Serbian Mathematical Olympiad

Solution:

The answer is $\left[\frac{3 n-7}{4}\right]$ for $n>3$, that is, 1 for $n=3$.

Triangles with two, one, or none of their sides being also a side of the $n$-gon ($n>3$) we call, in order, ears, thin, and thick triangles. Let there be $a$ thick triangles, $b$ thin ones, and $c$ ears in the division. The number of sides of the $n$-gon that they occupy is $b+2c=n$. On the other hand, the total number of triangles is $a+b+c=n-2$. From these two relations we obtain $c=a+2$.

Since there are at most $\left[\frac{n-1}{2}\right]$ distinct triangles that are not thick, the total number $N$ of non-congruent triangles in the division is no greater than $\frac{n-1}{2}+a$. On the other hand, in such a division there are $a+2$ ears, and all ears are congruent, so $N \leqslant n-2-(a+1)=n-a-3$. Adding these gives $2N \leqslant \left(\frac{n-1}{2}+a\right)+(n-a-3)=\frac{3n-7}{2}$, i.e. $N \leqslant \left[\frac{3n-7}{4}\right]$.

Finally, by drawing the diagonals $A_{0}A_{2i}$ and $A_{2i-2}A_{2i}\left(1 \leqslant i \leqslant \left[\frac{n}{4}\right]\right)$ and $A_{0}A_{j}$ $\left(2\left[\frac{n}{4}\right]<j \leqslant n-2\right)$ we obtain an example with exactly $\left[\frac{n-1}{2}\right]+\left[\frac{n}{4}\right]-1=\left[\frac{3n-7}{4}\right]$ non-congruent triangles.