# The 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^{[1]} 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` = `v` − `e` + `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.^{[2]}

We now make two observations. The first is that `n``f` = 2`e`, which is
obtained by noticing that each face in the polyhedron has `n` edges
around it, but each edge has two incident faces, so `n``f` double
counts the number of edges. The second is that `μ``v` = 2`e`, 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 `μ``v` also double counts the number of edges. With
these, we can write `v` − `e` + `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.^{[3]}

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`≥ 6Then, 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.

^{[1]}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.

^{[2]}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.

^{[3]}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.