Source: IMO-2025-P5
Category: Operations and Strategies

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

- 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.

Solution:
The threshold is $\lambda = \frac{1}{\sqrt{2}}$. Alice has a winning strategy for $\lambda > \frac{1}{\sqrt{2}}$, Bazza has a winning strategy for $\lambda < \frac{1}{\sqrt{2}}$, and neither player can force a win when $\lambda = \frac{1}{\sqrt{2}}$.

First, consider Alice when $\lambda \ge \frac{1}{\sqrt{2}}$. She plays
\[
x_{2i+1}=0
\]
on every odd turn until she decides to finish the game.
At time $n=2k+1$, we have
\[
x_1+x_2+\cdots+x_{2k} = x_2+x_4+\cdots+x_{2k}.
\]
By the Cauchy-Schwarz inequality,
\[
x_2+x_4+\cdots+x_{2k}
\le k\sqrt{\frac{x_2^2+x_4^2+\cdots+x_{2k}^2}{k}}.
\]
Because Alice plays $0$ on odd turns, we have $x_2^2+x_4^2+\cdots+x_{2k}^2 = x_1^2+x_2^2+\cdots+x_{2k}^2$. From Bazza's constraint on turn $2k$, this sum of squares is at most $2k$. Thus,
\[
x_2+x_4+\cdots+x_{2k} \le k\sqrt{\frac{2k}{k}} = \sqrt{2}k.
\]
Hence, the legal interval for Alice's next move is bounded below by $0$ and above by $\lambda(2k+1) - (x_1+\cdots+x_{2k})$, meaning she can pick any
\[
x_{2k+1} \in \bigl[0,\lambda(2k+1)-\sqrt{2}k\bigr].
\]
If $\lambda \ge \frac{1}{\sqrt{2}}$, this interval is always nonempty since $\lambda(2k+1)-\sqrt{2}k \ge \frac{1}{\sqrt{2}}(2k+1)-\sqrt{2}k = \frac{1}{\sqrt{2}} > 0$. Thus, Alice can always play $0$ to avoid losing forever.

If in fact $\lambda > \frac{1}{\sqrt{2}}$, Alice can choose $k$ large enough such that
\[
\sqrt{2}k < \lambda(2k+1)-\sqrt{2k+2}.
\]
Then on move $2k+1$, the upper bound of her legal interval is strictly greater than $\sqrt{2k+2}$, allowing Alice to choose
\[
x_{2k+1} > \sqrt{2k+2}.
\]
At the next step, Bazza would need to choose $x_{2k+2}$ such that
\[
x_1^2+\cdots+x_{2k+1}^2 + x_{2k+2}^2 \le 2k+2,
\]
but since $x_{2k+1}^2 > 2k+2$, this is impossible. Thus, Bazza has no legal move, and Alice wins for $\lambda > \frac{1}{\sqrt{2}}$.

Now, consider Bazza when $\lambda \le \frac{1}{\sqrt{2}}$. He plays on every even turn the largest possible value,
\[
x_{2i+2}=\sqrt{2-x_{2i+1}^2}.
\]
This keeps each pair $(x_{2i+1},x_{2i+2})$ contributing exactly $2$ to the quadratic budget, so $x_1^2 + \cdots + x_{2i+2}^2 = 2(i+1)$, which perfectly satisfies Bazza's condition for any even turn.
To estimate Alice's move $x_{2k+1}$, we use her constraint:
\[
\lambda(2k+1) \ge x_1+x_2+\cdots+x_{2k+1}.
\]
Grouping the terms in pairs gives
\[
\lambda(2k+1)
\ge \bigl(x_1+\sqrt{2-x_1^2}\bigr)+\cdots+\bigl(x_{2k-1}+\sqrt{2-x_{2k-1}^2}\bigr)+x_{2k+1}.
\]
For every $t\in[0,\sqrt{2}]$, we have $(t+\sqrt{2-t^2})^2 = 2 + 2t\sqrt{2-t^2} \ge 2$, which implies
\[
t+\sqrt{2-t^2} \ge \sqrt{2}.
\]
Applying this to each pair yields
\[
\lambda(2k+1) \ge \sqrt{2}k + x_{2k+1}.
\]
Therefore,
\[
x_{2k+1} \le \lambda(2k+1)-\sqrt{2}k.
\]
When $\lambda \le \frac{1}{\sqrt{2}}$, this upper bound is at most $\frac{1}{\sqrt{2}}(2k+1) - \sqrt{2}k = \frac{1}{\sqrt{2}} < \sqrt{2}$. Thus, Alice is forced to play $x_{2k+1} \le \sqrt{2}$, ensuring that Bazza can indeed play
\[
x_{2k+2}=\sqrt{2-x_{2k+1}^2}
\]
and never gets stuck.

If moreover $\lambda < \frac{1}{\sqrt{2}}$, then for all sufficiently large $k$,
\[
\lambda(2k+1)-\sqrt{2}k < 0.
\]
Since Alice must choose $x_{2k+1} \ge 0$, she eventually has no legal move at all. Hence, Bazza wins for $\lambda < \frac{1}{\sqrt{2}}$.

At the boundary value $\lambda = \frac{1}{\sqrt{2}}$, Alice's strategy of playing $0$ shows she can survive forever, while Bazza's maximal strategy also shows he can survive forever. Thus, neither player can force a win at the critical value.

Answer: Alice wins for $\lambda > 1/\sqrt2$; Bazza wins for $\lambda < 1/\sqrt2$; neither can force a win at $\lambda = 1/\sqrt2$

Grading guidelines:
(Partial) 1. Identified the correct threshold value \(\lambda=1/\sqrt2\) (e.g. by optimizing a Cauchy–Schwarz type inequality or by analyzing one full odd-even pair). 2. Gave a correct strategy for one player in one direction, with a real supporting estimate (e.g. Alice plays \(x_{2i+1}=0\) and shows \(x_2+x_4+\cdots+x_{2k}\le \sqrt2k\); or Bazza plays \(x_{2i+2}=\sqrt{2-x_{2i+1}^2}\) and derives \(x_{2i+1}+x_{2i+2}\ge \sqrt2\)). 3. Correctly proved at least one of the strict win statements: Alice wins for all \(\lambda>1/\sqrt2\) (by eventually choosing \(x_{2k+1}^2>2k+2\)); or Bazza wins for all \(\lambda<1/\sqrt2\) (by forcing \(\lambda(2k+1)-\sqrt2k<0\) for large \(k\)). (Almost) 1. Correctly handled both strict regimes \(\lambda>1/\sqrt2\) and \(\lambda<1/\sqrt2\) with valid strategies and inequalities, but did not correctly treat the boundary case \(\lambda=1/\sqrt2\) (e.g. claimed one player wins there, or omitted showing neither can force a win). 2. Gave essentially the full solution including the critical case, but left a localized gap in the key estimate (e.g. misapplied Cauchy–Schwarz in bounding \(x_2+\cdots+x_{2k}\), or did not justify \(t+\sqrt{2-t^2}\ge \sqrt2\) on the needed domain). 3. Solution is almost complete, but made minor mistakes which are not negligible.