Sine Theorem

A circle with center at point $O$ lies on the hypotenuse $A C$ of a right triangle $A B C$, touching its legs $A B$ and $B C$. Find $A C$, if it is known that $A M=\frac{20}{9}, A N: M N=6: 1$, where $M$ is the point of tangency of $A B$ with the circle, and $N$ is the point of intersection of the circle with $A C$, located between points $A$ and $O$.
Denote $\angle A O M=\angle A C B=\alpha$. Express the angles of triangle $A M N$ in terms of $\alpha$ and use the Law of Sines.

## Solution

Denote $\angle A O M=\angle A C B=\alpha$. Then

$$
\angle M A C=90^{\circ}-\alpha, \angle A M N=\frac{\alpha}{2}
$$

(angle between the tangent $A M$ and the chord $M N$). By the Law of Sines in triangle $A M N$, we find that $\frac{\sin \frac{\alpha}{2}}{\cos \alpha}=6$.

From this equation, it follows that

$$
\sin \frac{\alpha}{2}=\frac{2}{3}, \cos \frac{\alpha}{2}=\frac{\sqrt{5}}{3}, \sin \alpha=\frac{4 \sqrt{5}}{9}
$$

From triangle $A M O$, we find that

$$
M O=\frac{A M}{\operatorname{tg} \alpha}=\frac{\sqrt{5}}{9}
$$

Therefore,

$$
A C=\frac{A B}{\sin \alpha}=\frac{A M+M B}{\sin \alpha}=\frac{A M+M O}{\sin \alpha}=\frac{9\left(\frac{20}{9}+\frac{\sqrt{5}}{9}\right)}{4 \sqrt{5}}=\sqrt{5}+\frac{1}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6ed8f1205731b80a0039g-29.jpg?height=394&width=680&top_left_y=0&top_left_x=695)

## Answer

$\sqrt{5}+\frac{1}{4}$