7. In triangle $A B C$, side $A C$ is equal to 6, and angle $A C B$ is $120^{\circ}$. Circle $\Omega$ with radius $\sqrt{3}$ touches sides $B C$ and $A C$ of triangle $A B C$ at points $K$ and $L$ respectively, and intersects side $A B$ at points $M$ and $N$ ($M$ lies between $A$ and $N$) such that segment $M K$ is parallel to $A C$. Find the lengths of segments $C L, M K, A B$ and the area of triangle $A N L$.
Answer: $C L=1, M K=3, A B=2 \sqrt{13}, S_{\triangle A L N}=\frac{125 \sqrt{3}}{52}$.

Solution. Let the center of the circle be denoted as $O$, the foot of the altitude from vertex $C$ of the triangle as $H$, and the angle $B A C$ as $\alpha$.

The center of the circle inscribed in an angle lies on the bisector of that angle, so $\angle O C L = \frac{1}{2} \angle A C B = 60^{\circ}$. Since the radius drawn to the point of tangency is perpendicular to the tangent, $C L = O L \operatorname{ctg} \angle O C L = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1$. By the sum of the angles in quadrilateral $O L C K$, we find that $\angle L O K = 60^{\circ}$.

Since $O L \perp A C$ and $M K \| A C$, then $O L \perp M K$. Triangle $M O K$ is isosceles, and the height in it is also the bisector, so $\angle M O K = 2 \angle L O K = 120^{\circ}$. Hence, $M K = 2 \cdot M O \cdot \sin 60^{\circ} = 3$. $\angle M O L = \frac{1}{2} \angle M O K = 60^{\circ}, M O = L O$, therefore, triangle $M O L$ is equilateral, $\angle M L O = 60^{\circ}, M L = \sqrt{3}$. Then $\angle A L M = \angle A L O - \angle M L O = 30^{\circ}$.

Consider triangle $A L M$. By the cosine theorem, we get $A M^{2} = A L^{2} + L M^{2} - 2 \cdot A L \cdot L M \cdot \frac{\sqrt{3}}{2} = 13$.

By the tangent-secant theorem, $A L^{2} = A M \cdot A N$, from which $A N = \frac{25}{\sqrt{13}}$. The triangles $A L M$ and $A L N$ share a common height from vertex $L$, so their areas are in the ratio of their bases, i.e., as $A M : A N$. Therefore, $S_{\triangle A L N} = S_{\triangle A L M} \cdot \frac{A M}{A N} = \frac{1}{2} \cdot A L \cdot L M \cdot \sin 30^{\circ} \cdot \frac{25}{13} = \frac{125 \sqrt{3}}{52}$.

Since $K M \| A C$, triangles $B K M$ and $B C A$ are similar, with the similarity ratio being $K M : C A = 3 : 6$. Hence, $\frac{B M}{B A} = \frac{1}{2}, \frac{B A - \sqrt{13}}{B A} = \frac{1}{2}, A B = 2 \sqrt{13}$.

## TICKET 6