Ivlev $\Phi$.

On a circle, $2 n+1$ points are marked, dividing it into equal arcs ($n \geq 2$). Two players take turns erasing one point each. If after a player's move, all triangles with vertices at the remaining marked points are obtuse, the player wins, and the game ends. Who will win with correct play: the player starting the game or their opponent?
Let's present a winning strategy for the second player. He will strive to achieve the following condition: if after his move, there are $2k+1 \geq 5$ points left, then on any semicircle, there are at least $k$ marked points.

At the beginning of the game, this condition is satisfied. Suppose that at some point, it is satisfied before the first player's move; we will number the remaining points in order

$A_{0}, \ldots, A_{2k}$. Suppose, for definiteness, that the first player removes point $A_{0}$ on his move. The triangle $A_{k+1} A_{1} A_{k+2}$ is acute, so the game is not over yet. It is sufficient for the second player to remove point $A_{k}$. Now, if no more than one point has been removed from some semicircle, then it still has at least $k-1$ points; otherwise, points $A_{0}$ and $A_{k}$ have been erased from it, so it still has either points $A_{1}, \ldots, A_{k-1}$ or points $A_{k+1}, \ldots, A_{2k}$; in any case, the required condition is satisfied for it.

Thus, if the first player does not lose earlier, then after $2n-4$ moves, five points $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}$ will remain on the board. Suppose, for definiteness, that the first player removes point $A_{0}$; then the acute triangle $A_{1} A_{3} A_{4}$ will still remain. The second player will remove $A_{4}$ on his last move. The remaining triangle $A_{1} A_{2} A_{3}$ will be obtuse (otherwise, there would be a semicircle containing only $A_{2}$ out of the five points), meaning the second player will win.

## Answer

Opponent.

Author: $\underline{\text { Zhenedarov R.G. }}$.

Can the set of all natural numbers be partitioned into non-intersecting finite subsets $A_{1}, A_{2}, A_{3}, \ldots$ such that for any natural number $k$, the sum of all numbers in the subset $A_{k}$ equals $k + 2013$?

## Solution

Suppose the desired partition exists. We will call a set $A_{k}$ large if it contains more than one element. Suppose the number of large sets is finite. Then there exists a number $t > 1$ such that each of the sets $A_{t}, A_{t+1}, A_{t+2}, \ldots$ consists of a single element. Thus, the union of the sets $A_{t}, A_{t+1}, A_{t+2}, \ldots$ is the set $\{t+2013, t+2014, \ldots\}$. This means that the union of the sets $A_{1}, A_{2}, \ldots, A_{t-1}$ coincides with the set $\{1, 2, \ldots, t+2012\}$. However, the sum of the elements in these sets is $2014 + 2015 + \ldots + (t+2012)$, which is less than the sum of the elements in the set $\{1, 2, \ldots, t+2012\}$. Contradiction. Therefore, there are infinitely many large sets.

The sum of the numbers in each of the sets $A_{1}, A_{2}, \ldots, A_{n}$ does not exceed $n + 2013$, so all their elements lie in the set $\{1, 2, \ldots, n + 2013\}$. On the other hand, since there are infinitely many large sets, there exists a number $n$ such that among the sets $A_{1}, A_{2}, \ldots, A_{n}$, at least 2014 are large. Then the union of all sets $A_{1}, A_{2}, \ldots, A_{n}$ contains at least $n + 2014$ elements. Contradiction.

## Answer

No.