Given a unit cube $A B C D A_1 B_1 C_1 D_1$, $M$ is the midpoint of $B B_1$. Find the angle and the distance between the lines $A_1 B$ and $C M$. In what ratio does the common perpendicular of these lines divide the segments $C M$ and $A_1 B$?
## Перевод на английский

## Перьетеั Тик:ки.

Let $M^{\prime}$ be the orthogonal projection of point $M$ onto the plane $B 1 C 1 D A$, which is perpendicular to the line $A 1 B$ and intersects it at point $K$ - the center of the square $A A 1 B 1 B$, and $C^{\prime}$ be the orthogonal projection of point $C$ onto this plane (Fig.1). Then $C^{\prime}$ is the center of the square $C C 1 D 1 D, C^{\prime} M^{\prime}$ is the orthogonal projection of the line $C M$ onto this plane, and $M^{\prime}$ lies on $A B 1$ and $M^{\prime}$ is the midpoint of $B 1 K$. The distance between the lines $A 1 B$ and $C M$ is equal to the distance from point $K$ to the line $C^{\prime} M^{\prime}$. Let $P$ be the foot of the perpendicular dropped from point $K$ to the line $C^{\prime} M^{\prime}$. Then $K P$ is the height of the right triangle $K M^{\prime} C^{\prime}$, drawn from the vertex of the right angle (Fig.2). Further, we find:

$$
C^{\prime} M^{\prime}=\sqrt{C^{\prime} K^{2}+M^{\prime} K^{2}}=\sqrt{1+\left(\frac{\sqrt{2}}{1}\right)^{2}}=\frac{3}{2 \sqrt{2}}
$$

$$
K P=\frac{K M^{\prime} K C}{\sigma^{\prime} M^{\prime}}=\frac{\sqrt{2}}{1} \cdot \frac{1}{2 \sqrt{2}}=\frac{1}{3}
$$

The angle $\alpha$ between the lines $A 1 B$ and $C M$ complements to $90^{\circ}$ the angle between the intersecting lines $C M$ and $C^{\prime} M^{\prime}$. Therefore,

$$
\sin \alpha=\cos \left(90^{\circ}-\alpha\right)=\frac{\sigma^{M^{\prime}} M^{\prime}}{C M}=\frac{\frac{3}{\sqrt[3]{2}}}{\frac{\Sigma^{2}}{2}}=\frac{3}{\sqrt{10}}
$$

Thus, $\alpha=\arcsin \frac{\sqrt[3]{\sqrt{10}}}{}=\arccos \frac{1}{\sqrt{10}}$. Let $X Y$ be the common perpendicular of the lines $A 1 B$ and $C M$ (point $X$ lies on $A 1 B, Y$ - on $C M$). Then $K X Y P$ is a rectangle (Fig.3), $P Y\left\|M M^{\prime}\right\| C C^{\prime} \| A 1 B$. Therefore,

$$
\frac{\sigma V}{Y M}=\frac{\sigma^{\prime} P}{P M^{\prime}}=\frac{\sigma^{\prime} K^{2}}{M M^{2}}=\frac{1}{\left(\frac{1}{4}\right)^{2}}=8
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-20.jpg?height=51&width=277&top_left_y=709&top_left_x=-1)

Let $\overrightarrow{A B}=\vec{z}, \overrightarrow{A D}=\overrightarrow{\boldsymbol{y}}, \overrightarrow{A A_{1}}=\overrightarrow{\boldsymbol{z}}, A B=x, A D=y, A A 1=z$, where $x=y=z=1$. Then

$$
\overrightarrow{A_{1} \vec{B}}=\vec{y}-\vec{z}, \overrightarrow{C M}=-\vec{y}+\frac{1}{2} \vec{z}
$$

Let $\alpha$ be the angle between the lines $A 1 B$ and $C M$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-20.jpg?height=121&width=606&top_left_y=954&top_left_x=726)

![](https://cdn.mathpix.com/cropped/2024_05_06_db48a70f246fb4a097d0g-20.jpg?height=111&width=500&top_left_y=1151&top_left_x=779)

Let $X Y$ be the common perpendicular of the lines $A 1 B$ and $C M$ (point $X$ lies on $A 1 B, Y-$ on $C M$), and $\overrightarrow{A_{1} \boldsymbol{X}}=\overrightarrow{A_{1} \vec{B}_{\text {}}}$ and $\overrightarrow{C Y}=\mu \overrightarrow{C M}$. Then

$$
\begin{gathered}
\overrightarrow{X Y}=\overrightarrow{X B}+\overrightarrow{B C}+\overrightarrow{C Y}=(1-) \overrightarrow{A_{1} \vec{B}}+\overrightarrow{B C}+\mu \overrightarrow{C M}= \\
=(1-)(\vec{z}-\vec{z})+\vec{y}+\mu\left(-\vec{y}+\frac{1}{2} \vec{z}\right)= \\
=(1-) \vec{z}+(1-\mu) \vec{y}+\left(+\frac{1}{2} \mu-1\right) \cdot \vec{z}
\end{gathered}
$$

Since $\overrightarrow{X Y} \perp{\overrightarrow{A_{1} B}}_{\text {и }}^{\overrightarrow{X Y}} \perp \overrightarrow{C M}$, then $\overrightarrow{X Y} \cdot \overrightarrow{A_{1} B}=0$ and $\overrightarrow{X Y} \cdot \overrightarrow{C M}=0$, or

$$
\begin{gathered}
\left((1-) \vec{z}+(1-\mu) \vec{y}+\left(+\frac{1}{2} \mu-1\right) \vec{z}\right)(\vec{x}-\vec{z})= \\
=(1-)-\left(+\frac{1}{2} \mu-1\right)=-2-\frac{1}{2} \mu+2=0 \\
\left((1-) \vec{z}+(1-\mu) \vec{y}+\left(+\frac{1}{2} \mu-1\right) \vec{z}\right)\left(-\vec{y}+\frac{1}{2} \bar{z}\right)= \\
=(\mu-1)+\frac{1}{2}\left(+\frac{1}{2} \mu-1\right)=\frac{1}{2}+\frac{5}{1} \mu-\frac{3}{2}=0
\end{gathered}
$$

From the system

$$
\left\{\begin{array}{l}
-2 \lambda-\frac{1}{2} \mu+2=0 \\
\frac{1}{2} \lambda+\frac{5}{1} \mu-\frac{3}{2}=0
\end{array}\right.
$$

we find that $=\frac{7}{9}, \mu=\frac{8}{9}$. Therefore,

$$
\overrightarrow{X \vec{Y}}=(1-) \vec{z}+(1-\mu) \vec{y}+\left(+\frac{1}{2} \mu-1\right) \vec{z}=\frac{2}{9 \vec{z}}+\frac{1}{9} \vec{y}+\frac{2}{9 \vec{z}}
$$

Thus,

$$
X Y=\sqrt{\overrightarrow{X Y^{2}}}=\sqrt{\left(\frac{2}{9} \vec{z}+\frac{1}{9} \vec{y}+\frac{2}{9} \vec{z}\right)^{2}}=\frac{\sqrt{1+1+1}}{9}=\frac{1}{3}
$$

$$
\frac{A_{1} X}{A_{1} B}==\frac{\frac{T}{9}}{9}, \frac{A_{1} X}{X B}=\frac{T}{2}, \frac{C Y}{C M}=\mu=\frac{8}{9}, \frac{C Y}{Y M}=8 .
$$

## Tretichn' :tshn:takі.

Suppose we already know that $\sin \alpha=\frac{3}{\sqrt{10}}$. Let $V$ be the volume of the tetrahedron $B A 1 M C$ with vertex $C$. Then

$$
V=\frac{1}{3} S_{\triangle M B A} 1 \cdot C B=\frac{1}{3} \cdot \frac{1}{1} \cdot 1=\frac{1}{12}
$$

On the other hand, if $d$ is the desired distance between the lines $B A 1$ and $C M$, then

$$
V=\frac{1}{6} B A 1 \cdot C M \cdot d \sin \alpha=\frac{1}{6} \cdot \sqrt{2} \cdot \frac{\sqrt{15}}{2} \cdot d \cdot \frac{\sqrt[3]{10}}{\sqrt{10}}=\frac{d}{1} .
$$

From the equation $\frac{d}{\frac{d}{1}}=\frac{1}{12}$, we find that $d=\frac{1}{3}$.

## Answer

$\arccos \frac{1}{\sqrt{10}} ; \frac{1}{3}, C Y: Y M=8: 1($ on the segment $C M) ; \frac{A_{1} X}{X B}=\frac{7}{2}$.