That's a beautiful proof! And yeah, it generalizes to higher dimensions. If you're in the corner of an N-dimensional hypercube, then the hypercube has 2n faces

That's a beautiful proof! And yeah, it generalizes to higher dimensions. If you're in the corner of an N-dimensional hypercube, then the hypercube has 2n faces of dimension n-1. You're touching half of these faces, leaving n faces to form pyramids out of. And just like in the 3D case, they're arranged symmetrically so they must all have the same volume. EDIT: got around to reading TFA and it's also the top answer, so there's a nice visualization. Diving deeper into the proof, say the hypercube is a unit hypercube, spanning from (0, ..n, 0) to (1, ..n, 1) -- where "..n" means "a sequence of length n". Then: Each corner is a point (b_1, ..n, b_n) where each b_i is 0 or 1. You're located at (0, ..n, 0). The hypercube has 2n faces, which can be represented as a pair (i, b) where 0<=i<n and b is 0 or 1. The face (i, b) touches the 2(n-1) corners whose i'th coordinate is equal to b. You touch half of those faces: specifically the faces (i, 0) for each 0<=i<n. You can draw a pyramid to the other half of the faces, since you don't touch them. These pyramids must all have the same volume, by the symmetry of coordinate permutations, which is an operation that preserves volume.