5. (20 points) A circle passes through the vertices $A$ and $C$ of an isosceles triangle $ABC (AB = BC)$ and intersects the sides $AB$ and $BC$ at points $M$ and $N$, respectively. $MK$, a chord of this circle, equal in length to $2 \sqrt{5}$, contains point $H$, which lies on $AC$ and is the foot of the altitude of triangle $ABC$. The line passing through point $C$ and perpendicular to $BC$ intersects the line $MN$ at point $L$. Find the radius of the circumcircle of triangle $MKL$, if $\cos \angle ABK = \frac{2}{3}$.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c1ab4103bc5f9777fe4g-09.jpg?height=643&width=694&top_left_y=1917&top_left_x=727)

Quadrilateral $A M N C$ is an isosceles trapezoid. $\triangle A M H = \Delta H N C$ - by two sides and the angle between them.

$$
\begin{gathered}
\angle A H M = \angle H M N = \angle M N H = \angle N H C = \angle C H K = 180^{\circ} - \angle N C K \\
\angle M A C = \angle A C N = \angle N C A
\end{gathered}
$$

$$
\angle K H B = \angle K H C + 90^{\circ} = 180^{\circ} - \angle N C K + 90^{\circ} = \angle K C L,
$$

$\angle A M K = \angle A C K = \angle H N C$ - as angles of equal triangles and as angles subtending the same arc. Therefore, $\triangle H N C$ is similar to $\triangle H C K$ and $\triangle B H C$ is similar to $\triangle N L C$ by two angles. $\frac{K C}{K H} = \frac{N C}{C H} = \frac{L C}{H B}$ and considering that $\angle K H B = \angle K C L$, we get the similarity of $\triangle K C L$ and $\triangle B H K$ $\Rightarrow \angle C L K = \angle H B K \Rightarrow \angle M B K = \angle M L K$, which means points $K, M, B, L$ lie on the same circle and $R = \frac{M K}{2 \sin \angle A B K} = \frac{2 \sqrt{5}}{2 \sqrt{5} / 3} = 3$.

Answer: 3.

| Criterion | Points |
| :--- | :---: |
| The problem is solved correctly. | 20 |
| It is proven that points $K, M, B, L$ lie on the same circle. | 17 |
| It is proven that $\triangle K C L$ is similar to $\triangle B H K$. | 15 |
| It is proven that $\triangle B H C$ is similar to $\triangle N L C$. | 12 |
| It is proven that $\triangle A M H = \Delta H N C$ or $\triangle H N C$ is similar to $\triangle H C K$. | 5 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |