# Euler’s formula and the five Platonic solids

Euler’s Formula

Euler’s formula is a statement about convex polyhedra, that is a solid whose surface consists of polygons, called its faces, such that any side of a face lies on precisely one other face, and such that for any two points on the solid, the straight line connecting them lies entirely within the solid. Each convex polyhedron carries with it certain data, for example the number of faces F, the number of edges E, and the number of vertices V. Euler’s formula gives a relationship between these data, namely

V-E+F=2

for all convex polyhedra. The idea that this specific alternating sum should remain constant no matter what convex polyhedron you feed it is not at all clear. We will try to prove it by induction, the obvious question being induction on what? To answer that we first make a cute observation. We can transfer these data relating to our polyhedron into some data about a connected graph in a simple way. To see how lets look at the example of a cube. If you imagine peering through one of the faces like a window and tracing out what you see you might end up with something like this: The important observation is that this is a connected plane graph which has the same number of edges and vertices as the cube, and one fewer face (we lose the face that we were peering through). We can see that in fact this is the case for all convex polyhedra. In this way we translate the data about our polyhedra $(V,E,F)$ into data about a connected graph $(v,e,f)$ where $V=v, E=e, F=f+1$. Therefore proving Euler’s formula is equivalent to proving the very similar statement that

v-e+f=1

for any connected plane graph. We do so by induction.

Proof: If we have one edge, there is only one possible connected plane graph, and also for two edges there is also only one possibility, and in both these one can verify the desired formula, in case $e=1$ we have $v-e+f=2-1+0=1$ and in case $e=2$ we have $v-e+f=3-2+0=1$. With three edges there are multiple possible graphs, but we don’t need to worry about that. Let’s prove the inductive step: suppose that the formula holds in when there are $n$ edges. Now suppose we have a graph $\Gamma$ with $n+1$ edges, $\nu$ vertices and $\mu$ faces.

If $\Gamma$ has $\mu >0$, delete an edge from a face to obtain a new graph $\Gamma'$ with $(v,e,f) = (\nu, n, \mu-1)$. Now by our inductive hypothesis we have that, since $\Gamma'$ has n edges, it satisfies the desired formula, hence $v-e+f=\nu - n + \mu -1 = \nu - (n+1) + \mu =1$, and this is exactly  what we needed to show.

If $\Gamma$ doesn’t have $\mu =0$ then it must have an end vertex (a vertex joined only by one edge). Then $\Gamma$ has $\nu$ vertices and $n+1$ edges. If we delete an end vertex and its edge then we obtain a new graph $\Gamma''$ with $\nu-1$ vertices and n edges. Hence by the inductive hypothesis $v-e+f=(\nu-1)-n+0=\nu - (n+1) =1$, which is again just what we wanted to show. And this complets the proof of Euler’s formula. $\Box$

The Five Platonic Solids

We say a polyhedron is regular if it is made up of one kind of regular polygon such that each vertex has the same number of edges. This allows us to define the face degree $p$, which is the number of sides each face has, and the vertex degree $q$ which is the number of edges meeting at each vertex. So we have the following data associated to regular polyhedron, $V, E, F, p, q$. We can now use Euler’s formula to prove the remarkable result that there are only 5 regular polyhedra, the so called Platonic solids. They are the tetrahedron, cube, octahedron, dodecahedron and icosahedron. We record the data in a table below: Theorem: These are the only regular polyhedra.

Proof: First observe the following two relations, $pF=2E$ $qV = 2E$

First, if we count all the edges that a face has, and do this for each face, we will count each edge exactly twice. Similarly for the second relation, if we count each edge meeting at a vertex, and do so for all vertices we will count each edge twice. We can now substitute $V = \frac{2E}{q}$ and $F=\frac{2E}{p}$ into Euler’s formula $V-E+F=2$ to obtain $\frac{2E}{q}-E+\frac{2E}{p} = 2$ which becomes $\frac{1}{q} + \frac{1}{p} = \frac{1}{2}+\frac{1}{E}$

From this equation we can deduce that the only possibilities for the pair $(p,q) \in \mathbb{N}^2$ are $(3,3), (3,4), (4,3), (3,5)$ and $(5,3)$. It is geometrically clear that these 5 pairs lead to the 5 Platonic solids listed above, and to no others! $\Box$

Dual Polyhedra

We can describe the Platonic solids by their coordinates:

Tetrahedron: (1,1,1), (1,-1,-1), (-1,-1,1), (-1,1,-1) Cube: (1,1,1), (1,1,-1), (-1,1,1), (1,-1,1), (1, -1, -1), (-1, 1, -1), (-1,-1,1), (-1,-1,-1) Octahedron: (1,0,0), (0,0,1), (0,1,0), (-1,0,0), (0,-1,0), (0,0,-1) Dodecahedron: (0 $\pm\phi^{-1}$ $\pm\phi$), ( $\pm\phi^{-1}$ $\pm\phi$,0), ( $\pm\phi$, 0, $\pm\phi^{-1}$), ( $\pm1$, $\pm1$, $\pm1$) Icosahedron: (1,0, $\phi$), (1,0,- $\phi$), (-1,0, $\phi$), (-1,0,- $\phi$), (0, $\phi$,1), (0, $\phi$,-1), (0,- $\phi$,1), (0, – $\phi$,-1), ( $\phi$,1,0), ( $\phi$,-1,0), (- $\phi$,1,0), (- $\phi$,-1,0) Given a Platonic solid, putting a vertex at the midpoint of each face gives the vertices of the dual polyhedron. It should not be too much of a leap to believe that this dual is itself a Platonic solid. Well, we have just classified the Platonic solids, and there are only five of them, so taking the dual of one doesn’t give a ‘new’ shape, rather it will give one of the five we already have. Given the data (V,E,F, p, q) of a Platonic solid, say P, what can we say about the associated data of its dual, P’? Well, each face in P gives a vertex in P’, by definition of the dual construction. Again, it is geometrically clear that each vertex corresponds to a face in the dual, so we get $F \leftrightarrow V$, we get that the number of edges remains the same, and that $p \leftrightarrow q$. This implies that (P’)’=P. Indeed looking at the table above we get the following dual pairs: cube $\leftrightarrow$ octahedron, dodecahedron $\leftrightarrow$ icosahedron, tetrahedron $\leftrightarrow$ tetrahedron. It can be fun to picture a dual P’ sitting inside the original shape P, and then sitting its dual P”=P sitting inside P’, and so on, getting smaller and smaller indefinitely! Having this image in mind one see that dual shapes share the same symmetry groups.