2006 AMC 10A Problem 24

The six vertices of the octahedron are the six face centers of the cube. Take the top and bottom face centers as the two apexes; the other four vertices, the centers of the four side faces, all lie in the horizontal plane halfway up the cube and form the common base of two square pyramids.

Base: the centers of two opposite side faces are $1$ apart (the cube's width), and these are opposite corners of the base square. So the square has diagonal $1$ and area $\tfrac12 \cdot 1 \cdot 1 = \tfrac12$. (Equivalently its side is $\tfrac{\sqrt2}{2}$.)

Height: each apex is the center of the top or bottom face, at distance $\tfrac12$ from the middle plane.

Each pyramid has volume $\tfrac13 \cdot \tfrac12 \cdot \tfrac12 = \tfrac{1}{12}$, and the octahedron is two of them:
$$2 \cdot \frac{1}{12} = \frac{1}{6}.$$

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

A regular octahedron is two square pyramids glued base to base, so its volume only needs the base square and the apex height. Inside a unit cube those come for free: the base diagonal is a cube edge and each height is half an edge. Choosing the decomposition that aligns with the cube's own symmetry avoids any trigonometry.

Coordinates: the octahedron is $|x - \tfrac12| + |y - \tfrac12| + |z - \tfrac12| \le \tfrac12$, whose volume is $\tfrac{4}{3} r^3$ with $r = \tfrac12$, giving $\tfrac{4}{3} \cdot \tfrac18 = \tfrac16$. Or use the regular-octahedron formula $\tfrac{\sqrt2}{3} a^3$ with edge $a = \tfrac{\sqrt2}{2}$.

• Taking the base square's side to be $\tfrac12$ (area $\tfrac14$, volume $\tfrac{1}{12}$) or using the full cube height $1$ as a pyramid height (volume $\tfrac13$, choice (D)).
• Forgetting the factor $\tfrac13$ in the pyramid formula, which gives $\tfrac12$ (choice (E)).