A polyhedron is a solid which has a surface that consists of a number of polygonal faces. A polygon is a planar figure that is bounded by a closed path consisting of a finite sequence of straight line segments. For example, a cube or a tetrahedron is a polyhedron, while a triangle or a rectangle is a polygon.

Now, the *Euler characteristic* for polyhedra is a function (more technically, a topological invariant) defined by

,

where and are the number of vertices, edges and faces, respectively, in a given polyhedron.

It turns out that for a simply connected polyhedron, . Therefore,

is *Euler’s formula* for a simply connected polyhedron. Prof David Eppstein has a webpage that lists no less than nineteen proofs of the above formula.

Okay, so what has Euler’s formula for polyhedra got to do with the Platonic solids? Well, *using* this formula we can show/prove that there are exactly **five** Platonic solids.

The proof below will essentially employ the common technique (in combinatorics) of counting some objects in two different ways and equating the answers in order to obtain some useful information.

So, suppose a regular polyhedron has vertices, edges and faces, each of which is an -sided regular polygon. Also, suppose edges meet at each vertex.

Let the “degree of each face” be equal to the the number of edges that belong to it. Also, let the “degree of each vertex” be equal to the number of edges meeting it.

Now, since each face has edges surrounding it, the sum of the degrees of all the faces is . But, since each edge belongs to exactly two faces, each edge is counted *twice *in the expression . Therefore, we have

Again, since each vertex has edges meeting it, the sum of the degrees of all the vertices is . But, since each edge meets exactly two vertices, each edge is ,again, counted *twice* in the expression . Therefore, we have

Now, using and in Euler’s formula, we eliminate and to obtain

,

which yields

Further, we note that a polygon must have *at least* three sides (else it’s not really a polygon!); therefore, In addition, it is easy to picture that *at least* three edges must meet a vertex (else a polyhedral angle at each vertex wouldn’t exist!); therefore, . But, note that and *both* cannot be greater than 3, for otherwise, the left hand side of equation cannot exceed whereas the right hand side of the equation exceeds .

So, finally, we have two cases to consider.

. Plugging this value of into equation , we find that the possible values of are or . This yields or , which correspond to the tetrahedron, cube and dodecahedron, respectively.

. Plugging this value of into equation , we find that the possible values of are again or . This yields or , which correspond to the tetrahedron, octahedron and icosahedron, respectively.

Hence, there exist no more than **five** platonic solids, *viz.* tetrahedron, cube, octahedron, dodecahedron and icosahedron. For each of those solids, the corresponding “(vertices, edges, faces)” triples are and , respectively.

I stole the picture below from here.

And this concludes our proof.

## 21 comments

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March 1, 2008 at 3:04 pm

John ArmstrongYou might want to be very careful around the Euler characteristic. Have you read Lakatos’

Proofs and Refutations? It pretty neatly dissects some of these terms, and some of those “proofs” to figure out what’s really being proved.March 1, 2008 at 4:40 pm

VishalWell, indeed, I read the first few pages of

Proof and Refutationsbefore sorta “losing” interest in it. Maybe I should get a copy from the library again! Thanks very much for the feedback.September 6, 2010 at 4:47 pm

Raul TorresI’m an amateur on this, and I have an idea without Euler´s Formula

We know an N-sided regular polygon, it has an internal angle: A = (1-2/N) x 180

Just let see a vertex of any regular polyhedron, on it, we can put together F faces, but F x A must be less than 360, otherwise we could not cut this shape on one single piece of paper.

Then F 3, and N > 3 , both integers, then.

If N = 3; F = 3, 4, 5

If N = 4; F = 3

If N = 5; F = 3

There are no more solutions.

¿what is your opinion about this?

If N = 5; F = 3

There are no more solutions.

¿What is your opinion about this?

September 6, 2010 at 4:55 pm

Raul TorresWith F > 3, and N > 3 , both integers, then.

If N = 3; F = 3, 4, 5

If N = 4; F = 3

If N = 5; F = 3

There are no more solutions.

¿What is your opinion about this?

September 6, 2010 at 9:28 pm

Raul Torres(I’m sorry for the inconvenient)

I’m an amateur on this, and I have an idea without Euler´s Formula

We know an N-sided regular polygon, it has an internal angle: A = (1-2/N) x 180

Just let see a vertex of any regular polyhedron, on it, we can put together F faces, but F x A must be less than 360, otherwise we could not cut this shape on one single piece of paper.

Then F 2, and N > 2 , both integers, then.

If N = 3; F = 3, 4, 5

If N = 4; F = 3

If N = 5; F = 3

There are no more solutions.

¿What is your opinion about this?

September 6, 2010 at 9:30 pm

Raul TorresOh my good…….

September 6, 2010 at 9:31 pm

Raul TorresI’m an amateur on this, and I have an idea without Euler´s Formula

We know an N-sided regular polygon, it has an internal angle: A = (1-2/N) x 180

Just let see a vertex of any regular polyhedron, on it, we can put together F faces, but F x A must be less than 360, otherwise we could not cut this shape on one single piece of paper.

September 6, 2010 at 9:31 pm

Raul TorresThen F < 2xN / (N-2)

September 6, 2010 at 9:32 pm

Raul TorresWith F > 2, and N > 2 , both integers, then.

If N = 3; F = 3, 4, 5

If N = 4; F = 3

If N = 5; F = 3

There are no more solutions.

¿What is your opinion about this?

October 25, 2010 at 2:01 pm

ashishI need atleast 10 solids satisfying euler’s formula.(with their names)

January 2, 2011 at 6:28 am

2010 in review « Todd and Vishal’s blog[…] Platonic Solids and Euler’s Formula for Polyhedra March 2008 10 comments 3 […]

December 4, 2011 at 9:46 pm

Lily DavisCan you say that in a simpler form? 🙂

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