On Thu, 12 Jun 2003, David Wilson wrote:
Yes, I knew that power-of-2-greater-than-4-gons could not have rational vertices. I was just observing that the equilateral oddgon theorem gets us most of the way there. That with the regular octogon theorem gets us home.
I'm not sure which theorem you're talking about. It's a rather trivial remark that a polygon is embeddable (after a scale-change) with rational vertices if and only if every triple of vertices defines a triangle whose angles have rational tangents. Here's the proof. If a triangle ABC has such angles, put A at (0,0) and B at (1,0). Then C is that the intersection of two lines through these points with rational slope, so has rational vertices. For a quadrilateral ABCD this will be true of both C and D, etc. John Conway