I have this question:
"Consider a convex polyhedron, all of whose faces are square or regular pentagons. Say there are m squares and n pentagons. Assume that each vertex lies on exactly 3 edges"
Then I want to show that $3V=2E$ ! I can draw a picture and understand why it is true but I was wondering whether there was a more "mathsy" way of going about it?
Thank you!
The last sentence in the original problem is the only one that matters. From each vertex you have 3 edges. If the edges would not be connected at both ends, you will have $3V$ edges. Since each edge is connected to two vertices, the number of edges is $E=\frac{3V}2$.
To check that this is true in other cases, see a tetrahedron, with $V=4$, $E=6$, or a triangular prism $V=6$, $E=9$.