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Difference between revisions of "2013 USAMO Problems/Problem 1"

(→‎Solution 4)
(→‎Solution 4)
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\end{align*}</cmath>
 
\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 quadrilateral <math>CQZ'P</math> is also cyclic.
+
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 Y'TZ'
 +
&=180^\circ-\angle RTQ\
 +
&=180^\circ-(180^\circ-\angle RAQ)\
 +
&=\angle RAQ\
 +
&=\alpha.
 +
\end{align*}</cmath>
 
(Solution in progress)
 
(Solution in progress)
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:42, 6 August 2020

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


courtesy v_enhance

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 can 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("$A$", (-3.65,6.49333333333334), N * labelscalefactor); dot((-7.61,-5)); label("$B$", (-7.556666666666688,-4.866666666666666), SW * labelscalefactor); dot((7.09,-5)); label("$C$", (7.15,-4.866666666666666), SE * labelscalefactor); dot((-2.958888888888898,-5)); label("$P$", (-2.9033333333333546,-4.866666666666666), S * labelscalefactor); dot((0.5968131669050584,1.8770271258031248)); label("$Q$", (0.6566666666666455,2.0133333333333376), NE * labelscalefactor); dot((-5.053354907372894,2.4694710603912564)); label("$R$", (-4.996666666666688,2.6), dir(165) * labelscalefactor); dot((-3.143912404905382,-2.142970212141873)); label("$Z'$", (-3.09,-2.04), dir(170) * labelscalefactor); dot((-3.413789986031826,2.0243286531799747)); label("$Y'$", (-3.3566666666666882,2.1333333333333377), NE * labelscalefactor); dot((-3.3284001481939356,0.7057864725120093)); label("$X'$", (-3.276666666666688,0.8133333333333368), dir(20) * labelscalefactor); dot((1.7922953932137468,0.6108747864253139)); label("$V$", (1.8433333333333121,0.72), NE * labelscalefactor); dot((-5.8024625203461,-5)); label("$U$", (-5.7433333333333545,-4.8933333333333335), S * labelscalefactor); dot((-0.10264330299819162,1.125351256231488)); label("$T$", (-0.05,1.2266666666666703), dir(105) * 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 Y'TZ' &=180^\circ-\angle RTQ\\ &=180^\circ-(180^\circ-\angle RAQ)\\ &=\angle RAQ\\ &=\alpha. \end{align*} (Solution in progress)

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