2006 AMC 10A Problem 24

The octahedron has six corners, one at the center of each face of the cube. Split it into two pyramids.

1. The split. The centers of the four side faces all sit halfway up the cube, so they lie in one flat horizontal square. The top-face center is above that square and the bottom-face center is below it. So the octahedron is two square pyramids glued along this middle square.
2. The middle square. The centers of two opposite side faces (say front and back) are $1$ apart, the width of the cube, and they are opposite corners of the middle square. So the square's diagonal is $1$. A square is a special rhombus, so its area is half the product of its diagonals: $\frac12 \times 1 \times 1 = \frac12$.
3. The height. The top-face center is $\frac12$ above the middle of the cube, and the bottom-face center is $\frac12$ below. Each pyramid has height $\frac12$.
4. Volume. A pyramid's volume is $\frac13 \times \text{base} \times \text{height}$:
$$\frac13 \times \frac12 \times \frac12 = \frac{1}{12}.$$
Two pyramids give $2 \times \frac{1}{12} = \frac16$.

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

A regular octahedron is two square pyramids stuck together base to base, so its volume only needs the base square and the height. Inside a cube both come for free: the base's diagonal is the cube's width and each height is half the cube. Cutting a solid along a line of symmetry often turns it into shapes you already know.

Cut the unit cube into $8$ small cubes of side $\frac12$, all meeting at the cube's center. Inside one small cube, the octahedron is a corner piece: a pyramid whose three edges from the cube's center go to three face centers, each of length $\frac12$ and each at right angles to the others. Its base is a right triangle with legs $\frac12$, area $\frac18$, and its height is $\frac12$, so its volume is $\frac13 \times \frac18 \times \frac12 = \frac{1}{48}$. Eight such pieces give $\frac{8}{48} = \frac16$.