# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_aadf511cc872419fdfecg-04.jpg?height=308&width=468&top_left_y=1825&top_left_x=197)

$\angle C L A = \angle C A L = \angle C M A = \alpha, L A = M A = A K \Rightarrow \triangle L M K$ is a right triangle with $\angle L M K = 90^{\circ}$.

Then $\angle A M K = 90^{\circ} - \alpha = \angle B A K = \angle B K A = \beta \Rightarrow$

$$
C B = 2 R = \frac{A B}{\sin \beta} = \frac{B K}{\sin \angle K} \Rightarrow K B = C B \cdot \sin \angle K = 4 \cdot \frac{\sqrt{6}}{4} = \sqrt{6}
$$

Answer: $\sqrt{6}$.

| Criterion | Points |
| :--- | :---: |
| The problem is solved correctly. | 20 |
| It is proven that $C B$ is the diameter. | 15 |
| It is proven that $\triangle L M K$ is a right triangle. | 10 |
| It is proven that either $\triangle L A M$ is isosceles, or $\triangle M A K$ is isosceles, or $\triangle A B K$ is isosceles. | 5 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |
5. (20 points) Given triangle $K L M$. A circle is drawn passing through point $M$, touching segment $L K$ at point $A$, which is its midpoint, and intersecting sides $M L$ and $M K$ at points $C$ and $B$, respectively, such that $C B=4$, point $C$ is equidistant from points $A$ and $L$, and $\cos \angle K=\frac{\sqrt{10}}{4}$. Find the length of segment $B K$.