Difference between revisions of "1994 IMO Problems/Problem 1"

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Let <math>a_1, a_2, \dots a_m</math> satisfy the given conditions. We will prove that for all <math>j, 1 \le j \le m,</math>
 
  
<cmath>a_j+a_{m-j+1} \ge n+1</cmath>
 
 
WLOG, let <math>a_1 < a_2 < \dots < a_m</math>. Assume that for some <math>j, 1 \le j \le m,</math>
 
 
<cmath> a_j + a_{m-j+1} \le n</cmath>
 
 
This implies, for each <math>i, 1 \le i \le j,</math>
 
<cmath> a_i + a_{m-j+1} \le n </cmath>
 
because <math>a_i \le a_j</math>
 
 
For each of these values of i, we must have <math>a_i + a_{m-j+1} = a_{k_i}</math> such that <math>a_{k_i}</math> is a member of the sequence for each <math>i</math>. Because <math>a_i > 0, a_{k_i} > a_{m-j+1}</math>.
 
Combining all of our conditions we have that each of <math>k_i</math> must be distinct integers such that
 
<cmath>m-j+1 < k_i \le m</cmath>
 
 
However, there are <math>j</math> distinct <math>k_i</math>, but only <math>j-1</math> integers satisfying the above inequality, so we have a contradiction. Our assumption that <math>a_j + a_{m-j+1} \le n</math> was false, so <math>a_j + a_{m-j+1} \ge n+1</math> for all <math>j</math> such that <math>1 \le j \le m</math> Summing these inequalities together for <math>1 \le j \le m</math> gives
 
<cmath> 2(a_1+a_2+ \dots a_m) \ge m(n+1) </cmath>
 
which rearranges to
 
<cmath> \frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2} </cmath>
 

Revision as of 23:38, 9 April 2021

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