6. [5 points] a) 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 2. Find the ratio of the radii of the circles $\omega_{2}$ and $\omega_{1}$.

b) Find these radii if it is additionally known that $B D=2$.
Answer: a) $\frac{R_{2}}{R_{1}}=\frac{3}{2} ;$ b) $R_{1}=\sqrt{\frac{2}{3}}, R_{2}=\sqrt{\frac{3}{2}}$.

Solution. a) Let $R_{1}, R_{2}$ be the radii of 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 $S_{B O_{1} C O_{2}}=$ ${ }_{2}^{2} S_{O_{1} B O_{2}}=2 \cdot \frac{1}{2} \cdot B O_{1} \cdot B O_{2} \cdot \sin \alpha=R_{1} R_{2} \sin \alpha$. 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, $\frac{S_{B O_{1} C O_{2}}}{S_{B O_{2}} E}=\frac{R_{1}}{R_{2} \cos \alpha}=2$, that is, $\frac{1}{2} R_{1}=R_{2} \cos \alpha$. Further, $A B=D P$, so $2 R_{1}=R_{2}+R_{2} \cos \alpha$, hence $2 R_{1}=R_{2}+\frac{1}{2} R_{1}$, and $\frac{R_{2}}{R_{1}}=\frac{3}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ffd4340bf000b94108d2g-30.jpg?height=534&width=780&top_left_y=178&top_left_x=638)

Fig. 3: Variant 13, Problem 6

b) From the right triangle $A B D$ we get $B D^{2}=A B^{2}+A D^{2}$, that is, $B D^{2}=4 R_{1}^{2}+$ $B P^{2}=4 R_{1}^{2}+\left(R_{2} \sin \alpha\right)^{2}=4 R_{1}^{2}+R_{2}^{2}-\left(R_{2} \cos \alpha\right)^{2}=4 R_{1}^{2}+R_{2}^{2}-\left(2 R_{1}-R_{2}\right)^{2}=4 R_{1} R_{2}$. Therefore, $\frac{R_{2}}{R_{1}}=\frac{3}{2}$ and $4 R_{1} R_{2}=2^{2}=4$. Hence, $R_{1}=\sqrt{\frac{2}{3}}, R_{2}=\sqrt{\frac{3}{2}}$.