93. Two spheres touch each other and the faces of a dihedral angle, the magnitude of which is $\alpha$. Let $A$ and $B$ be the points of contact of these spheres with the faces ( $A$ and $B$ belong to different spheres and different faces). In what ratio is the segment $A B$ divided by the points of intersection with the surfaces of these spheres?
93. Let $A_{1}$ and $B_{1}$ be the other two points of tangency, $R$ and $r$ be the radii of the circles. In the trapezoid $A A_{1} B B_{1}$, we find the bases: $\left|A A_{1}\right|=$ $=2 R \cos \frac{\alpha}{2},\left|B B_{1}\right|=2 r \cos \frac{\alpha}{2}$ and the lateral sides $\left|A B_{1}\right|=$ $=\left|A_{1} B\right|=2 \sqrt{R r}$, after which we determine the diagonals $|A B|=\left|A_{1} B_{1}\right|=$ $=2 \sqrt{R r\left(1+\cos ^{2} \frac{\alpha}{2}\right)} \cdot$ If the tangent passing through $A$ and $A_{1}$ intersects $A B$ at point $K$, then $\left|A_{1} B\right|^{2}=|B K| \cdot|B A|$, from which

$$
|B K|=\frac{2 \sqrt{R r}}{\sqrt{1+\cos ^{2} \frac{\alpha}{2}}}=\frac{|A B|}{1+\cos ^{2} \frac{\alpha}{2}}, \quad|A K|=\frac{|A B| \cos ^{2} \frac{\alpha}{2}}{1+\cos ^{2} \frac{\alpha}{2}}
$$

Similarly, the other parts into which the segment $A B$ is divided are found. Answer: The segment $A B$ is divided in the ratio

$$
\cos ^{2} \frac{\alpha}{2}: \sin ^{2} \frac{\alpha}{2}: \cos ^{2} \frac{\alpha}{2} .
$$