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

Line 1: | Line 1: | ||

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

− | + | <math>\frac{a_1+a_2+...+a_m}{m} \le \frac{n+1}{2}</math>. | |

− | |||

==Solution== | ==Solution== |

## Revision as of 23:40, 9 April 2021

Let and be two positive integers. Let , , , be different numbers from the set such that for any two indices and with and , there exists an index such that . Show that .

## Solution

Let satisfy the given conditions. We will prove that for all

WLOG, let . Assume that for some

This implies, for each because

For each of these values of i, we must have such that is a member of the sequence for each . Because . Combining all of our conditions we have that each of must be distinct integers such that

However, there are distinct , but only integers satisfying the above inequality, so we have a contradiction. Our assumption that was false, so for all such that Summing these inequalities together for gives which rearranges to