A regular tetrahedron has edge length $6$. A regular octahedron has its six vertices at the midpoints of the six edges of the tetrahedron. What is the volume of the octahedron?
Cutting the tetrahedron through the midpoints of the edges at each vertex removes four small tetrahedra with edge $3$ and leaves exactly this octahedron.
A regular tetrahedron of edge $a$ has volume $\frac{a^3\sqrt2}{12}$: the large one has $\frac{216\sqrt2}{12}=18\sqrt2$, each small one has $\frac{27\sqrt2}{12}=\frac{9\sqrt2}4$.
The octahedron's volume is $18\sqrt2-4\cdot\frac{9\sqrt2}4=\boxed{9\sqrt2}$.
