4.5. Two tangent spheres are inscribed in a dihedral angle of magnitude $2 \alpha$. Let $A$ be the point of tangency of the first sphere with the first face, and $B$ be the point of tangency of the second sphere with the second face. In what ratio is the segment $A B$ divided by the points of intersection with these spheres?

## § 2. Tangents to Spheres
4.5. Let $a$ and $b$ be the radii of the spheres, $A_{1}$ and $B_{\mathrm{i}}$ be the points of their tangency with the faces of the angle. The sides of the trapezoid $A A, B B_{1}$ can be easily calculated: $A B_{1}=A_{1} B=2 \sqrt{a b}$ (Problem 4.1), $A A_{1}=$
$=2 a \cos \alpha$ and $B B_{1}=2 b \cos \alpha$. The square of the height of this trapezoid is $4 a b-(b-a)^{2} \cos ^{2} \alpha$, and the square of the diagonal is $4 a b-$ $-(b-a)^{2} \cos ^{2} \alpha+(a+b)^{2} \cos ^{2} \alpha=4 a b\left(1+\cos ^{2} \alpha\right)$. If a sphere passing through points $A$ and $A_{1}$ intersects the segment $A B$ at point $K$, then $B K=B A_{1}^{2} / B A,=2 \sqrt{a b} / \sqrt{1+\cos ^{2} \alpha}=$

![](https://cdn.mathpix.com/cropped/2024_05_21_3f2bc5f5a1af26514cfag-071.jpg?height=460&width=305&top_left_y=320&top_left_x=183)

Fig. 35 $=A B^{\prime}\left(1+\cos ^{2} \alpha\right)$ and $A K=A B \cos ^{2} \alpha /$ $/\left(1+\cos ^{2} \alpha\right)$. Similarly, the lengths of the segments into which $A B$ is divided by the intersection with the second sphere are found. In the end, we get that the segment $A B$ is divided in the ratio $\cos ^{2} \alpha: \sin ^{2} \alpha: \cos ^{2} \alpha$.