2012 AIME II Problems/Problem 15

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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 $CD=\frac{21}{8}$, $BD=\frac{35}{8}$, and use the Stewart's Theorem to find $AD=\frac{15}{8}$. Use Power of the Point to find $DE=\frac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \frac{\pi} {3}$, hence $\angle BAD = \frac{\pi}{3}$ 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'll do 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

Solution 3

[asy] size(6cm); pair E,X,B,C,A,D,M,F,R,I; real z=sqrt(3)*14/3; real y=2*sqrt(3)/21; real x=224*sqrt(3)/57; E=(z,0); X=(0,0); D=(sqrt(3)*7/6,-7/8); M=(sqrt(3)*7/6,0); B=z/2*dir(60); C=z/2*dir(300); A=(y,-8/7); F=(x,-sqrt(3)*x/4); R=circumcenter(A,B,C); I=circumcenter(M,E,F); draw(E--X); draw(A--E); draw(A--B); draw(A--C); draw(B--C); draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label("$A$",A,dir(220)); label("$B$",B,dir(110)); label("$C$",C,dir(250)); label("$D$",D,dir(60)); label("$E$",E,dir(0)); label("$F$",F,dir(315)); label("$X$",X,dir(180)); [/asy] First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\triangle ABC$ is $\frac{7\sqrt{3}}{3}$.

Since $DE$ is the diameter of circle $\gamma$, $\angle DFE$ is $90^\circ$. Extending $DF$ to intersect circle $\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\triangle ABC$ (since $\angle DFE$ is $90^\circ$). Therefore, $XE=\frac{14\sqrt{3}}{3}$.

Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem,




Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain:


\[a^2+2ab = \frac{5341}{192}.\]

By Power of a Point, $ab=BD \cdot DC=735/64=2205/192$, so

\[a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}\]


Since $a=XD$, $XD=\frac{7\sqrt{19}}{8\sqrt{3}}$.

Because $\angle EXF$ and $\angle EAF$ intercept the same arc in circle $\omega$ and the same goes for $\angle XFA$ and $\angle XEA$, $\angle EXF\cong\angle EAF$ and $\angle XFA\cong\angle XEA$. Therefore, $\triangle XDE\sim\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional,



\[AF \cdot \sqrt{19} = 30\]

\[AF = \frac{30}{\sqrt{19}}.\]

However, the problem asks for $AF^2$, so $AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}$.

-Solution by TheBoomBox77

Solution 4

It can be verified with law of cosines that $\angle BAC=120^\circ.$ Also, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\angle BEC=60.$ Thus $CBE$ is equilateral. Notice now that $\angle CFB=\angle BEC= 60.$ But $\angle DFE=90$ so $FD$ bisects $\angle BFC.$ Thus, $\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.$

Let $BF=5a, CF=3a.$ By law of cosines on $BFC$ we find $a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.$ But by ptolemy on $BFCA$, $15a+15a=7*AF,$ so $AF= \frac{30}{\sqrt{19}},$ so $AF^2=\frac{900}{19}$ and the answer is $900+19=\boxed{919}$


See Also

2012 AIME II (ProblemsAnswer KeyResources)
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