Problem 5. Inside triangle ABC, two points are given. The distances from one of them to the lines AB, BC, and AC are 1, 3, and 15 cm, respectively, and from the other - 4, 5, and 11 cm. Find the radius of the circle inscribed in triangle ABC.
Answer: 7 cm. First solution. Let $M_{1}$ and $M_{2}$ be the first and second given points, and let point $O$ be such that point $M_{2}$ is the midpoint of segment $O M_{1}$. Drop perpendiculars $M_{1} N_{1}, M_{2} N_{2}$, and $O N_{3}$ to line $A B$. Then segment $M_{2} N_{2}$ will be the midline of trapezoid $O N_{3} N_{1} M_{1}$. Therefore, $M_{2} N_{2}=\left(O N_{3}+M_{1} N_{1}\right) / 2 \Rightarrow 4=\left(O N_{3}+1\right) / 2$, from which $O N_{3}=7$. Similarly, we find that the perpendiculars $O M_{3}$ and $O K_{3}$ from point $O$ to lines $B C$ and $A C$ respectively are also equal to 7. Thus, point $O$ is the center of the inscribed circle of triangle $A B C$, which has a radius of 7. Second solution. Let $A B=c, B C=a$, $A C=b$. Then $3 a+15 b+c=5 a+11 b+4 c=r(a+b+c)=2 S_{A B C}$. From the first equality, we get $a=(4 b-3 c) / 2$. Substituting the found $a$ into the second equality $5 a+11 b+4 c=r(a+b+c)$, we find $r=7$.

- For an answer without justification - 0 points.
Problem 6. On the board, there are five "equations" of the form $x^{2}+\ldots x+\ldots=0$. Two players take turns writing natural numbers from 1 to 10 in place of the ellipses, with each number being used only once. The game ends when all numbers are written. The player who makes the first move wants to have as few equations with two distinct roots as possible at the end, while his opponent wants to have as many as possible. What is the best result the first player can achieve regardless of the second player's moves?

Answer. The first player can ensure that there are no more than three equations with two distinct roots on the board. This result cannot be improved. Solution. To create two equations without roots, the first player needs to use the first two moves to write the smallest of the numbers not yet written in the coefficient of $x$ in an "equation" where no numbers have been written yet. This is possible because if no more than two pairs of moves have been made, then at least one coefficient has been written in no more than four "equations". Let $x^{2}+p x+q=0$ be one of the equations obtained at the end of the game, where the coefficient $p$ was written by the first player on one of the first two moves. Then, obviously, $p \leq 3$ and $q \geq p+1$, from which $p^{2}-4 q \geq p^{2}-4(p+1)=p(p-4)-40$.

- For an answer without justification - 0 points. If there is a strategy for only one of the players, the solution is scored out of 4 points: 2 points for describing the strategy and 2 points for justifying it.


## SOURCES AND AUTHORS OF THE PROBLEMS

XII Tournament of Mathematical Flash Battles "League of Discoveries", Kazan, 2022: 6-4, 7-4.

Based on problems from Kostroma olympiads: 7-2, 9-2.

District-city round of the Chelyabinsk Region Olympiad 1999/2000: 9-1.

District-city round of the Chelyabinsk Region Olympiad 1998/1999: 11-5.

Third Kostroma City Tournament of Mathematical Battles, 1998 (numerical data changed): $9-3$.

Tatarstan, Olympiad for 6th grade, final round, 2014: 10-1.

Udmurtia, district round, 1979: 11-1

Folklore: $5-4=6.3,11-4$.

All other problems were composed by I.S. Rubanov specifically for this olympiad.