Let $\mathcal{P}$ be a regular 17 -gon; we draw in the $\left(\begin{array}{c}17 \\ 2\end{array}\right)$ diagonals and sides of $\mathcal{P}$ and paint each side or diagonal one of eight different colors. Suppose that there is no triangle (with vertices among vertices of $\mathcal{P}$ ) whose three edges all have the same color. What is the maximum possible number of triangles, all of whose edges have different colors?
We approach the problem by finding the minimum number of triangles with a pair of edges of the same color; we call such triangles "isosceles". We can count the number of such triangles by instead considering:

The number of isosceles triangles is equal to the number of pairs of adjacent edges of the same color.

So, if we let $n_{i, v}$ denote the number of edges of the ith color touching vertex $v$, the number of isosceles triangles can be written as

$$
\sum_{v \text { vertex }} \sum_{i=1}^{8}\left(\begin{array}{c}
n_{i, v} \\
2
\end{array}\right)
$$

Since $\sum_{i=1}^{8} n_{i, v}=16$, by Jensen's Inequality we actually have

$$
\sum_{v \text { vertex }} \sum_{i=1}^{8}\left(\begin{array}{c}
n_{i, v} \\
2
\end{array}\right) \geq \sum_{v \text { vertex }} 8\left(\begin{array}{c}
16 / 8 \\
2
\end{array}\right)=17 \cdot 8=136
$$

So, the number of triangles with all different colors is at most $\left(\begin{array}{c}17 \\ 3\end{array}\right)-17 \cdot 8=544$. We leave the construction of a maximal example as an exercise to the reader.