2007 USAMO Problems/Problem 3

Problem

Let $S$ be a set containing $n^2+n-1$ elements, for some positive integer $n$. Suppose that the $n$-element subsets of $S$ are partitioned into two classes. Prove that there are at least $n$ pairwise disjoint sets in the same class.

Solution

Call an $n+1$-element subset of $S$ separable if it has a subset in each class of the partition. We recursively build a set $Q$ of disjoint separable subsets of $S$: begin with $Q$ empty and at each step if there is a separable subset which is disjoint from all sets in $Q$ add that set to $Q$. The process terminates when every separable subset intersects a set in $Q$. Let $T$ be the set of elements in $S$ which are not in any set in $Q$. We claim that one class contains every $n$-element subset of $T$.

Suppose that $a_1, a_2, \ldots a_k$ are elements of $T$. Denote by $A_i$ the set $\left\{a_i, a_{i+1}, \ldots a_{i+n-1}\right\}$. Note that for each $i$, $A_i \cup A_{i+1}$ is not separable, so that $A_i$ and $A_{i+1}$ are in the same class. But then $A_i$ is in the same class for each $1 \leq i \leq k - n + 1$ — in particular, $A_1$ and $A_{k-n+1}$ are in the same class. But for any two sets we may construct such a sequence with $A_1$ equal to one and $A_{k-n+1}$ equal to the other.

We are now ready to construct our $n$ disjoint sets. Suppose that $|Q| = q$. Then $|T| = (n+1)(n-q) - 1 \geq n(n-q)$, so we may select $n - q$ disjoint $n$-element subsets of $T$. Then for each of the $q$ sets in $Q$, we may select a subset which is in the same class as all the subsets of $T$, for a total of $n$ disjoint sets.

See also

2007 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All USAMO Problems and Solutions

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