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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2012 AIME II Problems/Problem 15

Problem 15

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