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Difference between revisions of "2013 USAMO Problems/Problem 1"
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==Problem==
 
==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>.
  
 
==Solution 1==
 
==Solution 1==
Line 81: Line 81:
 
In this solution, all lengths and angles are directed.
 
In this solution, all lengths and angles are directed.
  
−
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>
+
Firstly, it is easy to see by that <math>\omega_A, \omega_B, \omega_C</math> concur at a point <math>M</math> (the Miquel point). 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 = 180^\circ - \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM</cmath>
 
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.
 
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.
  
−
courtesy v_enhance
+
 
 +
courtesy v_enhance, minor clarification by integralarefun
 +
----
 +
 
 +
==Solution 2==
 +
 
 +
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]
 +
Refer to the Diagram link.
 +
 
 +
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:
 +
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath>
 +
In addition, we have
 +
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath>
 +
Now, by the Ratio Lemma, we have
 +
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.
 +
The proof is complete.
 +
 
 +
==Solution 3==
 +
Use directed angles modulo <math>\pi</math>.
 +
 
 +
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math>
 +
 
 +
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath>
 +
 
 +
Now, it follows that (now not using directed angles)
 +
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath>
 +
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.
 +
 
 +
==Solution 4==
 +
We will use some construction arguments to solve the problem. Let <math>\angle BAC=\alpha,</math> <math>\angle ABC=\beta,</math> <math>\angle ACB=\gamma,</math> and let <math>\angle APB=\theta.</math> We construct lines through the points <math>Q,</math> and <math>R</math> that intersect with <math>\triangle ABC</math> at the points <math>Q</math> and <math>R,</math> respectively, and that intersect each other at <math>T.</math> We will construct these lines such that <math>\angle CQV=\angle ARV=\theta.</math>
 +
 
 +
 
 +
Now we let the intersections of <math>AP</math> with <math>RV</math> and <math>QU</math> be <math>Y'</math> and <math>Z',</math> respectively. This construction is as follows.
 +
<asy>
 +
import graph; size(12cm);
 +
real labelscalefactor = 1.9;
 +
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
 +
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);
 +
 
 +
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);
 +
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5));
 +
draw((-7.61,-5)--(7.09,-5));
 +
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669));
 +
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5));
 +
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139));
 +
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));
 +
dot((-3.6988888888888977,6.426666666666669));
 +
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor);
 +
dot((-7.61,-5));
 +
label("$B$", (-7.61,-5), SW * labelscalefactor);
 +
dot((7.09,-5));
 +
label("$C$", (7.09,-5), SE * labelscalefactor);
 +
dot((-2.958888888888898,-5));
 +
label("$P$", (-2.958888888888898,-5), S * labelscalefactor);
 +
dot((0.5968131669050584,1.8770271258031248));
 +
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor);
 +
dot((-5.053354907372894,2.4694710603912564));
 +
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor);
 +
dot((-3.143912404905382,-2.142970212141873));
 +
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor);
 +
dot((-3.413789986031826,2.0243286531799747));
 +
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor);
 +
dot((-3.3284001481939356,0.7057864725120093));
 +
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor);
 +
dot((1.7922953932137468,0.6108747864253139));
 +
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor);
 +
dot((-5.8024625203461,-5));
 +
label("$U$", (-5.8024625203461,-5), S * labelscalefactor);
 +
dot((-0.10264330299819162,1.125351256231488));
 +
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor);
 +
</asy>
 +
 
 +
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,
 +
<cmath>\begin{align*}
 +
\angle BRY'+\angle BPY'
 +
&=180^\circ-\theta+\theta\\
 +
&=180^\circ.
 +
\end{align*}</cmath>
 +
 
 +
Since the opposite angles of quadrilateral <math>RY'PB</math> add up to <math>180^\circ,</math> it must be cyclic. Similarly, we can also show that quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic.
 +
 
 +
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,
 +
<cmath>Y'=\omega_B\cap AP</cmath>
 +
and that
 +
<cmath>Z'=\omega_C\cap AP.</cmath>
 +
 
 +
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.
 +
 
 +
Since quadrilateral <math>AQTR</math> is also cyclic, we have,
 +
<cmath>\begin{align*}
 +
\angle Y'TZ'
 +
&=180^\circ-\angle RTQ\\
 +
&=180^\circ-(180^\circ-\angle RAQ)\\
 +
&=\angle RAQ\\
 +
&=\alpha.
 +
\end{align*}</cmath>
 +
 
 +
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,
 +
<cmath>\begin{align*}
 +
\angle TY'Z'
 +
&=180^\circ-\angle RY'P\\
 +
&=180^\circ-(180^\circ-\angle RBP)\\
 +
&=\angle RBP\\
 +
&=\beta,
 +
\end{align*}</cmath>
 +
and,
 +
<cmath>\begin{align*}
 +
\angle Y'Z'T
 +
&=180^\circ-\angle PZ'Q\\
 +
&=180^\circ-(180^\circ-\angle PCQ)\\
 +
&=\angle PCQ\\
 +
&=\gamma.
 +
\end{align*}</cmath>
 +
 
 +
Since these three angles are of <math>\triangle TY'Z',</math> and they are equal to corresponding angles of <math>\triangle ABC,</math> by AA similarity, we know that <math>\triangle TY'Z'\sim \triangle ABC.</math>
 +
 
 +
We now consider the point <math>X=\omega_c\cap AC.</math> We know that the points <math>A,</math> <math>Q,</math> <math>T,</math> and <math>R</math> are concyclic. Hence, the points <math>A,</math> <math>T,</math> <math>X,</math> and <math>R</math> must also be concyclic.
 +
 
 +
Hence, quadrilateral <math>AQTX</math> is cyclic.
 +
 
 +
<asy>
 +
import graph; size(12cm);
 +
real labelscalefactor = 1.9;
 +
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
 +
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451);
 +
 
 +
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);
 +
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5));
 +
draw((-7.61,-5)--(7.09,-5));
 +
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669));
 +
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5));
 +
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139));
 +
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));
 +
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));
 +
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));
 +
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));
 +
dot((-3.6988888888888977,6.426666666666669));
 +
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor);
 +
dot((-7.61,-5));
 +
label("$B$", (-7.61,-5), SW * labelscalefactor);
 +
dot((7.09,-5));
 +
label("$C$", (7.09,-5), SE * labelscalefactor);
 +
dot((-2.958888888888898,-5));
 +
label("$P$", (-2.958888888888898,-5), S * labelscalefactor);
 +
dot((0.5968131669050584,1.8770271258031248));
 +
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor);
 +
dot((-5.053354907372894,2.4694710603912564));
 +
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor);
 +
dot((-3.143912404905382,-2.142970212141873));
 +
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor);
 +
dot((-3.413789986031826,2.0243286531799747));
 +
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor);
 +
dot((-3.3284001481939356,0.7057864725120093));
 +
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor);
 +
dot((1.7922953932137468,0.6108747864253139));
 +
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor);
 +
dot((-5.8024625203461,-5));
 +
label("$U$", (-5.8024625203461,-5), S * labelscalefactor);
 +
dot((-0.10264330299819162,1.125351256231488));
 +
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor);
 +
</asy>
 +
 
 +
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,
 +
<cmath>\begin{align*}
 +
\angle AXT
 +
&=\angle ART\\
 +
&=\theta.
 +
\end{align*}</cmath>
 +
 
 +
Consider by this result, we can deduce that the homothety that maps <math>ABC</math> to <math>TY'Z'</math> will map <math>P</math> to <math>X.</math> Hence, we have that,
 +
<cmath>Y'X/XZ'=BP/PC.</cmath>
 +
 
 +
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,
 +
<cmath>YX/XZ=BP/PC,</cmath>
 +
 
 +
as required.
 +
 
 +
== Solution 5 (Simple Rotational Homotethy)==
 +
<asy>
 +
/* DRAGON 0.0.9.6 */
 +
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;
 +
/* 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(P--M, rgb(0.0,0.2,0.4));
 +
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(Z--M, rgb(0.0,0.2,0.4));
 +
draw(X--M, rgb(0.0,0.2,0.4));
 +
draw(Y--M, rgb(0.0,0.2,0.4));
 +
draw(C--Z, rgb(0.0,0.2,0.4));
 +
draw(B--Y, rgb(0.0,0.2,0.4));
 +
/* 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);
 +
/* 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);
 +
</asy>
 +
 
 +
We begin again by noting that the three circumcircles intersect at point <math>M</math> by Miquel's theorem. In addition, we state that the angle <math>\angle MQC = \alpha</math>, hence <math>\angle MPC = \angle MZC = 180 - \alpha</math>, as well as <math>\angle AQM</math>, from which follows that <math>\angle ARM = \alpha</math>, so <math>\angle BRM = 180 - \alpha</math>, and <math>\angle BPM = \alpha</math>.
 +
We shall prove that the triangles <math>\triangle MZC</math> and <math>\triangle XMP</math> and <math>\triangle YMB</math> are similar, which will imply a rotational homotethy with angle <math>\angle APB</math> about the point <math>M</math>, that takes <math>Y,X,Z</math> to <math>B,P,C</math>, thus proving the problem. (In essence, just imagine we rotate <math>YMZX</math> around M and lengthen things out and get <math>MCPB</math> - the ratios will remain identical.)
 +
 
 +
We do this by angle chasing. Denote angle <math>\angle ZMP = \beta</math>. From the angles labeled before, we now know <math>\angle ZPB = \alpha-\beta = \angle ZMC</math>. In addition, <math>\angle ZPM = \angle ZCM</math>. So the angles in triangle <math>\triangle ZMC</math>  are <math>\alpha-\beta</math>, <math>\beta</math> and thus <math>180-\alpha</math>, with <math>\alpha-\beta</math> at the point <math>M</math>. In addition, <math>\angle YRM = \angle YPM = \beta</math>, so <math>\angle ARY = \alpha-\beta</math>, so <math>\angle YMB = \alpha - \beta</math>. Since <math>\angle YBM = \angle YPM = \beta</math>, the triangle <math>\triangle YMB</math> has angles <math>\alpha-\beta, \beta, 180-\alpha</math> also, with <math>\alpha-\beta</math> at <math>M</math>. Finally, In triangle <math>\triangle XPM</math>, we already know the angle <math>\angle XPM</math> to be <math>\beta</math>; we also can find that <math>\angle AQM = 180-\alpha</math>, so <math>\angle AXM = \alpha</math>, so <math>\angle MXP = 180-\alpha</math>, so <math>\angle XMP = \alpha-\beta</math>. Thus, the three triangles are similar have a common point <math>M</math>, which proves that there is a rotational homotethy around <math>M</math> that maps <math>Y, X, Z</math> to <math>B, P, C</math> as desired.
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''Solution by SimilarTriangle.''
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==See Also==
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{{USAMO newbox|year=2013|before=First Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 22:38, 3 January 2025

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 $M$ (the Miquel point). 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 \[\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = 180^\circ - \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM\] Therefore, $\frac{XD}{XP} = \frac{PB}{PM}$. Analogously we find that $\frac{XE}{XP} = \frac{PC}{PM}$ and we are done.


courtesy v_enhance, minor clarification by integralarefun

Solution 2

Diagram Refer to the Diagram link.

By Miquel's Theorem, there exists a point at which $\omega_A, \omega_B, \omega_C$ intersect. We denote this point by $M.$ Now, we angle chase: \[\angle YMX = 180^{\circ} - \angle YXM - \angle XYM\]\[= 180^{\circ} - \angle AXM - \angle PYM\]\[= \left(180^{\circ} - \angle ARM\right) - \angle PRM\]\[= \angle BRM - \angle PRM\]\[= \angle BRP = \angle BMP.\] In addition, we have \[\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX\]\[= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP\]\[= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP\]\[= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP\]\[= \angle BPM - \angle MCP\]\[= 180^{\circ} - \angle MPC - \angle MCP\]\[= \angle CMP.\] Now, by the Ratio Lemma, we have \[\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}\]\[= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}\] (by the Law of Sines in $\triangle MZY$)\[= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}\]\[= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}\]\[= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}\] (by the Law of Sines in $\triangle MBC$)\[= \frac{PB}{PC}\] by the Ratio Lemma. The proof is complete.

Solution 3

Use directed angles modulo $\pi$.

Lemma. $\angle{XRY} \equiv \angle{XQZ}.$

Proof. \[\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.\]

Now, it follows that (now not using directed angles) \[\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}\] using the facts that $ARY$ and $APB$, $AQZ$ and $APC$ are similar triangles, and that $\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}$ equals twice the circumradius of the circumcircle of $AQR$.

Solution 4

We will use some construction arguments to solve the problem. Let $\angle BAC=\alpha,$ $\angle ABC=\beta,$ $\angle ACB=\gamma,$ and let $\angle APB=\theta.$ We construct lines through the points $Q,$ and $R$ that intersect with $\triangle ABC$ at the points $Q$ and $R,$ respectively, and that intersect each other at $T.$ We will construct these lines such that $\angle CQV=\angle ARV=\theta.$


Now we let the intersections of $AP$ with $RV$ and $QU$ be $Y'$ and $Z',$ respectively. This construction is as follows. [asy] import graph; size(12cm); real labelscalefactor = 1.9; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); draw((-7.61,-5)--(7.09,-5)); draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); dot((-3.6988888888888977,6.426666666666669)); label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); dot((-7.61,-5)); label("$B$", (-7.61,-5), SW * labelscalefactor); dot((7.09,-5)); label("$C$", (7.09,-5), SE * labelscalefactor); dot((-2.958888888888898,-5)); label("$P$", (-2.958888888888898,-5), S * labelscalefactor); dot((0.5968131669050584,1.8770271258031248)); label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); dot((-5.053354907372894,2.4694710603912564)); label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); dot((-3.143912404905382,-2.142970212141873)); label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); dot((-3.413789986031826,2.0243286531799747)); label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); dot((-3.3284001481939356,0.7057864725120093)); label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); dot((1.7922953932137468,0.6108747864253139)); label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); dot((-5.8024625203461,-5)); label("$U$", (-5.8024625203461,-5), S * labelscalefactor); dot((-0.10264330299819162,1.125351256231488)); label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); [/asy]

We know that $\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.$ Hence, we have, \begin{align*} \angle BRY'+\angle BPY' &=180^\circ-\theta+\theta\\ &=180^\circ. \end{align*}

Since the opposite angles of quadrilateral $RY'PB$ add up to $180^\circ,$ it must be cyclic. Similarly, we can also show that quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic.

Since points $Y'$ and $Z'$ lie on $AP,$ we know that, \[Y'=\omega_B\cap AP\] and that \[Z'=\omega_C\cap AP.\]

Hence, the points $Y'$ and $Z'$ coincide with the given points $Y$ and $Z,$ respectively.

Since quadrilateral $AQTR$ is also cyclic, we have, \begin{align*} \angle Y'TZ' &=180^\circ-\angle RTQ\\ &=180^\circ-(180^\circ-\angle RAQ)\\ &=\angle RAQ\\ &=\alpha. \end{align*}

Similarly, since quadrilaterals $CQZ'P,$ and $AQTR$ are also cyclic, we have, \begin{align*} \angle TY'Z' &=180^\circ-\angle RY'P\\ &=180^\circ-(180^\circ-\angle RBP)\\ &=\angle RBP\\ &=\beta, \end{align*} and, \begin{align*} \angle Y'Z'T &=180^\circ-\angle PZ'Q\\ &=180^\circ-(180^\circ-\angle PCQ)\\ &=\angle PCQ\\ &=\gamma. \end{align*}

Since these three angles are of $\triangle TY'Z',$ and they are equal to corresponding angles of $\triangle ABC,$ by AA similarity, we know that $\triangle TY'Z'\sim \triangle ABC.$

We now consider the point $X=\omega_c\cap AC.$ We know that the points $A,$ $Q,$ $T,$ and $R$ are concyclic. Hence, the points $A,$ $T,$ $X,$ and $R$ must also be concyclic.

Hence, quadrilateral $AQTX$ is cyclic.

[asy] import graph; size(12cm); real labelscalefactor = 1.9; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle); draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); draw((-7.61,-5)--(7.09,-5)); draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488)); draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564)); draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488)); dot((-3.6988888888888977,6.426666666666669)); label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); dot((-7.61,-5)); label("$B$", (-7.61,-5), SW * labelscalefactor); dot((7.09,-5)); label("$C$", (7.09,-5), SE * labelscalefactor); dot((-2.958888888888898,-5)); label("$P$", (-2.958888888888898,-5), S * labelscalefactor); dot((0.5968131669050584,1.8770271258031248)); label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); dot((-5.053354907372894,2.4694710603912564)); label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); dot((-3.143912404905382,-2.142970212141873)); label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); dot((-3.413789986031826,2.0243286531799747)); label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); dot((-3.3284001481939356,0.7057864725120093)); label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); dot((1.7922953932137468,0.6108747864253139)); label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); dot((-5.8024625203461,-5)); label("$U$", (-5.8024625203461,-5), S * labelscalefactor); dot((-0.10264330299819162,1.125351256231488)); label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); [/asy]

Since the angles $\angle ART$ and $\angle AXT$ are inscribed in the same arc $\overarc{AT},$ we have, \begin{align*} \angle AXT &=\angle ART\\ &=\theta. \end{align*}

Consider by this result, we can deduce that the homothety that maps $ABC$ to $TY'Z'$ will map $P$ to $X.$ Hence, we have that, \[Y'X/XZ'=BP/PC.\]

Since $Y'=Y$ and $Z'=Z$ hence, \[YX/XZ=BP/PC,\]

as required.

Solution 5 (Simple Rotational Homotethy)

[asy] /* DRAGON 0.0.9.6 */ 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; /* 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(P--M, rgb(0.0,0.2,0.4)); 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(Z--M, rgb(0.0,0.2,0.4)); draw(X--M, rgb(0.0,0.2,0.4)); draw(Y--M, rgb(0.0,0.2,0.4)); draw(C--Z, rgb(0.0,0.2,0.4)); draw(B--Y, rgb(0.0,0.2,0.4)); /* 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); /* 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); [/asy]

We begin again by noting that the three circumcircles intersect at point $M$ by Miquel's theorem. In addition, we state that the angle $\angle MQC = \alpha$, hence $\angle MPC = \angle MZC = 180 - \alpha$, as well as $\angle AQM$, from which follows that $\angle ARM = \alpha$, so $\angle BRM = 180 - \alpha$, and $\angle BPM = \alpha$. We shall prove that the triangles $\triangle MZC$ and $\triangle XMP$ and $\triangle YMB$ are similar, which will imply a rotational homotethy with angle $\angle APB$ about the point $M$, that takes $Y,X,Z$ to $B,P,C$, thus proving the problem. (In essence, just imagine we rotate $YMZX$ around M and lengthen things out and get $MCPB$ - the ratios will remain identical.)

We do this by angle chasing. Denote angle $\angle ZMP = \beta$. From the angles labeled before, we now know $\angle ZPB = \alpha-\beta = \angle ZMC$. In addition, $\angle ZPM = \angle ZCM$. So the angles in triangle $\triangle ZMC$ are $\alpha-\beta$, $\beta$ and thus $180-\alpha$, with $\alpha-\beta$ at the point $M$. In addition, $\angle YRM = \angle YPM = \beta$, so $\angle ARY = \alpha-\beta$, so $\angle YMB = \alpha - \beta$. Since $\angle YBM = \angle YPM = \beta$, the triangle $\triangle YMB$ has angles $\alpha-\beta, \beta, 180-\alpha$ also, with $\alpha-\beta$ at $M$. Finally, In triangle $\triangle XPM$, we already know the angle $\angle XPM$ to be $\beta$; we also can find that $\angle AQM = 180-\alpha$, so $\angle AXM = \alpha$, so $\angle MXP = 180-\alpha$, so $\angle XMP = \alpha-\beta$. Thus, the three triangles are similar have a common point $M$, which proves that there is a rotational homotethy around $M$ that maps $Y, X, Z$ to $B, P, C$ as desired.

Solution by SimilarTriangle.

See Also

2013 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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