4. The base of the pyramid is a parallelogram. A plane is drawn through one side of the base and the midline of the opposite lateral face. Into what ratio does this plane divide the volume of the pyramid?
Solution. Let $A_{1} A B C D$ be the given pyramid (with base $A B C D$), then $A B C D A_{1} B_{1} C_{1} D_{1}$ is a parallelepiped (with edges $B B_{1}, C C_{1}, D D_{1}$), $V$ is the volume of this parallelepiped, and $M$ and $N$ are the midpoints of segments $A_{1} D$ and $A_{1} C$ (diagram on the right). Then

$$
\begin{aligned}
V_{A_{1} A B C D} & =\frac{1}{3} V \\
V_{A_{1} A B N M} & =V_{D_{1} A_{1} A B}-V_{D_{1} A_{1} M N} \\
& =\frac{1}{6} V-\frac{1}{24} V=\frac{3}{24} V=\frac{3}{8} V_{A_{1} A B C D}
\end{aligned}
$$

According to this, the plane $A B N M$ divides the volume of the pyramid $A_{1} A B C D$ in the ratio 5:3.

![](https://cdn.mathpix.com/cropped/2024_06_05_17a2a9aba963f6a4e511g-2.jpg?height=407&width=407&top_left_y=1176&top_left_x=1075)

## 4th Year