Why does a convex polyhedron being vertex-, edge-, and face-transitive imply that it is a Platonic solid?

Solution 1:

Let a vertex have degree $\rho$ and a face have $s$ edges and note that $$\rho V=2E=sF\tag 1$$

Let $G$ be the group of symmetries. Let $G_v,G_e\text { and }G_f$ be the stabiliser of a vertex $v$, edge $e$ and face $f$, respectively. Then $$|G|=|G_v||V|=|G_e||E|=|G_f||F|\tag 2$$ If $G_e$ has a non-identity symmetry then it is a rotation of $180^o$ about the mid-point of the edge. Comparing (1) and (2) we then have $|G_f|=s$ and the face is regular.

Otherwise $|g_e|=1$. Again comparing (1) and (2) we have $|G_v|=\frac{\rho}{2} \text { and } |G_f|=\frac{s}{2}$. Therefore $\rho\ge 4$ and $s\ge 4$ and then Euler's formula gives $$\frac{E}{2}+\frac{E}{2}\ge E+2.$$

If $G$ includes reflections

The only 'extra' case is $|G_v|=\rho, |G_e|=2, |G_f|=s$, where both $G_v$ and $G_f$ contain reflections. However, the orders of the two groups, $\rho$ and $s$, are then both even and we still have $\rho\ge 4$ and $s\ge 4$.

Solution 2:

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.