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

(Added solution 5.)
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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 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, so AXM = \alpha, so MXP = 180-\alpha, so 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.  
+
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.  
  
 
<i>Solution by SimilarTriangle.
 
<i>Solution by SimilarTriangle.
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:26, 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.

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