Existence of a monochromatic equilateral triangle on a 3-colour plane **Given a $3$-colour plane, is it possible to give a geometrical representation to prove the existence of a monochromatic equilateral triangle i.e. an equilateral triangle all whose vertices are of same colour?** For a $2$-colour plane, we can prove the existence of such an equilateral triangle with the help of a regular hexagon. For a $3$-colour plane, if no $2$ points at a fixed distance $d$ are of same colour, then also we can again prove the existence of such an equilateral triangle with the help of a regular hexagon. But what if we don't have the above restriction, can we give an **appropriate geometry** to answer the question? All I know is that there will always exist a line segment of fixed length whose end points are of same colour.

This can be done with two applications of van der Waerden's theorem. First, there must be ten points, equally spaced, of one colour (say colour 1). Draw a large equilateral triangle with this base, giving 45 more points above it, each one the tip of an equilateral triangle whose other vertices have colour 1. So, none of these 45 points are colour 1.

In particular, the nine points just above are equally spaced and of colours 2 and 3 only, so there are three equally spaced points of one colour (by van der Waerden's again). From here, the argument in your second comment finishes the problem.

It might be worth noting your approach is ultimately similar to my second paragraph here - the existence of three points of the same colour is shown first, then a triangle placed on it.