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Saudi Arabian Mathematical Competitions, 2015

Geometry Difficulty 4.6 AIME Prove it Saudi Arabia

Let $ABC$ be a triangle, with $AB < AC$ , $D$ the foot of the altitude from $A$ , $M$ the midpoint of $BC$ , and $B'$ the symmetric of $B$ with respect to $D$ . The perpendicular line to $BC$ at $B'$ intersects $AC$ at point $P$ . Prove that if $BP$ and $AM$ are perpendicular then triangle $ABC$ is right-angled.

Solution

Let $E$ be the intersection point of $AD$ and $BP$ . Because $AD$ is perpendicular to $BM$ and $BP$ is perpendicular to $AM$ , the point $E$ is the orthocenter of triangle $ABM$ and therefore $ME$ is perpendicular to $AB$ .

Because point $D$ is the midpoint of the segment $BB'$ and $DE$ and $B'P$ are parallel, we deduce that point $E$ is the midpoint of segment $BP$ . But $M$ is the midpoint of segment $BC$ . We deduce that lines $ME$ and $BC$ are parallel.

It follows that $AB$ and $AC$ are perpendicular and triangle $ABC$ is right-angled.

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