1
$\begingroup$

I have this problem in my homework but it doesn't seem quite right to me:

Every face of a convex polyhedron has at least $5$ vertices, and every vertex has degree $3$. Prove that if the number of vertices is $n$, then the number of edges is at most $5(n-2)/3$.

I was thinking that If the number of vertices of the polyhedron is $n$, and every vertex has degree $3$, then the number of edges should be $3n/2$. Am I interpreting the question wrongly or is there an error in my reasoning? Thanks so much!

  • 1
    hmm.. sorry I'm a little confused. If the number of vertices is $n$, then the number of degrees is $3n$. since each face has at least $5$ vertices then number of faces is at most $3n/5$. Then using Euler's formula the number of edges would be less than $8n/5 -2$. This is incorrect.. and still I'm not sure why it is not possible to calculate the number of edges directly from the number of vertices and degrees? Thanks a lot for your time!2019-01-15

1 Answers 1

2

You are right. The sum of degrees, is twice the number of edges:

$\sum_{v \in V(G)} \deg(v) = 2 \cdot |E(G)|$

So you get

$3n = 2m \Rightarrow m = \frac 32 n$

which holds when $n > 20$. So you can consider the case where $n \leq 20$ separately.

  • 0
    That would be a possibility yes.2012-05-02