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 $m+n$ .

Quick Solution using Olympiad Terms

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 $900+19=\boxed{919}$ .

- Emathmaster

Side Note: You might be wondering what the motivation for this solution is. Most of the people who've done EGMO Chapter 8 should recognize this as problem 8.32 (2009 Russian Olympiad) with the computational finish afterwards. Now if you haven't done this, but still know what inversion is, here's the motivation. We'd see that it's kinda hard to angle chase, and if we could, it would still be a bit hard to apply (you could use trig, but it won't be so clean most likely). If you give up after realizing that angle chasing won't work, you'd likely go in a similar approach to Solution 1 (below) or maybe be a bit more insightful and go with the elementary solution above.

Finally, we notice there's circles! Classic setup for inversion! Since we're involving an angle-bisector, the first thing that comes to mind is a force overlaid inversion described in Lemma 8.16 of EGMO (where we invert with radius $\sqrt{AB \cdot AC}$ and center $A$ , then reflect over the $A$ -angle bisector, which fixes $B, C$ ). We try applying this to the problem, and it's fruitful - we end up with this solution. -MSC

Solution 1

Use the angle bisector theorem to find $CD=\tfrac{21}{8}$ , $BD=\tfrac{35}{8}$ , and use Stewart's Theorem to find $AD=\tfrac{15}{8}$ . Use Power of Point $D$ to find $DE=\tfrac{49}{8}$ , and so $AE=8$ . Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$ , hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$ . $[asy] size(150); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); draw(omega^^A--B--C--cycle^^gamma); draw(pic, A--E--F--cycle, gray); add(pic); dot("$W$",circumcenter(A,B,C),dir(180)); label("$\gamma$",gamma,dir(180)); [/asy]$ In triangle $AEF$ , let $X$ be the foot of the altitude from $A$ ; then $EF=EX+XF$ , where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$ , we get $\begin{align}\tag{1} EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{align}$ Note $\angle AFE = \angle ACE$ , and the Law of Cosines in $\triangle ACE$ gives $\cos \angle ACE = -\tfrac 17$ . Also, $\angle AEF = \angle DEF$ , and $\angle DFE = \tfrac{\pi}{2}$ ( $DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} = \tfrac{8}{49}\cdot EF$ .

Plugging in all our values into equation $(1)$ , we get: $EF = \frac{64}{49} EF -\frac{1}{7} AF \quad \Longrightarrow \quad EF = \frac{7}{15} AF.$ The Law of Cosines in $\triangle AEF$ , with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives $8^2 = AF^2 + \frac{49}{225} AF^2 + \frac 2{15} AF^2 = \frac{225+49+30}{225}\cdot AF^2$ Thus $AF^2 = \tfrac{900}{19}$ . The answer is $\boxed{919}$ .

Solution 2

Let $a = BC$ , $b = CA$ , $c = AB$ for convenience. Let $M$ be the midpoint of segment $BC$ . We claim that $\angle MAD=\angle DAF$ .

$\textit{Proof}$ . Since $AE$ is the angle bisector, it follows that $EB = EC$ and consequently $EM\perp BC$ . Therefore, $M\in \gamma$ . Now let $X = FD\cap \omega$ . Since $EX$ is a diameter, $X$ lies on the perpendicular bisector of $BC$ ; hence $E$ , $M$ , $X$ are collinear. From $\angle DAG = \angle DMX = 90^\circ$ , quadrilateral $ADMX$ is cyclic. Therefore, $\angle MAD = \angle MXD$ . But $\angle MXD$ and $\angle EAF$ are both subtended by arc $EF$ in $\omega$ , so they are equal. Thus $\angle MAD=\angle DAF$ , as claimed. $[asy] size(200); defaultpen(fontsize(10pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220)); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); draw(A--B--F--cycle, black+1); [/asy]$ As a result, $\angle CAM = \angle FAB$ . Combined with $\angle BFA=\angle MCA$ , we get $\triangle ABF\sim\triangle AMC$ and therefore $\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}$ By Stewart's Theorem on $\triangle ABC$ (with cevian $AM$ ), we get $AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},$ so $AF^2 = \tfrac{900}{19}$ , so the answer is $900+19=\boxed{919}$ .

-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 $\angle BEC=60^\circ.$ 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

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search