# 2012 AIME II Problems/Problem 15

## 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 the angle bisector theorem to find $CD=21/8$, $BD=35/8$, and use the Stewart's Theorem to find $AD=15/8$. Use Power of the Point to find $DE=49/8$, and so $AE=8$. Use law of cosines to find $\angle CAD = \pi /3$, hence $\angle BAD = \pi /3$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$.

I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines:

$AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot cos \angle AFE.$ (1)

$AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot cos \angle AEF.$ Adding these two and simplifying we get:

$EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF$ (2). Ah, but $\angle AFE = \angle ACE$ (since $F$ lies on $\omega$), and we can find $cos \angle ACE$ using the law of cosines:

$AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE$, and plugging in $AE = 8, AC = 3, BE = BC = 7,$ we get $cos \angle ACE = -1/7 = cos \angle AFE$.

Also, $\angle AEF = \angle DEF$, and $\angle DFE = \pi/2$ (since $F$ is on the circle $\gamma$ with diameter $DE$), so $cos \angle AEF = EF/DE = 8 \cdot EF/49$.

Plugging in all our values into equation (2), we get:

$EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}$, or $EF = \frac{7}{15} \cdot AF$.

Finally, we plug this into equation (1), yielding:

$8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}$. Thus,

$64 = \frac{AF^2}{225} \cdot (225+49+30),$ or $AF^2 = \frac{900}{19}.$ The answer is $919$.

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