2011 AMC 10A Problem 24

Label the cube's vertices by coordinates $0$ or $1$. A regular tetrahedron on these vertices must use four vertices that are pairwise joined by face diagonals (length $\sqrt2$), so its vertices are the four with an even number of $1$s, or the four with an odd number. Call these $T_1 = \{(0,0,0), (1,1,0), (1,0,1), (0,1,1)\}$ and $T_2$ its complement.

Volume of one tetrahedron. $T_1$ is the cube with four corners sliced off at the vertices of $T_2$. The corner at $(1,0,0)$, for instance, is the pyramid with three mutually perpendicular unit edges, volume $\frac{1}{6}$. So
$$[T_1] = 1 - 4 \cdot \frac{1}{6} = \frac{1}{3}.$$

How $T_2$ cuts $T_1$. On every face of the cube, one diagonal is an edge of $T_1$ and the other is an edge of $T_2$; they cross at the face center. Take the face of $T_2$ through $(1,0,0), (0,1,0), (0,0,1)$, the plane $x + y + z = 1$. It contains $(\tfrac12, \tfrac12, 0)$, $(\tfrac12, 0, \tfrac12)$, $(0, \tfrac12, \tfrac12)$, the midpoints of the three edges of $T_1$ that meet at $(0,0,0)$. Hence this face of $T_2$ shaves off the corner of $T_1$ at $(0,0,0)$, and that corner is a tetrahedron similar to $T_1$ with ratio $\tfrac12$, so its volume is $\left(\tfrac12\right)^3 \cdot \tfrac13 = \tfrac{1}{24}$.

By symmetry the other three faces of $T_2$ do the same at the other three vertices of $T_1$, and these four corner pieces are disjoint. Everything left of $T_1$ lies inside $T_2$, so
$$[T_1 \cap T_2] = \frac{1}{3} - 4 \cdot \frac{1}{24} = \frac{1}{6}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{1}{6}}$.

The two inscribed tetrahedra are mirror images through the cube's center, and each face plane of one passes through the midpoints of three edges of the other. "Cut a tetrahedron at the midpoints of its edges" always removes four half-size copies (half the volume) and leaves a regular octahedron. Coordinates turn all of these facts into one-line checks such as $x + y + z = 1$ at $(\tfrac12, \tfrac12, 0)$.

Identify the intersection directly: it is the regular octahedron whose six vertices are the centers of the cube's faces. Split it along the horizontal plane $z = \tfrac12$ into two square pyramids. The base is the square through the four side-face centers, with side $\frac{\sqrt2}{2}$ and area $\tfrac12$; each height is $\tfrac12$. Volume: $2 \cdot \tfrac13 \cdot \tfrac12 \cdot \tfrac12 = \tfrac16$.

• Using the wrong tetrahedron volume: each tetrahedron is the cube minus four corner pyramids of volume 1/6, so its volume is 1/3, not 1/6 or 1/2.
• Guessing an answer with $\sqrt2$ or $\sqrt3$ because the tetrahedra have edge $\sqrt2$; the intersection's vertices are face centers with rational coordinates, so its volume is rational.
• Assuming the intersection is a smaller cube or tetrahedron rather than an octahedron, or computing the octahedron with base side $1$ instead of $\frac{\sqrt2}{2}$.