6. [4 points] Two parallel lines $\ell_{1}$ and $\ell_{2}$ touch the circle $\omega_{1}$ with center $O_{1}$ at points $A$ and $B$ respectively. The circle $\omega_{2}$ with center $O_{2}$ touches the line $\ell_{1}$ at point $D$, intersects the line $\ell_{2}$ at points $B$ and $E$, and intersects the circle $\omega_{1}$ again at point $C$ (with point $O_{2}$ lying between the lines $\ell_{1}$ and $\ell_{2}$). It is known that the ratio of the area of quadrilateral $B O_{1} C O_{2}$ to the area of triangle $O_{2} B E$ is $\frac{6}{5}$. Find the ratio of the radii of the circles $\omega_{2}$ and $\omega_{1}$.
Answer: $\frac{R_{2}}{R_{1}}=\frac{7}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_734640e0e89abd1c02f1g-15.jpg?height=542&width=780&top_left_y=2082&top_left_x=638)

Fig. 3: Option 10, Problem 6

Solution. a) Let $R_{1}, R_{2}$ be the radii of the circles $\omega_{1}$, $\omega_{2}$ respectively, $\angle O_{1} B O_{2}=\alpha$, and the lines $D O_{2}$ and $\ell_{2}$ intersect at point $P$. Then, from the condition of tangency, $\angle O_{2} B E=\frac{\pi}{2}-\alpha$, and $\angle B O_{2} P=\frac{\pi}{2}-\angle O_{2} B E=\alpha$. Triangles $B O_{1} O_{2}$ and $C O_{1} O_{2}$ are equal by three sides, so $S_{B O_{1} C O_{2}}=2 S_{O_{1} B O_{2}}=2 \cdot \frac{1}{2} \cdot B O_{1} \cdot B O_{2} \cdot \sin \alpha=R_{1} R_{2} \sin \alpha$. The area of triangle $B O_{2} E$ is $\frac{1}{2} O_{2} P \cdot B E=\frac{1}{2} R_{2} \cos \alpha \cdot 2 R_{2} \sin \alpha=R_{2}^{2} \sin \alpha \cos \alpha$. Therefore, $\frac{6}{5}=\frac{S_{B O_{1} O_{2}}}{S_{B O_{2} E}}=\frac{R_{1}}{R_{2} \cos \alpha}$, which means $\frac{5}{6} R_{1}=R_{2} \cos \alpha$. Further, $A B=D P$, so $2 R_{1}=R_{2}+R_{2} \cos \alpha$, hence $2 R_{1}=R_{2}+\frac{5}{6} R_{1}$, and $\frac{R_{2}}{R_{1}}=\frac{7}{6}$.