Platonic Solids

Question

It´s a theorem that there exist only five platonic solids ( up to similarity). I was searching some proofs of this, but I could not. I want to see some proof of this, specially one that uses principally group theory.

Here´s the definition of Platonic solid Wikipedia Platonic solids

Answer 1

From Regular Polytopes by Coxeter, let a spherical graph have $N_0$ vertices, $N_1$ edges, and $N_2$ faces. Euler's formula reads $$ N_0 - N_1 + N_2 = 2. \; \; \; \; \; (1.61) $$ Now, suppose our graph is regular, each face has $p$ sides, each vertex has $q$ surrounding faces. Then both $p,q \geq 3.$ Next, $$ q N_0 = 2 N_1 = p N_2 . \; \; \; \; \; (1.71) $$ Put them together, $$ \frac{1}{N_1} = \frac{1}{p} + \frac{1}{q} - \frac{1}{2}. \; \; \; \; \; (1.72) $$ As $N_1$ is positive, and $p,q \geq 3,$ the possible solutions to $$ \frac{1}{p} + \frac{1}{q} > \frac{1}{2} $$ are $$ \{p,q\} =\{3,3\}, \; \; \{3,4\}, \; \; \{4,3\}, \; \; \{3,5\}, \; \; \{5,3\}. $$
Note that the spherical dual graph to $\{p,q\}$ is $\{q,p\},$ while the tetrahedron $\{3,3\}$ is self dual.

Answer 2

The proof I know doesn't use group theory. At each vertex, there must be the same number $\ge 3$ of the same regular polygon meeting, and the angles must add to less than $180^{\circ}$. The only possibilities are three, four, or five triangles, three squares, or three pentagons. Then observe that each of these generates only one solid by Euler's formula for the number of faces and regularity forcing where each face goes.

Answer 3

Here's a group theoretic proof that one can use plus some euclidean geometry.

Let $X$ be a regular platonic solid. Now I believe you can let a finite subgroup of $SO_3$ act on $X$ by acting on the faces, edges or vertices. Now to each face you can draw a line perpendicular to that face, and let $SO_3$ act on that line instead. We will call such a line a pole. Similarly one can let $SO_3$ act on the poles above an edge and a vertex. So now we have that $SO_3$ is acting on the set of poles associated to a face,edge or vertex. If $p$ is a pole above a vertex, let

$$|G_p| = r_p = \text{number of faces that meet at a vertex}.$$

If $p'$ is a pole above an edge, let

$$|G_{p'}| = r_{p'} = \text{number of faces that meet at an edge} = 2.$$

Finally if $p''$ is a pole above a face, let

$$|G_{p''}| = r_{p''} = \text{number of sides a face has} = n.$$

Now it is not hard to show (I can provide a proof of this) that

$$\sum_{\text{over all poles $p,p'$ or $p''$}}(r_p - 1) = 2|G| - 2 $$

where $G$ is the group of rotational symmetries of $X$. In fact the proof of the formula above is group theoretic: One looks at the order of group elements and orders of stabilisers.

Now at the same time we know that the number of poles is equal to $V + F + E$, where $V$ is the number of vertices, $F$ the number of faces and $E$ the number of edges. If you use this information and plug it into our formula above, we have that

$$kV + nF + 2E - (V + F + E) = 2|G| - 2.$$

At the same time, the Orbit - Stabiliser Theorem gives us that $kV = nF = 2E = |G|$. Hence

$$\begin{eqnarray} 3nF - V - F - E &=& 2|G| - 2 \\ \\ \implies \frac{3nF}{|G|} - \frac{1}{k} - \frac{1}{n} - \frac{1}{2} &=& 2 - \frac{2}{|G|}. \end{eqnarray}$$

However $3nF/|G| = 3$, so that upon simplifying we have

$$\begin{eqnarray} \frac{1}{2} + \frac{2}{|G|} &=& \frac{1}{k} + \frac{1}{n} \\ \implies \frac{1}{k} + \frac{1}{n} &>& 2. \end{eqnarray}$$

The task now is reduced to finding integers that satisfy that inequality above. Since $k,n \geq 3$, the only possible integer solutions are $k=3, n=3$ or $k=3,n=4$ or $k=3, n=5$ or $k=4,n=3$ or $k=5,n=3.$

In the first case for example, we have a regular polyhedron made out of an equilateral triangle, with 3 faces (made out of equilateral triangles) meeting at a vertex. The tetrahedron does satisfy these requirements, but as Mariano has noted above it remains to check that the tetrahedron is the only one that satisfies this. Similarly one has to check that the other 4 cases only give the octahedron, cube, dodecahedron and icosahedron. I leave this to you to check!

$\textbf{Edit:}$ Mariano has told me that the proof that there is a finite subgroup of $SO_3$ acting on $X$ is not trivial.

Answer 4

This is another proof which is not based on group theory. You may see it as an expanded version of Ross' answer, so read this if you need more details on one of the steps there.

Suppose you have a patonic solid where each face is a regular $n$-gon and at each corner $m$ faces meet. Then you have several constraints:

1. $m,n\in\mathbb N$ (since they obviously have to be non-gnegative integers)
2. $n>2$ (to have at least triangles not line segments)
3. $m>2$ (since for $m=2$ the “solid” would be a pair of polygons glued back to back)
4. $m\cdot\frac{n-2}{n}\cdot180°<360°$, which means the inner angles at a corner add up to less than one whole turn. You need this to ensure that if you place the faces next to one another, they form a convex cone instead of a flat or even overlapping (or hyperbolic) surface. Laying out the faces in the plane leaves a gap, and closing that gap produces the cone.
   This condition can be rewritten to $m<\frac{2n}{n-2}$.

For $n=6$ this last formula would result in $m<\frac{12}4=3$ which together with $m>2$ leaves no valid integer solutions. For $n>6$ this only gets worse. So you can conclude that $n<6$. Then you can start enumerating things:

• $n=3\implies2<m<6\implies m\in\{3,4,5\}$ (tetrahedron, octahedron, icosahedron)
• $n=4\implies2<m<4\implies m=3$ (cube)
• $n=5\implies2<m<\tfrac{10}3\implies m=3$ (dodecahedron)

So there are only very few possible combinations of $m$ and $n$ which satisfy all of the given constraints. Each of them can indeed be realized as a platonic solid, demonstrating that these neccessary conditions are also sufficient. Furthermore, each solid is unique up to similarity, since $m$ and $n$ tell you exactly how to combine the faces according to the rules (in terms of convexity and symmetry) of a platonic solid.