# The five Platonic solids Figure 1. Paper models of all five Platonic solids.

I’ve been fascinated by polyhedra since I was really young. My mom had this book called Polyhedron Models for the Classroom by Magnus Wenniger which gave patterns for building models of all the regular and semiregular polyhedra as well as a short history of the pursuit to discover all of them. I thought it was really cool, and I’ve made many models of the Platonic and Archimedean solids through the years. Unlike polygons, there are only finitely many Platonic solids. Why is that? In this page, I’ll give a proof that there are at most five Platonic solids.

The picture to the right shows a set of models of all five Platonic solids. From left to right they are the tetrahedron, the dodecahedron, the cube (or hexahedron), the icosahedron, and the octahedron, and they are each named for their respective number of faces. These forms have been known for thousands of years, and were named after Plato who, for better or for worse, said that each polyhedron corresponded to a different classical element (earth, air, fire, and water; he left the dodecahedron to the gods). They really should have been named after Theaetetus, a contemporary of Plato, who first proved that there were exactly five regular convex polyhedra. Or even Euclid, who completed the Elements with constructions of each of the solids.

I haven’t been clear about what I mean by a regular polyhedron, so I give the following definition:

Definition. A regular polyhedron is a polyhedron whose faces are congruent regular polygons such that each vertex has the same number of incident faces.

For a given regular convex polyhedron, we will let n be the number of sides of each of its polygonal faces and μ be the number of faces around each vertex (where μ stands for monodromy). Out of our familiarity with regular polygons, we immediately know that n ≥ 3. I’ll appeal to our geometric intuition to see that μ ≥ 3 as well (polyhedra must have some actual volume, which is analogous to the bound on n for polygons).

Let v, e, and f be the numbers of vertices, edges, and faces, respectively of the polyhedron. The quantity χ = ve + f is known as the Euler characteristic, and for convex polyhedra, χ is always 2, a fact which has been proven so many times I’m not going to show it here.

We now make two observations. The first is that nf = 2e, which is obtained by noticing that each face in the polyhedron has n edges around it, but each edge has two incident faces, so nf double counts the number of edges. The second is that μv = 2e, which is obtained by noticing that each vertex has μ incident edges (separating each of the μ faces), and each edge has two incident vertices, so μv also double counts the number of edges. With these, we can write ve + f = 2 in terms of n, μ, and e as follows:

 1 n
+
 1 μ
=
 1 2
+
 1 e

I’m told that diophantine equations which are sums of reciprocals involve deep number theory, which is something I’d like to understand.

Since e is positive, this equation reduces to the following inequality:

 1 n
+
 1 μ
>
 1 2

Let’s now just do a case analysis to find all the solutions to this inequality. We’ll go through each possible value of n, starting with 3 since n ≥ 3. For the following, recall that μ ≥ 3.

Case I. n = 3
In this case, μ < 6. When n = 3 and μ = 3, this is a tetrahedron, when n = 3 and μ = 4, this is an octahedron, and when n = 3 and μ = 5, this is an icosahedron.
Case II. n = 4
In this case, μ < 4, so the only solution is n = 4 and μ = 3, which is a hexahedron (cube).
Case III. n = 5
In this case, μ < 10/3, so we have n = 5 and μ = 3, which is the dodecahedron.
Case IV. n ≥ 6

Then, since μ ≥ 3,

 1 n
+
 1 μ
≤
 1 6
+
 1 3
=
 1 2
,

so there is no solution.

We can verify that each of these solutions have corresponding integral values for v, e, and f, which we compute in the following table:

 n μ v e f name 3 3 4 6 4 tetrahedron 3 4 6 12 8 octahedron 3 5 12 30 20 icosahedron 4 3 8 12 6 hexahedron (cube) 5 3 20 30 12 dodecahedron

Therefore, we have shown that there cannot be more than five Platonic solids!

Unfortunately, we have only proven that there are at most five regular convex polyhedra, and no attempt has been made to show that each of these solutions correspond to an actual regular polyhedron, despite the suggestive images of paper models and my attempts identify the solutions with them. I’ll leave this as an exercise for the reader. Or, you can read Book XIII of the Elements (and probably most of the books leading up to it) to get constructions of each of them.

 According to a review of Polyhedron Models for the Classroom on Amazon, the book is an excerpt from the book Polyhedron Models by the same author.

 See Nineteen Proofs of Euler’s Theorem.

One method of proof which isn’t listed uses algebraic topology. First you show you can replace each face with triangles, and this does not change χ. This gives a simplicial complex which is homeomorphic to a 2-sphere. There is an exercise of Hatcher which says that alternating sums of the degrees of the homology groups of a space are constant up to homotopy equivalence. For a 2-sphere, this alternating sum is 1 − 0 + 1 = 2. Since for simplicial complexes the alternating sum is the Euler characteristic, this shows that χ = 2.

 The page Egyptian Fractions on MathWorld deals with some equations of this form. I think it also has something to do with Lie something-or-another.