[ Intersecting lines, angle between them ]

The height $A A_1$ of the rectangular parallelepiped $A B C D A_1 B_1 C_1 D_1$ is twice as large as each side of the base. Find the angle between the lines $B D_1$ and $A M$, where $M$ is the point of intersection of the diagonals of the face $D C C_1 D_1$.
Let $A B=x, A D=y, A A 1=z, \overrightarrow{A B}=\vec{x}, \overrightarrow{A D}=\vec{y}, \overrightarrow{A A_{1}}=\vec{z}, x=y=a, z=2 a$. Then

$$
\overrightarrow{B D_{1}}=-\vec{x}+\vec{y}+\vec{z}, \overrightarrow{A M}=\frac{1}{2} \vec{x}+\vec{y}+\frac{1}{2} \vec{z}
$$

If $\alpha-$ is the desired angle, then

$$
\begin{aligned}
& \cos \alpha=\frac{\overrightarrow{B D_{1}} \cdot \overrightarrow{A M}}{B D_{1} \cdot A M}=\frac{(-\vec{x}+\vec{y}+\vec{z})\left(\frac{1}{2} \vec{x}+\vec{y}+\frac{1}{2} \vec{z}\right)}{\sqrt{\alpha^{2}+\alpha^{2}+4 \alpha^{2}} \cdot \sqrt{\frac{1}{4} \alpha^{2}+\alpha^{2}+\alpha^{2}}}= \\
& =\frac{-\frac{1}{2} \vec{z} \cdot \vec{z}+\vec{y} \cdot \vec{y}+\frac{1}{2} \vec{z} \cdot \vec{z}}{a^{2} \sqrt{6} \cdot \frac{3}{2}}=\frac{-\frac{1}{2} a^{2}+a^{2}+2 a^{2}}{a^{2} \sqrt{6} \cdot \frac{3}{2}}=\frac{5}{3 \sqrt{6}} .
\end{aligned}
$$

Therefore, $\alpha=\arccos \frac{5}{3 \sqrt{6}}$.

## Answer

$\arccos \frac{5}{3 \sqrt{6}}$