where can I find a proof that truncated octahedron tiles Euclidean 3D space?
Truncated octahedron tiles 3D space. Proof?
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$\begingroup$
geometry
tiling
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2You mean references for the [bitruncated cubic honeycomb](http://en.wikipedia.org/wiki/Bitruncated_cubic_honeycomb)? Sadly, [there doesn't seem to be an online version of Grünbaum's paper](http://www.uccs.edu/~geombina/1994.html#2)... – 2010-12-04
1 Answers
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To tile space, center them at lattice points and at the centers of lattice cubes (i.e. lattice points plus $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$). The eight faces of the octahedron face the neighboring 8 vertices of the "other" lattice (the one relatively shifted by $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$), while the six truncated corners face the six neighbor vertices on the "same" lattice.
Most sources (often crystallographic) consider this to be obvious enough that it needs no proof.