Olya and Tolya are playing a game on $[0,1]$ segment. In the beginning it is white. In the first round Tolya chooses a number $0 \leq l \leq 1$, and then Olya chooses a subsegment of $[0,1]$ of length $l$ and recolors every its point to the opposite color(white to black, black to white). In the next round players change roles, etc. The game lasts $2024$ rounds. Let $L$ be the sum of length of white segments after the end of the game. If $L > \frac{1}{2}$ Olya wins, otherwise Tolya wins. Which player has a strategy to guarantee his win? A. Naradzetski
Problem
Source: Belarus TST 2024
Tags: games, combinatorics
YaoAOPS
18.07.2024 13:37
I think the solution to this problem depends on how you define "subsegment" and length. Notably, if $L \ne \frac12$, then note that Olya can choose $l \in \{0, 1\}$ on the $2024$th round so as to guarentee to a win. This raises questions to how $L$ is chosen.
TYERI
22.07.2024 11:20
After finite amount of rounds [0,1] would be split into finite amount of of black/white subsegments, so I think that L would be well-defined.
Olya wins.
Denote by $L_k$ the sum of lengths of white segments after $k$ rounds. Also for any moment denote by endsegments two (possibly matching) segments of partition that touch $0$ and $1$. It it important to note that if $0 < l < 1$, then any possible choice of subsegment recolors at most one endsegment, and for $l \in \{0, 1\}$ it recolors none/both of them.
Lemma. Olya can keep the following invariant:
For even $k$-s $L_k \neq \frac{1}{2}$ and at least one endsegment is white.
For odd $k$-s $L_k \neq \frac{1}{2}$ or both endsegments are white.
Proof. We will proceed using the induction by $k$. For the intial configuration conditions are fullfilled. Now fix $k$ and suppose that on previous step Olya was successful.
If $k$ is odd, then $L_{k-1} \neq \frac{1}{2}$, at least one endsegment is white, and it is Tolya's choice. If they choose $0$ or $1$, then $L_k - \frac{1}{2}$ is still nonzero. For $0 < l < 1$ Olya can either keep both endsegments white (by choosing subsegment inside) or recolor one black endsegment .
If $k$ is even, then $L_{k-1} \neq \frac{1}{2}$ or both endsegments are white. In the former case Olya can pick $l \in \{0,1\}$ in a way to keep disbalance and leave at least one white segment, so assume the latter case. Denote the minimum of lengths of endsegments by $a$. What happens if Olya chooses $l = 1 - b$ for $0 < b < a$? Then any possible placing of segment of such length would cover all black part of the partition and all but two pieces of endsegments of total length $b$.
So, such choice would guarantee $L_k$ being equal to $1 - L_{k-1} + b$. Therefore, Olya just need to choose $b$ such that this quantity is not $\frac{1}{2}$ (because Tolya cannot recolor more than one endsegment). $\blacksquare$
Now Olya proceed the described procedure for the first $2023$ rounds. If $L_{2023} \neq \frac{1}{2}$, then Olya can pick $0$ or $1$ to win. If that is not the case, then Olya is guaranteed that both endsegments are white. Then (in the same manner as above) choise $l = 1 - \varepsilon $ for $\varepsilon > 0$ small enough gives $L_{2024} = 1/2 + \varepsilon$.