Competition problem — Combinatorics

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Combinatorics Difficulty 5.1 AIME, harder Prove it Romania

In a square of side length $60$ , $121$ distinct points are given. Show that among them there exists three points which are vertices of a triangle with an area not exceeding $30$ .

Solution

Divide the square into $60$ rectangles $5 \times 12$ . By the pigeonhole principle, there are three points among the given ones inside one of the rectangles. The area of this triangle does not exceed half of the area of the rectangle, that is $5 \cdot 12/2 = 30$ , as needed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.