A recent local math olympiad problem showed how to tile a regular hexagon and octagon using rhombi (with each polygon edge coinciding with a rhombus edge), and asked "What is the number of rhombi needed for such a tiling of a 2n-gon ?"

This wasn't hard to answer when the given tiling method was generalized to the 2n-gon.  But implicit in the problem was the fact that *any* such rhombus tiling of a 2n-gon uses the same number of rhombi.

I've since found the answer proved in passing in a book, but y'all, or y'some, may find this an interesting puzzle.  (Also, I suspect there must be a more straightforward proof than the one I happened upon.)

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.

--Dan