Through the midpoint of the edge $A B$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, a plane is drawn parallel to the lines $B D_{1}$ and $A_{1} C_{1}$.

1) In what ratio does this plane divide the diagonal $D B_{1}$?
2) Find the area of the resulting section.
1) Let $M$ be the midpoint of $AB$. The plane of the face $ABCD$ has a common point $M$ with the cutting plane and passes through the line $AC$, which is parallel to the cutting plane (since $AC \| A_{1}C_{1}$). Therefore, the line of intersection of these planes passes through the point $M$ and is parallel to $AC$. If $N$ is the point of intersection of this line with the edge $BC$, then $MN$ is the midline of the triangle $ABC$.

Let $F$ be the point of intersection of the lines $MN$ and $BD$. The plane of the rectangle $BB_{1}D_{1}D$ has a common point $F$ with the cutting plane and passes through the line $BD_{1}$, which is parallel to the cutting plane. Therefore, the line of intersection of these planes passes through the point $F$ and is parallel to $BD_{1}$. Let this line intersect the edge $DD_{1}$ at point $K$, and the diagonal $DB_{1}$ at point $T$. Then $T$ is the point of intersection of the cutting plane with the diagonal $DB_{1}$.

Let $G$ be the point of intersection of the diagonals of the square $ABCD$, and $O$ be the point of intersection of the diagonals of the cube. Since $F$ is the midpoint of the segment $BG$, $DF / BD = 3 / 4$, and since $FT \| BO$, then $DT / DO = DF / DB = 3 / 4$. Therefore,

$$
DT / DB_{1} = DT / (2 \cdot DO) = (DT / DO) / 2 = 3 / 8
$$

Thus, $DT / B_{1} = 3 / 5$.

2) It is clear that the angle $\boldsymbol{\alpha}$ between the cutting plane and the plane of the base $A_{1}B_{1}C_{1}D_{1}$ is equal to the angle $BD_{1}B_{1}$. Then $\cos \alpha = B_{1}D_{1} / BD_{1} = \sqrt{2 / 3}$.

Let $M_{1}$ and $N_{1}$ be the orthogonal projections of the points $M$ and $N$ onto the plane of the base $A_{1}B_{1}C_{1}D_{1}$. Then $M_{1}$ and $N_{1}$ are the midpoints of the segments $A_{1}B_{1}$ and $B_{1}C_{1}$, respectively, and the pentagon $A_{1}M_{1}N_{1}C_{1}D_{1}$ is the orthogonal projection of the pentagon of the section onto the plane of the base $A_{1}B_{1}C_{1}D_{1}$.

Since $M_{1}N_{1}$ is the midline of the triangle $A_{1}B_{1}C_{1}$,

$$
\begin{aligned}
& S\left(M_{1}B_{1}N_{1}\right) = \frac{1}{4} S\left(A_{1}B_{1}C_{1}\right) = \frac{1}{8} S\left(A_{1}B_{1}C_{1}D_{1}\right), \text{and} \\
& S\left(A_{1}M_{1}N_{1}C_{1}D_{1}\right) = (7 / 8) \cdot S\left(A_{1}B_{1}C_{1}D_{1}\right) = 7a^{2} / 8
\end{aligned}
$$

Therefore, the desired area is

$$
S\left(A_{1}M_{1}N_{1}C_{1}D_{1}\right) / \cos \alpha = \left(7a^{2} / 8\right) \cdot \sqrt{3 / 2} = 7a^{2} \sqrt{6} / 16
$$

## Answer

$$
3 / 5 ; 7a^{2} \sqrt{6} / 16
$$