In parallelogram $A B C D$, the acute angle $B A D$ is equal to $\alpha$. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the centers of the circumcircles of triangles $D A B, D A C, D B C, A B C$ respectively. Find the ratio of the area of quadrilateral $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ to the area of parallelogram $A B C D$.
Draw four perpendicular bisectors to the sides of a parallelogram. Their intersection forms a parallelogram $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$.

## Solution

Consider the quadrilateral whose vertices are the points of intersection of the four perpendicular bisectors of the sides of the given parallelogram. This is also a parallelogram, and its vertices are points $O_{1}, O_{2}, O_{3}$, and $O_{4}$. Let $O$ be the common center of these parallelograms.

Let the acute angle between the diagonals of parallelogram $A B C D$ be $\varphi$. Then the acute angle between the diagonals of parallelogram $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3} \mathrm{O}_{4}$ is also $\varphi$ (acute angles with correspondingly perpendicular sides are equal).

Consider the circle centered at $O_{1}$, circumscribed around triangle $A B D$. The central angle $B O_{1} D$ is twice the inscribed angle $B A D$, so

$$
\angle B O_{1} O=\frac{1}{2} \angle B O_{1} D=\angle B A D=\alpha
$$

From the right triangle $B O_{1} O$, we find that $\frac{O O_{1}}{B O}=\operatorname{ctg} \alpha$. Similarly, we find that $\frac{O O_{2}}{A O}=\operatorname{ctg} \alpha$. Therefore,

$$
\begin{gathered}
\frac{S_{O_{1} O_{2} O_{3} O_{4}}}{S_{A B C D}}=\frac{\frac{1}{2} O_{1} O_{3} \cdot O_{2} O_{4} \sin \varphi}{\frac{1}{2} B D \cdot A C \sin \varphi}= \\
=\frac{O_{1} O_{3}}{B D} \cdot \frac{O_{2} O_{4}}{A C}=\frac{O_{1} O}{B O} \cdot \frac{O O_{2}}{A O}=\operatorname{ctg} \alpha \cdot \operatorname{ctg} \alpha=\operatorname{ctg}^{2} \alpha
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-50.jpg?height=468&width=809&top_left_y=1138&top_left_x=636)

## Answer

$\operatorname{ctg}^{2} \alpha$.

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