1994 IMO Problems/Problem 1

Let $m$ and $n$ be two positive integers. Let $a_1$ , $a_2$ , $\ldots$ , $a_m$ be $m$ different numbers from the set $\{1, 2,\ldots, n\}$ such that for any two indices $i$ and $j$ with $1\leq i \leq j \leq m$ and $a_i + a_j \leq n$ , there exists an index $k$ such that $a_i + a_j = a_k$ . Show that $\frac{a_1+a_2+...+a_m}{m} \ge \frac{n+1}{2}$ .

Solution

Let $a_1, a_2, \dots a_m$ satisfy the given conditions. We will prove that for all $j, 1 \le j \le m,$

$a_j+a_{m-j+1} \ge n+1$

WLOG, let $a_1 < a_2 < \dots < a_m$ . Assume that for some $j, 1 \le j \le m,$

$a_j + a_{m-j+1} \le n$

This implies, for each $i, 1 \le i \le j,$ $a_i + a_{m-j+1} \le n$ because $a_i \le a_j$

For each of these values of i, we must have $a_i + a_{m-j+1} = a_{k_i}$ such that $a_{k_i}$ is a member of the sequence for each $i$ . Because $a_i > 0, a_{k_i} > a_{m-j+1}$ . Combining all of our conditions we have that each of $k_i$ must be distinct integers such that $m-j+1 < k_i \le m$

However, there are $j$ distinct $k_i$ , but only $j-1$ integers satisfying the above inequality, so we have a contradiction. Our assumption that $a_j + a_{m-j+1} \le n$ was false, so $a_j + a_{m-j+1} \ge n+1$ for all $j$ such that $1 \le j \le m$ Summing these inequalities together for $1 \le j \le m$ gives $2(a_1+a_2+ \dots a_m) \ge m(n+1)$ which rearranges to $\frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2}$

1994 IMO (Problems) • Resources
Preceded by First Question	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
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1994 IMO Problems/Problem 1

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