6. Two circles touch each other and the sides of triangle $A B C$. The first circle with radius $\frac{1}{18}$ touches sides $A B$ and $A C$ at points $L$ and $K$, the second circle with radius $\frac{2}{9}$ touches sides $A C$ and $B C$ at points $N$ and $M$. Find the area of triangle $A B C$, if $A L=\frac{1}{9}, C M=\frac{1}{6}$ (20 points)
Solution: $a=\frac{1}{9}, \quad b=\frac{1}{6}, \quad r=\frac{1}{18}, \quad R=\frac{2}{9}$ $K N=2 \sqrt{r R}=\frac{2}{9}, \quad A C=a+K N+b=\frac{1}{2}$

$\operatorname{tg} \frac{A}{2}=\frac{r}{a}, \quad \operatorname{tg} \frac{C}{2}=\frac{R}{b}$

$\cos ^{2} \frac{A}{2}=\frac{a^{2}}{a^{2}+r^{2}}, \quad \cos ^{2} \frac{C}{2}=\frac{b^{2}}{b^{2}+R^{2}}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_c995faa73b19f5c426fdg-04.jpg?height=713&width=865&top_left_y=1865&top_left_x=1121)

Fig. 1
$\cos A=2 \cos ^{2} \frac{A}{2}-1=\frac{2 a^{2}}{a^{2}+r^{2}}-1=\frac{a^{2}-r^{2}}{a^{2}+r^{2}}=\frac{3}{5}$,

$\cos C=2 \cos ^{2} \frac{C}{2}-1=\frac{2 b^{2}}{b^{2}+R^{2}}-1=\frac{b^{2}-R^{2}}{b^{2}+R^{2}}=-\frac{7}{25}$,

$\sin A=\frac{2 a r}{a^{2}+r^{2}}=\frac{4}{5}, \quad \sin C=\frac{2 b R}{b^{2}+R^{2}}=\frac{24}{25}, \quad \sin B=\sin (A+C)=\sin A \cos C+\cos A \sin C=\frac{44}{125}$.

Sine theorem: $\frac{A B}{\sin C}=\frac{A C}{\sin B}$.

$A B=\frac{A C \sin C}{\sin B}, \quad S_{A B C}=\frac{1}{2} A B \cdot A C \sin A=\frac{A C^{2} \sin A \sin C}{2 \sin B}=\frac{4 \cdot 24 \cdot 125}{8 \cdot 5 \cdot 25 \cdot 44}=\frac{3}{11} . \quad$ Answer: $\frac{3}{11}$.

## Solution for Variant № 10-1