Two circles touch each other at $C$; draw a common tangent to the two circles, which touches the circles at points $A$ and $B$. Calculate the sides of the triangle $ABC$, if the radii of the circles ($R$ and $r$) are given.
Let's draw a line parallel to $AB$ from $O$, which intersects $O_{1}B$ at $D$.

From the right triangle $O O_{1} D$:

$$
\overline{O D}^{2}=\overline{A B}^{2}=(R+r)^{2}+(R-r)^{2}
$$

from which

$$
A B=2 \sqrt{R r}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e3118c055f072373a25cg-1.jpg?height=309&width=461&top_left_y=453&top_left_x=821)

From the triangle $O_{1} B C$:

$$
B C=2 R \sin \frac{\alpha}{2}=2 R \sqrt{\frac{1-\cos \alpha}{2}}
$$

But from the triangle $O O_{1} D$:

$$
\cos \alpha=\frac{R-r}{R+r}
$$

so

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

From the triangle $A O C$:

$$
\begin{gathered}
A C=2 r \sin \frac{180^{\circ}-\alpha}{2}=2 r \cos \frac{\alpha}{2}=2 r \sqrt{\frac{1+\cos \alpha}{2}} \\
A C=2 r \sqrt{\frac{R}{R+r}}
\end{gathered}
$$

(Aladár Koós, Budapest.)

The problem was also solved by: Barna D., Bella I., Bobál S., Devecis M., Döme B., Freibauer E., Goldziher K., Juvancz I., Kiss A., Kohn B., Krausz B., Gy. Krisztián, R. Obláth, J. Prohászka, Z. Rehberger, G. Sasvári, Ö. Spitzer, Arthur Weisz, J. Weisz.