Croatian Junior Mathematical Olympiad

The numbers 8, 10, 11 and 12 are not good, while 7 and 9 are good.

We first show that even numbers are not good. Let us assume $n$ is a good number and consider a fixed vertex $A$ of a regular $n$-gon coloured in the required way. For each vertex $B$ different from $A$, there is a unique vertex $C$ such that the sides of the triangle $ABC$ have the same colour. Hence, we may partition all vertices other than $A$ into pairs. This shows that $n$ is odd. From this we conclude that 8, 10 and 12 are not good numbers.

Next, we show that numbers of the form $n=3k+2$ are not good. Again, let us assume $n$ is good and for a regular $n$-gon coloured in the required way, let $t$ be the number of triangles $ABC$ with all sides of the same colour. Each such triangle has exactly three pairs of vertices, while for each pair of vertices of the $n$-gon there is a unique triangle with all the sides of the same colour. This shows that $$3t=\frac{n(n-1)}{2}.$$ Hence, 3 divides $n$ or $n-1$. This shows that 11 (and, again, 8) are not good.

To show that 7 and 9 are good numbers we construct examples. In the example we denote vertices of the $n$-gon by numbers 1, 2, ..., $n$ and write down triples $(a,b,c)$ representing triangles that have the sides of the same colour. Note that there should be $t=\frac{n(n-1)}{6}$ triples, and for each pair of numbers $a,b\in \{1,2,\dots ,n\}$ there should be exactly one triple in which the pair of number occurs together.

For $n=7$ we have the following $t=7$ triples: $(1,2,3)$, $(1,4,5)$, $(1,6,7)$, $(2,4,6)$, $(2,5,7)$, $(3,4,7)$ and $(3,5,6)$.

For $n=9$ we have the following $t=12$ triples: $(1,2,3)$, $(4,5,6)$, $(7,8,9)$, $(1,4,7)$, $(2,5,8)$, $(3,6,9)$, $(1,5,9)$, $(2,6,7)$, $(3,4,8)$, $(1,6,8)$, $(2,4,7)$ and $(3,5,7)$.