Imagine that a painter went down to a mathematical plane and colored all of the points on that plane one of three colors.

Prove that there exist two points on this plane, exactly one meter apart, that have the same color.

(In reply to Proof by contradiction by Brian Smith)

Good job, Brian! I was trying to think of something similar to this, but you beat me to the punch.

Obviously, you can't have four distinct points equidistant from one another in a two-dimensional plane (I think someone mentioned a regular tetrahedron for a 3-D case).

The approach I was trying for involved a large set of equilateral triangles, squares, circles, soccer balls, trying to find a set of points at a unit distance from each other that would be impossible to give three colors without two adjacent points being the same. That approach, as DuCk prematurely ventured, may well be impossible without overlapping the shapes. What I can do (so as not to feel totally dumb) is refine Brian's answer geometrically.

Start with an isoceles triangle with legs of length √3 and a base of unit length (or 1 meter, for this specific problem). Draw the perpendicular bisector of each leg, to a segment of unit length which is bisected by the leg. Both pairs of endpoints of these two segments form an equilateral triangle with the peak point of the large triangle and with its respective base point. Therefore, as Brian said, each set of endpoints needs to be a different color, as they are endpoints of a 1m segment, and each pair is comprised of the two colors that are different than the apex of the large isosceles triangle. Therefore, each segment forces its respective base point to be the same color as the apex. But, since the base of that triangle is of unit length, they cannot be the same color. So, somewhere in those 7 points are two that are 1 meter apart and have to be the same color.

Posted by DJ on 2003-05-01 12:46:00