AIME 2012 Number 15

Geometry Level 5

Triangle $ABC$ is inscribed in circle $\omega$ with $AB = 5$ , $BC = 7$ , and $AC = 3$ . The bisector of angle $A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$ . Let $\gamma$ be the circle with diameter $DE$ . Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$ . Then $AF^2 = \frac mn$ , where m and n are relatively prime positive integers. Find $m + n$ .

The answer is 919.

1 solution

Alan Yan
Dec 25, 2015

Consider inversion centered at $A$ with radius $\sqrt{bc}$ .

We get that $D$ and $E$ are inverses and since $\angle DFE = 90^{\circ}$ , we have that $\angle D'F'E' = 90^{\circ}$ and since the circumcircle gets inverted to $B'C'$ , we have that $F'$ is the foot of the perpendicular of $D'$ to $B'C'$ . Since $D'$ is the midpoint of arc $B'C'$ of circle $A'B'C'$ , we have that $F'$ is the midpoint of $B'C'$ . We can use Stewart's theorem to find $AF' = \frac{\sqrt{19}}{2}$ . We have $AF = \frac{bc}{AF'} = \frac{30}{\sqrt{19}} \implies AF^2 = \frac{900}{19}$ .

AIME 2012 Number 15

The answer is 919.

1 solution

0 pending reports