12.392 In the sector $P O Q$ of radius $R$ with a central angle $\alpha$, a rectangle is inscribed; two of its vertices lie on the arc of the sector, and the other two on the radii $P O$ and $P Q$. Find the area of the rectangle, if the acute angle between its diagonals is $\beta$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-282.jpg?height=269&width=345&top_left_y=1283&top_left_x=73)

Let the rectangle $ABCD$ be inscribed in the given sector $POQ$, and let $N$ be the midpoint of side $AD$. Then $ON \perp AD$ (as a radius passing through the midpoint of a chord) $\Rightarrow BM = MC$, where $M$ is the intersection point of $ON$ with $BC \Rightarrow \triangle OMB = \triangle OMC \Rightarrow$

$$
\Rightarrow \angle BOM = \frac{\alpha}{2} \Rightarrow OM = BM \cdot \operatorname{ctg} \frac{\alpha}{2} = \frac{1}{2} BC \cdot \operatorname{ctg} \frac{\alpha}{2}
$$

$$
\begin{gathered}
ON = OM + MN = \frac{BC}{2} \cdot \operatorname{ctg} \frac{\alpha}{2} + AB \Rightarrow R^2 = OA^2 = AN^2 + ON^2 = \\
= \left(\frac{BC}{2}\right)^2 + \left(\frac{BC}{2} \cdot \operatorname{ctg} \frac{\alpha}{2} + AB\right)^2
\end{gathered}
$$

Now let $K$ be the intersection point of the diagonals $AC$ and $BD$. Since $\angle CKD = \angle CAD + \angle BDA = 2 \angle CAD$, then $\angle CAD = \frac{\beta}{2}$ or $\angle CAD = 90^\circ - \frac{\beta}{2}$ (depending on whether $\angle CKD$ is acute or obtuse). Let $\gamma = \angle CAD$. Then

$$
\begin{gathered}
AB = CD = BC \cdot \operatorname{tg} \gamma \Rightarrow 4R^2 = BC^2 + \left(BC \cdot \operatorname{ctg} \frac{\alpha}{2} + 2BC \cdot \operatorname{tg} \gamma\right)^2 \Rightarrow \\
\Rightarrow BC^2 = \frac{4R^2}{1 + \left(\operatorname{ctg} \frac{\alpha}{2} + 2 \operatorname{tg} \gamma\right)^2}
\end{gathered}
$$

But $S_{ABCD} = AB \cdot BC = BC^2 \cdot \operatorname{tg} \gamma \Rightarrow S_{ABCD} = \frac{4R^2 \cdot \operatorname{tg} \gamma}{1 + \left(\operatorname{ctg} \frac{\alpha}{2} + 2 \operatorname{tg} \gamma\right)^2}$.

Considering that $\gamma = \frac{\beta}{2}$ or $\gamma = 90^\circ - \frac{\beta}{2}$, we have

Answer: $\frac{4R^2 \cdot \operatorname{tg} \frac{\beta}{2}}{1 + \left(\operatorname{ctg} \frac{\alpha}{2} + 2 \operatorname{tg} \frac{\beta}{2}\right)^2}$ or $\frac{4R^2 \cdot \operatorname{ctg} \frac{\beta}{2}}{1 + \left(\operatorname{ctg} \frac{\alpha}{2} + 2 \operatorname{ctg} \frac{\beta}{2}\right)^2}$.