Truncated octahedron

As suggested by its name, a truncated octahedron is constructed from a regular octahedron by cutting off all vertices. This resulting polyhedron has six squares and eight hexagons, leaving out six square pyramids.

The Cartesian coordinates of the vertices of a truncated octahedron with edge length 1 are all permutations of^[2]$${\displaystyle {\bigl (}\pm {\sqrt {2}},\pm {\tfrac {\sqrt {2}}{2}},0{\bigr )}.}$$The truncated octahedron is the result of the regular octahedron's edges being expanded. The edges are separated and pushed away in the direction of two-fold rotational symmetry axes passing through the midpoint of opposite edges in the octahedral symmetry. With the distance between the closest parallel edges that is the regular octahedron's edge-length, connecting the endpoints of each edge to form hexagons and triangles, the highlight shows that the third Johnson solid, the triangular cupola, coincides with the hexagonal faces.^[3]

Loeb (1986) constructs a truncated octahedron by attaching eight cubes to each face.^[4] In this hinge, eight segments on each half-cube form hexagons, and the hexagons bisect the cubes. It thus can fold inward and outward, forming a truncated octahedron and a cube, respectively.^[5]

Setting the edge length of the regular octahedron equal to ${\displaystyle 3a}$, it follows that the length of each edge of a square pyramid (to be removed) is ${\displaystyle a}$ (the square pyramid has four equilateral triangles as faces, the first Johnson solid). The volume of each of these equilateral square pyramids is ${\textstyle {\tfrac {\sqrt {2}}{6}}a^{3}}$. Because six of them are removed by truncation, the volume ${\displaystyle V}$ of the truncated octahedron is given by^[6]$${\displaystyle V={\frac {\sqrt {2}}{3}}(3a)^{3}-6\cdot {\frac {\sqrt {2}}{6}}a^{3}=8{\sqrt {2}}a^{3}\approx 11.3137a^{3}.}$$The surface area ${\displaystyle A}$ of this truncated octahedron can be obtained by summing all polygons' area, six squares and eight hexagons:^[6]$${\displaystyle A=(6+12{\sqrt {3}})a^{2}\approx 26.7846a^{2}.}$$

The truncated octahedron is one of the thirteen Archimedean solids. In other words, it is a highly symmetric, semi-regular polyhedron with two or more different regular polygonal faces meeting at a vertex.^[7] The dual polyhedron of a truncated octahedron is the tetrakis hexahedron. They both have the same three-dimensional symmetry group as the regular octahedron does, the octahedral symmetry ${\displaystyle \mathrm {O} _{\mathrm {h} }}$.^[8] Each vertex is surrounded by a square and two hexagons, so its vertex figure is ${\displaystyle 4\cdot 6^{2}}$.^[9]

A truncated octahedron has two different dihedral angles, angles between two polygonal faces. The angle between square and hexagonal faces is ${\textstyle \arccos(-1/{\sqrt {3}})\approx 125.26^{\circ }}$, and the angle between two adjacent hexagonal faces is ${\textstyle \arccos(-1/3)\approx 109.47^{\circ }}$.^[10]

Fourteen-sided Chinese dice

Similar shape, octahedron truncated down to the inscribed sphere

The first Brillouin zone of the FCC lattice, showing symmetry labels for high symmetry lines and points.

The structure of the faujasite framework

The truncated octahedron has been used as a fourteen-sided die, dating back to China in Warring States period, although a cube was also used as well.^[18]

The truncated octahedron appears in the structure in the framework of a faujasite-type of zeolite crystals. The structure consists of sodalite cages that resemble truncated octahedra, connecting each other by hexagonal prisms^[19]

In solid-state physics, the first Brillouin zone of the face-centered cubic lattice is a truncated octahedron.^[20]

The shape of a truncated octahedron appears as the Wigner–Seitz cell of a sodium. The background for this discovery dates back to Eugene Wigner and Frederick Seitz's proposal on the Wigner–Seitz cell's application to condensed matter physics on solving the Schrödinger equation for free electrons in sodium.^[21][22]

The truncated octahedron (in fact, the generalized truncated octahedron) appears in the error analysis of quantization index modulation (QIM) in conjunction with repetition coding.^[23]

Second and third genus toroids

The truncated octahedron can be dissected into a central octahedron, surrounded by 8 triangular cupolae on each face, and 6 square pyramids above the vertices.^[24] Removing the central octahedron and 2 or 4 triangular cupolae creates two Stewart toroids, with dihedral and tetrahedral symmetry.

It is possible to slice a tesseract by a hyperplane so that its sliced cross-section is a truncated octahedron.^[25]

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

Jungle gym nets often include truncated octahedra.

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  sculpture in Bonn
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  Rubik's Cube variant
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  model made with Polydron construction set
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  Pyrite crystal
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  Boleite crystal

Quick facts Vertices, Edges ...

Truncated octahedral graph
3-fold symmetric Schlegel diagram
Vertices	24
Edges	36
Automorphisms	48
Chromatic number	2
Book thickness	3
Queue number	2
Properties	Cubic, Hamiltonian, regular, zero-symmetric
Table of graphs and parameters

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In the mathematical field of graph theory, a truncated octahedral graph is the graph of vertices and edges of the truncated octahedron. It has 24 vertices and 36 edges, and is a cubic Archimedean graph.^[26] It has book thickness 3 and queue number 2.^[27]

As a Hamiltonian cubic graph, it can be represented by LCF notation in multiple ways: [3, −7, 7, −3]⁶, [5, −11, 11, 7, 5, −5, −7, −11, 11, −5, −7, 7]², and [−11, 5, −3, −7, −9, 3, −5, 5, −3, 9, 7, 3, −5, 11, −3, 7, 5, −7, −9, 9, 7, −5, −7, 3].^[28]

More information LCF [3, −7, 7, −3]6, LCF [5, −7, 7, −5]6 ...

LCF [3, −7, 7, −3]6	LCF [5, −7, 7, −5]6	Configuration
		\ v1 v2 e1 e2 e3 v1 12 * 1 0 2 v2 * 12 1 2 0 e1 1 1 12 * * e2 0 2 * 12 * e3 2 0 * * 12

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• DNA nanotechnology
• Skew apeirohedron