On the edges $D A, D B$ and $D C$ of the pyramid $A B C D$, points $K, L$ and $M$ are taken respectively, such that $D K=\frac{1}{2} D A, D L=\frac{2}{5} D B$ and $D M=\frac{3}{4} D C, G-$ is the point of intersection of the medians of triangle $A B C$. In what ratio does the plane $K L M$ divide the segment $D G ?$
Let $V_{A B C D}=V$. Then

$$
V_{D K L M}=\frac{D K}{D A} \cdot \frac{D L}{D B} \cdot \frac{D M}{D C} \cdot V_{A B C D}=\frac{1}{2} \cdot \frac{2}{5 \cdot \frac{3}{4}} \cdot V=\frac{3}{20} V
$$

Since $G$ is the point of intersection of the medians of triangle $A B C$, triangles $A G B, A G C$, and $B G C$ are equal in area, therefore

$$
V_{D A G B}=V_{D A G C}=V_{D B G C}=\frac{1}{3} V_{A B C D}
$$

Let $\frac{D O}{D G}=x$. Suppose $O$ is the point of intersection of segment $D G$ with the plane $K L M$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_1f91526468b1b4d32730g-21.jpg?height=116&width=877&top_left_y=1314&top_left_x=593)

$$
\begin{aligned}
& V_{D M L O}=\frac{D M}{D C} \cdot \frac{D L}{D B} \cdot \frac{D O}{D G} \cdot \frac{1}{3} V_{A B C D}=\frac{\frac{3}{4}}{4} \cdot \frac{2}{5} \cdot x \cdot \frac{1}{3} V=\frac{1}{10} x V, \\
& V_{D K M O}=\frac{D K}{D A} \cdot \frac{D M}{D C} \cdot \frac{D O}{D G} \cdot \frac{1}{3} V_{A B C D}=\frac{1}{2} \cdot \frac{3}{4} \cdot x \cdot \frac{1}{3} V=\frac{1}{3} x V,
\end{aligned}
$$

The condition for points $K, L, M$, and $O$ to lie in the same plane is equivalent to the equality

$$
V_{D K L O}+V_{D M L O}+V_{D K M O}=V_{D K M L}
$$

or

$$
\frac{1}{15} x V+\frac{1}{10} x V+\frac{1}{8} x V=\frac{3}{20} V
$$

From this, we find that $\frac{D Q}{D G}=x=\frac{18}{85}$. Therefore, $\frac{D O}{O G}=\frac{18}{17}$.

## Answer

$18: 17$.