Find the number of triangular faces in a polyhedron, that has only triangles and fourgons as faces.

Question

A polyhedron has only triangles and fourgons as faces, each of its vertices has degree four, and the polygon has four fourgons. How many triangles does it have?

Answer 1

Let $q$ be the number of quadrilateral faces, and $t$ the number of triangular faces.

If you were to count the total number of corners the faces have, how many would that be (expressed in terms of $q$ and $t$)? You are also given that the vertices have degree 4, so now express the total number of corners that the faces have in terms of $v$ (the number of vertices).

The above counts the number of corners in two different ways, and this gives you an equation that expresses $v$ in terms of $q$ and $t$.

You can do a similar thing by counting the sides of the faces in two different ways to create an equation that expresses $e$ (the number of edges) in terms of $q$ and $t$.

It is trivial to express $f$ (the total number of faces) in terms of $q$ and $t$.

So now every variable in Euler's identity, $v-e+f=2$, can be substituted by something involving only $q$ and $t$.

It turns out that you don't even need to know the value of $q$ - it drops out of the equation - so you then have an equation for $t$ alone, and can deduce its value.

Answer 2

Use Euler's identity for planar graphs to set up system of equations.