2009 USAMO Problems/Problem 2

Problem

Let $n$ be a positive integer. Determine the size of the largest subset of $\{ - n, - n + 1, \ldots , n - 1, n\}$ which does not contain three elements $a, b, c$ (not necessarily distinct) satisfying $a + b + c = 0$ .

Solution 1

Let $S$ be a subset of $\{-n,-n+1,\dots,n-1,n\}$ of largest size satisfying $a+b+c\neq 0$ for all $a,b,c\in S$ . First, observe that $0\notin S$ . Next note that $|S|\geq \lceil n/2\rceil$ , by observing that the set of all the odd numbers in $\{-n,-n+1,\dots,n-1,n\}$ works. To prove that $|S|\leq \lceil n/2\rceil$ , it suffices to only consider even $n$ , because the statement for $2k$ implies the statement for $2k-1$ as well. So from here forth, assume $n$ is even.

For any two sets $A$ and $B$ , denote by $A+B$ the set $\{a+b\mid a\in A,b\in B\}$ , and by $-A$ the set $\{-a\mid a\in A\}$ . Also, let $A_+$ denote $A\cap\{1,2,\dots\}$ and $A_-$ denote $A\cap\{-1,-2,\dots\}$ . First, we present a lemma:

Lemma 1: Let $A$ and $B$ be two sets of integers. Then $|A+B|\geq|A|+|B|-1$ .

Proof: Write $A=\{a_1,\dots,a_n\}$ and $B=\{b_1,\dots,b_m\}$ where $a_1<\dots<a_n$ and $b_1<\dots<b_m$ . Then $a_1+b_1,a_1+b_2,\dots,a_1+b_m,a_2+b_m,\dots,a_n+b_m$ is a strictly increasing sequence of $n+m-1$ integers in $A+B$ .

Now, we consider two cases:

Case 1: One of $n,-n$ is not in $S$ . Without loss of generality, suppose $-n\notin S$ . Let $U=\{-n+1,-n+2,\dots,n-2,n-1,n\}$ (a set with $2n$ elements), so that $S\subseteq U$ by our assumption. Now, the condition that $a+b+c\neq 0$ for all $a,b,c\in S$ implies that $-S\cap(S_++S_-)=\emptyset$ . Since any element of $S_++S_-$ has absolute value at most $n-1$ , we have $S_++S_-\subseteq \{-n+1,-n+2,\dots,n-2,n-1\}\subseteq U$ . It follows that $S_++S_-\subseteq U\setminus -S$ , so $|S_++S_-|\leq 2n-|S|$ . However, by Lemma 1, we also have $|S_++S_-|\geq |S_+|+|S_-|-1=|S|-1$ . Therefore, we must have $|S|-1\leq 2n-|S|$ , or $2|S|\leq 2n+1$ , or $|S|\leq n$ .

Case 2: Both $n$ and $-n$ are in $S$ . Then $n/2$ and $-n/2$ are not in $S$ , and at most one of each of the pairs $\{1,n-1\},\{2,n-2\},\ldots,\{\frac n2-1,\frac n2+1\}$ and their negatives are in $S$ . This means $S$ contains at most $2+(n-2)=n$ elements.

Thus we have proved that $|S|\leq n=\lceil n/2\rceil$ for even $n$ , and we are done.

Solution 2

Let $Z_n$ be the set of subsets satisfying the $a+b+c \neq 0$ condition for $n$ , and let $s_n$ be the largest size of a set in $Z_n$ . Let $n = 2k$ if $n$ is even, and $n = 2k-1$ if $n$ is odd. We note that $s_n \ge 2k$ due to the following constuction: $\{-2k+1, -2k + 3, \ldots, 2k-3, 2k-1\}$ or all of the odd numbers in the set. Then the sum of any three will be odd and thus nonzero.

Lemma 1: $s_{n+1} \le s_{n} + 2$ . If $M_{n+1} \in Z_{n+1}, |M_{n+1}| = s_{n+1}$ , then we note that $M_{n+1} \setminus \{-2n+1, 2n-1\} \in Z_n$ , so $|M_{n+1}| - 2 \le s_n$ . $\square$

Lemma 2: $s_{2k} = s_{2k-1}$ . Suppose, for sake of contradiction, that $M_{2k} \in Z_{n+1}$ and $M_{2k} > s_{2k-1}$ . Remove $2k,-2k$ from $M_{2k}$ , and partition the rest of the elements into two sets $P_{2k-1}, Q_{2k-1}$ , where $P$ and $Q$ contain all of the positive and negative elements of $M_{2k}$ , respectively. (obviously $0 \not \in M_{2k}$ , because $0+0+0 = 0$ ). WLOG, suppose $|P_{2k-1}| \ge |Q_{2k-1}|$ . Then $|P_{2k-1}| + |Q_{2k-1}| \ge s_{2k-1} - 1 \ge 2k - 1$ . We now show the following two sub-results:

Sub-lemma (A): if $|P_{2k-1}| \ge k$ , $-2k \not \in M_{2k}$ [and similar for $Q$ ]; and

Sub-lemma (B): we cannot have both $|P_{2k-1}| + |Q_{2k-1}| \ge 2k$ and $|Q_{2k-1}| < k$ simultaneously hold.

This is sufficient, because the only two elements that may be in $M_{2k}$ that are not in $P_{2k-1}, Q_{2k-1}$ are $2k$ and $-2k$ ; for $M_{2k} > s_{2k-1}$ , we must either have $|P_{2k-1}| + |Q_{2k-1}| = s_{2k-1} - 1 \ge 2k-1$ and both $2k,-2k \in M_{2k}$ [but by pigeonhole $|P_{2k-1}| \ge k$ , see sub-lemma (A)], or $|P_{2k-1}| + |Q_{2k-1}| = s_{2k-1} \ge 2k$ , and $2k \in M_{2k}$ , in which case by (A) we must have $|Q_{2k-1}| < k$ , violating (B).

(A): Partition $\{1,2,\ldots,2k-1\}$ into the $k$ sets $\{k\}, \{1,2k-1\}, \{2, 2k-2\}, \cdots, \{k-1, k+1\}$ . Because $(k+\delta) + (k-\delta) - 2k = 0$ , then if any of those sets are within $P_{2k-1}$ , $-2k \not \in M_{2k}$ . But by Pigeonhole at most $k-1$ elements may be in $P_{2k-1}$ , contradiction.

(B): We prove this statement with another induction. We see that the statement easily holds true for $k=1$ or $2$ , so suppose it is true for $k-1$ , but [for sake of contradiction] false for $k$ . Let $P_{2k-3} = P_{2k-1} \setminus \{2k-2, 2k-1\}$ , and similarly for $Q_{2k-3}$ . Again WLOG $|P_{2k-3}| \ge |Q_{2k-3}|$ . Then we have $|P_{2k-1}| + |Q_{2k-1}| \le |P_{2k-3}| + |Q_{2k-3}| + 4$ .

If $|P_{2k-3}| + |Q_{2k-3}| \ge 2k-2$ , then by inductive hypothesis, we must have $|P_{2k-3}| = |Q_{2k-3}| = k-1$ . But (A) implies that we cannot add $2k-2$ or $-2k+2$ . So to satisfy $|P_{2k-1}| = |Q_{2k-1}| \ge 2k$ we must have $2k-1, -2k+1$ added, but then $|P_{2k-1}| = |Q_{2k-1}| = k$ contradiction.

If $|P_{2k-3}| + |Q_{2k-3}| = 2k-3$ , then at least three of $\{2k-2, 2k-1, -2k+2, -2k + 1\}$ added. But $|P_{2k-3}| \ge k-1$ , and by (A) we have that $-2k+2$ cannot be added. If $|P_{2k-3}| \ge k$ , then another grouping similar to (A) shows that $-2k+1$ canno be added, contradiction. So $|P_{2k-3}| = k-1$ , $|Q_{2k-3}| = k-2$ , and adding the three remaining elements gives $|P_{2k-1}| = |Q_{2k-1}| = k$ contradiction.

If $|P_{2k-3}| + |Q_{2k-3}| = 2k-4$ , then all four of $\{2k-2, 2k-1, -2k+2, -2k + 1\}$ must be added, and furthermore $|P_{2k-3}| > |Q_{2k-3}|$ . Then $|P_{2k-3}| \ge k-1$ , and by previous paragraph we cannot add $-2k+2$ . $\square$

So we have $\begin{cases} s_1 &= 2 \\ s_{2k} &= s_{2k-1} \\ s_{2k-1} &\le s_{2k-2} + 2 \\ \end{cases}$ and by induction, that $s_n \le \boxed{2\left\lceil \frac n2 \right \rceil} = 2k$ , which we showed is achievable above.

2009 USAMO (Problems • Resources)
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2009 USAMO Problems/Problem 2

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Problem

Solution 1

Solution 2

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