Given a tetrahedron $A B C D$. In what ratio does the plane passing through the points of intersection of the medians of the faces $A B C$, $A B D$, and $B C D$ divide the edge $B D$?
Let $M, K$ and $N$ be the points of intersection of the medians of triangles $ABC, ABD$ and $BCD$ respectively; $CE, CF$ and $BG$ be the medians of triangles $BCD, ABC$ and $ABD$ respectively. Then

$$
\frac{CN}{CE}=\frac{CM}{CH}=\frac{2}{3}
$$

Therefore, $MN \| EF$. This means that the line $MN$ is parallel to the plane $ABD$. The plane $KMN$ passes through the line $MN$, which is parallel to the plane $ABD$, and intersects this plane along some line $a$ passing through point $K$. Thus, the line $a$ is parallel to the lines $MN, EF$ and $AD$. Let the line $a$ intersect the edge $BD$ at point $P$. Then

$$
\frac{BP'}{BD}=\frac{BK}{BG}=\frac{2}{3}
$$

Therefore, $\frac{\partial P^{P}}{BP'}=\frac{1}{2}$.

## Answer

$1: 2$, counting from point $D$.