1. On each edge of a tetrahedron, a plane parallel to the opposite edge is placed.

a) What polyhedron defines the part of space bounded by these planes?

b) Find the ratio of the volume of the resulting polyhedron to the volume of the tetrahedron.
Solution. a) The tetrahedron has three pairs of opposite edges, so, three pairs of parallel planes are drawn, i.e., the body bounded by the planes is a parallelepiped. The edges of the tetrahedron are diagonals of the faces of the parallelepiped, so the parallelepiped is composed of the given tetrahedron and four pyramids with bases half the base of the parallelepiped and heights equal to the height of the parallelepiped.

b) Let $V$ be the volume of the parallelepiped, $H$ its height, $B$ the area of its base, $V_{\mathrm{T}}$ the volume of the tetrahedron, and let $V_{\circ}$ be the volume of one of the remaining pyramids. Then

$$
\begin{aligned}
& V=B H, V_{\circ}=\frac{1}{3} \frac{B}{2} H=\frac{B H}{6} \\
& V=V_{\mathrm{T}}+4 V_{\circ}=V_{\mathrm{T}} \frac{2}{3} V, V_{\mathrm{T}}=\frac{1}{3} V
\end{aligned}
$$

i.e., $V: V_{\mathrm{T}}=3: 1$.

![](https://cdn.mathpix.com/cropped/2024_06_05_0970dfed9f8278f60f58g-4.jpg?height=444&width=466&top_left_y=374&top_left_x=1014)