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Harvard-MIT Math Tournament

Geometry Difficulty 4.5 AIME Prove it United States

Problem:

Points $A$ , $B$ , $C$ in the plane satisfy $\overline{A B} = 2002$ , $\overline{A C} = 9999$ . The circles with diameters $A B$ and $A C$ intersect at $A$ and $D$ . If $\overline{A D} = 37$ , what is the shortest distance from point $A$ to line $B C$ ?

Solution

Solution:

$\angle A D B = \angle A D C = \pi / 2$ since $D$ lies on the circles with $A B$ and $A C$ as diameters, so $D$ is the foot of the perpendicular from $A$ to line $B C$ , and the answer is the given $37$ .

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