At 01:49 AM 6/11/03, David Wilson wrote:
Prove the following:
If a polygon in R^2 has an odd number of equal-length sides, it has a nonrational vertex (not in Q^2).
I just wanted to know if there was a proof very different from my own.
If there is an equilateral polygon with vertices in Q^2, it can be scaled so that its vertices are in Z^2. If such a polygon has sides of even length, they can be halved (as any Pythagorean triangle with an even hypotenuse has even legs). Thus the sides can be taken odd, which means the parity of the sum of the coordinates changes between each consecutive pair of vertices. (Odd hypotenuse implies one odd leg, one even leg.) Which is impossible in a circuit with an odd number of edges. -- Fred W. Helenius <fredh@ix.netcom.com>