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 Using Only Elementary Geometry Methods

Let $P$ be the unique point on circle $\omega$ such that $\bigtriangleup BCP$ is equilateral. Point $E$ must be the point on $\omega$ such that $\angle PED$ is right. We extend line $ED$ to hit circle $\omega$ again at $X$ .

Claim 1: $X$ is the midpoint of minor arc $\overarc{BC}$ . Proof: Notice that $\angle PEX$ and $\angle PAX$ are subtended by the same arc, thus they are both equal to 90 degrees.

Now it is obvious that $\bigtriangleup PED \sim \bigtriangleup ADX$ . After this, we length chase with similar triangles and Pythag, which yields $AP = 8$ , $EX = \frac{56}{\sqrt{57}}$ , $AX = \frac{2}{\sqrt{3}}$ , $PE = \frac{14}{\sqrt{19}}$ , and $PX = \frac{14}{\sqrt{3}}$ . We can tie this all together with Ptolemy's theorem for cyclic quadrilaterals, and we find that $AE = \frac{30}{\sqrt{19}}$ . Thus, $AE^2 = \frac{900}{19}$ , and the answer is $\boxed{919}$ -jj_ca888

Quick Solution in Terms of Olympiad Terms (Similar to above)

Take a force-overlaid inversion about $A$ and note $D$ and $E$ map to each other. As $DE$ was originally the diameter of $\gamma$ , $DE$ is still the diameter of $\gamma$ . Thus $\gamma$ is preserved. Note that the midpoint $M$ of $BC$ lies on $\gamma$ , and $BC$ and $\omega$ are swapped. Thus points $F$ and $M$ map to each other, and are isogonal. It follows that $AF$ is a symmedian of $\triangle{ABC}$ , or that $ABFC$ is harmonic. Then $(AB)(FC)=(BF)(CA)$ , and thus we can let $BF=5x, CF=3x$ for some $x$ . By the LoC, it is easy to see $\angle{BAC}=120^\circ$ so $(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49$ . Solving gives $x^2=\frac{49}{19}$ , from which by Ptolemy's we see $AF=\frac{30}{\sqrt{19}}$ . We conclude the answer is $900+19=\boxed{919}$ , as desired.

- Emathmaster

Solution 1

Use the angle bisector theorem to find $CD=\frac{21}{8}$ , $BD=\frac{35}{8}$ , and use 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,

$x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}$

and

$x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.$

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

$(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}$

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

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

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

$a^2=\frac{931}{192}.$

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,

$\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}$

$\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}$

$AF \cdot \sqrt{19} = 30$

$AF = \frac{30}{\sqrt{19}}.$

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 $\angle BFC=\angle BFE= 60.$ 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 $900+19=\boxed{919}$

~abacadaea

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