Counting faces of each type in a rhombicosidodecahedron If I know that, in a rhombicosidodecahedron, at every vertex one triangle, one pentagon, and two squares meet, then how can I compute the number of faces and edges that are needed to build it?
There must be some mathematically way to show that there are exactly $x$ triangles, $y$ squares, and $z$ pentagons.
 A: Let $v$ be the number of vertices, $t$ the number of triangles, $p$ the number of pentagons, and $s$ the number of squares.
Each vertex touches one triangle, one pentagon, and two squares.
So we know $v = 3t = 5p = \frac{4}{2} s$
From every vertex there are four edges. But each edge is connected to two vertices. So the number of edges $e$ satisfies $4v = 2e$. Similarly we can count the edges by the number of edges per face, so we get $2e = 4s+3t+5p$
The number of faces is $f=s+t+p$.
As we are looking for a convex (and therefore characteristic $\chi = 2$) we can use eulers formula which says $v-e+f = \chi = 2$.
This is a system of linear equatios which can now be solved.

EDIT: Regaring the comment of @DavidK that there are not enough equatiosn: Yes there are enough equations, there are 7 of them and they are idependent. Let's go somewhat more in depth:
Let's rewrite this sytem as matrix equation. The equations from top to bottom are in the order in which they occur in this post. It is quite obvious that the matrix has full rank and therefore is invertible:
$$\begin{pmatrix}
  &-3&  & 1&  &  &  \\
  &  &-5& 1&  &  &  \\
-2&  &  & 1&  &  &  \\
-4&-3&-5&  & 2&  &  \\
-1&-1&-1&  &  & 1&  \\
  &  &  &  &  &  &1\\
  &  &  & 1&-1& 1&-1\\
\end{pmatrix}
\begin{pmatrix} s\\ t\\p \\ v\\ e\\ f\\ \chi \end{pmatrix}
= \begin{pmatrix}0\\0\\0\\0\\0\\2\\0\end{pmatrix}$$
I used matlab to solve this system with following code:
A=[0,-3, 0,1, 0,0, 0;
   0, 0,-5,1, 0,0, 0;
  -2, 0, 0,1, 0,0, 0;
  -4,-3,-5,0, 2,0, 0;
  -1,-1,-1,0, 0,1, 0;
   0, 0, 0,0, 0,0, 1;
   0, 0, 0,1,-1,1,-1];
b= [0;0;0;0;0;2;0];
y=A\b

And we get the solution $(s,t,p,v,e,f,\chi) = (30,20,12,60,120,62,2)$ which exactly corresponds to the data from wikipedia.
A: Hint to get you started (but probably not enough to finish the problem): use Euler's formula $V - E + F = 2$.
For faces:
$$
F = x + y + z .
$$
For vertices 
$$
4V = 3x + 4y + 5z .
$$
You should be able to write the equation for $2E$ since every edge is on two faces.
