# For certain pairs of Platonic Solids, are the edge-centers and face-centers equivalent?

Notice that a Tetrahedron has 6 edges, and a Cube has 6 faces. So lets draw points at the center of those 6 edges, and at the center of the 6 faces.

If we project these points onto a unit sphere, do they turn out to be the same?

This is the chart I'm looking at. It's not easy to eyeball, but I suspect the answer is no.

I also have the same question for other pairs. E.g., the Cube has 12 edges and the Dodecahedron has 12 faces. Do their centers coincide on a unit sphere? And what about the Octahedron edges vs the Dodecahedron faces?

I think those are the only possibilities. Can't really talk about the edges of the Dodecahedron or Icosahedron because there are 30. No Platonic Solid has 30 faces.

(Note: I didn't bother with vertexes because the dual of one Platonic Solid will swap the vertexes and faces, even with the Tetrahedron despite being a self-dual.)

• The cube & octahedron are dual to one another (in relation to their face centers.) The edge centers of a cube and an octahedron give the same configuration of $12$ points that correspond to the cuboctahedron (not a dodecahedron, as you suggest) See en.wikipedia.org/wiki/Cuboctahedron Similarly the edge centers of a dodechedron or icosahedron give an Icosidodecahedron en.wikipedia.org/wiki/Icosidodecahedron Jul 8 '19 at 21:32
• You can draw a tetrahedron whose vertices are half those of the cube, so that each edge of the former is a diagonal of a face of the latter. Consequently, the tetrahedron's edge-centers are indeed arranged identically to the cube's face-centers.
– Blue
Jul 8 '19 at 21:37
• You can explore this (and other) questions using Polyhedronisme, which applies various operations to polyhedra. The "dual" operation on polyhedron $P$ creates a polyhedron with vertices at the centers of the faces of $P$; the "ambo" (aka, "truncation-to-midpoint") operation creates a polyhedron with vertices at the midpoints of edges of $P$. Observe that dual-cube and ambo-tetrahedron are both octahedra; on the other hand, dual-dodecahedron is an icosahedron, while ambo-cube is a cuboctahedron.
– Blue
Jul 8 '19 at 23:17
• But it is possible to orient the squares of the cuboctahedron in such a consistent way, that the diametral folds of those produce the overall edge structure of an icosahedron, at least within its combinatorical sense (the square diametrals clearly are longer here than the sides of the cuboctahedral triangles). Jul 9 '19 at 6:19