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. ----- Original Message ----- From: John Conway To: math-fun Sent: Wednesday, June 11, 2003 9:45 AM Subject: Re: [math-fun] What's your proof? On Wed, 11 Jun 2003, David Wilson wrote:
With the additional step, Fred's proof coincides with mine.
So, there is no such thing as an equilateral oddgon with rational vertices.
Here's a quick application:
If a regular polygon has rational vertices, it's a power-of-2-gon.
Yes, since in fact it's a 4-gon, and 4 is indeed a power of 2. John Conway _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun