The base of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ is a square $A B C D$. Find the greatest possible value of the angle between the line $B D_{1}$ and the plane $B D C_{1}$.

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Consider the section $B D C_{1}$ of the given parallelepiped (see the left figure). Let $O$ be the point of intersection of the diagonals of the square $A B C D$. Since triangle $D C_{1} B$ is isosceles, $C_{1} O$ is its height. Let $P$ be the point of intersection of the diagonals of the parallelepiped, and $P Q$ be the perpendicular to the plane $B D C_{1}$. Clearly, point $P$ is equidistant from the vertices of triangle $D C_{1} B$, so $Q$ is the center of the circumscribed circle around it. Since this triangle is acute and isosceles, point $Q$ lies on the segment $C_{1} O$. The angle between the line $B D_{1}$ and the plane $B D C_{1}$ is the acute angle $\alpha$ between $\left(B D_{1}\right)$ and its orthogonal projection onto $\left(B D C_{1}\right)$, which is angle $P B Q$.
![](https://cdn.mathpix.com/cropped/2024_05_06_dd7345f2b5ef58b5189dg-20.jpg?height=410&width=826&top_left_y=332&top_left_x=615)

Let $A B=a, A A_{1}=b$. Then $P B=\frac{\sqrt{2 a^{2}+b^{2}}}{2}$. The length of $P Q$ can be found by considering, for example, the rectangle $C C_{1} O_{1} O$, where $O_{1}$ is the point of intersection of the diagonals of the square $A_{1} B_{1} C_{1} D_{1}$ (see the right figure). From the similarity of the right triangles $P O Q$ and $C_{1} O O_{1}$, we get that $\frac{P Q}{C_{1} O_{1}}=\frac{P O}{C_{1} O}$. Since $C_{1} O_{1}=\frac{a \sqrt{2}}{2}, C_{1} O=\sqrt{\frac{a^{2}}{2}+b^{2}}$, then $P Q=\frac{\frac{b}{2} \cdot \frac{a \sqrt{2}}{2}}{\sqrt{\frac{a^{2}}{2}+b^{2}}}=\frac{a b}{2 \sqrt{a^{2}+2 b^{2}}}$. Therefore, $\sin \alpha=\frac{P Q}{P B}=\frac{a b}{\sqrt{\left(a^{2}+2 b^{2}\right)\left(2 a^{2}+b^{2}\right)}}=\frac{a b}{\sqrt{2 a^{4}+2 b^{4}+5 a^{2} b^{2}}}=$ $\frac{1}{\sqrt{2\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\right)+5}} \leq \frac{1}{\sqrt{2 \cdot 2+5}}=\frac{1}{3}$.

Thus, the maximum value of $\alpha$ is $\arcsin 1 / 3$. It is achieved when $a=b$, that is, when the given parallelepiped is a cube.

## Answer

$\arcsin 1 / 3$.