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Problem 1120

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Combinatorics Difficulty 5.1 Prove it Romanian Mathematical Olympiad · 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 $30$ .

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

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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