A convex polyhedron has $f_n$ faces with $n$ edges and $v_n$ vertices at which $n$ edges meet.

Question

This problem is from "Notes on Geometry" by E. Rees.

A convex polyhedron has $f_n$ faces with $n$ edges and $v_n$ vertices at which $n$ edges meet.

Show that

• $\sum nf_n=\sum nv_n$
• $\sum f_{2n+1}$ is even
• $v_3+f_3 >0$

I would like to get some help with this problem. I am trying to use Euler's Formula (i.e. $V+F-E=2$).

What I can imagine is separating all faces and counting the edges of each polygon will give us $\sum nf_n$. And the total count of vertices of each polygon is $\sum nv_n$. But, I wonder if this is a general fact for any convex polyhedron. If this is right, then $E=\frac{\sum nf_n}{2}$ as edges overlap as a pair. And, we can rewrite the formula as $\sum f_n + \sum v_n – \frac{\sum nf_n}{2}=2$. This is because $F=\sum f_n$, $V= \sum v_n$.

Is this the right approach?

Any kind of help would be appreciated a lot!

Answer 1

Yes, your method is the correct one.

The first bullet point is thus proved with each sum equalling $2E$.

HINTS

Consider the equation $\sum nf_n \equiv 2E $ modulo $2$.

If $v_3+f_3 =0$ then $\sum nv_n \ge 4V$ and so $2E\ge V$......