2012 AIME II Problems/Problem 15

Revision as of 00:15, 18 April 2012 by JHW (talk | contribs)

Problem 15

Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. 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 $m+n$.

Solution

Use angle bisector theorem to find $CD=21/8$, $BD=35/8$, and $AD=15/8$. Use Power of the Point to find $ED=49/8$, and so $AE=8$. Use law of cosines to find \angle CAD = \pi/3 }

See Also

2012 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions