HMMT February

Geometry Difficulty 4.5 AIME Find the answer

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 $B C$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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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