I have a question with my homework. I'm pretty sure I have to use Euler's formula to solve it, but I'm kind of stuck on how to use the formula to solve this problem.
I started to draw the problem on paper, but it only made me more confused. Is there anyone that can give me some insight on maybe a good starting point?
Question: "Suppose you have a solid having only triangles and squares as faces. Suppose that the degree of each vertex is 4. Prove that the solid must contain exactly 8 triangles."
Let $m$ be the number of triangles and $n$ the number of squares. The total number of vertices is $3m+4n$ and note that in this each vertex is counted four times since the degree at each vertex is 4. If $V$ is the number of vertices of the solid, we have $4V = 3m+4n$. The number of edges $E$ is given by $3m+4n = 2E$ since each edge will be counted twice, and the number of faces is $m+n$. By Euler's formula, we have \begin{align*} V- E + F = 2 \\ \frac{3m+4n}{4} - \frac{3m+4n}{2} + m+n = 2 \\ \frac{m}{4} = 2 \end{align*} Hence $m=8$. Thus the number of triangles is 8.