Track / Stage 5 / 338 of 400 #938 of 1964

Problem 938

AIME late

Combinatorics Difficulty 5.8 Prove it

$5.2 n$ real numbers are placed at $2 n$ distinct positions on a circle. It is known that the sum of these $2 n$ numbers is positive. Prove: there exists one of these positions such that: starting from this position (inclusive), the sum of the numbers at the next $n$ consecutive positions in both the clockwise and counterclockwise directions is positive.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

5. Let the $2n$ numbers in clockwise order be $a_{1}, a_{2}$ , $\cdots, a_{2 n}$ , with their sum $S>0$ .
Let $S_{i}=a_{i}+a_{i+1}+\cdots+a_{i+n-1}$ , where $a_{2 n+i}=a_{i}, S_{2 n+i}=S_{i}$ .
It suffices to prove that there exists $i$ such that $S_{i}$ and $S_{i+1-n}$ are both greater than 0. Since $S_{i}+S_{n+i}=S>0$ , there is a positive $S_{i}$ .
If all $S_{i}$ are positive, the conclusion is obviously true. Otherwise, there exists $i$ such that $S_{i}>0, S_{i+1} \leqslant 0$ . Therefore,
$S_{i+1-n}=S-S_{i+1}>0 .$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.