Burden of Proof Activity

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:

• lines may not cross each other as they move from dot to dot
• every dot on your page must be connected to every other dot through a sequence of lines
• every region must topologically be a disk with no holes

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).

Since n is positive, and n times k/p -k/2 +1 multiply to give the positive number 2, then

k/p -k/2 +1 > 0.

k/p + 1 > k/2.

Multiply both sides by 2/k which is positive.

2/p + 2/k > 1

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.