TITLE: Painting the plane, and finding points one unit apart
QUESTION [37 upvotes]: An old (rather easy) contest problem reads as follows:

Each point in a plane is painted one of two colors. Prove that there exist two points exactly one unit apart that are the same color.

This proof can be easily written by constructing an equilateral triangle of side length $1$ unit and asserting that it is impossible for the colors of all three vertices to be pairwise unequal.
However, I was curious about the trickier problem

Each point in a plane is painted one of three colors. Do there exist two points exactly one unit apart that are the same color?

...now, if this happened in $3$-space, I could construct a tetrahedron... but I can't do this in $2$-space. Does this not work with three colors, or is the proof just more complicated? If it doesn't work, how can I construct a counterexample?

REPLY [3 votes]: You can alternatively construct two equilateral triangles ABC and BDC with side length 1 unit, such that they share the basis BC (see image). 

Now consider the three sets G, R, and Y, where G contains all the green points, R all the red points and Y all the yellow points (these are the three colors I've used). Let w.l.o.g. A $\in$ G. If B $\in$ G or C $\in$ G, we are done with the proof. We will assume therefore that B $\notin$ G and C $\notin$ G. Now if B and C have the same color, we would be also done with the proof, so we'll assume w.l.o.g. that B $\in$ R and C $\in$ Y. 
Again, if D $\in$ R or D $\in$ Y, we could complete our prove, because one of the 1-unit-segments BD and CD would connect two points with the same color, so we'll let D $\in$ G.

Construct now a circle centered at A (colored green) with radius $\sqrt{3}$. Since the altitude of an equilateral triangle with side 1 is $\frac{\sqrt{3}}{2}$, point D would lie on the circumference. Move now the point D (colored green too) around the circumference, preserving the two-equilateral-triangles-construction.
Note, that if any point P lying on the circumference is colored red or yellow, either the one unit segment PB or PC would join two points with the same color, one unit apart, and we could conclude our proof, so we'll let all points lying on the circumference be colored green. With this configuration, you can find a one unit chord connecting two points on the circumference colored green (do you see why?).
Q.E.D
PS: Sorry for my spelling and grammar mistakes. How can I attach the images correctly?