2011 AMC 10A Problem 24

Put the cube on a grid so that every corner has coordinates that are each $0$ or $1$, like $(1, 0, 1)$.

1. Find the two tetrahedra. In a regular tetrahedron all six edges are equal. Among cube corners, the only way to get four points all the same distance apart is to use face diagonals (length $\sqrt2$). That happens for the four corners with an even number of $1$s:
$$T_1:\ (0,0,0),\ (1,1,0),\ (1,0,1),\ (0,1,1),$$
and for the other four corners, with an odd number of $1$s:
$$T_2:\ (1,0,0),\ (0,1,0),\ (0,0,1),\ (1,1,1).$$
2. Volume of one tetrahedron. $T_1$ is what's left of the cube after slicing off the four corners at the points of $T_2$. For example, the slice at $(1,0,0)$ is a small pyramid with three edges of length $1$ meeting at right angles. Its volume is $\frac{1}{3} \times (\text{base } \tfrac{1}{2}) \times (\text{height } 1) = \frac{1}{6}$. So
$$\text{volume of } T_1 = 1 - 4 \times \frac{1}{6} = \frac{1}{3}.$$
3. How $T_2$ cuts $T_1$. Look at the face of $T_2$ through $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. Every point on it has coordinates adding to $1$. Now look at the corner $(0,0,0)$ of $T_1$ and its three edges, going to $(1,1,0)$, $(1,0,1)$, $(0,1,1)$. Their midpoints are $(\tfrac12,\tfrac12,0)$, $(\tfrac12,0,\tfrac12)$, $(0,\tfrac12,\tfrac12)$, and each has coordinates adding to $1$. So this face of $T_2$ slices through the midpoints of those three edges.
4. The piece cut off. That slice removes the corner of $T_1$ at $(0,0,0)$. The piece removed is a smaller tetrahedron of the same shape, with every length half as long. Halving every length makes the volume $\tfrac12 \times \tfrac12 \times \tfrac12 = \tfrac18$ as big, so the piece has volume $\tfrac18 \times \tfrac13 = \tfrac{1}{24}$.
5. All four corners. By symmetry, each of the four faces of $T_2$ cuts off one corner of $T_1$ in the same way, and the four pieces don't overlap. What remains of $T_1$ is inside $T_2$. So the overlap has volume
$$\frac{1}{3} - 4 \times \frac{1}{24} = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}.$$

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

The two tetrahedra are mirror images of each other through the center of the cube, and each face of one passes through the midpoints of three edges of the other. Cutting a tetrahedron through the midpoints of its edges always removes four half-size copies (each $\tfrac18$ of the volume, so half the volume in total). Coordinates make every step a quick check, such as "these coordinates add to $1$".

Recognise the overlap directly. It is an octahedron (two square pyramids glued base to base) whose six corners are the centers of the cube's six faces.

• Cut it by the flat plane halfway up the cube. The cross-section is the square joining the centers of the four side faces. Its corners are like $(\tfrac12, 0, \tfrac12)$ and $(1, \tfrac12, \tfrac12)$, so its side is $\sqrt{\tfrac14 + \tfrac14} = \tfrac{\sqrt2}{2}$ and its area is $\tfrac12$.
• Each pyramid has this square as its base and height $\tfrac12$ (up to the top face center, or down to the bottom one). Volume of each: $\frac13 \times \frac12 \times \frac12 = \frac{1}{12}$.
• Two pyramids: $2 \times \frac{1}{12} = \frac16$.