5. A regular $n$-gon is divided into triangles with $n-3$ diagonals having no common interior points. At most how many incongruent triangles can occur among these triangles?

(Dušan Djukić)
5. The answer is $\left[\frac{3 n-7}{4}\right]$ for $n>3$, and 1 for $n=3$.

Fro $n>3$, we call a triangle in a triangulation external, thin or fat if the sides it shares with the polygon are two, one, or none, respectively. Let there be $a$ fat, $b$ thin and $c$ external triangles. The number of sides of the $n$-gon they cover is $b+2 c=n$. On the other hand, there are $n-2=a+b+c$ triangles in total, so these two relations give us $c=a+2$.

Among triangles that are not fat, there can be at most $\left[\frac{n-1}{2}\right]$ incongruent ones, so the total number $N$ of incongruent triangles is at most $\frac{n^{2}-1}{2}+a$. On the other hand, there are also $a+2$ external triangles, all of which are congruent, so $N \leqslant$ $n-2-(a+1)=n-a-3$. Addition yields $2 N \leqslant\left(\frac{n-1}{2}+a\right)+(n-a-3)=\frac{3 n-7}{2}$, i.e. $N \leqslant\left[\frac{3 n-7}{4}\right]$.

Finally, note that drawing diagonals $A_{0} A_{2 i}$ and $A_{2 i-2} A_{2 i}\left(1 \leqslant i \leqslant\left[\frac{n}{4}\right]\right)$ and $A_{0} A_{j}$ $\left(2\left[\frac{n}{4}\right]<j \leqslant n-2\right)$ yields a triangulation with $\left[\frac{n-1}{2}\right]+\left[\frac{n}{4}\right]-1=\left[\frac{3 n-7}{4}\right]$ incongruent triangles.