5. 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 $A C, K C=1, A L=6$. Find $\angle A C B$, the lengths of segments $M K, A B$, and the area of triangle $C M N$.
Answer: $\angle A C B=\frac{2 \pi}{3}, M K=3, A B=\frac{7 \sqrt{21}}{4}, S_{\triangle C M N}=\frac{5 \sqrt{3}}{4}$.

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 $\alpha$.

Since the radius drawn to the point of tangency is perpendicular to the tangent, $\operatorname{to} \operatorname{tg} \angle O C K=$ $\frac{O K}{C K}=\sqrt{3}$ and $\angle O C K=60^{\circ}$. The center of the circle inscribed in an angle lies on the bisector of that angle, so $\angle A C B=2 \angle O C K=120^{\circ}$. Then, by the sum of the angles of the 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, the height in it is also the bisector, so $\angle M O K=2 \angle L O K=120^{\circ}$. Hence, $M K=2 \cdot M O \cdot \sin 60^{\circ}=3$.

(c) MIPT, 2018
$\angle M O L=\frac{1}{2} \angle M O K=60^{\circ}, M O=L O$, 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 $A M^{2}=A L^{2}+L M^{2}-2$. $A L \cdot L M \cdot \frac{\sqrt{3}}{2}=21$. By the sine theorem, $\frac{A M}{\sin 30^{\circ}}=\frac{L M}{\sin \alpha}$, from which $\sin \alpha=\frac{1}{2 \sqrt{7}}$. Since angle $\alpha$ lies opposite the smaller side of the triangle, it is acute. From triangle $A C H$, we get $C H=A C \sin \alpha=\frac{\sqrt{7}}{2}$

By the tangent and secant theorem, $A L^{2}=A M \cdot A N$, from which $A N=\frac{36}{\sqrt{21}}, M N=A N-A M=\frac{15}{\sqrt{21}}$. Then the area of triangle $C M N$ is $\frac{1}{2} \cdot C H \cdot M N=\frac{5 \sqrt{3}}{4}$.

Since $K M \| A C$, triangles $B K M$ and $B C A$ are similar, with the similarity ratio being $K M: C A=3: 7$. Therefore, $\frac{B M}{B A}=\frac{3}{7}, \frac{B A-\sqrt{21}}{B A}=\frac{3}{7}, A B=\frac{7 \sqrt{21}}{4}$.