**Exercise: ***Show that any deltahedron has an even number of
faces.*

**Answer: **

In a deltahedron with *E* edges and *F* faces, *E* =
(3/2) *F*. So *F* must be even, as *E* is an integer.

To derive the equation, imagine a pile of loose triangles, before they
are joined together. As each triangle has three sides, there are three
times as many triangle-edges as triangles. When we join the triangles together,
two triangle-edges become one polyhedron edge, giving *E* = (3/2)
*F*.

**Further:**

You can show in a similar manner that in any polyhedron, the total number of odd-edged faces is even. In other words, if you add the number of triangles, plus the number of pentagons, plus the number of heptagons, ... , you get an even number.