There is a little well-known problem in geometry about equilateral triangles. Prove that it is impossible to have an equilateral triangle in the plane whose vertices are rational coordinates (a rational coordinate is a point (x,y) where x and y are rational numbers).

I found a one line proof of this problem that in fact generalizes this classic problem.

The problem is to find what kind of polygons are able to appear in the plane with vertices as rational coordinates. If you play with this a little bit you should realize that squares easily appear with rational coordinates. It turns out that only squares are the only such polygons that have this property.

## Wednesday, February 23, 2011

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And the one-line proof is...

ReplyDelete"And the one-line proof is... "

ReplyDeleteWe can assume the polygon has integer coordinates, then Pick's theorem would say $A=\frac{B}{2} + I - 1$, i.e. A is rational, but the only regular polygons with rational area with integer sides is a square.

The proof would need a second line to say that an equilateral triangle with integer sides cannot have rational area.

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