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Difference between revisions of "1994 IMO Problems/Problem 1"
 
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which rearranges to
 
which rearranges to
 
<cmath> \frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2} </cmath>
 
<cmath> \frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2} </cmath>
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==See Also==
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{{IMO box|year=1994|before=First Question|num-a=2}}

Latest revision as of 00:28, 22 November 2023

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 \[\frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2}\]


See Also

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