10.222. Circles of radii $R$ and $r$ touch each other externally. The lateral sides of an isosceles triangle are their common tangents, and the base touches the larger of the circles. Find the base of the triangle.
Solution.

$O$ and $O_{1}$ are the centers of the given circles, $\triangle A B C$ is the triangle mentioned in the problem, $A C=B C$ (Fig. 10.32). $C K$ is the altitude of $\triangle A B C, O$ and $O_{1}$ lie on $C K . F$ and $E$ are the points of tangency of the circles with centers $O_{1}$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0730.jpg?height=632&width=507&top_left_y=80&top_left_x=752)

Fig. 10.32 $O$ respectively with the side $A C$ of the triangle. $O E=O K=R . O_{1} F=r$.

Since $\triangle A K C \sim \triangle O E C$ (by two angles: $\angle A K C=\angle O E C=90^{\circ}$, the angle at vertex $C$ is common), then $\angle C A K=\angle C O E$. Since the circle with center $O$ is tangent to the sides of $\angle C A K$, then $A O$ is its bisector. Let $\angle C O E=\alpha$. Then $\angle O A K=\frac{\alpha}{2}$. Draw $O_{1} N \| A C$. Then $O_{1} N \perp O E, E N=O_{1} F=r$, $O N=O E-E N=R-r, \quad O O_{1}=R+r$. From $\quad \triangle O_{1} N O \quad\left(\angle O_{1} N O=90^{\circ}\right):$ $\cos \alpha=\frac{O N}{O O_{1}}=\frac{R-r}{R+r} \cdot$ From $\triangle A K O\left(\angle A K O=90^{\circ}\right):$

$A K=O K \operatorname{ctg} \angle O A K=R \cdot \operatorname{ctg} \frac{\alpha}{2}=R \frac{1+\cos \alpha}{\sin \alpha}=R \frac{1+\frac{R-r}{R+r}}{\sqrt{1-\left(\frac{R-r}{R+r}\right)^{2}}}=$

$=R \frac{R+r+R-r}{(R+r) \sqrt{\frac{(R+r)^{2}-(R-r)^{2}}{(R+r)^{2}}}}=\frac{2 R^{2}}{\sqrt{4 R r}}=\frac{R^{2}}{\sqrt{R r}}$.

$A B=2 A K=\frac{2 R^{2}}{\sqrt{R r}}$.

Answer: $\frac{2 R^{2}}{\sqrt{R r}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0731.jpg?height=430&width=483&top_left_y=7&top_left_x=128)

Fig. 10.33

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0731.jpg?height=325&width=535&top_left_y=76&top_left_x=661)

Fig. 10.34