[ Inscribed and Circumscribed Circles ]

In an isosceles triangle $A B C$ with base $A C$, a circle is inscribed, touching the lateral side $A B$ at point $M$. From point $M$, a perpendicular $M L$ is dropped to side $A C$. Find the measure of angle $C$, given that the area of triangle $A B C$ is 1, and the area of quadrilateral $L M B C$ is $s$.

#
Triangle $A M L$ is similar to triangle $A B K$.

## Solution

Let the required angle be $\alpha$. Let $K$ be the point of tangency of the inscribed circle with side $A C$. Then $K$ is the midpoint of $A C$. Triangles $A M L$ and $A B K$ are similar with a ratio of ${ }^{A M} /{ }_{A B}={ }^{A K} /_{A B}=\cos \alpha$. Therefore, $1-s=$ $S_{A M L}=\cos ^{2} \alpha S_{A B K}=1 / 2 \cos ^{2} \alpha$. Consequently, $\cos ^{2} \alpha=2(1-s)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_65daf5394c369f17193bg-16.jpg?height=351&width=460&top_left_y=842&top_left_x=799)

## Answer

$\arccos \sqrt{2(1-s)}$.