## [ Properties and characteristics of the tangent

A right triangle $ABC$ has a perimeter of 54, and the leg $AC$ is greater than 10. A circle with a radius of 6, whose center lies on the leg $BC$, is tangent to the lines $AB$ and $AC$. Find the area of triangle $ABC$.

#
Let $O$ be the center of the given circle. Denote $A C=x$ and express the tangent of angle $O A C$ in terms of $x$.

## Solution

Let $O$ be the center of the given circle, $M$ be the point of tangency of the circle with the line $A B$. Denote $A C=A M=x, \angle B A C=$ $\alpha$. Then

$$
\begin{gathered}
\angle O A C=\frac{\alpha}{2}, \operatorname{tg} \frac{\alpha}{2}=\frac{O C}{A C}=\frac{6}{x} \\
O B=\frac{O M}{\cos \angle M O B}=\frac{O M}{\cos \alpha}=\frac{6\left(1+\operatorname{tg}^{2} \frac{\alpha}{2}\right)}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{6\left(x^{2}+36\right)}{x^{2}-36}
\end{gathered}
$$

$$
B M=O M \cdot \operatorname{tg} \alpha=\frac{72 x}{x^{2}-36}
$$

Since the perimeter of triangle $A B C$ is 54, then $A C+A M+C O+O B+B M=54$, or

$$
2 x+6+\frac{6\left(x^{2}+36\right)}{x^{2}-36}+\frac{72 x}{x^{2}-36}
$$

or

$$
x^{3}-21 x^{2}+27 \cdot 36=0
$$

Divide both sides of this equation by 27 and let $t=\frac{x}{3}$. We get the equation

$$
t^{3}-7 t^{2}+36=0
$$

Its roots are: (-2), 6, 3. Since $x=3 t>10$, then $x=18$. Then

$$
\operatorname{tg} \frac{\alpha}{2}=\frac{1}{3}, \cos \alpha=\frac{4}{5}, O B=\frac{15}{2}, B C=6+\frac{15}{2}=\frac{27}{2}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} \cdot A C \cdot B C=\frac{1}{2} \cdot 18 \cdot \frac{27}{2}=\frac{243}{2}
$$

## Answer

$\frac{243}{2}$.