Heard a long time ago, in 1990s.

(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:

(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.

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?

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