For $k=1,2,3$, let $a_{k}$ be the number of isosceles triangles whose vertices contain exactly $k$ colours. Suppose on the contrary that $a_{3}=0$. Let $b, c, d$ be the number of vertices of the three different colours respectively. We now count the number of pairs $(\triangle, E)$ where $\triangle$ is an isosceles triangle and $E$ is a side of $\triangle$ whose endpoints are of different colours. On the one hand, since we have assumed $a_{3}=0$, each triangle in the pair must contain exactly two colours, and hence each triangle contributes twice. Thus the number of pairs is $2 a_{2}$. On the other hand, if we pick any two vertices $A, B$ of distinct colours, then there are three isosceles triangles having these as vertices, two when $A B$ is not the base and one when $A B$ is the base since $n$ is odd. Note that the three triangles are all distinct as $(n, 3)=1$. In this way, we count the number of pairs to be $3(b c+c d+d b)$. However, note that $2 a_{2}$ is even while $3(b c+c d+d b)$ is odd, as each of $b, c, d$ is. This yields a contradiction and hence $a_{3} \geqslant 1$. Comment. A slightly stronger version of this problem is to replace the condition $(n, 6)=1$ by $n$ being odd (where equilateral triangles are regarded as isosceles triangles). In that case, the only difference in the proof is that by fixing any two vertices $A, B$, one can find exactly one or three isosceles triangles having these as vertices. But since only parity is concerned in the solution, the proof goes the same way. The condition that there is an odd number of vertices of each colour is necessary, as can be seen from the following example. Consider $n=25$ and we label the vertices $A_{0}, A_{1}, \ldots, A_{24}$. Suppose colour 1 is used for $A_{0}$, colour 2 is used for $A_{5}, A_{10}, A_{15}, A_{20}$, while colour 3 is used for the remaining vertices. Then any isosceles triangle having colours 1 and 2 must contain $A_{0}$ and one of $A_{5}, A_{10}, A_{15}, A_{20}$. Clearly, the third vertex must have index which is a multiple of 5 so it is not of colour 3.