I would address this question from the point of view of group theory. Namely, if one starts to classify finite subgroups $G$ of $\mathrm{SO}(3)$ (say, first looking at thier poles – points where an rotation axis intersects the sphere $S^2$), then it's not difficult to prove by orbit counting that there are following possibilities for finite subgroups of $\mathrm{SO}(3)$:
- cyclic
- dihedral
- some group $\mathbf T$ of order 12 with three types of pole orbits of sizes 4, 6 and 4 (hence with stabilisers of sizes 3, 2 and 3);
- some group $\mathbf O$ of order 24 with three types of pole orbits of sizes 8, 12 and 6 (hence with stabilisers of sizes 3, 2 and 4);
- some group $\mathbf I$ of order 60 with three types of pole orbits of sizes 12, 30 and 20 (hence with stabilisers of sizes 5, 2 and 3).
Of course, the tricky point is to check that $\mathbf T$, $\mathbf O$ and $\mathbf I$ exist without appealing to the existence of Platonic solids :) This can be done in at least two ways:
- using explicit presentations of them as described, for instance, in § 57 of the book G. A. Miller, H. F. Blichfeldt, L. E. Dickson, Theory and applications of finite groups, Dover, New York, 1916; for instance, for the tetrahedron it reads $s_1^3=s_2^3 = (s_1s_2)^2 = 1$ which is easily realisable by rotations;
- or (amazingly!) using Riemann surfaces.
Now, if we take any orbit whose stabiliser has size more than 2, (meaning an orbit of size 4 for $\mathbf T$, an orbit of size 8 or 6 for $\mathbf O$, an orbit of size 12 or 20 for $\mathbf I$), then its points will define vertices of a regular polytope (whose edges can be defined as connecting a vertex with nearest vertices).
Each vertex can be rotated by $G$ to another one by construction, and the stabiliser of each vertex consists of rotations around it, which clearly have to permute edges going out of this vertex. Now, the sizes of stabilisers guarantee that there are exactly as many edges going out of each vertex as the order of the stabiliser, and therefore every edge can be rotated to every other edge by an element of $G$. As $G$ clearly preserves faces of our polyhedron, they are forced to be regular polygons. An inspection of orders of stabilisers thus gives the list of Platonic solids.
Notice that the “exceptional” orbits have stabilisers of order 2, so the above construction doesn't work there because there's not enough rotations around these vertices (however, we were only concerned with existence of Platonic solids anyway).
This is a more elaborate version of my comment below the question.
The first two observations are:
- if $P$ is edge-transitive, then all its edges are of the same length.
- if $P$ is vertex-transitive, then $P$ is inscribed, that is, all its vertices are one a common sphere.
The next observation is the following: if $P$ has both properties, then also all its faces must have both properties. This is obvious for the edge lengths. For being inscribed consider the following image:
All vertices of $P$ are on a sphere $S$. If $f$ is a face of $P$ and $E$ is the plane that contains $f$, then $S\cap E$ is a circle that contains all the vertices of $f$. In other words, $f$ is inscribed as well.
So all faces of $P$ are inscribed and have all edges of the same length.
The final observation is that the faces must then be regular polygons.
This is easiest seen via an explicit construction:
If the radius of the circle and the edge lengths are fixed, then placing a single edge in the circle inductively determines all other edges as shown in the figure. That is, the inscribed polygon with this edge length is uniquely determined. But a regular polygon has this property, and so the face must be this regular polygon.
Best Answer
The possible ways to put polygons together to form a sphere-like object are constrained by Euler's formula $V - E + F = 2$ (where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number of faces). Equivalently you can think of this as a statement about planar graphs.
Suppose we use $f_3$ triangles, $f_4$ squares, $f_5$ pentagons, etc. Every edge meets exactly two faces, and an edge of type $f_n$ meets $n$ faces, so let's double-count the number of pairs of an edge and a face next to it: on the one hand, this is $2E$, and on the other hand, this is
$$3f_3 + 4f_4 + 5f_5 + ...$$
Plugging this into Euler's formula gives $V - \frac{f_3 + 2f_4 + 3f_5 + ...}{2} = 2$. If in addition the polyhedron is convex and the polygons are regular, there are constraints on the faces that can meet at each vertex coming from the fact that the angles must sum up to less than $360^{\circ}$. (This is one way to prove the classification of Platonic solids.) For example, at most $5$ faces can meet at each vertex if we allow arbitrary faces; this means $3f_3 + 4f_4 + ... \le 5V$. (If you really want to, you can allow six triangles to touch at one point, but I would just count this as a hexagon.) If we don't allow triangles, exactly $3$ faces meet at each vertex; this means $4f_4 + 5f_5 + ... = 3V$.
Here is an application in chemistry: a fullerene is a certain type of molecule made from carbon atoms. (One of these, the buckyball, looks just like a soccer ball.) It gives a convex polyhedron in which each face is either a regular pentagon or hexagon. This gives $V - \frac{3f_5 + 4f_6}{2} = 2$ on the one hand, and $3V = 5f_5 + 6f_6$ on the other. Together these equations give $f_5 = 12$ and $V - 2f_6 = 20$; in other words, any fullerene must have exactly twelve pentagons (Twelve Pentagon theorem for fullerene).
(Hexagons are special. One way to interpret this result is that an infinite plane can be tiled with hexagons, so hexagons correspond to zero curvature, whereas since pentagons have a smaller angle at each vertex they correspond to positive curvature. What the above statement says, roughly, is that the total amount of curvature is a constant. This is a simple form of the Gauss-Bonnet theorem, which is closely related to Euler's formula.)
Here are some other things you can prove, again under the assumptions of convexity and regularity:
This is already most of the way to the classification of Platonic solids. If you're interested in learning more about Euler's formula, I highly recommend David Richeson's Euler's Gem. Extremely well-written and informative. You might also enjoy David Eppstein's Nineteen ways to prove Euler's formula.