Brazilian Mathematical Olympiad

Consider the planar graph which vertices are the vertices of the triangles and edges are the sides of the triangles. Let $V$, $E$ and $F$ be the number of vertices, edges and faces of such graph and $d$ be the degree of each vertex. Note that $n = F - 1$. Then $E = \frac{3F}{2} = \frac{dV}{2} \iff V = \frac{2E}{d}$ and $F = \frac{2E}{3}$. By Euler's theorem, $V - E + F = 2 \iff \frac{1}{d} + \frac{1}{3} = \frac{1}{2E} + \frac{1}{2} > \frac{1}{2} \implies \frac{1}{d} > \frac{1}{6} \iff d < 6$. It's not hard to see that $d \ge 3$ and that each value for $d$ determines $V$, $E$ and $F$. Indeed $d = 3 \implies F = 4$; $d = 4 \implies F = 8$; $d = 5 \implies F = 20$. The following examples show that the possible values for $n$ are indeed $3$, $7$ and $19$.