In a circle with a radius of 4 cm and center at point $O$, two diameters $A B$ and $C D$ are drawn such that the angle $\angle A O C=\frac{\pi}{9}$. From point $M$, lying on the circle and distinct from points $A, B, C$, and $D$, perpendiculars $M Q$ and $M P$ are drawn to diameters $A B$ and $C D$ respectively (point $Q$ lies on $A B$, and point $P$ on $C D$) such that $\angle M P Q=\frac{2 \pi}{9}$. Find the area of triangle $M P Q$.
Prove that points $P$ and $Q$ lie on a circle with diameter $O M$. If $r$ is the radius of this circle, then

$$
S_{\triangle \mathrm{MPQ}}=2 r^{2} \sin \angle Q M P \sin \angle M P Q \sin \angle M Q P
$$

## Solution

First, we will prove that the area $S$ of a triangle can be calculated using the formula:

$$
S=2 R^{2} \sin \alpha \sin \beta \sin \gamma
$$

where $\alpha, \beta, \gamma$ are the angles of the triangle, and $R$ is the radius of its circumscribed circle.

Indeed, if $b$ and $c$ are the sides of the triangle opposite the angles $\beta$ and $\gamma$, respectively, then

$$
S=\frac{1}{2} \cdot b c \cdot \sin \alpha=\frac{1}{2} \cdot 2 R \sin \beta \cdot 2 R \sin \gamma \cdot \sin \alpha=2 R^{2} \sin \alpha \sin \beta \sin \gamma
$$

which is what we needed to prove.

From the condition of the given problem, it follows that the segment $O M$ is seen from points $P$ and $Q$ at a right angle. This means that points $P$ and $Q$ lie on a circle with diameter $O M$. If $r$ is the radius of this circle, then $r=\frac{1}{2} O M=2$.

By the theorem of inscribed angles subtending the same arc,

$$
\angle Q M P=\angle Q O P=\angle A O C=\frac{\pi}{9}
$$

therefore,

$$
\angle M Q P=\pi-\angle Q M P-\angle M P Q=\pi-\frac{\pi}{9}-\frac{2 \pi}{9}=\pi-\frac{\pi}{3}
$$

By the previously proven formula,

$$
S_{\triangle \mathrm{MPQ}}=2 r^{2} \sin \angle Q M P \sin \angle M P Q \sin \angle M Q P=8 \sin \frac{\pi}{9} \sin \frac{2 \pi}{9} \sin \frac{\pi}{3}=4 \sqrt{3} \sin \frac{\pi}{9} \sin \frac{2 \pi}{9}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-40.jpg?height=771&width=826&top_left_y=975&top_left_x=619)

Answer

$S=4 \sqrt{3} \sin \frac{\pi}{9} \sin \frac{2 \pi}{9}$.

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