Two circles with radii $R$ and $r(R>r)$ touch externally at point $C$. A common external tangent $A B$ is drawn to them, where $A$ and $B$ are the points of tangency. Find the sides of triangle $A B C$.
Drop a perpendicular from the center of the smaller circle to the radius of the larger circle, extended to the point of tangency with the common external tangent. Find the sides and the cosine of the acute angle of the resulting right triangle.

## Solution

Let $O_{1}$ and $O_{2}$ be the centers of the circles with radii $r$ and $R$, $A$ be a point on the first circle, $B$ be on the second, and $K$ be the projection of point $O_{1}$ onto $B O_{2}$. Then

$$
\begin{gathered}
K O_{2}=B O_{2}-A O_{1}=R-r, O_{1} O_{2}=r+R \\
A B=K O_{1}=\sqrt{O_{1} O_{2}^{2}-K O_{2}^{2}}=\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \sqrt{R r}
\end{gathered}
$$

Let $\angle A B C=\frac{1}{2} \angle B O_{2} C=\alpha$. Then

$$
\cos 2 \alpha=\frac{K O_{2}}{O_{1} O_{2}}=\frac{R-r}{R+r}, \cos \alpha=\sqrt{\frac{1+\cos 2 \alpha}{2}}=\sqrt{\frac{1+\frac{R-r}{R+r}}{2}}=\sqrt{\frac{R}{R+r}}
$$

$$
B C=A B \cos \alpha=2 \sqrt{R r} \sqrt{\frac{R}{R+r}}=2 R \sqrt{\frac{r}{R+r}}
$$

Similarly, we find $A C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-52.jpg?height=426&width=604&top_left_y=236&top_left_x=727)

## Answer

$2 \sqrt{R r}, 2 r \sqrt{\frac{R}{R+r}}, 2 R \sqrt{\frac{r}{R+r}}$.

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