Proving that there are only five Platonic solids using spherical geometry

Question

In 100 Great Problems of Elementary Mathematics by Dorrie, it is proved that there are only five possible tessellations of the sphere using congruent regular (spherical) polygons: $4$ regular triangles, $6$ regular squares, $8$ regular triangles, $20$ regular triangles, and $12$ regular pentagons.

This is great, but the author then proceeds to say that if we connect all the vertices of the spherical polygons using straight lines, we obtain five regular solids: tetrahedron, octahedron, cube, icosahedron, and dodecahedron. Hence, there are only $5$ Platonic solids. $Q.E.D$

I feel rather uneasy about this claim. Am I missing something here? How do we know that the sphere tessellations necessarily correspond to the Platonic solids? How can we prove for sure that there doesn't exist a Platonic solid which can't be constructed by connecting the vertices of all the possible sphere tessellations?

Answer 1

It will take a bit of thought to realize the result.

First fact: one uses is the sum of the angles around a vertex. Projected to the circle it's obvious - its $360^\circ$, and one should realize that non-projected has to be less than that. This means that the polygons will have corners of less than $120^\circ$.

Second fact: the angle sum of a polygon is $(N-2)180^\circ$, and on a regular polygon the angles are therefore $(1-2/N)180^\circ$. This makes it impossible to have polygons with more than $5$ corners.

First and second fact puts a limit on how many polygons can meet at a vertex. For pentagons you can only meet $3$ polygons at a vertex since the vertex angle is $108^\circ$ and a fourth would exceed a sum of $360^\circ$. For squares you would have the same even though a fourth square would only result in a sum of $360^\circ$. For triangles on the other hand you can have up to $5$ joining at a vertex, for the same reason as the square you can't have $6$ even though it would only add up to $360^\circ$.

After that it's just a matter of verifying the possible candidates, and they are all possible. You have with triangles: three joining at vertices means tetrahedron, four means octahedron, and five means icosahedron. For squares and pentagons there's only one possibility and that will mean the cube and the dodecahedron.

The same reasoning can be done using spherical projection of course, but then you use the fact that the angle sum at vertices is constant $360^\circ$ while the angle sum of a polygon is larger than $(N-2)180^\circ$.

Answer 2

By symmetry, a regular polyhedron has all vertices the same distance from its centre, i.e. they all lie on a sphere known as its circumsphere. Imagine placing a wire model of such a polyhedron inside a translucent circumsphere, with a light source at the centre. The polyhedron has straight sides but their shadows on the sphere create a spherical tiling. This creates a one-to-one correspondence between regular spherical tilings and regular polyhedra: for every tiling you can join the vertices with straight wires, for every polyhedron you can project it onto its circumsphere.