22. The diagonals of quadrilateral $A B C D$ intersect at point $M$, the angle between them is $\alpha$. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the circumcircles of triangles $A B M, B C M, C D M$, $D A M$ respectively. Determine the ratio of the areas of quadrilateral $A B C D$ and $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$.
22. Note that $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is a parallelogram with angles $\alpha$ and $\pi-\alpha\left(O_{1} O_{4} \perp A C\right.$ and $O_{2} O_{3} \perp A C$, hence, $O_{1} O_{4} O_{2} O_{3}$ and so on). If $K$ is the midpoint of $A M$ and $L$ is the midpoint of $M C$, then $\left|O_{3} O_{4}\right|=\frac{|K L|}{\sin \alpha}=$ $=\frac{|A C|}{2 \sin \alpha}$. Similarly, $\left|O_{2} O_{3}\right|=\frac{B D}{2 \sin \alpha}$, therefore,

$$
S_{O_{1} O_{2} O_{3} O_{4}}=\frac{|A B| \cdot|B D| \sin \alpha}{4 \sin ^{2} \alpha}=\frac{S_{A B C D}}{2 \sin ^{2} \alpha}
$$

Answer: $2 \sin ^{2} \alpha$.