The volume of a parallelepiped is $V$. Find the volume of the polyhedron whose vertices are the centers of the faces of the given parallelepiped.

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Let $S$ be the area of the base of the parallelepiped, and $h$ be the height of the parallelepiped. The polyhedron, whose vertices are the centers of the faces of the given parallelepiped, consists of two quadrilateral pyramids with a common base, the area of which is half the area of the base of the parallelepiped, and the height of each pyramid is half the height of the parallelepiped. Then, if $v$ is the desired volume of the polyhedron, we have

$$
v=2 \cdot \frac{1}{3} \cdot \frac{1}{2} S \cdot \frac{1}{2} h=\frac{1}{6} S h=\frac{1}{6} \mathrm{~V}
$$

## Answer

$\frac{1}{6} V$.