2015 AMC 10B Problem 17

The box is $5 \times 4 \times 3$. Stand it so the $5 \times 4$ faces are the top and bottom, making the height $3$.

1. The four side faces. The center of each side face is halfway up, at height $\frac32$. So all four of these centers lie in one flat layer that cuts the box through the middle.
2. The shape they make. Look down on that middle layer. It is a $5 \times 4$ rectangle, and the four side-face centers are the midpoints of its four sides. Joining them gives a diamond (a rhombus) whose diagonals are $5$ and $4$.
3. Area of the diamond. The diamond fills exactly half of the rectangle (each corner triangle cut off matches a triangle inside the diamond). So its area is
$$\frac12 \cdot 5 \cdot 4 = 10.$$
4. The top and bottom corners. The centers of the top and bottom faces sit straight above and below the middle of the diamond, each $\frac32$ away from it.
5. Two pyramids. The octahedron is two pyramids glued on the diamond: one pointing up, one pointing down. A pyramid has volume $\frac13 \times \text{base area} \times \text{height}$, so
$$V = 2 \cdot \frac13 \cdot 10 \cdot \frac32 = 10.$$

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

Joining the face centers of any box gives this same shape: a diamond in the middle layer with a point above and a point below. Working it out in general gives one sixth of the box's volume, and here $\frac{5 \cdot 4 \cdot 3}{6} = 10$. When a solid is the same above and below a flat slice, cut it there and look for pyramids.