# Truncated octahedron

Truncated octahedron Rank3
Notation
Bowers style acronymToe
Coxeter diagramo4x3x (     )
Conway notationtO
Stewart notationK3
Elements
Faces6 squares, 8 hexagons
Edges12+24
Vertices24
Vertex figureIsosceles triangle, edge lengths 2, 3, 3 Measures (edge length 1)
Circumradius${\frac {\sqrt {10}}{2}}\approx 1.58113$ Volume$8{\sqrt {2}}\approx 11.31371$ Dihedral angles6–4: $\arccos \left(-{\frac {\sqrt {3}}{3}}\right)\approx 125.26439^{\circ }$ 6–6: $\arccos \left(-{\frac {1}{3}}\right)\approx 109.47122^{\circ }$ Central density1
Number of external pieces14
Level of complexity3
Related polytopes
ArmyToe
RegimentToe
DualTetrakis hexahedron
ConjugateNone
Abstract & topological properties
Flag count144
Euler characteristic2
SurfaceSphere
OrientableYes
Genus0
Properties
SymmetryB3, order 48
ConvexYes
NatureTame

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

• $\left(\pm {\sqrt {2}},\,\pm {\frac {\sqrt {2}}{2}},\,0\right)$ .

## 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 (B2 axial, square-first)
• xxux3xuxx&#xt (A2 axial, hexagon-first)
• Qqo xux4ooq&#zx (B2×A1 symmetry)
• xu(wx)(wx)ux oq(oQ)(oQ)qo&#xt (K2 axial, edge-first)
• xu(xd)ux xu(dx)ux&#xt (square-first when seen as rectangle)

## Semi-uniform variant

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 $(a^{3}+6a^{2}b+12ab^{2}+5b^{3}){\frac {\sqrt {2}}{3}}$ .
• $\left(\pm (a+b){\frac {\sqrt {2}}{2}},\,\pm {\frac {b{\sqrt {2}}}{2}},\,0\right)$ .