$\left[\begin{array}{l}\text { B } \quad \text { Inscribed and Circumscribed Circles } \\ \text { [Trigonometric Ratios in a Right Triangle] }\end{array}\right.$

In triangle $ABC$, a circle with center $O$ is inscribed. The line $BO$ intersects this circle at points $M$ and $N$, and the segments $AO$ and $CO$ intersect the circle at points $P$ and $Q$, respectively. Find the ratio of the areas of triangles $MNP$ and $MQN$, if $\angle A=\alpha, \angle C=\gamma$.
Express the distance from points $Q$ and $P$ to the line $B O$ in terms of the radius of the circle.

## Solution

Let $R$ be the radius of the circle, $B D$ be the angle bisector of triangle $A B C$, and $K$ and $F$ be the projections of points $P$ and $Q$ onto the line $B O$. Then,

$$
\angle P O K = \angle B A O + \angle A B D = \frac{\alpha}{2} + \frac{\pi - \alpha - \gamma}{2} = \frac{\pi}{2} - \frac{\gamma}{2}
$$

Therefore,

$$
P K = P O \sin \angle P O K = P O \sin \left(\frac{\pi}{2} - \frac{\gamma}{2}\right) = R \cos \frac{\gamma}{2}
$$

Similarly, $Q F = R \cos \frac{\alpha}{2}$. Thus,

$$
\frac{S_{\triangle M N P}}{S_{\triangle M N Q}} = \frac{P K}{Q F} = \frac{\cos \frac{\gamma}{2}}{\cos \frac{\alpha}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-01.jpg?height=480&width=1054&top_left_y=1575&top_left_x=502)

Answer

$\frac{\cos \frac{\gamma}{2}}{\cos \frac{\alpha}{2}} \cdot$