# Euler formula

In a convex polyhedron, number of faces F, number of vertices S and number of edges A
are related by :

For instance a cube has 6 faces, 8 vertices and 12 edges : 8+6=12+2

### Proof

By induction.
Let's consider an

**opened** convex polyhedral area with a border being a (plane or not) polygonal line.

The relation to prove becomes :
S+F=A+1.

This formula is true in the case of only one face (a polygon) where F=1 and S=A.

Adding a polygon with m sides and m vertices, touching the area border at p common edges,
the result will have p+1 common vertices and F becomes F'=F+1, S becomes S'=S+m-(p+1) and A becomes A'=A+m-p,

Hence F'+S'-A'=F+S-A=1.

We can then end the proof by closing the polyhedron by a last face
giving F'=F+1, A'=A and S'=S that is what we wanted.

### Applications

- Regular Polyhedrons

If all faces are similar, with p>2 sides, and all vertices are similar, with q edges, we have :

2A=pF=qS hence A(2/p+2/q-1)=2 or A=2pq/(2p+2q-pq) should be a positive integer.

p+q>pq/2 gives q<2p/(p-2) function which is decreasing for p>2.
Its maximal value is then for p=3 hence q<6.

Hence (p,q)={3,4,5} results into a finite number of possible cases, and finally the only ones where A is an integer :
- p=3 q=3 : tetrahedron 4 triangular faces
- p=3 q=4 : octahedron with 8 triangular faces
- p=3 q=5 : icosahedron with 20 triangular faces
- p=4 q=3 : hexahedron with 6 square faces (or cube)
- p=5 q=3 : dodecahedron with 12 pentagonal faces

- Football ball

### Polyhedrons with holes

The relation is no more true for Polyhedrons with holes, or made of several connected parts.