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Truncated Octahedron

\begin{figure}\BoxedEPSF{Truncated_Octahedron_net.epsf scaled 600}\end{figure}

An Archimedean Solid, also known as the Mecon, whose Dual Polyhedron is the Tetrakis Hexahedron. It is also Uniform Polyhedron $U_8$ and has Schläfli Symbol t$\{3,4\}$ and Wythoff Symbol $2\,4\,\vert\,3$. The faces of the truncated octahedron are $8\{6\}+6\{4\}$. The truncated octahedron has the $O_h$ Octahedral Group of symmetries.


The solid can be formed from an Octahedron via Truncation by removing six Square Pyramids, each with edge slant height $a=s/3$ and height $h$, where $s$ is the side length of the original Octahedron. From the above diagram, the height and base area of the Square Pyramid are

$\displaystyle h$ $\textstyle =$ $\displaystyle \sqrt{a^2-d^2}={\textstyle{1\over 2}}\sqrt{2}\,a$ (1)
$\displaystyle A_b$ $\textstyle =$ $\displaystyle a^2.$ (2)

The Volume of the truncated octahedron is then given by the Volume of the Octahedron
V_{\rm octahedron}={\textstyle{1\over 3}}\sqrt{2}\,s^3=9\sqrt{2}\,a^3
\end{displaymath} (3)

minus six times the volume of the Square Pyramid,
V=V_{\rm octahedron}-6({\textstyle{1\over 3}}A_b h)=(9\sqrt{2}-\sqrt{2})a^3=8\sqrt{2}\,a^3.
\end{displaymath} (4)

The truncated octahedron is a Space-Filling Polyhedron. The Inradius, Midradius, and Circumradius for $a=1$ are

$\displaystyle r$ $\textstyle =$ $\displaystyle {\textstyle{9\over 20}}\sqrt{10} \approx 1.42302$ (5)
$\displaystyle \rho$ $\textstyle =$ $\displaystyle {\textstyle{3\over 2}} = 1.5$ (6)
$\displaystyle R$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\sqrt{10} \approx 1.58114.$ (7)

See also Octahedron, Square Pyramid, Truncation


Coxeter, H. S. M. Regular Polytopes, 3rd ed. New York: Dover, pp. 29-30 and 257, 1973.

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© 1996-9 Eric W. Weisstein