Two Special Archimedeans

If you divide the edges of the Cube or Octahedron in half and truncate, something very interesting happens: the resulting solid is the same in either case! Likewise, when the Dodecahedron and Icosahedron are truncated in this way the same solid results.

These solids strongly indicate the special relationship that exists between the Cube and Octahedron (and between the Dodecahedron and the Icosahedron). They are called a "dual-pair"; the Cube and Octahedron have the same number of edges, and the number of faces of one is the same as the number of vertices of the other (the same is true for the Dodecahedron and the Icosahedron).

This duality is even more clearly seen in the solid which is a compound of a Cube and Octahedron, and the solid which is a compound of a Dodecahedron and Icosahedron. Note that every vertex of one solid is above a face of the other, so the number of vertices of one must be equal to the number of faces of the other. Also, every edge of one solid is crossed by an edge of the other, so the number of edges in each must be the same.

Cuboctahedron (3,4,3,4)

Icosidodecahedron (3,5,3,5)

Here is a good thought exercise for you: what happens when you truncate the Tetrahedron in this way? That is, what solid results if you divide the edges of a Tetrahedron in half, and then cut off the vertices at these points? (Hint: it is the dual of the Tetrahedron)

Truncated Platonic Solids
The Rhombi-Archimedeans

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I welcome your comments & questions!