112. In the plane $P$, there is an isosceles triangle $ABC (|AB| = |BC| = l, |AC| = 2a)$. A sphere of radius $r$ touches the plane $P$ at point $B$. Two skew lines pass through points $A$ and $C$ and are tangent to the sphere. The angle between each of these lines and the plane $P$ is $\alpha$. Find the distance between these lines.
112. Let $M$ and $N$ be the points of tangency of the tangents drawn from $A$ and $B$, and $M_{1}$ and $N_{1}$ be the projections of points $M$ and $N$ onto the plane $ABC$ (Fig. 19, $a$; the figure shows one of the two equivalent cases of the arrangement of the tangents, in which the tangents intersect, in the other two cases these tangents lie in the same plane).

![](https://cdn.mathpix.com/cropped/2024_05_21_9b03e22d2b8ca67aa653g-072.jpg?height=414&width=616&top_left_y=1172&top_left_x=318)

b)

Fig. 19.

It is not difficult to find: $|A M| = |C N| = l, |M M_{1}| = |N N_{1}| = l \sin \alpha, |A M_{1}| = |C N_{1}| = l \cos \alpha$. Let's find $|B M_{1}|$ and $|B N_{1}|$ (Fig. 19, $b$; $O$ is the center of the sphere, $O L \parallel B M_{1}$)

$$
\begin{aligned}
& |B N_{1}| = |B M_{1}| = |O L| = \sqrt{r^{2} - (l \sin \alpha - r)^{2}} = \\
& = \sqrt{2 r l \sin \alpha - l^{2} \sin^{2} \alpha}
\end{aligned}
$$

When rotating around point $B$ by the angle $\varphi = \widehat{A B C}$, point $A$ will move to $C$, and $M_{1}$ to $N_{1}$, hence the triangles $B M_{1} N_{1}$ and $B A C$ are similar,

$$
|M N| = |M_{1} N_{1}| = |B M_{1}| \frac{|A C|}{|A B|} = \frac{2 a}{l} \sqrt{2 r l \sin \alpha - l^{2} \sin^{2} \alpha}
$$

$\triangle M_{1} B N_{1}$ is obtained from $\triangle A B C$ by rotating around $B$ by the angle $\gamma = \widehat{A B M_{1}}$ followed by a homothety. Therefore, the angle between $M_{1} N_{1}$ and $A C$ is $\gamma$, and since $M_{1} N_{1}$ and $M N$ are parallel, the angle between $M N$ and $A C$ is also $\gamma$.

From $\triangle B M_{1} A$ we find

$\cos \gamma = \frac{2 r l \sin \alpha - l^{2} \sin^{2} \alpha + l^{2} - l^{2} \cos^{2} \alpha}{2 l \sqrt{2 r l \sin \alpha - l^{2} \sin^{2} \alpha}} = \frac{r \sin \alpha}{\sqrt{2 r l \sin \alpha - l^{2} \sin^{2} \alpha}}$.

Then

$$
\sin \gamma = \frac{\sqrt{2 r l \sin \alpha - (l^{2} + r^{2}) \sin^{2} \alpha}}{\sqrt{2 r l \sin \alpha - l^{2} \sin^{2} \alpha}}
$$

Now, using the obtained values for $|M N|, |M M_{1}|$, and $\sin \gamma$, the volume of the pyramid $A C M N$ is:

$$
\begin{aligned}
& V_{A C M N} = \frac{1}{6} |A C| \cdot |M N| \cdot |M M_{1}| \sin \gamma = \\
& = \frac{2 a^{2} \sin \alpha}{3} \sqrt{2 r l \sin \alpha - (l^{4} + r^{2}) \sin^{2} \alpha}
\end{aligned}
$$

Now, let's take point $P$ such that $M_{1} N_{1} C P$ is a parallelogram, which means $M N C P$ is also a parallelogram. Let $\beta$ be the angle between $A M$ and $C N$, then $\beta = \widehat{A M P}$. But $\triangle A B M_{1}$ is obtained from $\triangle C B N_{1}$ by rotating around $B$ clockwise by the angle $\varphi = \widehat{A B C}$. Therefore, the angle between $A M_{1}$ and $C N_{1}$ is $\varphi$, and thus $\widehat{A M_{1}} P = \varphi$. That is, triangles $A M_{1} P$ and $A B C$ are similar. From this similarity, we find $|A P| = 2 a \cos \alpha$. The angle $\beta$ is equal to the angle $\widehat{A M P}$, and $\triangle A M P$ is isosceles, where $|A M| = |M P| = l, |A P| = 2 a \cos \alpha$. Therefore,

$$
\begin{gathered}
\sin \frac{\beta}{2} = \frac{a \cos \alpha}{l} \\
\sin \beta = 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2} = \frac{2 a \cos \alpha \sqrt{l^{2} - a^{2} \cos^{2} \alpha}}{l^{2}}
\end{gathered}
$$

Let's express the volume of the pyramid $A C M N$ in another way,

$$
V_{A C M N} = \frac{1}{6} |A M| \cdot |C N| \cdot \sin \beta = \frac{1}{3} a x \cos \alpha \sqrt{l^{2} - a^{2} \cos^{2} \alpha},
$$

where $x$ is the desired distance. Comparing this formula with the equality

(1), we get

$$
x = \frac{2 a \tan \alpha \sqrt{2 r l \sin \alpha - (l^{2} + r^{2}) \sin^{2} \alpha}}{\sqrt{l^{2} - a^{2} \cos^{2} \alpha}}
$$