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Difference between revisions of "2013 USAMO Problems/Problem 1"
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==Problem==
 
In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively.  Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively.  Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math>
 
In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively.  Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively.  Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math>
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 +
==Solution 1==
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<asy>
 +
/* DRAGON 0.0.9.6
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Homemade Script by v_Enhance. */
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import olympiad;
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import cse5;
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size(11cm);
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real lsf=0.8000;
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real lisf=2011.0;
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defaultpen(fontsize(10pt));
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/* Initialize Objects */
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pair A = (-1.0, 3.0);
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pair B = (-3.0, -3.0);
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pair C = (4.0, -3.0);
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pair P = (-0.6698198198198195, -3.0);
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pair Q = (1.1406465288818244, 0.43122416534181074);
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pair R = (-1.6269590345062048, 1.119122896481385);
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path w_A = circumcircle(A,Q,R);
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path w_B = circumcircle(B,P,R);
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path w_C = circumcircle(P,Q,C);
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pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));
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pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));
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pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));
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pair X = (2)*(foot(O_A,A,P))-A;
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pair Y = (2)*(foot(O_B,A,P))-P;
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pair Z = (2)*(foot(O_C,A,P))-P;
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pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;
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pair D = (2)*(foot(O_B,X,M))-M;
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pair E = (2)*(foot(O_C,X,M))-M;
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/* Draw objects */
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draw(A--B, rgb(0.6,0.6,0.0));
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draw(B--C, rgb(0.6,0.6,0.0));
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draw(C--A, rgb(0.6,0.6,0.0));
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draw(w_A, rgb(0.4,0.4,0.0));
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draw(w_B, rgb(0.4,0.4,0.0));
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draw(w_C, rgb(0.4,0.4,0.0));
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draw(A--P, rgb(0.0,0.2,0.4));
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draw(D--E, rgb(0.0,0.2,0.4));
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draw(P--D, rgb(0.0,0.2,0.4));
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draw(P--E, rgb(0.0,0.2,0.4));
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draw(P--M, rgb(0.4,0.2,0.0));
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draw(R--M, rgb(0.4,0.2,0.0));
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draw(Q--M, rgb(0.4,0.2,0.0));
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draw(B--M, rgb(0.0,0.2,0.4));
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draw(C--M, rgb(0.0,0.2,0.4));
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draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));
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draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));
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draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));
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draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));
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/* Place dots on each point */
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dot(A);
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dot(B);
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dot(C);
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dot(P);
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dot(Q);
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dot(R);
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dot(X);
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dot(Y);
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dot(Z);
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dot(M);
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dot(D);
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dot(E);
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/* Label points */
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label("$A$", A, lsf * dir(110));
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label("$B$", B, lsf * unit(B-M));
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label("$C$", C, lsf * unit(C-M));
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label("$P$", P, lsf * unit(P-M) * 1.8);
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label("$Q$", Q, lsf * dir(90) * 1.6);
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label("$R$", R, lsf * unit(R-M) * 2);
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label("$X$", X, lsf * dir(-60) * 2);
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label("$Y$", Y, lsf * dir(45));
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label("$Z$", Z, lsf * dir(5));
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label("$M$", M, lsf * dir(M-P)*2);
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label("$D$", D, lsf * dir(150));
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label("$E$", E, lsf * dir(5));</asy>
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In this solution, all lengths and angles are directed.
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 +
Firstly, it is easy to see by that <math>\omega_A, \omega_B, \omega_C</math> concur at a point <math>M</math>. Let <math>XM</math> meet <math>\omega_B, \omega_C</math> again at <math>D</math> and <math>E</math>, respectively. Then by Power of a Point, we have <cmath>XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP</cmath> Thusly <cmath>\frac{XY}{XZ} = \frac{XD}{XE}</cmath> But we claim that <math>\triangle XDP \sim \triangle PBM</math>. Indeed, <cmath>\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM</cmath> and <cmath>\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM</cmath>
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Therefore, <math>\frac{XD}{XP} = \frac{PB}{PM}</math>. Analogously we find that <math>\frac{XE}{XP} = \frac{PC}{PM}</math> and we are done.
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courtesy v_enhance

Revision as of 19:19, 11 May 2013

Problem

In triangle $ABC$, points $P,Q,R$ lie on sides $BC,CA,AB$ respectively. Let $\omega_A$, $\omega_B$, $\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, $CPQ$, respectively. Given the fact that segment $AP$ intersects $\omega_A$, $\omega_B$, $\omega_C$ again at $X,Y,Z$ respectively, prove that $YX/XZ=BP/PC$

Solution 1

[asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair A = (-1.0, 3.0); pair B = (-3.0, -3.0); pair C = (4.0, -3.0); pair P = (-0.6698198198198195, -3.0); pair Q = (1.1406465288818244, 0.43122416534181074); pair R = (-1.6269590345062048, 1.119122896481385); path w_A = circumcircle(A,Q,R); path w_B = circumcircle(B,P,R); path w_C = circumcircle(P,Q,C); pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5)); pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5)); pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5)); pair X = (2)*(foot(O_A,A,P))-A; pair Y = (2)*(foot(O_B,A,P))-P; pair Z = (2)*(foot(O_C,A,P))-P; pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P; pair D = (2)*(foot(O_B,X,M))-M; pair E = (2)*(foot(O_C,X,M))-M; /* Draw objects */ draw(A--B, rgb(0.6,0.6,0.0)); draw(B--C, rgb(0.6,0.6,0.0)); draw(C--A, rgb(0.6,0.6,0.0)); draw(w_A, rgb(0.4,0.4,0.0)); draw(w_B, rgb(0.4,0.4,0.0)); draw(w_C, rgb(0.4,0.4,0.0)); draw(A--P, rgb(0.0,0.2,0.4)); draw(D--E, rgb(0.0,0.2,0.4)); draw(P--D, rgb(0.0,0.2,0.4)); draw(P--E, rgb(0.0,0.2,0.4)); draw(P--M, rgb(0.4,0.2,0.0)); draw(R--M, rgb(0.4,0.2,0.0)); draw(Q--M, rgb(0.4,0.2,0.0)); draw(B--M, rgb(0.0,0.2,0.4)); draw(C--M, rgb(0.0,0.2,0.4)); draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8)); draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8)); /* Place dots on each point */ dot(A); dot(B); dot(C); dot(P); dot(Q); dot(R); dot(X); dot(Y); dot(Z); dot(M); dot(D); dot(E); /* Label points */ label("$A$", A, lsf * dir(110)); label("$B$", B, lsf * unit(B-M)); label("$C$", C, lsf * unit(C-M)); label("$P$", P, lsf * unit(P-M) * 1.8); label("$Q$", Q, lsf * dir(90) * 1.6); label("$R$", R, lsf * unit(R-M) * 2); label("$X$", X, lsf * dir(-60) * 2); label("$Y$", Y, lsf * dir(45)); label("$Z$", Z, lsf * dir(5)); label("$M$", M, lsf * dir(M-P)*2); label("$D$", D, lsf * dir(150)); label("$E$", E, lsf * dir(5));[/asy]

In this solution, all lengths and angles are directed.

Firstly, it is easy to see by that $\omega_A, \omega_B, \omega_C$ concur at a point $M$. Let $XM$ meet $\omega_B, \omega_C$ again at $D$ and $E$, respectively. Then by Power of a Point, we have \[XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP\] Thusly \[\frac{XY}{XZ} = \frac{XD}{XE}\] But we claim that $\triangle XDP \sim \triangle PBM$. Indeed, \[\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM\] and \[\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM\] Therefore, $\frac{XD}{XP} = \frac{PB}{PM}$. Analogously we find that $\frac{XE}{XP} = \frac{PC}{PM}$ and we are done.

courtesy v_enhance

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