Difference between revisions of "2012 AIME II Problems/Problem 15"

Revision as of 12:27, 3 September 2024

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: $EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.$ The Law of Cosines in $\triangle AEF$ , with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives $8^2 = AF^2 + \tfrac{49}{225} AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2$ Thus $AF^2 = \frac{900}{19}$ . The answer is $\boxed{919}$ . ~Shen Kislay Kai

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 $\angle EFX=90^\circ$ , $EX$ is a diameter, so $X$ 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. $[asy] size(175); defaultpen(fontsize(10pt)); picture pic; pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); pair M=MP("M",midpoint(B--C),dir(220)); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, gray); draw(pic, A--M--C--cycle^^A--B--F--cycle); draw(A--M, royalblue); dot("$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

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 $AE=8$ . Then use the Extended Law of Sine to find that the length of the circumradius of $\triangle ABC$ is $\tfrac{7\sqrt{3}}{3}$ . $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP("B",origin,dir(180)); C=MP("C",(7,0),dir(0)); A=MP("A",IP(CR(B,5),CR(C,3)),N); D=MP("D",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP("E",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP("F",OP(omega,gamma),SE); pair X=MP("X",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70)); [/asy]$ 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 $\omega$ (since $\angle DFE$ is $90^\circ$ ). Therefore, $XE=\tfrac{14\sqrt{3}}{3}$ .

Let $EF=x$ , $XD=u$ , and $DF=v$ . Then $XE^2-XF^2=EF^2=DE^2-DF^2$ , so we get $(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}$ which simplifies to $u^2+2uv = \frac{5341}{192}.$ By Power of Point $D$ , $uv=BD \cdot DC=735/64$ . Combining with above, we get $XD^2=u^2=\frac{931}{192}.$ Note that $\triangle XDE\sim \triangle ADF$ and the ratio of similarity is $\rho = AD : XD = \tfrac{15}{8}:u$ . Then $AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R$ and $AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.$ The answer is $900+19=\boxed{919}$ .

-Solution by TheBoomBox77

Solution 4

Use Law of Cosines in $\triangle ABC$ to get $\angle BAC=120^\circ$ . Because $AE$ bisects $\angle A$ , $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\angle BEC=60^\circ.$ Thus $\triangle BEC$ is equilateral. Notice now that $\angle BFC=\angle BFE= 60^\circ.$ But $\angle DFE=90^\circ$ so $FD$ bisects $\angle BFC.$ Thus, $\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\triangle BFC$ to get $25k^2+9k^2-15k^2 = 49 \qquad \Longrightarrow\qquad k=\frac 7{\sqrt{19}}$ Use Ptolemy's Theorem on $BFCA$ , to get $15k+15k=7\cdot AF, \qquad \Longrightarrow\qquad AF= \frac{30}{\sqrt{19}},$ so $AF^2=\frac{900}{19}$ and the answer is $900+19=919$

~Shen Kislay Kai

Solution 5

Denote $AB = c, BC = a, AC = b, \angle A = 2 \alpha.$ Let M be midpoint BC. Let $\theta$ be the circle centered at $A$ with radius $\sqrt{AB \cdot AC} =\sqrt{bc}.$

We calculate the length of some segments. The median $AM = \sqrt{\frac {b^2}{2} + \frac {c^2}{2} - \frac {a^2}{4}}.$ The bisector $AD = \frac {2 b c \cos \alpha}{b+c}.$ One can use Stewart's Theorem in both cases.

$AD$ is bisector of $\angle A \implies BD = \frac {a c}{b + c}, CD = \frac {a b}{b + c} \implies$ $BD \cdot CD = \frac {a^2 bc }{(b+c)^2}.$ We use Power of Point $D$ and get $AD \cdot DE = BD \cdot CD.$ $AE = AD + DE = AD + \frac {BD \cdot CD}{AD},$ $AE =\frac {2 b c \cos \alpha}{b+c} + \frac {a^2 bc \cdot (b+c) }{(b+c)^2 \cdot 2 b c \cos \alpha} =$ $= \frac {b c \cos^2 \alpha + a^2}{2(b+c)\cos \alpha} =\frac {4bc \cos^2 \alpha + b^2 +c^2 -2 b c \cos 2\alpha}{2(b+c) \cos \alpha} = \frac {b+c}{2} \implies AD \cdot AE = 2 bc \cos \alpha.$ We consider the inversion with respect $\theta.$

$B$ swap $B' \implies AB' = AC, B' \in AB \implies B'$ is symmetric to $C$ with respect to $AE.$

$C$ swap $C' \implies AC' = AB, C'$ lies on line $AC \implies C'$ is symmetric to $B$ with respect to $AE.$

$BC^2 = AB^2 + AC^2 + AB \cdot BC \implies \alpha = 60^\circ \implies AD \cdot AE = bc \implies D$ swap $E.$

Points $D$ and $E$ lies on $\Gamma \implies \Gamma$ swap $\Gamma.$

$DE$ is diameter $\Gamma, \angle DME = 90^\circ \implies M \in \Gamma.$ Therefore $M$ is crosspoint of $BC$ and $\Gamma.$

Let $\Omega$ be circumcircle $AB'C'. \Omega$ is image of line $BC.$ Point $M$ maps into $M' \implies M' = \Gamma \cap \Omega.$

Points $A, B',$ and $C'$ are symmetric to $A, C,$ and $B,$ respectively.

Point $M'$ lies on $\Gamma$ which is symmetric with respect to $AE$ and on $\Omega$ which is symmetric to $\omega$ with respect to $AE \implies$

$M'$ is symmetric $F$ with respect to $AE \implies AM' = AF.$

We use Power of Point $A$ and get $AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}} = \frac {30}{\sqrt{19}} \implies \boxed{\textbf{919}}.$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution by mop 2024

https://youtu.be/mIFUuY4ybeg

~r00tsOfUnity

@@ Line 2: / Line 2: @@
 Triangle <math>ABC</math> is inscribed in circle <math>\omega</math> with <math>AB=5</math>, <math>BC=7</math>, and <math>AC=3</math>. The bisector of angle <math>A</math> meets side <math>\overline{BC}</math> at <math>D</math> and circle <math>\omega</math> at a second point <math>E</math>. Let <math>\gamma</math> be the circle with diameter <math>\overline{DE}</math>. Circles <math>\omega</math> and <math>\gamma</math> meet at <math>E</math> and a second point <math>F</math>. Then <math>AF^2 = \frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
-== Solution 1==
+==Quick Solution using Olympiad Terms ==
-Use the angle bisector theorem to find <math>CD=\frac{21}{8}</math>, <math>BD=\frac{35}{8}</math>, and use the Stewart's Theorem to find <math>AD=\frac{15}{8}</math>.  Use Power of the Point to find <math>DE=\frac{49}{8}</math>, and so <math>AE=8</math>.  Use law of cosines to find <math>\angle CAD = \frac{\pi} {3}</math>, hence <math>\angle BAD = \frac{\pi}{3}</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>.
-I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:
+Take a force-overlaid inversion about <math>A</math> and note <math>D</math> and <math>E</math> map to each other. As <math>DE</math> was originally the diameter of <math>\gamma</math>, <math>DE</math> is still the diameter of <math>\gamma</math>. Thus <math>\gamma</math> is preserved. Note that the midpoint <math>M</math> of <math>BC</math> lies on <math>\gamma</math>, and <math>BC</math> and <math>\omega</math> are swapped. Thus points <math>F</math> and <math>M</math> map to each other, and are isogonal. It follows that <math>AF</math> is a symmedian of <math>\triangle{ABC}</math>, or that <math>ABFC</math> is harmonic. Then <math>(AB)(FC)=(BF)(CA)</math>, and thus we can let <math>BF=5x, CF=3x</math> for some <math>x</math>. By the LoC, it is easy to see <math>\angle{BAC}=120^\circ</math> so <math>(5x)^2+(3x)^2-2\cos{60^\circ}(5x)(3x)=49</math>. Solving gives <math>x^2=\frac{49}{19}</math>, from which by Ptolemy's we see <math>AF=\frac{30}{\sqrt{19}}</math>. We conclude the answer is <math>900+19=\boxed{919}</math>.
-<math>AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.</math> (1)
+'''- Emathmaster'''
-<math>AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.</math>  Adding these two and simplifying we get:
+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.
-<math>EF = AF \cdot \cos \angle AFE + AE \cdot \cos \angle AEF</math> (2).  Ah, but <math>\angle AFE = \angle ACE</math> (since <math>F</math> lies on <math>\omega</math>), and we can find <math>cos \angle ACE</math> using the law of cosines:
+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 <math>\sqrt{AB \cdot AC}</math> and center <math>A</math>, then reflect over the <math>A</math>-angle bisector, which fixes <math>B, C</math>). We try applying this to the problem, and it's fruitful - we end up with this solution.
+-MSC
-<math>AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot \cos \angle ACE</math>, and plugging in <math>AE = 8, AC = 3, BE = BC = 7,</math> we get <math>\cos \angle ACE = -1/7 = \cos \angle AFE</math>.
+== Solution 1==
+Use the angle bisector theorem to find <math>CD=\tfrac{21}{8}</math>, <math>BD=\tfrac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\tfrac{15}{8}</math>. Use Power of Point <math>D</math> to find <math>DE=\tfrac{49}{8}</math>, and so <math>AE=8</math>. Use law of cosines to find <math>\angle CAD = \tfrac{\pi} {3}</math>, hence <math>\angle BAD = \tfrac{\pi}{3}</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>.
+<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 <math>AEF</math>, let <math>X</math> be the foot of the altitude from <math>A</math>; then <math>EF=EX+XF</math>, where we use signed lengths. Writing <math>EX=AE \cdot \cos \angle AEF</math> and <math>XF=AF \cdot \cos \angle AFE</math>, we get
+<cmath>\begin{align}\tag{1}
+    EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE.
+\end{align}</cmath>
+Note <math>\angle AFE = \angle ACE</math>, and the Law of Cosines in  <math>\triangle ACE</math> gives  <math>\cos \angle ACE = -\tfrac 17</math>.
+Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \tfrac{\pi}{2}</math> (<math>DE</math> is a diameter), so <math>\cos \angle AEF = \tfrac{EF}{DE} =  \tfrac{8}{49}\cdot EF</math>.
-Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \pi/2</math> (since <math>F</math> is on the circle <math>\gamma</math> with diameter <math>DE</math>), so <math>\cos \angle AEF = EF/DE = 8 \cdot EF/49</math>.
+Plugging in all our values into equation <math>(1)</math>, we get:
+<cmath>EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.</cmath>
+The Law of Cosines in  <math>\triangle AEF</math>, with <math>EF=\tfrac 7{15}AF</math> and <math>\cos\angle AFE = -\tfrac 17</math> gives
+<cmath>8^2 = AF^2 + \tfrac{49}{225}  AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2</cmath>
+Thus <math>AF^2 = \frac{900}{19}</math>. The answer is <math>\boxed{919}</math>.
+~Shen Kislay Kai
-Plugging in all our values into equation (2), we get:
+== Solution 2==
-<math>EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}</math>, or <math>EF = \frac{7}{15} \cdot AF</math>.
+Let <math>a = BC</math>, <math>b = CA</math>, <math>c = AB</math> for convenience. Let <math>M</math> be the midpoint of segment <math>BC</math>. We claim that <math>\angle MAD=\angle DAF</math>.
-Finally, we plug this into equation (1), yielding:
+<math>\textit{Proof}</math>. Since <math>AE</math> is the angle bisector, it follows that <math>EB = EC</math> and consequently <math>EM\perp BC</math>. Therefore, <math>M\in \gamma</math>. Now let <math>X = FD\cap \omega</math>. Since <math>\angle EFX=90^\circ</math>, <math>EX</math> is a diameter, so <math>X</math> lies on the perpendicular bisector of <math>BC</math>; hence <math>E</math>, <math>M</math>, <math>X</math> are collinear. From <math>\angle DAG = \angle DMX = 90^\circ</math>, quadrilateral <math>ADMX</math> is cyclic. Therefore, <math>\angle MAD = \angle MXD</math>. But <math>\angle MXD</math> and <math>\angle EAF</math> are both subtended by arc <math>EF</math> in <math>\omega</math>, so they are equal. Thus <math>\angle MAD=\angle DAF</math>, as claimed.
+<asy>
+size(175);
+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));
-<math>8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot \frac{-1}{7}</math>.  Thus,
-<math>64 = \frac{AF^2}{225} \cdot (225+49+30),</math> or <math>AF^2 = \frac{900}{19}.</math>  The answer is <math>\boxed{919}</math>.
+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);
-== Solution 2==
+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);
-Let <math>a = BC</math>, <math>b = CA</math>, <math>c = AB</math> for convenience. We claim that <math>AF</math> is a symmedian. Indeed, let <math>M</math> be the midpoint of segment <math>BC</math>. Since <math>\angle EAB=\angle EAC</math>, it follows that <math>EB = EC</math> and consequently <math>EM\perp BC</math>. Therefore, <math>M\in \gamma</math>. Now let <math>G = FD\cap \omega</math>. Since <math>EG</math> is a diameter, <math>G</math> lies on the perpendicular bisector of <math>BC</math>; hence <math>E</math>, <math>M</math>, <math>G</math> are collinear. From <math>\angle DAG = \angle DMG = 90</math>, it immediately follows that quadrilateral <math>ADMG</math> is cyclic. Therefore, <math>\angle MAD = \angle MGD=\angle EAF</math>, implying that <math>AF</math> is a symmedian, as claimed.
+</asy>
+As a result, <math>\angle CAM = \angle FAB</math>. Combined with <math>\angle BFA=\angle MCA</math>, we get <math>\triangle ABF\sim\triangle AMC</math> and therefore<cmath>\frac c{AM}=\frac {AF}b\qquad \Longrightarrow \qquad AF^2=\frac{b^2c^2}{AM^2} = \frac{15^2}{AM^2}</cmath>
-The rest is standard; here's a quick way to finish. From above, quadrilateral <math>ABFC</math> is harmonic, so <math>\dfrac{FB}{FC}=\dfrac{AB}{BC}=\dfrac{c}{a}</math>. In conjunction with <math>\triangle ABF\sim\triangle AMC</math>, it follows that <math>AF^2=\dfrac{b^2c^2}{AM^2}=\dfrac{4b^2c^2}{2b^2+2c^2-a^2}</math>. (Notice that this holds for all triangles <math>ABC</math>.) To finish, substitute <math>a = 7</math>, <math>b=3</math>, <math>c=5</math> to obtain <math>AF^2=\dfrac{900}{19}\implies 900+19=\boxed{919}</math> as before.
+By Stewart's Theorem on <math>\triangle ABC</math> (with cevian <math>AM</math>), we get <cmath>AM^2 = \tfrac 12 (b^2+c^2)-\tfrac 14 a^2 = \tfrac{19}{4},</cmath> so <math>AF^2 = \tfrac{900}{19}</math>, so the answer is <math>900+19=\boxed{919}</math>.
 '''-Solution by thecmd999'''
 ==Solution 3==
+Use the angle bisector theorem to find <math>CD=\tfrac{21}{8}</math>, <math>BD=\tfrac{35}{8}</math>, and use Stewart's Theorem to find <math>AD=\tfrac{15}{8}</math>. Use Power of Point <math>D</math> to find <math>DE=\tfrac{49}{8}</math>, and so <math>AE=8</math>. Then use the Extended Law of Sine to find that the length of the circumradius of <math>\triangle ABC</math> is <math>\tfrac{7\sqrt{3}}{3}</math>.
 <asy>
-size(6cm);
+size(175);
-pair E,X,B,C,A,D,M,F,R,I;
+defaultpen(fontsize(9pt));
-real z=sqrt(3)*14/3;
+pair A,B,C,D,E,F,W;
-real y=2*sqrt(3)/21;
+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);
-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 <math>BD=35/8</math> and <math>CD=21/8</math>, and use [[Stewart's Theorem]] to find that <math>AD=15/8</math>. Then use [[Power of a Point Theorem|Power of a Point]] to find that <math>DE=49/8</math>. Then use the [[Circumradius|circumradius of a triangle]] formula to find that the length of the circumradius of <math>\triangle ABC</math> is <math>\frac{7\sqrt{3}}{3}</math>.
-Since <math>DE</math> is the diameter of circle <math>\gamma</math>, <math>\angle DFE</math> is <math>90^\circ</math>. Extending <math>DF</math> to intersect circle <math>\omega</math> at <math>X</math>, we find that <math>XE</math> is the diameter of the circumcircle of <math>\triangle ABC</math> (since <math>\angle DFE</math> is <math>90^\circ</math>). Therefore, <math>XE=\frac{14\sqrt{3}}{3}</math>.
+draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue);
-Let <math>EF=x</math>, <math>XD=a</math>, and <math>DF=b</math>. Then, by the [[Pythagorean Theorem]],
+dot("$W$",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label("$\gamma$",gamma,dir(180)); label("$u$",X--D,dir(60)); label("$v$",D--F,dir(70));
+</asy>
+Since <math>DE</math> is the diameter of circle <math>\gamma</math>, <math>\angle DFE</math> is <math>90^\circ</math>. Extending <math>DF</math> to intersect circle <math>\omega</math> at <math>X</math>, we find that <math>XE</math> is the diameter of <math>\omega</math> (since <math>\angle DFE</math> is <math>90^\circ</math>). Therefore, <math>XE=\tfrac{14\sqrt{3}}{3}</math>.
-<cmath>x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}</cmath>
+Let <math>EF=x</math>, <math>XD=u</math>, and <math>DF=v</math>. Then <math>XE^2-XF^2=EF^2=DE^2-DF^2</math>, so we get
+<cmath>(u+v)^2-v^2=\frac{196}{3}-\frac{2401}{64}</cmath>
+which simplifies to
+<cmath>u^2+2uv = \frac{5341}{192}.</cmath>
+By Power of Point <math>D</math>, <math>uv=BD \cdot DC=735/64</math>. Combining with above, we get
+<cmath>XD^2=u^2=\frac{931}{192}.</cmath>
+Note that <math>\triangle XDE\sim \triangle ADF</math> and the ratio of similarity is <math>\rho = AD : XD = \tfrac{15}{8}:u</math>. Then <math>AF=\rho\cdot XE = \tfrac{15}{8u}\cdot R</math> and <cmath>AF^2 = \frac{225}{64}\cdot \frac{R^2}{u^2} = \frac{900}{19}.</cmath>
+The answer is <math>900+19=\boxed{919}</math>.
-and
+'''-Solution by TheBoomBox77'''
-<cmath>x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.</cmath>
+==Solution 4==
+Use Law of Cosines in <math>\triangle ABC</math> to get <math>\angle BAC=120^\circ</math>. Because <math>AE</math> bisects <math>\angle A</math>, <math>E</math> is the midpoint of major arc <math>BC</math> so <math>BE=CE,</math> and <math>\angle BEC=60^\circ.</math> Thus <math>\triangle BEC</math> is equilateral. Notice now that <math>\angle BFC=\angle BFE= 60^\circ.</math> But <math>\angle DFE=90^\circ</math> so <math>FD</math> bisects <math>\angle BFC.</math> Thus, <cmath>\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.</cmath>
+Let <math>BF=5k, CF=3k.</math> Use Law of Cosines on <math>\triangle BFC</math> to get<cmath>25k^2+9k^2-15k^2 = 49 \qquad \Longrightarrow\qquad k=\frac 7{\sqrt{19}}</cmath> Use Ptolemy's Theorem on <math>BFCA</math>, to get <cmath>15k+15k=7\cdot AF, \qquad \Longrightarrow\qquad AF= \frac{30}{\sqrt{19}},</cmath> so <math>AF^2=\frac{900}{19}</math> and the answer is <math>900+19=919</math>
-Subtracting the first equation from the second, the <math>x^2</math> term cancels out and we obtain:
+~Shen Kislay Kai
-<cmath>(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}</cmath>
+==Solution 5==
+[[File:2012 AIME II 15a.png|500px|right]]
+Denote <math>AB = c, BC = a, AC = b, \angle A = 2 \alpha.</math>
+Let M be midpoint BC. Let <math>\theta</math> be the circle centered at <math>A</math> with radius <math>\sqrt{AB \cdot AC} =\sqrt{bc}.</math>
-<cmath>a^2+2ab = \frac{5341}{192}.</cmath>
+We calculate the length of some segments.
+The median <math>AM = \sqrt{\frac {b^2}{2} + \frac {c^2}{2} -  \frac {a^2}{4}}.</math>
+The bisector <math>AD = \frac {2 b c \cos \alpha}{b+c}.</math> One can use Stewart's Theorem in both cases.
+<math>AD</math> is bisector of <math>\angle A \implies BD = \frac {a c}{b + c}, CD = \frac {a b}{b + c} \implies</math>
+<cmath>BD \cdot CD = \frac {a^2 bc }{(b+c)^2}.</cmath>
+We use Power of Point <math>D</math> and get <math>AD \cdot DE = BD \cdot CD. </math>
+<cmath>AE = AD + DE = AD + \frac {BD \cdot CD}{AD},</cmath>
+<cmath>AE =\frac {2 b c \cos \alpha}{b+c} + \frac {a^2 bc \cdot (b+c) }{(b+c)^2 \cdot 2 b c \cos \alpha} =</cmath>
+<cmath>=  \frac {b c \cos^2 \alpha + a^2}{2(b+c)\cos \alpha} =\frac {4bc \cos^2 \alpha  + b^2 +c^2 -2 b c \cos 2\alpha}{2(b+c) \cos \alpha} = \frac {b+c}{2} \implies AD \cdot AE = 2 bc \cos \alpha.</cmath>
+We consider the inversion with respect <math>\theta.</math>
-By Power of a Point, <math>ab=BD \cdot DC=735/64=2205/192</math>, so
+<math>B</math> swap <math>B' \implies AB' = AC, B'  \in AB \implies B'</math> is symmetric to <math>C</math> with respect to <math>AE.</math>
-<cmath>a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}</cmath>
+<math>C</math> swap <math>C' \implies AC' = AB, C'</math> lies on line <math>AC \implies C'</math> is symmetric to <math>B</math> with respect to <math>AE.</math>
-<cmath>a^2=\frac{931}{192}.</cmath>
+<math>BC^2 = AB^2 + AC^2 + AB \cdot BC \implies \alpha = 60^\circ \implies AD \cdot AE = bc \implies D</math> swap <math>E.</math>
-Since <math>a=XD</math>, <math>XD=\frac{7\sqrt{19}}{8\sqrt{3}}</math>.
+Points <math>D</math> and <math>E</math> lies on <math>\Gamma \implies \Gamma </math> swap <math>\Gamma.</math>
-Because <math>\angle EXF</math> and <math>\angle EAF</math> intercept the same arc in circle <math>\omega</math> and the same goes for <math>\angle XFA</math> and <math>\angle XEA</math>, <math>\angle EXF\cong\angle EAF</math> and <math>\angle XFA\cong\angle XEA</math>. Therefore, <math>\triangle XDE\sim\triangle ADF</math> by [[Similarity (geometry)|AA Similarity]]. Since side lengths in similar triangles are proportional,
+<math>DE</math> is diameter <math>\Gamma, \angle DME = 90^\circ \implies  M \in \Gamma.</math> Therefore <math>M</math> is crosspoint of <math>BC</math> and <math>\Gamma.</math>
-<cmath>\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}</cmath>
+Let <math>\Omega</math> be circumcircle <math>AB'C'.  \Omega</math> is image of line <math>BC.</math>
+Point <math>M</math> maps into <math>M' \implies M' = \Gamma \cap \Omega.</math>
-<cmath>\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}</cmath>
+Points <math>A, B',</math> and <math>C'</math> are symmetric to <math>A, C,</math> and <math>B,</math> respectively.
-<cmath>AF \cdot \sqrt{19} = 30</cmath>
+Point <math>M'</math> lies on <math>\Gamma</math> which is symmetric with respect to <math>AE</math> and on <math>\Omega</math> which is symmetric to <math>\omega</math> with respect to <math>AE  \implies</math>
-<cmath>AF = \frac{30}{\sqrt{19}}.</cmath>
+<i><b><math>M'</math> is symmetric <math>F</math> with respect to <math>AE \implies AM' = AF.</math></b></i>
-However, the problem asks for <math>AF^2</math>, so <math>AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}</math>.
+We use Power of Point <math>A</math> and get
+<cmath>AF = AM' = \frac {AD \cdot AE}{AM} = \frac {4b c}{\sqrt{2 b^2 + 2 c^2 – a^2}} = \frac {4 \cdot 3 \cdot 5}{\sqrt{ 50 + 18 – 49}}
+= \frac {30}{\sqrt{19}} \implies \boxed{\textbf{919}}.</cmath>
-'''-Solution by TheBoomBox77'''
+'''vladimir.shelomovskii@gmail.com, vvsss'''
-==Solution 4==
-It can be verified with law of cosines that <math>\angle BAC=120^\circ.</math> Also, <math>E</math> is the midpoint of major arc <math>BC</math> so <math>BE=CE,</math> and <math>\angle BEC=60.</math> Thus <math>CBE</math> is equilateral. Notice now that <math>\angle CEB=\angle BFE= 60.</math> But <math>\angle DFE=90</math> so <math>FD</math> bisects <math>\angle BFC.</math> Thus, <math>\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.</math>
-Let <math>BF=5a, CF=3a.</math> By law of cosines on <math>BFC</math> we find <math>a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.</math> But by ptolemy on <math>BFCA</math>, <math>15a+15a=7*AF,</math> so <math>AF= \frac{30}{\sqrt{19}},</math> so <math>AF^2=\frac{900}{19}</math> and the answer is <math>900+19=\boxed{919}</math>
+==Video Solution by mop 2024==
+https://youtu.be/mIFUuY4ybeg
-~abacadaea
+~r00tsOfUnity
 == See Also ==
 {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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