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.
Tesselate the plane with equilateral triangles of side length = 1 meter.
Label the three corners of one triangle R, G and B (as, if one were the same we'd already have a match--we're trying to show we can't keep producing lacks of matches forever). This is the first triangle.
In the adjacent triangle that also includes the R-G side, label the remaining vertex B. This is the second triangle. In the triangle adjacent to the first triangle and sharing the R-B side, label the remaining vertex G, so as to avoid any matching colors at 1 meter so far. Finally in the triangle adjacent to the second triangle and sharing its R-B side, label the remaining vertex, G.
There are now 3 vertices labelled G that are at the vertices of an equilateral triangle whose sides are each sqrt(3) as they are at tips of back-to-back equilateral triangles whose side lengths are 1, and so each of those altitudes is sqrt(3)/2.
Draw circles of radius 1 around each of these G vertices. They form a 3-lobed figure where the tips of the lobes are also sqrt(3) meters apart. As the points are all 1 meter from one G point or another, all points on this 3-lobed figure are R or B.
In particular consider the arcs on the counterclockwise side of the centerline of each lobe. If points are chosen equally far out on each of these arcs an equilateral triangle is formed. This equilateral triangle can be any size from zero-length sides, to the full sqrt(3)-length sides. Choose the one with sides of length 1 meter. As there are now only 2 colors to choose from, R and B, and 3 vertices to this triangle, two of them must have the same color.
Posted by Charlie
on 2003-05-01 08:13:04