Truncated octahedron

The truncated octahedron or toe is one of the 13 Archimedean solids. It consists of 6 squares and 8 ditrigons. Each vertex joins one square and two hexagons. As the name suggests, it can be obtained by the truncation of the octahedron. It is also the omnitruncate of the tetrahedral family.

It is the only Archimedean solid that can tile 3D space by itself. This results in the bitruncated cubic honeycomb.

It can be alternated into the icosahedron after all edge lengths are made equal.

It is the 4th-order permutohedron.

Vertex coordinates
A truncated octahedron of edge length 1 has vertex coordinates given by all permutations of
 * $$\left(±\sqrt2,\,±\frac{\sqrt2}{2},\,0\right).$$

Representations
A truncated octahedron has the following Coxeter diagrams:


 * o4x3x (full symmetry)
 * x3x3x (A3 symmetry, as great rhombitetratetrahedron)
 * s4x3x (as hexagon-alternated great rhombicuboctahedron)
 * xuxux4ooqoo&#xt (BC2 axial, square-first)
 * xxux3xuxx&#xt (A2 axial, hexagon-first)
 * Qqo xux4ooq&#zx (BC2×A1 symmetry)
 * xu(wx)(wx)ux oq(oQ)(oQ)qo&#xt (A1×A1 axial, edge-first)
 * xu(xd)ux xu(dx)ux&#xt (square-first when seen as rectangle)

Semi-uniform variant


The truncated octahedron has a semi-uniform variant of the form o4y3x that maintains its full symmetry. This variant has 6 squares of size y and 8 ditrigons as faces.

With edges of length a (between two ditrigons) and b (between a ditrigon and a square), its circumradius is given by $$\frac{\sqrt{2a^2+4b^2+4ab}}{2}$$ and its volume is given by $$(a^3+6a^2b+12ab^2+5b^3)\frac{\sqrt2}{3}$$.

Generally, alternating these polyhedra gives a pyritohedral icosahedron.

It has coordinates given by all permutations of:


 * $$\left(±(a+b)\frac{\sqrt2}{2},\,±b\frac{\sqrt2}{2},\,0\right).$$