Problem 6. In an acute-angled triangle $A B C$ with side $A C=1$, the altitude $B H$ is drawn, in triangle $B H C$ - the bisector $C L$, in triangle $B L C$ - the median $B M$. The line $A L$ intersects side $B C$ at point $K$, and $\angle B H K=\angle M H C=15^{\circ}$. Find the area of quadrilateral $K L H M$.
Answer: $\frac{3(2-\sqrt{3})}{8}$.

Solution. Triangle $C H L$ is a right triangle, so its vertices lie on a circle with the center at the midpoint of the hypotenuse - point $M$. Then $M H = M C$ and $\angle L H K = \angle M H C = \angle M C H = \angle M C K = 15^{\circ}$. Therefore, point $K$ also lies on the same circle. Then $\angle L K C = 90^{\circ}$ as an inscribed angle subtending the diameter, and $A K$ is also an altitude of triangle $A B C$. Therefore, $L$ is the orthocenter of triangle $A B C$, and thus ray $C L$ contains both the altitude and the angle bisector of this triangle, making the triangle isosceles ($A C = B C = 1$). From this, it follows that $C K = C H$ and $\triangle K L C = \triangle H L C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_06261581b98c3717a5ffg-2.jpg?height=788&width=856&top_left_y=1071&top_left_x=223)

Since the median divides the area of the triangle in half,

$S(K L H M) = S(K L M) + S(H L M) = \frac{1}{2} S(K L C) + \frac{1}{2} S(H L C) = S(K L C) = \frac{1}{2} \cdot K C \cdot L C \cdot \sin 15^{\circ} = \frac{1}{2} \cdot K C \cdot \frac{K C}{\cos 15^{\circ}} \cdot \sin 15^{\circ} = \frac{1}{2} \cdot (A C \cos 30^{\circ})^2 \cdot \tan 15^{\circ} = \frac{3(2-\sqrt{3})}{8}$.[^0]

Answer: a) The second player; b) 1 point.

Solution. Note that the first player does not score any points after their first move, while the second player can guarantee 1 point for each move. Indeed, it is sufficient for the second player to choose an already occupied point with a number that has an unoccupied neighboring point and place a different number there. Thus, the second player can score at least 100 points with correct play. There are 199 segments in total, so the second player always wins with correct play. Starting from their second move, the first player can also guarantee 1 point per move (by placing the same number in a neighboring unoccupied point). Thus, with correct play, the first player can guarantee 99 points.

[^0]: Problem 7. On the number line, 200 points are marked with coordinates $1, 2, \ldots, 200$. Two players take turns placing the number "0" or "1" in any of the still unoccupied points. When all possible moves are made, 199 segments $[1; 2], [2; 3], [3; 4], \ldots, [199; 200]$ are considered, and points are scored for each segment: if the endpoints of the segment contain the same number, the first player scores 1 point; if they are different, the second player scores 1 point.

a) Which player can ensure a higher total score regardless of the opponent's moves?

b) What is the maximum advantage one player can have over the other (assuming each player plays optimally)?