Example 78. A triangular pyramid is cut by a plane into two polyhedra. We will find the ratio of the volumes of these polyhedra, given that the cutting plane divides the edges converging at one vertex of the pyramid in the ratio $1: 2, 1: 2, 2: 1$, counting from this vertex.

Construction of the image. Let the quadrilateral $S A B C$ with its diagonals be the image of the given pyramid (Fig. 87), $\triangle P Q R$ - the image of the given section. This image is complete, and what is quite curious, no parameter has been spent on it.

Given: $\quad S A B C$ - pyramid, $\quad S$ - vertex, $\quad P \in[S A]$, $Q \in[A B]$, $\quad R \in[A C]$, $\quad|A Q|:|Q B|=1: 2,|A R|:|R C|=1: 2$, $|A P|:|P S|=2: 1,(P Q R)$ - cutting plane.

To find: $V_{1}: V_{2}$ - the ratio of the volumes of the resulting polyhedra.
Solution. Let $V$ be the volume of the pyramid $SABC$, and $V_{1}$ the volume of the pyramid $PAQR$. Then $V_{1}=V-V_{2}$.

Construct $[SO]$ and assume that $[SO]$ is the image of the height of the pyramid $SABC$ (thus, two parameters are used). Construct $(AO)$ and $[PM] \|[SO]$. Then $M \in (AO)$. To simplify the calculations, let $|AB|=a, |AC|=b$, $|AS|=c, |SO|=H$.

Then $|AQ|=\frac{1}{3} a, |AR|=\frac{1}{3} b, |AP|=\frac{2}{3} c$, and from the similarity of triangles $APM$ and $ASO$ we have: $|PM|=\frac{2}{3} H$. Now let's calculate $V$ and $V_{2}$.

We have:

$$
\begin{aligned}
V & =\frac{1}{3} S_{\triangle ABC} \cdot H=\frac{1}{6} ab \sin \widehat{BAC} \cdot H \\
V_{2} & =\frac{1}{3} S_{\triangle AQR} \cdot \frac{2}{3} H=\frac{1}{81} ab \sin \widehat{BAC} \cdot H
\end{aligned}
$$

Then

$$
V_{1}=V-V_{2}=\frac{25}{162} abH \sin \widehat{BAC}
$$

Thus,

$$
\frac{V_{1}}{V_{2}}=\frac{\frac{25}{162} abH \sin \widehat{BAC}}{\frac{1}{81} abH \sin \widehat{BAC}}=\frac{25}{2}
$$

Investigation. By the problem's meaning, $V > V_{2}$. Therefore, $V_{1}=V-V_{2}>0$ and, consequently, $\frac{V_{1}}{V_{2}}>1$. The found value of $\frac{V_{1}}{V_{2}}$ satisfies this inequality.

Thus, $V_{1}: V_{2}=25: 2$.

## Problems for Independent Solving

244. The base of a parallelepiped is a rhombus $ABCD$, with side length $a$ and an acute angle of $60^{\circ}$. Find the volume of the parallelepiped if the lateral edge is equal to $a$, and $\widehat{A_{1}AB}=\widehat{A_{1}AD}=45^{\circ}$.
245. Each edge of the parallelepiped is equal to $a$. Each of the three planar angles at one of the vertices of the parallelepiped is equal to $2\alpha$. Find the volume of the parallelepiped.
246.