A convex polyhedron has exactly $2$ decagonal faces and in each of its vertices has exactly $4$ edges coming together. Prove that this polyhedron has at least $20$ triangular faces.
Let's call:
- v = number of vertices;
- e = number of edges;
- f = number of faces.
From eulers equation I know that: $v - e + f = 2$
We know also that: $2e = 4v$ since there are $2$ vertices incident with $1$ egde and $4$ edges coming together in $1$ vertice.
But how to prove that there are at least $20$ triangular faces knowing that there are exactly $2$ decagonal faces? I don't even know anything about the other possible faces of that polyhedron.
Let's say the polyhedron has $V$ vertices, $E$ edges and $F$ faces in total. We know that $2$ of these are decagons; say $F_3$ are triangles and $F'$ are neither triangles nor decagons (so $F=2+F_3+F'$). Importantly, these other faces have at least $4$ sides.
Not only do we have $2E=4V$, as you say, by counting the vertices on each face we also have $2\cdot 10+3F_3+4F' \le 4V$ (this is the key observation).
Now, by Euler's formula, $$\begin{align} F+V&=E+2 \\ F+V&=2V+2 \\ F&=V+2 \\ F&\ge \frac14 \left[20+3F_3+4F'\right]+2 \\ 2+F_3+F' &\ge 5+\frac34 F_3 + F'+2 \\ \frac14 F_3 &\ge 5 \end{align}$$
and so $F_3 \ge 20$, as required.