Difference between revisions of "2007 USAMO Problems/Problem 3"

Latest revision as of 14:36, 2 October 2024

Problem

(András Gyárfás) 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

Solution 1

Claim: If we have instead $k(n+1)-1$ elements, then we must have k disjoint subsets in a class. We proceed by Induction

Base Case: $k=1$ is trivial.

Case 1: If there exists an $n+1$ element subset $X$ s.t. $X$ has 2 n-element subsets in different classes. Consider the $(k-1)(n+1)-1$ elements that are not in $X$ , from the Induction hypothesis, we have $k-1$ disjoint subsets in one of the classes. Those $k-1$ subsets paired with a subset of $X$ in the same class forms $k$ disjoint subsets of size n.

Case 2: All $n+1$ element subsets have all of their n-element subsets in one class. Look at any 2 n-element subsets $a_1$ and $a_p$ . Then we can create a path $a_1,a_2,...,a_p$ s.t. any 2 consecutive terms belong to the same n+1-element subset.(i.e. we change 1 element each time.) Thus, $a_1$ and $a_p$ are in the same class and we see that all subsets are in one class. Obviously we can find $k$ disjoint subsets of size n.

The question is asking for the case when we substitute $k=n$ which we have proved.

Solution 2

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.

Solution 3

In order to apply induction, we generalize the result to be proved so that it reads as follows:

Proposition. If the $n$ -element subsets of a set $S$ with $(n + 1)m - 1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.

Proof. Fix $n$ and proceed by induction on $m$ . The case of $m = 1$ is trivial. Assume $m > 1$ and that the proposition is true for $m - 1$ . Let $\mathcal{P}$ be the partition of the $n$ -element subsets into two classes. If all the $n$ -element subsets belong to the same class, the result is obvious. Otherwise select two $n$ -element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A\cap B| = n - 1$ . (If $|A\cap B| = k < n - 1$ , then build $C$ from $B$ by replacing some element not in $A\cap B$ with an element of $A$ not already in $B$ . Then $|A\cap C| = k + 1$ and $|B\cap C| = n - 1$ and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A\cup B$ from $S$ , there are $(n + 1)(m - 1) - 1$ elements left. On this set the partition induced by $\mathcal{P}$ has, by the inductive hypothesis, $m - 1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\blacksquare$

Remark: The value $n^2 + n - 1$ is sharp. A set $S$ with $n^2 + n - 2$ elements can be split into a set $A$ with $n^2 - 1$ elements and a set $B$ of $n - 1$ elements. Let one class consist of all $n$ -element subsets of $A$ and the other consist of all $n$ -element subsets that intersect $B$ . Then neither class contains $n$ pairwise disjoint sets.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 16: / Line 16: @@
 Look at any 2 n-element subsets <math>a_1</math> and <math>a_p</math>. Then we can create a path <math>a_1,a_2,...,a_p</math> s.t. any 2 consecutive terms belong to the same n+1-element subset.(i.e. we change 1 element each time.) Thus, <math>a_1</math> and <math>a_p</math> are in the same class and we see that all subsets are in one class. Obviously we can find <math>k</math> disjoint subsets of size n.
-(Correct me if I'm wrong) Note: Because <math>k</math><math>(n+1)</math>-1><math>k</math><math>(n+1)</math>, we can group the first n+1 elements, then the next <math>n+1</math> elements, and we have <math>k</math> of these groupings. Then, we assign each of the 2 classes to have one <math>n</math>-element group from each of these <math>k</math> groupings, so that their elements don't repeat. This means that they're disjoint!
 The question is asking for the case when we substitute <math>k=n</math> which we have proved.
-~Brackie1331
 === Solution 2 ===

2007 USAMO (Problems • Resources)
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