Suppose you have an infinite plane, and each point on the plane has been arbitrarily painted one of two colors.

Prove that there exists an equilateral triangle whose vertices are all the same color.

What is the fewest number of points needed to prove this?

The MOST you need is a seven point proof, because you won't be able to place the eighth.

Start with 3 adjacent, and then start adding so that you don't have three in a triangle. The best way ends up being having a diagonal row of one color and a diagonal row of the other. So you have 7 points. Zig-zag causes problems around 5 points. HOw's that?

Posted by Lawrence on 2003-08-30 20:42:11