Through the midpoints $M$ and $N$ of the edges $A A_1$ and $C_1 D_1$ of the parallelepiped $A B C D A_1 B_1 C_1 D_1$, a plane parallel to the diagonal $B D$ of the base is drawn. Construct the section of the parallelepiped by this plane. In what ratio does it divide the diagonal $A_1 C$?
Continue the median line $N K$ of triangle $B 1 C 1 D 1$ until it intersects with line $A 1 B 1$ at point $X$, and with line $A 1 D 1$ at point $Y$ (Fig.1). Let segments $M X$ and $B B 1$ intersect at point $L$, and segments $M Y$ and $D D 1$ intersect at point $P$. Then the pentagon $M L K N P$ is the desired section, since the cutting plane contains the line $K N$, which is parallel to $D 1 B 1$.

Let $A 1 C 1$ and $N K$ intersect at point $Q$. Then the cutting plane and the plane of the diagonal section $A C C 1 A 1$ intersect along the line $M Q$, and the point $O$ of intersection of the lines $A 1 C$ and $M Q$, lying in the plane of the diagonal section, is the desired point of intersection of the line $A 1 C$ with the cutting plane.

Consider the parallelogram $A C C 1 A 1$ (Fig.2). Let $A C=4 x$. Extend $Q M$ until it intersects with line $A C$ at point $T$. From the equality of triangles $A M T$ and $A 1 M Q$, it follows that

$$
A T=A 1 Q=\frac{3}{4} A 1 C 1=\frac{3}{4} \cdot 4 x=3 x
$$

Then $C T=A T+A C=3 x+4 x=7 x$, and since triangles $A 1 O Q$ and $C O T$ are similar, we have

$$
\frac{O A_{1}}{O C}=\frac{A_{1} Q}{C T}=\frac{3 I}{7 I}=\frac{3}{7}
$$

## Answer

3:7, measured from vertex $A 1$.

## Problem