Ex. 105. Two circles are constructed on a plane such that each passes through the center of the other. $P$ and $Q$ are the two points of their intersection. A line through point $P$ intersects the first circle at point $A$ and the second circle at point $B$, with $P$ lying between $A$ and $B$. A line through point $Q$, parallel to $AB$, intersects the first circle at point $D$ and the second circle at point $C$. Find the area of quadrilateral $ABCD$, given that $AP=2, PB=1$.
Ex. 105. Answer: $\frac{9 \sqrt{3}}{2}$. Solution. Let the center of the first circle be $O_{1}$, the center of the second circle be $O_{2}$, the angle $\angle A O_{1} P=\alpha, \angle D O_{1} Q=\beta$. Note that the arc $\mathrm{PO}_{2} Q$ is equal to the arc $A D$ and is equal to $120^{\circ}$. Therefore, it is true that $\alpha+\beta=\frac{2 \pi}{3}$. Draw the line $P Q$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_21_91ff4e46083e03d62f1eg-38.jpg?height=443&width=554&top_left_y=755&top_left_x=751)

The area of the parallelogram $A B C D$ consists of two areas of identical trapezoids $A P Q D$ and $P B C Q$. The area of trapezoid $A P Q D$, in turn, consists of four areas of triangles with vertices at point $O_{1}$. Thus, $S_{A B C D}=$

$=2\left(S_{A P O_{1}}+S_{P Q O_{1}}+S_{Q D O_{1}}+S_{D A O_{1}}\right)=2 \cdot \frac{1}{2} r^{2}\left(\sin \alpha+\sin 120^{\circ}+\sin \beta+\sin 120^{\circ}\right)=$ $=r^{2}\left(\sqrt{3}+2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}\right)=r^{2} \sqrt{3}\left(1+\cos \frac{\alpha-\beta}{2}\right)=r^{2} \sqrt{3}\left(1+\cos \left(\alpha-\frac{\pi}{3}\right)\right)$. From the conditions of the problem, it follows that $2=2 r \sin \frac{\alpha}{2}$ and $1=2 r \sin \frac{\beta}{2}$, from which we get that $2 \sin \frac{\beta}{2}=\sin \frac{\alpha}{2}$. Considering that $\frac{\beta}{2}=\frac{\pi}{3}-\frac{\alpha}{2}$, rewrite the equation as $2 \sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right)=\sin \frac{\alpha}{2} \Rightarrow \sqrt{3} \cos \frac{\alpha}{2}=2 \sin \frac{\alpha}{2} \Rightarrow \operatorname{tg} \frac{\alpha}{2}=\frac{\sqrt{3}}{2} \Rightarrow$ $\sin \frac{\alpha}{2}=\frac{\sqrt{3}}{\sqrt{7}}, \cos \frac{\alpha}{2}=\frac{2}{\sqrt{7}}, r=\frac{\sqrt{7}}{\sqrt{3}}$. From this, it follows that $\sin \alpha=\frac{4 \sqrt{3}}{7}, \cos \alpha=$ $\frac{1}{7}, \cos \left(\alpha-\frac{\pi}{3}\right)=\frac{13}{14}$. Thus, $\quad S_{A B C D}=\frac{7}{3} \sqrt{3}\left(1+\frac{13}{14}\right)=\frac{9 \sqrt{3}}{2}$.