# Archimedean and Catalan solids: picking favorites

All five Platonic solids are perfect in every way; I cannot justify putting any one of them above others. Tetrahedron has the virtue of being self-dual: taking the centers of its faces as new vertices, we get another tetrahedron. It’s also the simplest of them all, but perhaps too simple.

Archimedean solids are another matter. Their isometry groups act transitively on vertices but not on faces. Which means the faces can vary in size and shape, though they are all regular polygons. Out of 13 Archimedean solids two stand out on the ground of being 50-50 interpolants between dual Platonic solids (and, as a result, being edge-transitive in addition to vertex-transitive).

Cuboctahedron is a 50-50 hybrid of Cube and Octahedron. Take either of two solids and truncate all vertices just enough for the new faces to meet at new vertices. Alternatively, takes the intersection of cube and octahedron, appropriately scaled. In coordinates $(x_1,x_2,x_3)$ a cuboctahedron is determined by inequalities $\forall i \ |x_i|\le 1$, $\sum |x_i|\le 2$. I’m going to take cuboctahedron as my Archimedean solid #1. It combines 8 triangles, 6 squares, and 4 hexagons (central section) into a shape of remarkable simplicity.

Similarly, Icosidodecahedron is the 50-50 hybrid of Icosahedron and Dodecahedron. Theoretically, it should be just as nice as cuboctahedron. Yet, I don’t find it nearly as appealing…

It’s either the disparity of face sizes, or of their number of sides (3 vs 5). Either way, Icosidodecahedron loses #2 spot to Truncated Icosahedron aka soccer ball (old design) aka buckyball:

This is also a hybrid, but it’s 2/3 of icosahedron and 1/3 of dodecahedron. The vertices of Icosahedron are truncated just enough so that triangular faces become regular hexagon: that is, 1/3 of sidelength is cut. The resulting combination of nearly equal hexagons and pentagons is both aesthetically pleasing and mathematically natural. Think of the standard hexagonal grid on the plane being forced to take spherical shape: some hexagons will become pentagons, allowing the surface to curve. Hence the structure of $C_{60}$.

The other asymmetric hybrid, 2/3 of dodecahedron and 1/3 of icosahedron, is so ugly that I don’t even want to show it here.

Unlike Platonic solids, the duals of Archimedean solids do not belong to the same class. Indeed, they form a dual class of 13 polyhedra known as Catalan solids. Given the perfect duality of the two classes, the gap between the times of Archimedes (-287 – -212) and Catalan (1814 – 1894) may appear surprising. The probable reason for neglect of Catalan solids is that their faces are not regular polygons. Yet, they beat their Archimedean counterparts in “real-world application” category: since their isometry group is face-transitive, they are equally likely to land on any face and thus can be used as game dice.

This is the dual of cuboctahedron… unremarkable, in my opinion. It does get an honorable mention for being space-tiling.

Having picked truncated icosahedron over icosidodecahedron, I should also prefer its dual, Pentakis Dodecahedron, to Rhombic Triacontahedron. But it seems that duality does not quite work this way: in my opinion, the latter solid looks better.

It’s the shape of these triangles (almost regular but not quite) that makes the pentakis dodecahedron look like some kind of a fake, a wannabe Platonic solid. In contrast, the rhombic triacontahedron does not pretend to be anything it’s not:

If you are going to be a rhombus, let your ratio of diagonals be the golden ratio. This is exactly what we see in the rhombic triacontahedron. But what clinches its #1 place among Catalan solid is the following. There are five cubes that can be inscribed into dodecahedron using some 8-subsets of the 20 vertices: [video]. The intersection of these five cubes is a rhombic triacontahedron: each of the 30 faces of the latter is a part of a face of some cube ($30=5\cdot 6$).