Supercali wrote:

We will use induction. Just notice that there exist 3 consecutive vertices with different colours.

The idea is correct but some more detail is required.

Let the three colours be red (R), blue (B), and green (G). Let the convex polygon be $A_1A_2 \cdots A_n$ . The colour of a vertex $v$ is denoted by $c(v)$ . Suppose some colour, say red, occurs only once as the colour of some vertex, say $c(A_n) = R$ . Then the vertices from $A_1$ to $A_{n-1}$ must be coloured alternately with the colours $B$ and $G$ . Thus the triangulation $A_nA_1A_2, A_nA_2A_3 , \cdots , A_nA_{n-2}A_{n-1}$ works.

Otherwise for each colour, there exist at least two vertices with that colour. Here, we show that Supercali's claim is true. Indeed, assume that it was not true. Let $c(A_1) = R$ and $c(A_2) = B$ , say. Then by our assumption, $c(A_3) = R$ , $c(A_4) = B,$ etc, and the whole colouring is determined by only red and blue. But this would lead to a contradiction, as the colour green occurred nowhere. Thus some vertex $v$ has neighbours of different colours. The triangle $T$ that they form has all three colours different. Separate this triangle, and the remaining polygon is $n-1$ -sided, and each colour occurs at least once in it ( this is why we need to make cases; we need each colour to occur at least once). Now this is induction.