IMO 2026 Problem 3

Let $n$ be a positive integer. Alice and Bob have a stick of length $1$ and want to divide it between themselves. Alice marks at most $n$ points on the stick, and then Bob marks at most $n$ points on the stick. The marked points are distinct. Then, the stick is cut at all marked points, creating a number of pieces. Afterwards, they take turns claiming any unclaimed piece of the stick, with Alice going first. Each player's goal is to maximize the total length of their own pieces. For each $n$, determine the largest value $c$ such that Alice may guarantee a total length of at least $c$, regardless of Bob's play.

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The answer is $\boxed{c=\frac{2^n}{2^{n+1}-1}}$.

Alice gets at least $\frac{2^n}{2^{n+1}-1}$:

Up to scaling factor $\frac{1}{2^{n+1}-1}$, Alice makes $n+1$ segments of lengths $1,2,4,\dots,2^n$. At least one of them is not further split by Bob. Regardless of how Bob plays, Alice gets at least half of $2^n,2^{n-1},\dots,2^{k+1}$, and the entire $2^k$. Hence Alice gets at least $2^n$, which we further divide by $2^{n+1}-1$.

Bob gets at least $\frac{2^n-1}{2^{n+1}-1}$:

Let the lengths of the $n+1$ segments made by Alice be $\ell_1,\dots,\ell_{n+1}$. If any of them is $0$, then Bob splits every segment into two equal halves and gets at least $\frac{1}{2}$. Otherwise, we claim that Bob gets at least $\frac{1-e}{2}$ where $$e=\min_{A,B\subseteq [n+1],A\ne B}\big|\sum_{i\in A}\ell_i-\sum_{i\in B}\ell_i\big|.$$ We make $A$ and $B$ disjoint by removing common elements. Say $|A|=a$ and $|B|=b$. There are exactly $n+1-a-b$ segments outside of $A\cup B$. Bob cuts once to get $e$, cuts $a-1+b-1$ times to get pairs of equal-lengthed segments, and then cuts $n+1-a-b$ times to split all other segments into two equal halves. Later Bob misses a segment of length at most $e$.

We have $e\le\frac{1}{2^{n+1}-1}$ because the multiset $L=\left\{\sum_{i\in A}\ell_i:A\subseteq[n+1]\right\}\subset[0,1]$ of size $2^{n+1}$ has two elements that differ by at most $\frac{1}{2^{n+1}-1}$. Hence, Bob gets at least $$\frac{1-e}{2}=\frac{2^n-1}{2^{n+1}-1}.$$