Be real!

Consider all sets of 100 real numbers r i r_i whose sum is positive.

Define t i j = r i + r j t_{ij} = r_i + r_j for i < j i < j . There are ( 100 2 ) = 4950 { 100 \choose 2 } = 4950 such pairs.

What is the maximum number of t i j t_{ij} which are non-positive?


The answer is 4851.

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1 solution

Luca Bernardelli
Apr 10, 2018

The maximum number of t_i_j which are non-positive is when all of the 100 chosen real numbers but 1 are negative. (The total sum could still be positive if the only positive number is bigger then the opposite of the sum of the negative ones):

  • r 1 , r 2 , . . . r 99 < 0 r_1,r_2, ... r_{99} < 0

  • r 100 > i = 1 99 r i r_{100} > -\sum\limits_{i=1}^{99} r_i

Then, the t i j 0 t_{ij}\le 0 are the ones between negative numbers, while the ones that involve r 100 r_{100} are positive.

4950 99 = 4851 4950 - 99 = \boxed{4851}

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