# Truncated octahedron

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Truncated octahedron
Truncated octahedron

Type Archimedean solid
Uniform polyhedron
Elements F = 14, E = 36, V = 24 (χ = 2)
Faces by sides 6{4}+8{6}
Schläfli symbol t0,1{3,4}
t0,1,2{3,3}
Wythoff symbol 2 4 | 3
3 3 2 |
Coxeter-Dynkin
Symmetry Oh, [4,3], (*432)
Th, [3,3] and (*332)
Dihedral Angle
References U08, C20, W7
Properties Semiregular convex zonohedron
permutohedron

Colored faces

4.6.6
(Vertex figure)

Tetrakis hexahedron
(dual polyhedron)

Net

In geometry, the truncated octahedron is an Archimedean solid. It has 14 faces (8 regular hexagonal and 6 square), 36 edges, and 24 vertices. Since each of its faces has point symmetry the truncated octahedron is a zonohedron.

If the original truncated octahedron has unit edge length, its dual tetrakis cube has edge lengths $\frac{9}{8}\sqrt{2}$ and $\frac{3}{2}\sqrt{2}$.

## Construction

A truncated octahedron is constructed from a regular octahedron with side length 3a by the removal of six right square pyramids, one from each point. These pyramids have both base side length (a) and lateral side length (e) of a, to form equilateral triangles. The base area is then a². Note that this shape is exactly similar to half an octahedron or Johnson solid J1.

From the properties of square pyramids, we can now find the slant height, s, and the height, h of the pyramid:

$h = \sqrt{e^2-\frac{1}{2}a^2}=\frac{\sqrt{2}}{2}a$
$s = \sqrt{h^2 + \frac{1}{4}a^2} = \sqrt{\frac{1}{2}a^2 + \frac{1}{4}a^2}=\frac{\sqrt{3}}{2}a$

The volume, V, of the pyramid is given by:

$V = \frac{1}{3}a^2h = \frac{\sqrt{2}}{6}a^3$

Because six pyramids are removed by truncation, there is a total lost volume of √2 a³.

## Coordinates and permutohedron

All permutations of (0, ±1, ±2) are Cartesian coordinates of the vertices of a truncated octahedron centered at the origin. The vertices are thus also the corners of 12 rectangles whose long edges are parallel to the coordinate axes.

The truncated octahedron can also be represented by even more symmetric coordinates in four dimensions: all permutations of (1,2,3,4) form the vertices of a truncated octahedron in the three-dimensional subspace x + y + z + w = 10. Therefore, the truncated octahedron is the permutohedron of order 4.

## Area and volume

The area A and the volume V of a truncated octahedron of edge length a are:

$A = (6+12\sqrt{3}) a^2 \approx 26.7846097a^2$
$V = 8\sqrt{2} a^3 \approx 11.3137085a^3.$

## Uniform colorings

There are two uniform colorings, with tetrahedral symmetry and octahedral symmetry:

Octahedral symmetry Tetrahedral symmetry
(Omnitruncated tetrahedron)

122 coloring
Wythoff: 2 4 | 3

123 coloring
Wythoff: 3 3 2 |

## Related polyhedra

The truncated octahedron exists within the set of truncated forms between a cube and octahedron:

 Cube Truncated cube cuboctahedron Truncated octahedron Octahedron

It also exists as the omnitruncate of the tetrahedron family:

 Tetrahedron Truncated tetrahedron Rectified tetrahedron Cantellated tetrahedron Omnitruncated tetrahedron Snub tetrahedron

## Tessellations

The truncated octahedron exists in three different convex uniform honeycombs (space-filling tessellations):

The cell-transitive bitruncated cubic honeycomb can also be seen as the Voronoi tessellation of the body-centred cubic lattice. The truncated octahedron is one of five three-dimensional primary parallelohedron.

## References

• Gaiha, P., and Guha, S. K. (1977). "Adjacent vertices on a permutohedron". SIAM Journal on Applied Mathematics 32 (2): 323–327. doi:10.1137/0132025.
• Alexandrov, A. D. (1958). Convex polyhedra. Berlin : Springer, cop.. pp. 539. doi:3-540-23158-7.

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