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 1

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 $\boxed{919}$ .

Solution 2

Let $a = BC$ , $b = CA$ , $c = AB$ for convenience. We claim that $AF$ is a symmedian. Indeed, let $M$ be the midpoint of segment $BC$ . Since $\angle EAB=\angle EAC$ , it follows that $EB = EC$ and consequently $EM\perp BC$ . Therefore, $M\in \gamma$ . Now let $G = FD\cap \omega$ . Since $EG$ is a diameter, $G$ lies on the perpendicular bisector of $BC$ ; hence $E$ , $M$ , $G$ are collinear. From $\angle DAG = \angle DMG = 90$ , it immediately follows that quadrilateral $ADMG$ is cyclic. Therefore, $\angle MAD = \angle MGD=\angle EAF$ , implying that $AF$ is a symmedian, as claimed.

The rest is standard; here's a quick way to finish. From above, quadrilateral $ABFC$ is harmonic, so $\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}$ . In conjunction with $\triangle ABF\sim\triangle AMC$ , it follows that $AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}$ . (Notice that this holds for all triangles $ABC$ .) To finish, substitute $a = 7$ , $b=3$ , $c=5$ to obtain $AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}$ as before.

-Solution by thecmd999

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2012 AIME II Problems/Problem 15

Contents

Problem 15

Solution 1

Solution 2

See Also