Croatian Mathematical Olympiad

Let $n$ be the number of smaller triangles, $t$ the number of points in the triangulation (including the vertices of the triangle), $d$ the number of segments (including the sides of the triangle) and $k$ the number of segments meeting in each point of the triangulation.

Obviously $t \cdot k = 2 \cdot d$ holds. Furthermore, $d$ segments are sides of $n + 1$ triangles, so $2d = 3(n + 1)$ since each segment is a side of exactly two triangles.

Finally, let us consider the sum of inner angles of smaller triangles. That sum is equal to $n \cdot 180^\circ$. On the other hand, in each of $t - 3$ points in the interior of the triangle that sum is equal to $360^\circ$, so when we add the angles of the big triangle we get that the sum of angles of the smaller triangles is equal $180^\circ + (t - 3) \cdot 360^\circ$. Hence $n \cdot 180^\circ = 180^\circ + (t - 3) \cdot 360^\circ$, i.e. $2t = n + 5$. From these equations, it follows that
$$n = \frac{5k - 6}{6 - k} = \frac{24}{6 - k} + 5.$$
So, $6 - k$ divides $24$ and the only possibilities for a positive integer $n$ are obtained if $k \in \{2, 3, 4, 5\}$, i.e. $n \in \{1, 3, 7, 19\}$. The maximal possible number of smaller triangles is $19$ and the following example shows that this can be achieved.