A regular tetrahedron has two vertices on the body diagonal of a cube with side length 12 . The other two vertices lie on one of the face diagonals not intersecting that body diagonal. Find the side length of the tetrahedron.
Solution: Let $A B C D$ be a tetrahedron of side $s$. We want to find the distance between two of its opposite sides. Let $E$ be the midpoint of $A D, F$ the midpoint of $B C$. Then $A E=s / 2$, $A F=s \sqrt{3} / 2$, and angle $A E F=90^{\circ}$. So the distance between the two opposite sides is $E F=$ $\sqrt{A F^{2}-A E^{2}}=\sqrt{3 s^{2} / 4-s^{2} / 4}=s / \sqrt{2}$.

Now we find the distance between a body diagonal and a face diagonal of a cube of side $a$. Let $O$ be the center of the cube and $P$ be the midpoint of the face diagonal. Then the plane containing $P$ and the body diagonal is perpendicular to the face diagonal. So the distance between the body and face diagonals is the distance between $P$ and the body diagonal, which is $\frac{a}{2} \sqrt{\frac{2}{3}}$ (the altitude from $P$ of right triangle $O P Q$, where $Q$ is the appropriate vertex of the cube). So now $\frac{s}{\sqrt{2}}=\frac{a}{2} \sqrt{\frac{2}{3}}$, thus $s=a / \sqrt{3}=12 / \sqrt{3}=4 \sqrt{3}$.