## [ Sections, developments, and skeletons (other) ] [ Volume of a tetrahedron and pyramid $]$

The base of the pyramid $P A B C D$ is a parallelogram $A B C D$. Points $K$ and $M$ are taken on the edges $A B$ and $P C$ respectively, such that $A K: K B = C M: M P = 1: 2$. In what ratio does the plane passing through points $K$ and $M$ and parallel to the line $B D$ divide the volume of the pyramid $P A B C D$?
The base plane $ABCD$ passes through the line $BD$, which is parallel to the cutting plane, and intersects the cutting plane along a line $l$ passing through point $K$ and parallel to $BD$. Let $L$ be the point of intersection of line $l$ with side $AD$ of parallelogram $ABCD$. Then, $\frac{AL}{LD} = \frac{AK}{KB} = \frac{1}{2}$. Let $F$ and $E$ be the points of intersection of line $l$ with the extensions of $BC$ and $CD$, respectively. Denote $BC = AD = a$. From the similarity of triangles $BKF$ and $AKL$, we find that

$$
BF = AL \cdot \frac{BK}{AK} = \frac{1}{3} a \cdot 2 = \frac{2}{3} a, \quad CF = BC + BF = a + \frac{2}{3} a = \frac{5}{3} a
$$

The planes of the faces $BPC$ and $APD$ pass through parallel lines $BC$ and $AD$ and have a common point $P$. Therefore, they intersect along some line $m$ parallel to lines $BC$ and $AD$. Consider the plane of the face $BPC$. Let line $FM$ intersect edge $PB$ at point $N$ and line $m$ at point $T$. From the similarity of triangles $PMT$ and $CMF$, we find that

$$
PT = CF \cdot \frac{PM}{MC} = \frac{5}{3} a \cdot 2 = \frac{10}{3} a
$$

$$
\frac{PN}{NB} = \frac{PM}{BF} = \frac{\frac{10}{3} a}{\frac{2}{3} a} = 5
$$

If $Q$ is the point of intersection of line $EM$ with edge $DP$, then similarly, we can prove that $\frac{PQ}{QD} = 5$. Let $h$ be the height of the given pyramid, $S$ be the area of its base, and $V$ be the volume. Then the height of the triangular pyramid $MCFE$ with vertex $M$ is $\frac{1}{3} h$, and the heights of the triangular pyramids $NBFK$ and $QDEL$ with vertices $N$ and $Q$ are $\frac{1}{6} h$. We have:

$$
\begin{gathered}
S_{\triangle AKL} = \frac{1}{9} S_{\triangle ABD} = \frac{1}{9} \cdot \frac{1}{2} S = \frac{1}{18} S, \\
S_{\triangle DEL} = S_{\triangle BFK} = 4 S_{\triangle AKL} = \frac{2}{9} S, \\
S_{\triangle CFE} = S_{ABCD} - S_{\triangle AKL} + 2 S_{\triangle BFK} = S - \frac{1}{18} S + \frac{4}{9} S = \frac{25}{18} S
\end{gathered}
$$

Thus,

$$
\begin{gathered}
V_{MCFE} = \frac{1}{3} S_{\triangle CFE} \cdot \frac{1}{3} h = \frac{1}{3} \cdot \frac{25}{18} S \cdot \frac{1}{3} h = \frac{25}{54} \cdot \frac{1}{3} S h = \frac{25}{54} V \\
V_{NBFK} = \frac{1}{3} S_{\triangle BFK} \cdot \frac{1}{6} h = \frac{1}{3} \cdot \frac{2}{9} S \cdot \frac{1}{6} h = \frac{1}{27} \cdot \frac{1}{3} S h = \frac{1}{27} V, \\
V_{CBKLDQMN} = V_{MCFE} - 2 V_{NBFK} = \frac{25}{54} V - \frac{2}{27} V = \frac{21}{54} V = \frac{7}{18} V .
\end{gathered}
$$

Therefore, the cutting plane divides the volume of the given pyramid in the ratio 11:7.

## Answer

$11: 7$