Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra? $\textbf{(A)}\ \frac{1}{12}\qquad$ $\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad$ $\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad$ $\textbf{(D)}\ \frac{1}{6}\qquad$ $\textbf{(E)}\ \frac{\sqrt{2}}{6}$
Solution
Two vertices of the cube are adjacent iff their coordinates differ in exactly one place. A regular tetrahedron inside the cube has all six edges equal, so they are face diagonals; hence its vertices are pairwise non-adjacent, i.e.\ they are the even-parity vertices $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$ ($T_1$) or the odd-parity vertices $(1,0,0),(0,1,0),(0,0,1),(1,1,1)$ ($T_2$). These are the only two (distinct) tetrahedra.
Describe $T_1$ by its four faces. The face opposite $(0,0,0)$ lies on $x+y+z=2$; the face opposite $(1,1,0)$ passes through $(0,0,0),(1,0,1),(0,1,1)$ and lies on $x+y-z=0$, and similarly for the other two. So $T_1$ is $x+y+z\le2$, $x+y-z\ge0$, $x-y+z\ge0$, $-x+y+z\ge0$.
$T_2$ is the image of $T_1$ under $(x,y,z)\mapsto(1-x,1-y,1-z)$, so it is $x+y+z\ge1$, $x+y-z\le1$, $x-y+z\le1$, $-x+y+z\le1$.
Shift to the cube's center: $u=x-\frac12,v=y-\frac12,w=z-\frac12$. Then the conditions for $T_1\cap T_2$ become $|u+v+w|\le\frac12$, $|u+v-w|\le\frac12$, $|u-v+w|\le\frac12$, $|-u+v+w|\le\frac12$.
These cover all eight sign patterns $\pm u\pm v\pm w$ (each pattern or its negative occurs), and the largest of them is $|u|+|v|+|w|$. So the intersection is $|u|+|v|+|w|\le\frac12$, a regular octahedron with vertices at the centers of the cube's faces.
It consists of two square pyramids with common base the square of side $\frac{\sqrt2}2$ (area $\frac12$) and height $\frac12$, so its volume is $2\cdot\frac13\cdot\frac12\cdot\frac12=\frac16$. The answer is (\textbf{D}).
A rectangular solid measures $1\times1\times2$. Two distinct tetrahedra have all their vertices among the eight vertices of the solid, and in each tetrahedron no two vertices are joined by an edge of the solid. What is the volume of the union of the two tetrahedra?
$\textbf{(A)}\ \frac{1}{12}\qquad$ $\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad$ $\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad$ $\textbf{(D)}\ \frac{1}{6}\qquad$ $\textbf{(E)}\ \frac{\sqrt{2}}{6}$ 
