# Homework Help: Number of edges of a convex polytope with n vertices

1. Dec 11, 2008

### squaremeplz

1. The problem statement, all variables and given/known data

What is the number of edges of a convex polytope with n vertices all of whose faces are triangles.

2. Relevant equations

# of faces + #of vertecies = # of edges + 2

3. The attempt at a solution

My reasoning is as follows:

n/3 + n = # of edges + 2

4n/3 - 2 = #of edges

This doesn't seem right though. Am I using the correct formula?

Last edited: Dec 11, 2008
2. Dec 11, 2008

### Dick

Right formula. Wrong logic. Each edge meets two faces. Each face meets three edges. What's the relation between edges and faces?

3. Dec 11, 2008

### squaremeplz

3/2 * faces = edges?

Is the result

3/2*n - 2 = # of edges by any chance?

I just did

3/2*1/3*n + n = e + 2

Last edited: Dec 11, 2008
4. Dec 11, 2008

### Dick

Yes. Three times the number of faces equals twice the number of edges. If you have to do this again pick some example like tetrahedron E=6, V=4, F=4. Octohedron E=12, V=6 and F=8. It's a good sanity check.

5. Dec 11, 2008

### squaremeplz

That's actually how I found it. I drew a little tetrahedron with 4 vertecies, 6 edges and 4 faces. I don't know why but i just couldn't wrap my mind around it without a drawing. Thanks again for your time.