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.
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 $0.7209$.
