10,11

A plane passes through the vertex $A$ of the triangular pyramid $S A B C$, bisects the median $S K$ of triangle $S A B$, and intersects the median $S L$ of triangle $S A C$ at a point $D$ such that $S D: D L=1: 2$. In what ratio does this plane divide the volume of the pyramid?
Let $P$ be the midpoint of the median $S K$ of triangle $S A B, M$ be the intersection point of lines $A P$ and $S B, N$ be the intersection point of lines $A D$ and $S C$ (Fig.1). Consider the plane of triangle $S A B$ (Fig.2). Draw a line through point $S$ parallel to $A B$. Let $T$ be the intersection point of this line with the extension of segment $A M$. From the equality of triangles $S P T$ and $K P A$, we find that $S T = A K = \frac{1}{2} A B$. From the similarity of triangles $S M T$ and $B M A$, it follows that

$$
\frac{S M}{M A B} = \frac{S \Gamma}{A B} = \frac{1}{2} \cdot \frac{A B}{A B} = \frac{1}{2}
$$

Therefore, $\frac{\frac{S M}{S B}}{} = \frac{1}{3}$. Consider the plane of triangle $S A C$ (Fig.3). Draw a line through point $S$ parallel to $A C$. Let $Q$ be the intersection point of this line with the extension of segment $A N$. From the similarity of triangles $S D Q$ and $L D A$, we find that $S Q = \frac{1}{2} A L = \frac{1}{1} A C$. From the similarity of triangles $S N Q$ and $C N A$, it follows that

$$
\frac{S N}{N C} = \frac{S Q}{A C} = \frac{1}{1} \cdot \frac{A C}{A C} = \frac{1}{1}
$$

Therefore, $\frac{5 N}{S C} = \frac{1}{5}$. Hence,

$$
\frac{V_{S A N N}}{Y_{A A B C}} = \frac{S A}{S A} \cdot \frac{5 M}{S B} \cdot \frac{S N}{S C} = \frac{1}{1} \cdot \frac{1}{3} \cdot \frac{1}{b} = \frac{1}{15}
$$

Thus, the cutting plane divides the volume of the pyramid in the ratio $1: 14$.

## Answer

$1: 14$