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Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Hong Kong

Given finitely many points in a plane, it is known that the area of the triangle formed by any three points of the set is less than $1$ . Show that all points of the set lie inside or on the boundary of a triangle with area less than $4$ .

Solution

Let $A$ , $B$ , $C$ be points in $S$ such that $[ABC]$ is maximized. Let $\ell_1$ be the line passing through $A$ which is parallel to $BC$ . If there exists a point $X \in S$ which lies on a different side of $\ell_1$ as $B$ , then the distance from $X$ to $BC$ is greater than that from $A$ to $BC$ . This yields the contradiction $[XBC] > [ABC]$ . Therefore, $X$ must lie on the same side of $\ell_1$ as $B$ .

Similarly, let $\ell_2$ be the line passing through $B$ which is parallel to $CA$ , and let $\ell_3$ be the line passing through $C$ which is parallel to $AB$ . Then any point in $S$ must lie inside the triangle $\triangle DEF$ bounded by $\ell_1$ , $\ell_2$ , $\ell_3$ . Note that $\triangle ABC$ is the median triangle of $\triangle DEF$ . So $[DEF] = 4[ABC] < 4$ . Thus, we can simply take $\triangle DEF$ .

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