In triangle $A B C$ with perimeter $2 p$, side $A C$ is equal to $a$, and the acute angle $A B C$ is $\boldsymbol{\alpha}$. The incircle of triangle $A B C$ with center $O$ touches side $B C$ at point $K$. Find the area of triangle $B O K$.

#
The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side of the triangle.

## Solution

Since

$$
B K=p-A C=p-a, O K=B K \operatorname{tg} \frac{\alpha}{2}=(p-a) \operatorname{tg} \frac{\alpha}{2}
$$

then,

$$
S_{\triangle \mathrm{BOK}}=\frac{1}{2} B K \cdot O K=\frac{1}{2}(p-a)^{2} \operatorname{tg} \frac{\alpha}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-11.jpg?height=383&width=734&top_left_y=806&top_left_x=659)

## Answer

$\frac{1}{2}(p-a)^{2} \operatorname{tg} \frac{\alpha}{2}$.