2015 AMC 10B Problem 17

The box is $5 \times 4 \times 3$. Put it with the $5 \times 4$ faces on top and bottom, so the height is $3$.

The centers of the four vertical side faces all sit at half-height, in the horizontal plane that cuts the box in the middle. In that plane they are the midpoints of the four sides of a $5 \times 4$ rectangle, so they form a rhombus whose diagonals are $5$ and $4$. Its area is
$$\frac{1}{2} \cdot 5 \cdot 4 = 10.$$

The remaining two vertices of the octahedron are the centers of the top and bottom faces, directly above and below the center of the rhombus at distance $\frac{3}{2}$ each.

So the octahedron is two pyramids sharing the rhombus as base:
$$V = 2 \cdot \frac{1}{3} \cdot 10 \cdot \frac{3}{2} = 10.$$

The answer is $\boxed{\textbf{(B)}\ 10}$.

Joining face centers of any box always produces this structure: a rhombus in the mid-plane plus an apex on each side. In general the volume is $2 \cdot \frac13 \cdot \frac{ab}{2} \cdot \frac{c}{2} = \frac{abc}{6}$, exactly one sixth of the box, here $\frac{60}{6} = 10$. When a solid is symmetric about a plane, slice it there and look for a pyramid.

• Assuming the octahedron is regular and using a regular-octahedron volume formula, or taking the rhombus area as the full 5 x 4 = 20.
• Using the full height $3$ for each pyramid instead of $\frac{3}{2}$, which doubles the answer to $20$.