On the edge \$AD\$ and diagonal \$A_1C\$ of the parallelepiped \$ABCDA_1B_1C_1D_1\$, points \$M\$ and \$N\$ are taken, respectively, such that the line \$MN\$ is parallel to the plane \$BDC_1\$ and \$AM:AD = 1:5\$. Find the ratio \$CN:CA_1\$.
Let $P$ be the center of parallelogram $ABCD$. Planes $BDC_1$ and $AA_1C_1C$ intersect along a line

![](https://cdn.mathpix.com/cropped/2024_05_06_2f46c6bf14fbb7f478d4g-20.jpg?height=66&width=2032&top_left_y=2651&top_left_x=-1)
\[
\frac{CP}{A_1C_1} = \frac{CP}{AC} = \frac{1}{2}.
\]
Since line $MN$ is parallel to plane $BDC_1$, this line lies in plane $\alpha$, passing through point $M$ parallel to plane $BDC_1$. By the theorem on the intersection of two parallel planes by a third, line $MK$ of the intersection of planes $ABCD$ and $\alpha$ is parallel to $BD$. Let point $K$ lie on line $AB$, and lines $MK$ and $AC$ intersect at point $E$. Then
\[
\frac{AE}{AP} = \frac{AM}{AD} = \frac{1}{5}, \quad \frac{AE}{AC} = \frac{1}{10}, \quad \frac{PC}{PE} = \frac{5}{4}.
\]
By the theorem on the intersection of two parallel planes by a third, planes $\alpha$ and $AA_1C_1C$ intersect along a line passing through point $E$ parallel to $PC_1$. Clearly, the intersection point of this line with line $CA_1$ is point $N$ (line $MN$ lies in a plane parallel to plane $BDC_1$).

Consider parallelogram $AA_1C_1C$. Since
\[
\frac{9}{5} \cdot \frac{1}{3} CA_1 = \frac{3}{5} CA_1,
\]
it follows that
\[
\frac{CN}{CA_1} = \frac{3}{5}.
\]

## Answer

$3: 5$.