5. A trapezoid $ABCD (AD \| BC)$ and a rectangle $A_{1}B_{1}C_{1}D_{1}$ are inscribed in a circle $\Omega$ with a radius of 10, such that $AC \| B_{1}D_{1}, BD \| A_{1}C_{1}$. Find the ratio of the areas of $ABCD$ and $A_{1}B_{1}C_{1}D_{1}$, given that $AD=16, BC=12$.
Answer: $\frac{49}{50}$ or $\frac{1}{2}$.

Solution. Draw a line through the center of the circle $O$ perpendicular to the bases of the trapezoid. Let it intersect $A D$ and $B C$ at points $N$ and $M$ respectively. Since a diameter perpendicular to a chord bisects the chord, $B M=M C=6, A N=N D=8$. By the Pythagorean theorem from triangles $O N D$ and $O M C$, we find that $O N=\sqrt{O D^{2}-D N^{2}}=6, O M=\sqrt{O C^{2}-M C^{2}}=8$. There are two cases.

1) Point $O$ does not lie on the segment $M N$. Then the height of the trapezoid is $M N=O M-O N=2$. Let $H$ be the foot of the perpendicular dropped from vertex $B$ to the base $A D$. Since the trapezoid inscribed in a circle is isosceles, $D H=\frac{A D+B C}{2}=14$. Then $B D=\sqrt{D H^{2}+B H^{2}}=\sqrt{200}=10 \sqrt{2}$. The area of any quadrilateral is half the product of the diagonals multiplied by the sine of the angle between them. Since the diagonals of the rectangle $A_{1} B_{1} C_{1} D_{1}$ are parallel to the diagonals of the trapezoid $A B C D$, the angle between them is equal to the angle between the diagonals of the trapezoid. Let this angle be $\psi$. Moreover, the diagonals of the rectangle inscribed in a circle are its diameters, so $A_{1} C_{1}=B_{1} D_{1}=20$. Then $S_{A B C D}=\frac{1}{2} A C \cdot B D \sin \psi=100 \sin \psi ; S_{A_{1} B_{1} C_{1} D_{1}}=\frac{1}{2} A_{1} C_{1} \cdot B_{1} D_{1} \sin \psi=$ $\frac{1}{2} \cdot 20 \cdot 20 \sin \psi=200 \sin \psi$. Therefore, the ratio of the areas is $\frac{100 \sin \psi}{200 \sin \psi}=\frac{1}{2}$.
2) Point $O$ lies on the segment $M N$. Then $M N=O N+O M=14$. Similarly to the first case, we find that $D H=14, B D=\sqrt{D H^{2}+B H^{2}}=14 \sqrt{2}, S_{A B C D}=\frac{1}{2} A C \cdot B D \sin \varphi=196 \sin \varphi ; S_{A_{1} B_{1} C_{1} D_{1}}=\frac{1}{2} A_{1} C_{1}$. $B_{1} D_{1} \sin \varphi=\frac{1}{2} \cdot 20 \cdot 20 \sin \varphi=200 \sin \varphi$, where $\varphi$ is the angle between the diagonals of the trapezoid. Hence, the ratio of the areas is $\frac{196 \sin \varphi}{200 \sin \varphi}=\frac{49}{50}$.