3. Two lines $l_{1}$ and $l_{2}$, the angle between which is $\alpha$, are common internal tangents to circles of radii $R_{1}$ and $R_{2}$. A common external tangent to the circles intersects the lines $l_{1}$ and $l_{2}$ at points $A$ and $B$ respectively. Find the length of the segment $A B$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_1ce6d950be05a9f52a95g-2.jpg?height=594&width=1105&top_left_y=1211&top_left_x=498)

The notations are shown in the figure. The external tangent is the line $A B$. Let the length of the segment $A B$ be denoted by $a$. Suppose the segments of the tangents drawn from points $A$ and $B$ are $x$ and $y$, respectively. Considering the given data, we find the lengths of the segments of the internal tangents drawn from the point of intersection $C$. $C F = C P = R_{1} \operatorname{ctg} \frac{\alpha}{2}, C E = C Q = R_{2} \operatorname{ctg} \frac{\alpha}{2}$. The segments of the tangents $A Q$ and $A B$, drawn to the second circle from point $A$, are equal to each other. Similarly, the segments $B P$ and $B K$ are equal. This leads to a system of two equations

$$
\left\{\begin{array}{l}
x + R_{1} \operatorname{ctg} \frac{\alpha}{2} + R_{2} \operatorname{ctg} \frac{\alpha}{2} = a + y \\
y + R_{1} \operatorname{ctg} \frac{\alpha}{2} + R_{2} \operatorname{ctg} \frac{\alpha}{2} = a + x
\end{array}\right.
$$

From the equations, it follows that $x = y$, hence, $a = \left(R_{1} + R_{2}\right) \operatorname{ctg} \frac{\alpha}{2}$.

Answer. $\left(R_{1} + R_{2}\right) \operatorname{ctg} \frac{\alpha}{2}$