IMO 2025 Problem 5

Alice and Bazza are playing the inekoalaty game, a two‑player game whose rules depend on a positive real number $\lambda$ which is known to both players.  On the $n$th turn of the game (starting with $n=1$) the following happens:

-If $n$ is odd, Alice chooses a nonnegative real number $x_n$ such that

$$x_1 + x_2 + \cdots + x_n \le \lambda n$$

-If $n$ is even, Bazza chooses a nonnegative real number $x_n$ such that

$$x_1^2 + x_2^2 + \cdots + x_n^2 \le n$$

If a player cannot choose a suitable $x_n$, the game ends and the other player wins.  If the game goes on forever, neither player wins.  All chosen numbers are known to both players.

Determine all values of $\lambda$ for which Alice has a winning strategy and all those for which Bazza has a winning strategy.

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For $\lambda>1/\sqrt{2}$, Alice wins; for $\lambda<1/\sqrt{2}$, Bazza wins; otherwise it is a tie if they both play optimally.

Bazza's strategy is to choose $x_n$ such that $x^2_{n-1}+x^2_n=2$ whenever possible.

If $\lambda\ge1/\sqrt{2}$, Alice keeps choosing $x_n=0$. The best Bazza can do to maximize the sum is to follow his strategy by choosing $x_n=\sqrt{2}$. After $n=2k$ rounds, Alice can choose at least $y=(n+1)\lambda-n/\sqrt{2}=\lambda+n(\lambda-1/\sqrt{2})\ge0$. Thus Alice does not lose. When $\lambda>1/\sqrt{2}$, $y^2>n+2$ for sufficiently large $n$. Hence Alice wins by switching from $0$ to $y$.

If $\lambda\le1/\sqrt{2}$, Bazza follows his strategy. The sum after Bazza's turn is at least $n/\sqrt{2}$, so Alice's choice cannot exceed $$y=(n+1)\lambda-n/\sqrt{2}=n(\lambda-1/\sqrt{2})+\lambda\le\lambda\le1/\sqrt{2}<\sqrt{2}.$$ Thus inductively Bazza can always follow his strategy and does not lose. If $\lambda<1/\sqrt{2}$ then $y<0$ for sufficiently large $n$, i.e., Alice cannot choose a suitable $x_n$ and Bazza wins.