A convex polyhedron has triangular and quadrilateral faces meeting four to a vertex. Prove the number of triangles is always the same; find the number

Solution 1:

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.