A convex polyhedron has triangular and quadrilateral faces, not necessarily regular, with exactly four faces meeting at each vertex. Prove that the number of triangular faces is always the same, and find this number.
I know that I have to use $V-E+F=2$, but I am not sure where to start with this. Is it something to do with the sum of the angles of the shapes?
Since four faces meet at each vertex, four edges do also and each edge is shared by two vertices. Thus
$E=2V$ Eq. 1
Next, each triangular face is shared by three vertices and each quadranglular face by four. Comparing that with each edge being shared by two vertices again, get
$E=(3/2)F_{\triangle}+2F_{\square}$ Eq. 2
From Eq. 1 we have $V=E/2$. So we may render the Euler characteristic:
$(E/2)+F_{\triangle}+F_{\square}-E=2$
$F_{\triangle}+F_{\square}-E/2=2$
$F_{\triangle}+F_{\square}-(1/2)((3/2)F_{\triangle}+2F_{\square})=2$
and if you simplify the last equation algebraically, $F_{\square}$ will drop out leaving only a single constant value for $F_{\triangle}$.
By similar logic, a polyhedron consisting of pentagons and hexagons meeting three to a vertex always has 12 pentagons.