Talk:Tetrakis hexahedron

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Polyhedra

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Dual polyhedron[edit]

I think the dual should be truncated octahedron rather than truncated cube.

Agreed, I corrected it. Tom Ruen 19:59, 13 June 2006 (UTC)[reply]

Verification[edit]

This section was removed until verified. SockPuppetForTomruen (talk) 02:51, 4 December 2009 (UTC)[reply]

Geometric proportions[edit]

The tetrakis hexahedron can be seen to contain 2 types of edges, the 12 edges of the cube, and 24 lateral edges connecting the cube to the apex of the augmented square pyramids on each face.

If its latereral edge lengths are $a$ , and the cubic edges are length 1, its surface area is $\scriptstyle {\frac {16}{3}}{\sqrt {5}}a^{2}$ and its volume is $\scriptstyle {\frac {32}{9}}a^{3}$ .

In order for the tetrakis hexahedron to be convex, it must be the case that $\scriptstyle a<{\frac {\sqrt {3}}{2}}$ ; when $\scriptstyle a={\frac {\sqrt {3}}{2}}$ , pairs of triangular faces become coplanar and the polyhedron degenerates to a rhombic dodecahedron.

Why do I sometimes see "tetrakishexahedron" (no space)? 4 T C 14:31, 29 January 2010 (UTC)[reply]