2. Maximum 15 points. On side AB of an equilateral triangle $\mathrm{ABC}$, a right triangle $\mathrm{A} H \mathrm{~B}$ is constructed ( $\mathrm{H}$ - the vertex of the right angle), such that $\angle \mathrm{HBA}=60^{\circ}$. Let point K lie on ray $\mathrm{BC}$ beyond point $\mathrm{C}$ and $\angle \mathrm{CAK}=15^{\circ}$. Find the angle between line HK and the median of triangle $\mathrm{AHB}$, drawn from vertex $\mathrm{H}$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_040ad17443cbde3a5398g-3.jpg?height=617&width=662&top_left_y=568&top_left_x=734)

Extend NB and NA beyond points B and A respectively (H-B-B1, H-A-A1)

$\angle \mathrm{B} 1 \mathrm{BC}=60^{\circ}$

$\angle$ KAA1 $=75^{\circ}$, so BK is the bisector of $\angle \mathrm{ABB} 1$

AK is the bisector of $\angle \mathrm{A} 1 \mathrm{AC}$, therefore, HK is the bisector of $\angle \mathrm{AHB}$, so we need to find the angle between the bisector and the median of triangle AHB, drawn from the vertex of the right angle. Answer: $15^{\circ}$

| Criteria | Score |
| :--- | :---: |
| The solution algorithm is provided, the drawing is correct, and the correct answer is obtained | 15 |
| The solution algorithm is provided, the drawing is correct, but due to an arithmetic error, the answer is incorrectly calculated or the answer is correct, but the logic of the proof is violated | 6 |
| The solution algorithm has significant logical errors, but some points of the proof are correct | 2 |
| The solution does not meet any of the presented criteria | 0 |