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

(→‎Quick Solution in Terms of Olympiad Terms (Similar to above))
(→‎Solution 6:)
 
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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>.
 
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 Using Only Elementary Geometry Methods ==
+
==Quick Solution using Olympiad Terms ==
−
Let <math>P</math> be the unique point on circle <math>\omega</math> such that <math>\bigtriangleup BCP</math> is equilateral. Point <math>E</math> must be the point on <math>\omega</math> such that <math>\angle PED</math> is right. We extend line <math>ED</math> to hit circle <math>\omega</math> again at <math>X</math>.
 
  
−
Claim 1: <math>X</math> is the midpoint of minor arc <math>\overarc{BC}</math>.  
+
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>.
−
Proof: Notice that <math>\angle PEX</math> and <math>\angle PAX</math> are subtended by the same arc, thus they are both equal to 90 degrees.
 
  
−
Now it is obvious that <math>\bigtriangleup PED \sim \bigtriangleup ADX</math>. After this, we length chase with similar triangles and Pythag, which yields <math>AP = 8</math>, <math>EX = \frac{56}{\sqrt{57}}</math>, <math>AX = \frac{2}{\sqrt{3}}</math>, <math>PE = \frac{14}{\sqrt{19}}</math>, and <math>PX = \frac{14}{\sqrt{3}}</math>. We can tie this all together with Ptolemy's theorem for cyclic quadrilaterals, and we find that <math>AE = \frac{30}{\sqrt{19}}</math>. Thus, <math>AE^2 = \frac{900}{19}</math>, and the answer is <math>\boxed{919}</math>
+
'''- Emathmaster'''
−
-jj_ca888
 
  
−
==Quick Solution in Terms of Olympiad Terms (Similar to above)==
+
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.
  
−
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 Ptomtely's we see <math>AF=\frac{30}{\sqrt{19}}</math>. We conclude the answer is <math>900+19=\boxed{919}</math>, as desired.
+
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
−
-Emathmaster
 
  
 
== Solution 1==
 
== Solution 1==
−
Use the angle bisector theorem to find <math>CD=\frac{21}{8}</math>, <math>BD=\frac{35}{8}</math>, and use 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>.
+
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>.
  
−
I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines:
+
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
  
−
<math>AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot \cos \angle AFE.</math> (1)
+
== Solution 2==
  
−
<math>AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot \cos \angle AEF.</math> Adding these two and simplifying we get:
+
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>.
  
−
<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:
+
<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>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>.
 
  
−
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>. 
+
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);
  
−
Plugging in all our values into equation (2), we get:
+
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);
−
<math>EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}</math>, or <math>EF = \frac{7}{15} \cdot AF</math>.
+
</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>
−
Finally, we plug this into equation (1), yielding:
+
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>.  
−
 
 
−
<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>.
 
−
 
 
−
== Solution 2==
 
−
 
 
−
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.
 
−
 
 
−
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.
 
  
 
'''-Solution by thecmd999'''
 
'''-Solution by thecmd999'''
  
 
==Solution 3==
 
==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>
 
<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);
+
draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue);
−
X=(0,0);
+
 
−
D=(sqrt(3)*7/6,-7/8);
+
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));
−
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>
 
</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 <math>\omega</math> (since <math>\angle DFE</math> is <math>90^\circ</math>). Therefore, <math>XE=\tfrac{14\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>.
+
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>.  
  
−
Let <math>EF=x</math>, <math>XD=a</math>, and <math>DF=b</math>. Then, by the [[Pythagorean Theorem]],
+
'''-Solution by TheBoomBox77'''
  
−
<cmath>x^2+b^2=\left(\frac{49}{8}\right)^2=\frac{2401}{64}</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>  
  
−
and
+
~Shen Kislay Kai
  
−
<cmath>x^2+(a+b)^2=\left(\frac{14\sqrt{3}}{3}\right)^2=\frac{196}{3}.</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>
  
−
Subtracting the first equation from the second, the <math>x^2</math> term cancels out and we obtain:
+
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>
  
−
<cmath>(a+b)^2-b^2=\frac{196}{3}-\frac{2401}{64}</cmath>
+
<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+2ab = \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>
  
−
By Power of a Point, <math>ab=BD \cdot DC=735/64=2205/192</math>, so
+
<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>
  
−
<cmath>a^2+2 \cdot \frac{2205}{192}=\frac{5341}{192}</cmath>
+
Points <math>D</math> and <math>E</math> lies on <math>\Gamma \implies \Gamma </math> swap <math>\Gamma.</math>
  
−
<cmath>a^2=\frac{931}{192}.</cmath>
+
<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>
  
−
Since <math>a=XD</math>, <math>XD=\frac{7\sqrt{19}}{8\sqrt{3}}</math>.
+
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>
  
−
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,
+
Points <math>A, B',</math> and <math>C'</math> are symmetric to <math>A, C,</math> and <math>B,</math> respectively.  
  
−
<cmath>\frac{AF}{\frac{15}{8}}=\frac{\frac{14\sqrt{3}}{3}}{\frac{7\sqrt{19}}{8\sqrt{3}}}</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>\frac{AF}{\frac{15}{8}}=\frac{16}{\sqrt{19}}</cmath>
+
<i><b><math>M'</math> is symmetric <math>F</math> with respect to <math>AE \implies AM' = AF.</math></b></i>
  
−
<cmath>AF \cdot \sqrt{19} = 30</cmath>
+
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>
  
−
<cmath>AF = \frac{30}{\sqrt{19}}.</cmath>
+
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
== Solution 6: ==
 +
To do this, we first define the intersection of <math>EF</math> and <math>BC</math> to be <math>K</math>.
  
−
However, the problem asks for <math>AF^2</math>, so <math>AF^2 = \frac{900}{19}\implies 900 + 19 = \boxed{919}</math>.
+
<b>Lemma 1:</b> <math>(K, C, D, B)</math> are harmonic. First of all, define the midpoint of <math>BC</math> to be <math>M</math>. Then, we have that angle FMD is <math>90</math> degrees, and as a result, <math>M</math> lies on this circle. By [[Power of a Point]], <math>(KD)(KM) = (KE)(KF) = (KB)(KC)</math>. As a result, <math>(K, C, D, B)</math> are harmonic from another famous harmonic lemma.
  
−
'''-Solution by TheBoomBox77'''
+
As a result, since <math>\angle EFD = 90^{\circ}</math>, by another Harmonic Lemma, <math>FD</math> is the angle bisector of <math>BFC</math>. Since <math>\frac{BD}{CD} = \frac{5}{3}</math> by [[angle bisector theorem]], <math>\frac{BF}{CF} = \frac{5}{3}</math>. Since <math>\angle BAC</math> is <math>120^{\circ}</math> by [[Law of Cosines]] (LOC), we can use LOC to finish off. Call <math>BF = 5a</math>, and <math>CF = 3a</math>, <math>(5a)^2+(3a)^2-15a^2 = 49</math>, so <math>a = \frac{7}{\sqrt{19}}</math>. We do [[Ptolemy's Theorem]] on <math>ABFC</math>. Our answer is:
 +
<cmath>\frac{(3)(35)+(5)(21)}{7\sqrt{19}} = \frac{30}{\sqrt{19}}.</cmath>
 +
As a result, the final answer is <math>\boxed{919}</math>.
 +
 
 +
-sepehr2010
  
−
==Solution 4==
+
Minor edits ~[[Zhenghua]]
−
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 BFC=\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 ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}
 
{{AIME box|year=2012|n=II|num-b=14|after=Last Problem}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:28, 17 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

2012 AIME II 15a.png

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

Solution 6:

To do this, we first define the intersection of $EF$ and $BC$ to be $K$.

Lemma 1: $(K, C, D, B)$ are harmonic. First of all, define the midpoint of $BC$ to be $M$. Then, we have that angle FMD is $90$ degrees, and as a result, $M$ lies on this circle. By Power of a Point, $(KD)(KM) = (KE)(KF) = (KB)(KC)$. As a result, $(K, C, D, B)$ are harmonic from another famous harmonic lemma.

As a result, since $\angle EFD = 90^{\circ}$, by another Harmonic Lemma, $FD$ is the angle bisector of $BFC$. Since $\frac{BD}{CD} = \frac{5}{3}$ by angle bisector theorem, $\frac{BF}{CF} = \frac{5}{3}$. Since $\angle BAC$ is $120^{\circ}$ by Law of Cosines (LOC), we can use LOC to finish off. Call $BF = 5a$, and $CF = 3a$, $(5a)^2+(3a)^2-15a^2 = 49$, so $a = \frac{7}{\sqrt{19}}$. We do Ptolemy's Theorem on $ABFC$. Our answer is: \[\frac{(3)(35)+(5)(21)}{7\sqrt{19}} = \frac{30}{\sqrt{19}}.\] As a result, the final answer is $\boxed{919}$.

-sepehr2010

Minor edits ~Zhenghua

Video Solution by mop 2024

https://youtu.be/mIFUuY4ybeg

~r00tsOfUnity

See Also

2012 AIME II (Problems • Answer Key • Resources)
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Last Problem
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All AIME Problems and Solutions

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