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Brazilian Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Brazil

$P$ is a fixed point in the plane. $A$ , $B$ , $C$ are points such that $PA = 3$ , $PB = 5$ , $PC = 7$ and the area $ABC$ is as large as possible. Show that $P$ must be the orthocenter of $ABC$ .

Solution

Consider all points $A'$ such that $PA' = 3$ . They lie on a circle with center $P$ . The area of $A'BC$ is $\frac{BC}{2}$ times the distance of $A'$ from $BC$ . That distance is maximal for $A'P$ perpendicular to $BC$ (because the distance is the distance of $P$ from $BC$ is $PA' \sin \theta$ , where $\theta$ is the angle between $A'P$ and $BC$ ). Hence $AP$ must be perpendicular to $BC$ . Similarly $BP$ must be perpendicular to $AC$ , so $P$ must be the orthocenter.

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