4
$\begingroup$

A Platonic Solid is defined to be a convex polyhedron where all the faces are congruent and regular, and the same number of faces meet at each vertex. An Archimedean Solid drops the requirement that all the faces have to be the same, but they must still all be regular, and each vertex must have the same arrangement of faces.

However, for Archimedean Solids, the pseudorhombicuboctahedron fits this definition, despite not being vertex-transitive (meaning that the rotation group of the solid does not act transitively on the vertices).

I was wondering: For Platonic Solids, is it equivalent to define them as convex polyhedra that are face, vertex, and edge-transitive (where for Archimedean solids, we drop the face-transitivty condition)? Face-transitivity forces all the faces to be congruent, edge-transitivity forces all the faces to be equilateral, and vertex-transitivity forces the same number (or arrangement, in terms of Archimedean solids). It's not immediately obvious to me that these conditions force the faces to be equiangular as well as equilateral...does it indeed follow, or is there a counterexample?

$\endgroup$
1
  • 1
    $\begingroup$ Most Archimedean solids are not even edge transitive, they only are bound to have edges of the same size. For example consider the truncated tetrahedron: it has edges between 2 hexagons as well as edges between a triangle and a hexagon. Those 2 edge types clearly never are exchangeable by whatever symmetry! $\endgroup$ Sep 20 '21 at 15:24
1
$\begingroup$

In fact the modern definition of regular polyhedron / Platonic solid is a polyhedron that is vertex, edge, and face transitive, so that definition is equivalent to the older one.

To see that the new definition doesn't include any further polyhedra, first we observe that, if in a polyhedron each face has the same number of sides and each vertex touches the same number of edges, then the graph of the polyhedron is isomorphic to the graph of one of the five Platonic solids. This follows from Euler's formula V - E + F = 2 and some case analysis. Three of the solids are composed of triangles, and of course if a triangle has equal sides it is regular. For a polyhedron isomorphic to the cube, if the side is not a square it is a rhombus with two different angles. The total number of large angles is 12, as is the total number of small angles. Since the vertices are transitive, each vertex must have the same number of large and small angles, meaning they would have to have 1.5 each, which is impossible. The same argument applies to the dodecahedron: if the angles of the pentagonal sides are not all the same, a particular angle must occur 12n times for n = 1,2,3,4. However, that same angle must occur some multiple of 20 times when you consider the vertices, so we again have an impossibility.

So, we conclude from the above that the faces of a regular polyhedron must be regular. So the polyhedron must be one of the five Platonic solids (using the old definition).

$\endgroup$
0
$\begingroup$

Being an Archimedean solid is not the same as being both vertex- and edge-transitive. While the Archimedean solids are all vertex-transitive (they are uniform polyhedra after all), most of them are not edge-transitive.

Observe that there are only nine edge-transitive (convex) polyhedra (five of them being regular), but there are more than nine Archimedean solids.

$\endgroup$
0
$\begingroup$

The main observation here is the following:

Being vertex transitive (and considering bounded solids only) requires the vertices to lie on sphere. Faces on the other hand are bound to be planar by definition, thence those are contained within a supporting plane. The intersection of the plane and the sphere (ball) simply provides a circle (disc). Now, using equisized edges only (being an Archimedean solid), it becomes evident that the edge circuit of each polygonal face has all its vertices on that very circle, they will be evenly spaced appart, so in effect those each happen to be regular polygons only.

In fact this even would apply for their non-convex counterparts as well, just that the faces then might become non-convex regular polygrams as well. But still all corner angles thereof would be forced to the same size.

Note that the same argument also applies to the other non-flat geometry, i.e. for the hyperbolic tilings. Again the cross-section of a planar face plane and the there supporting hyperboloid defines a circle. - In contrast this argument would break down within any flat geometry (euclidean space of arbitrary dimension). So for instance there is a vertex- edge-, and face-transitive tiling, which still isn't a regular tiling of the plane: the rhombic tiling (of non-square rhombs).

--- rk

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.