[ Touching Spheres

Two equal spheres $S 1$ and $S 2$ touch each other, and, in addition, each sphere touches both faces $P$ and $Q$ of a dihedral angle of size $2 \alpha$. Sphere $S 1$ touches face $P$ at point $A$, and sphere $S 2$ touches face $Q$ at point $B$. In what ratio is the segment $A B$ divided by the spheres

#
Let $O_1$ and $O_2$ be the centers of spheres $S_1$ and $S_2$ respectively (Fig. 1), $R$ be the radii of the spheres, $H$ be the point of tangency of sphere $S_1$ with plane $Q$, $K$ be the point of tangency of sphere $S_2$ with plane $P$, $M$ and $N$ be the points of intersection of spheres $S_1$ and $S_2$ with segment $AB$ other than $A$ and $B$ respectively. Radii $O_1A$ and $O_1H$ of sphere $S_1$ are perpendicular to the planes $P$ and $Q$ respectively. Let the plane of intersecting lines $O_1A$ and $O_1H$ intersect the edge of the given dihedral angle at point $C$. Then $ACH$ is the linear angle of the given dihedral angle, so $\angle ACH = 2\alpha$. Similarly, if the plane $O_2BK$ intersects the edge of the dihedral angle at point $D$, then $\angle BDK = 2\alpha$. The plane $ACH$ passes through the perpendicular $O_1H$ to the plane $Q$, so these planes are perpendicular, and the perpendicular $AF$ dropped from point $A$ to the plane $Q$ lies in the plane $ACH$. Consider the quadrilateral $ACHO_1$ (Fig. 2), in which

$$
O_1A = O_1H = R, \angle O_1AC = \angle O_1HC = 90^\circ, \angle ACH = 2\alpha, CA = CH.
$$

From the right triangles $O_1CH$ and $AFC$, we find that

$$
CH = O_1H \operatorname{ctg} \angle O_1CH = R \operatorname{ctg} \alpha, \quad CF = AC \cos \angle ACH = CH \cos 2\alpha = R \operatorname{ctg} \alpha \cos 2\alpha,
$$

$$
AF = AC \sin \angle ACH = R \operatorname{ctg} \alpha \sin 2\alpha = 2R \alpha \cos 2\alpha.
$$

Then

$$
FH = CH - CF = R \operatorname{ctg} \alpha - R \operatorname{ctg} \alpha \cos 2\alpha = R \operatorname{ctg} \alpha (1 - \cos 2\alpha) = R \operatorname{ctg} \alpha \cdot 2 \sin 2\alpha = 2R \sin \alpha \cos \alpha.
$$

Note that $CHBD$ and $O_1O_2BH$ are rectangles, so $HB = O_1O_2 = 2R$. From the right triangles $BHF$ and $AFB$, we find that

$$
\begin{gathered}
BF = \sqrt{BH^2 + FH^2} = \sqrt{4R^2 + 4R \sin^2 \alpha \cos^2 \alpha} = 2R \sqrt{1 + \sin^2 \alpha \cos^2 \alpha} \\
AB = \sqrt{BF^2 + AF^2} = \sqrt{4R^2 + 4R \sin^2 \alpha \cos^2 \alpha + 4R^2 \alpha \cos^4 \alpha} = \\
= 2R \sqrt{1 + \sin^2 \alpha \cos^2 \alpha + \cos^4 \alpha} = 2R \sqrt{1 + \cos^2 \alpha (\sin^2 \alpha + \cos^2 \alpha)} = 2R \sqrt{1 + \cos^2 \alpha}.
\end{gathered}
$$

Let $G$ be the projection of point $N$ onto the plane $Q$. Then point $G$ lies on segment $BF$. Consider the right trapezoid $NO_2BG$, in which $O_2N = O_2B = R$. Denote $\angle NBG = \angle ABF = \beta$. If $E$ is the midpoint of the base $BN$ of the isosceles triangle $BO_2N$, then $\angle BO_2E = \angle NBG = \beta$, so

$$
BE = O_2B \sin \beta = R \cdot \frac{AF}{AB} = R \cdot \frac{2R \cos^2 \alpha}{2R \sqrt{1 + \cos^2 \alpha}} = \frac{R \cos^2 \alpha}{\sqrt{1 + \cos^2 \alpha}}
$$

Thus,

$$
AM = BN = 2BE = \frac{2R \cos^2 \alpha}{\sqrt{1 + \cos^2 \alpha}}
$$

$$
MN = AB - 2AM = 2R \sqrt{1 + \cos^2 \alpha} - \frac{4R \cos^2 \alpha}{\sqrt{1 + \cos^2 \alpha}} = \frac{2R \sin^2 \alpha}{\sqrt{1 + \cos^2 \alpha}}
$$

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_786127613a7b2f99b094g-34.jpg?height=262&width=618&top_left_y=172&top_left_x=720)

## Answer

$1: \operatorname{ctg} 2\alpha: 1$.