5. (20 points) In a right triangle $A B C\left(\angle C=90^{\circ}\right)$ with side $B C=a$, points $M$ and $N$ are the midpoints of sides $A B$ and $B C$, respectively. The bisector of $\angle A$ intersects line $M N$ at point $L$. Find the radius of the circumcircle of triangle $A C L$, if angle $\angle C B L=\alpha$.
Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_aadf511cc872419fdfecg-09.jpg?height=474&width=511&top_left_y=1174&top_left_x=841)

Draw the line $B L$, intersecting $A C$ at point $X$. Then, in $\triangle A B X$, the bisector $A L$ is also a median, and therefore, an altitude. Thus, $\angle A L B=90^{\circ}$ and points $A, B, L, C$ lie on the same circle. Consequently, $\angle C B L=\angle C A L$, as they subtend the same arc and the radius of the circumcircle of $\triangle C A L$ is equal to the radius of the circumcircle of $\triangle C A B$ and is equal to half of $A B$. Therefore, $R=\frac{A B}{2}=\frac{a}{2 \sin 2 \alpha}$. Answer: $\frac{a}{2 \sin 2 \alpha}$.

| Criterion | Points |
| :--- | :---: |
| The problem is solved correctly. | 20 |
| It is proven that $\angle C B L=\angle C A L$. | 15 |
| It is proven that points $A, B, L, C$ lie on the same circle. | 10 |
| It is proven that $\angle A L B=90^{0}$. | 5 |
| The solution does not meet any of the criteria listed above. | 0 |
| Maximum score | 20 |