross-section allows a square to be fitted entirely inside it whose dimensions are larger than the sides of the original cube. This insight allows us to solve an old puzzle by Henry Dudeney:

“I had two solid cubes of lead, one very slightly larger than the other, just as shown in the illustration. Through one of them I cut a hole (without destroying the continuity of its four sides) so that the other cube could be passed right through it. On weighing them afterwards it was found that the larger cube was still the heavier of the two! How was this possible?”

(d) It is easy to find sequences of 27 distinct small cubes such that pairs of adjacent small cubes have common faces. However, it is not possible for such a sequence to enjoy the additional property that the first and the last small cubes also have a face in common.

Explanation: Color the 1×1×1 cubes either black or white such that any horizontal or vertical 3×3×1 slice looks like a chessboard. In other words, if two 1×1×1 cubes share a face, they are of opposite color. If we visit 27 small cubes in any sequence such that any pair of adjacent small cubes have common faces, then the first and the last cubes must have identical color and therefore will not have a common face.

Previous Puzzle: Coin Toss Guess

Alice and Bob are playing a game. They are teammates, so they will win or lose together. Before the game starts, they can talk to each other and agree on a strategy.

When the game starts, Alice and Bob go into separate soundproof rooms — they cannot communicate with each other in any way. They each flip a coin and note whether it came up Heads or Tails. (No funny business allowed — it has to be an honest coin flip and they have to tell the truth later about how it came out.) Now Alice writes down a guess as to the result of Bob’s coin flip; and Bob likewise writes down a guess as to Alice’s flip.

If either or both of the written-down guesses turns out to be correct, then Alice and Bob both win as a team. But if both written-down guesses are wrong, then they both lose.

Can you think of a strategy Alice and Bob can use that is guaranteed to win every time?

Next Puzzle: Truchet Tilings

An 8x8 square grid has to be covered with isosceles triangular tiles with two tiles per square. Tiles come in two colors: black and white. Such tilings are called Truchet tilings. A tiling is said to be "fine" if no two tiles sharing an edge have the same color. How many "fine" Truchet tilings are there?