Are triangulations of polygons 3-colourable?

Question

Let us take a polygon and divide it into triangles in an arbitrary manner. When playing with such triangulations, I was not able to generate any which were not 3-colourable.

Is it true in general that they are 3-colourable? If not, can someone give a counterexample?

Here are two examples to illustrate clearly what I mean by "triangulation" and "colouring".

Answer 1

I found Brooks's theorem, which says:

For any connected undirected graph $G$ with maximum degree $\Delta$, the chromatic number of $G$ is at most $\Delta$ unless $G$ is a complete graph or an odd cycle, in which case the chromatic number is $\Delta+1$.

A triangle can have at most 3 neighbours, so the maximum degree is 3.

Let us look at the exceptions mentioned by the theorem:

(Odd) cycles have max-degree 2, so they are still 3-colourable.

Complete graphs with more than 4 vertices are not planar, thus not realizable as a triangulation. The complete graph with 4 vertices cannot be realized as a triangulation either because if two neighbours of a triangle are adjacent, they cannot be also adjacent to the third neighbour. Complete graphs with less than 4 vertices are 3-colourable.

Thus it is true that any triangulation of a polygon is 3-colourable.