Truncated cube, cuboctahedron and regular tetrahedron

Problem

A cube $ABCDEFGH$ has diagonal $AG=6$. a) Find the edge of the cube. b) All $8$ vertices are cut off by joining the midpoints of the edges meeting at each vertex: find the volume of the remaining solid. c) Alternatively, vertices $A,C,F,H$ are cut along the face diagonals: prove that a regular tetrahedron is obtained and find its volume.

Solution

a) The cube’s diagonal is $a\sqrt3$, so $a=\dfrac{6}{\sqrt3}=2\sqrt3\approx3{,}464$. b) Each cut removes a corner tetrahedron with three perpendicular edges of length $\dfrac{a}{2}=\sqrt3$, of volume $\dfrac{1}{6}(\sqrt3)^{3}=\dfrac{\sqrt3}{2}$. With $8$ vertices we remove $4\sqrt3$; since $V_{\text{cube}}=a^{3}=24\sqrt3$, there remains $24\sqrt3-4\sqrt3=20\sqrt3\approx34{,}64$ (a cuboctahedron). c) The vertices $B,D,E,G$ form a tetrahedron whose edge equals the face diagonal $a\sqrt2=2\sqrt6$; it is regular and has volume $\dfrac{V_{\text{cube}}}{3}=8\sqrt3\approx13{,}86$. $\boxed{\,a=2\sqrt3\approx3{,}46,\quad V_{\text{cubott}}=20\sqrt3\approx34{,}64,\quad V_{\text{tetr}}=8\sqrt3\approx13{,}86\,}$