PUZZLE: Given a rhombus tiling of a 2n-gon (so that each polygon edge is a rhombus edge), show that the number of rhombi is determined by n.
Hmm. I don't think you need rhombi here, right? This is just a statement about any tiling by parallelograms. But you need to be a little careful that you're discussing *convex* 2n-gons -- otherwise, when you place a single tile down in your 2n-gon, you have a new (concave) 2n-gon which you can cover with one fewer tile :-). Okay, spoiler below. Or maybe spoiler above, if the idea that it's true in parallelogram generality gives it away... Fundamentally, parallelograms propagate edges: if you have an edge of your original 2n-gon with a given length and slope, then the parallelogram that uses that edge will leave you with a translation of that edge, and so on -- the only way you can ever have a tiling is if you eventually reach a matching edge, all the way across the 2n-gon. So in particular, the edges of the original 2n-gon come in matched pairs. Now if the 2n-gon is convex, then the parallelo-path from one edge to its mate has to cross the path from any other edge to its mate, since the slopes of the edges determine their cyclic order. Each one of those crossings is exactly one parallelogram. So there are n-choose-2 parallelograms, and their shapes (including orientation!) are exactly given by taking pairs of the original n distinct types of edges. Cool stuff! --Michael Kleber kleber@brandeis.edu