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NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Geometry Difficulty 5.0 AIME, harder Prove it Romania

Consider two equilateral triangles $ABC$ and $MNP$ with $AB \parallel MN$ , $BC \parallel NP$ and $CA \parallel PM$ , intersecting over a convex hexagon. The distances between the pairs of parallel sides do not exceed $1$ . Show that at least one of the triangles has the side length less than or equal to $\sqrt{3}$ .

Solution

Let $P$ be an interior point to the hexagon, therefore also interior to the triangles. Denote by $a$ , respectively $b$ , the lengths of the sides of the two triangles. The sum of the distances from $P$ to the sides of an equilateral triangle is equal to the altitude of the triangle, hence the sum of the distances from $P$ to the lines $AB$ , $BC$ , $CA$ , $MN$ , $NP$ , $PM$ is $(a + b) \frac{\sqrt{3}}{2}$ .

On the other hand, the sum of the distances from $P$ to the parallel lines $AB$ and $MN$ is precisely the distance between those lines, hence at most $1$ . It follows that $(a + b) \frac{\sqrt{3}}{2} \le 3$ , so $a + b \le 2\sqrt{3}$ , whence $a \le \sqrt{3}$ or $b \le \sqrt{3}$ , which is what was asked to be proved.

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