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If four equal size spheres form close-packed tetrahedron, what is the fraction of space that is occupied/unoccupied by the spheres in the tetrahedron.

Similarly, if six equal size spheres form close-packed octahedron, what is the fraction of space that is occupied/unoccupied by the spheres in the octahedron.

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    $\begingroup$ This is a good question (the calculations can be unpleasant if you haven't done this sort of thing before), but not really on topic here since mathoverflow is focused on research questions. I'm going to post an answer, since I happen to know what these fractions are, but for more information or further discussion I'd recommend math.stackexchange.com. $\endgroup$
    – Henry Cohn
    Oct 3, 2013 at 13:49

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For the tetrahedron, this calculation comes up in the Rogers sphere packing bound. The answer is $\sqrt{18}(\cos^{-1}(1/3) - \pi/3)$, which is roughly $0.7796$.

For the octahedron, one can calculate the answer using the fact that the face-centered cubic packing is made up of octahedral and tetrahedral cells, with twice as many tetrahedra as octahedra. This means the face-centered cubic packing density $\pi/\sqrt{18}$ satisfies $\pi/\sqrt{18} = D_T/3 + 2D_O/3$, where $D_T$ is the tetrahedron density and $D_O$ is the octahedron density. (Note that we must weight the cells by volume, not just number of cells, and a regular octahedron has four times the volume of a regular tetrahedron with the same edge length.) Solving for $D_O$, we get $3\pi \sqrt{2}/4 - 3 \sqrt{2} \cos^{-1}(1/3)/2$, which is roughly $0.7209$.

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  • $\begingroup$ (In hindsight the argument with the face-centered cubic packing is not really simpler than computing $D_O$ directly, but it's how I happened to compute it.) $\endgroup$
    – Henry Cohn
    Oct 3, 2013 at 14:14

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