One may think at first that one can construct an infinite number of regular polyhedra in three-dimensions, just as we could construct an infinite number of regular polygons in two-dimensions. However, this does not turn out to be the case. There are only five regular polyhedra, but why?

**V-E+F Experiment**

Draw a few dots on a piece of paper

Connect the dots with lines, subject to the following rules:

Compute Vertices (V) - Edges (E) + Faces Separated by Edges (F)

[Do not forget to count the outside as a region for F too.]

For the plane and the sphere, the answer is always 2. So, if you are a farmer who wants to fence off 4 (interior) pastures together with 55 sections of fence, you can calculate exactly how many fenceposts you need, no matter how you arrange the fences.

Euler's formula gives us that

V - E + F = 2

If we have a regular polyhedra with *n*
faces that are polygons with *k* sides, and if *p* faces are
touching at each vertex, then Euler's formula can be written in a different
form.

How many vertices do we have? Well the polygons have *k* sides,
so they have *k* vertices too. There are *n* different polygons,
so that makes *n k * vertices, but we have to take into consideration
the fact that when *p* faces touch, they do so only at 1 point:

V= *nk/p*

How about the number of edges? Similarly, there are
*nk* edges on the polygons, but we are double counting because
each polygon edge meets one other polygon edge, so we divide by 2.

E = *nk/2*

2 = V -E + F = *nk/p* -*nk/2* +*n* =
n(*k/p* -*k/2* +1).

Add *k/2* to both sides.

In the plane, a regular polygon with non-zero area must have
at least 3 sides, so *k* is at least 3.

How many polygons meet at a point? 1 polygon doesn't close up in 3-D.
2 flat polygons meeting at a vertex
don't close up in 3-D unless they overlap each other,
but then this isn't a polyhedron.
So there must be at least 3 polygons meeting at a vertex, and so
*p* must be at least 3.

Now we combine *k* and *p* at least 3 with our formula
2/*k* + 2/*p* > 1

k=3 and p=3

k=3 and p=4

k=3 and p=5

k=4 and p=3

k=5 and p=3

are the only numbers that work. So now we know that these are the only possibilities. They are all realized - tetrahedron, octahedron, icosahedron, cube, and dodecahedron [build nets and present information about remembering the name as well as the vertices, edges, and faces]

On the
sphere, we can have polygons that are
closed lunes, formed by 2 great circles. These
have positive area, so *k* can also be 2, and *p* can
be any number [these are the hosohedrons].

Or, we can have just 2
spherical polygons touching, where *p = 2* - how can we
do this? By taking the vertices along a great circle. Then one of the
polygons is the great circle plus the
upper hemisphere, and the other is the great circle plus the lower.
Now *k* can be any number.