2009 USAMO Problems/Problem 2

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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 $a + b + c = 0$.


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 $s_{2k} > s_{2k-1}$. 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.

See also

2009 USAMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All USAMO Problems and Solutions

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