Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.7 AIME Prove it Romania

On a circle are written several real numbers, of positive sum. Let $S$ be the largest and $s$ the least of the sums of consecutive numbers on the circle. Prove that $S + s > 0$ .

Solution

Denote by $T > 0$ the total sum of the numbers around the circle. Clearly $S \ge T > 0$ . If $s \ge 0$ , we are done. If $s < 0$ , the sum of the numbers which are not terms of $s$ is equal to $T - s$ . Since $S \ge T - s$ , then $S + s \ge T > 0$ , as needed.

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