Edit: clearly since this is going to be the contest thread, I'll post my solution here.

Did anyone else not use the Miquel Point for this problem? I just used LOS like four times. Will post proof later

Darn is the Miquel point citable? It was quick to prove but still...

robinpark you have two similar triangles, angle YRX = ZQX and then bash with the LOS right?

Not at all, you don't even need LOS, you can just have two similar triangles, one of which is the spiral similarity unique to the miguel point

Or once you have the similar triangles you're done because similar triangles means ratios which is all you needed (see tenniskidperson's solution on the other thread).

Whoops, didn't notice any of the similar triangles and bashed everything with LOS. Ended up trying to prove 2 angles were equal, but I had 5 minutes because #2 EVIL.

bary bashing?

Anyone else get trolled when your line AP went straight through where the circles were concurrent?

willwang123 wrote: Anyone else get trolled when your line AP went straight through where the circles were concurrent? Yup, until I drew a larger diagram

For some reason every time I drew the diagram the Miquel point, $B$, and $X$ were collinear :O But I didn't use it in my proof so yay. Also this took me like two hours darn. And USAMO #2/3 too hard QQ

V_enhance must be very happy with this problem. lolol, yeah my first thought was "this is baryable so I will get >=7 today". Then equality occured. -- v_E

I did miquel's theorem and then spiral similarity

Me too.

Hmmm I said "By Miquel's Pivot Theorem" but hopefully that's no big deal.

This looks pretty long, but it actually took me longer to draw the diagram (so not used to using compasses!) than to do it (I bashed it "live" i.e. writing it down on official solution sheet as I went in under 30 min).

ssilwa wrote: V_enhance must be very happy with this problem. darn I actually didn't bary this, even though it was pretty clearly baryable in less than 40 minutes. [asy][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][/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.

By Miquel, $\omega_A$, $\omega_B$, $\omega_C$ concur a single point, which we denote $O$. Then we have \[\angle PBO = \angle OYZ, \quad \angle PCO = \angle YZO, \quad \angle OPC = \angle ORB = \angle OXA,\] implying $\triangle BOC \sim \triangle YOZ$ with corresponding cevians $OP$ and $OX$, giving the desired. $\blacksquare$

first, by miquel point $\omega_A,\omega_B,$ and $\omega_C$ intersect at a point. we will denote it as point $O$. using spiral similarity, we get that triangles $OZY$ and $OCB$ are similar, so $\frac{YZ}{BC}=\frac{OZ}{OC}$ we can also get that triangles $XOY$ and $POB$ are similar from angles, so $\frac{XZ}{PC}=\frac{OZ}{OC}$ therefore, $\frac{XY}{BP}=\frac{OZ}{OC}=\frac{XZ}{PC}$

Let the Miquel point be $O$. Note $\angle{Y}=\angle{B}; \angle{X}=\angle{C}, \triangle{OYZ}\sim \triangle{OBC}$. Moreover, $\angle{AXO}=\angle{ARO}=\angle{OPB}, \triangle{OYX}\sim \triangle{OBP}$, a same reason yields $\triangle{OXZ}\sim \triangle{OPC}$. Thus, $\frac{OY}{OB}=\frac{YZ}{BX}=\frac{YX}{BP}=\frac{XZ}{PC}$, completed the proof.

Mwahaha, coaxiality is real! First, notice $(ARQ), (CQP), (BPR)$ concur by angles at the intersection of any two. Let this point be $T$. Then it follows $(XTP)$ is tangent to $BC$ since $\angle TPC = \pi - \angle TQC = \angle AQT = \pi - \angle AXT = \angle TXP$. Now, by the legendary coaxiality lemma, we have $\frac{XY}{XZ} = \frac{pow_{(BPR)}(X)}{pow_{(CPQ)}(X)} = \frac{pow_{(BPR)}(P)}{pow_{(CPQ)}(P)} = \frac{BP}{PC}$, finishing as desired.

First, we have $\omega_A$, $\omega_B$ and $\omega_C$ concurring at the Miquel point of $\triangle ABC$, $K$. Then we can angle chase to find $\measuredangle{KYZ} = \measuredangle{KYP} = \measuredangle{KBP}$ and similarly $\measuredangle{KZY} = \measuredangle{KCB}$, so $\triangle KYZ \sim \triangle KBC$. Then since $\measuredangle{KXA} = \measuredangle{KRA} = \measuredangle{KRB} = \measuredangle{KPB}$ so $\triangle KXY \sim \triangle KPB$ gives us $YX/XZ = BP/PC$ as desired.

Denote by $\measuredangle$ an angle modulo $\pi$. Let $M$ denote the Miquel point of $\triangle ABC$. Observe the spiral similarity at $M$ sending $YZ \stackrel{M}{\mapsto} BC \implies \triangle MYZ \sim \triangle MBC$. Also observe that as \[\measuredangle MPB = \measuredangle MRA = \measuredangle MXA \text{ and } \measuredangle PBA = \measuredangle PYM,\]we have that $\triangle MYX \sim \triangle MBP$. Hence \[\frac{YZ}{BC} = \frac{MY}{MB} = \frac{YX}{BP} \implies \frac{YZ}{BC} = \frac{YX}{BP} = \frac{YZ - YX}{BC - BP} = \frac{XZ}{PC} \implies \frac{YX}{XZ} = \frac{BP}{PC}, \]as desired. $\blacksquare$

Let $M$ be the miquel point of $\omega_A$, $\omega_B$ and $\omega_C$. Claim: $\triangle YXM \sim \triangle BPM$. Proof: We have $\angle YXM = \angle AXM = \angle ARM = \angle BPM$ and $\angle XYM = \angle PYM = \angle PBM$. Similarly, $\triangle ZXM \sim \triangle CPM$, so the desired ratios follow.

Is spiral similarity necessary? Draw the Miquel point $M$. Through angle chasing, we get $\triangle MYX \sim \triangle MBP$ and $\triangle MYZ \sim \triangle MBC$ with the same scale factor $MY/MB$. So, we are done.

Assume $AYXZP$ is the order, let $M$ denote miquel point, $\angle MZP = 180- \angle MBP$, $\angle MCP = \angle MYX$ so $\triangle YMZ ~ \triangle MCB$, $\angle MXY = \angle MRB = \angle MPC$, thus $\triangle MPC ~ \triangle MXY$ so the spiral similarity sending $YZ$ to $CB$ is the same as the spiral similarity sending $XY$ to $PC$ which suffices.

Let $M = w_A \cap w_B \cap w_C$ which exists by Miquels.

, and similarly $\triangle MXZ \sim \triangle MPC$. By the two similarities we have $\frac{YX}{BP} = \frac{XD}{DP}$ and $\frac{XD}{DP} = \frac{XZ}{CP}$, done after rearrangement

Let $M$ be the miquel point of $\triangle PQR$ wrt $\triangle ABC$. Claim:$\triangle MZX \sim \triangle MCP$ and similarly $\triangle MYX \sim \triangle MBP$ \begin{align*} \measuredangle MZX & \equiv \measuredangle MZP \\ & \stackrel{\omega_{C}}{=} MCP\\ \end{align*}And \begin{align*} \measuredangle MXZ & \equiv \measuredangle MXA \\ & \stackrel{\omega_{A}}{=} MQA\\ & \stackrel{\omega_{C}}{=} MPC\\ \end{align*}Hence by AA similarity criterion. We get that $\triangle MZX \sim \triangle MCP$ Similarly $\triangle MYX \sim \triangle MBP$. $\blacksquare$ Now by the similarities we get \begin{align*} & \frac{XY}{XM} = \frac{BP}{MP} & \dots (1)\\ & \frac{XZ}{XM} = \frac{CP}{MP} & \dots (2) \end{align*}Dividing (1) by (2) we get \[ \frac{YX}{XZ} = \frac{BP}{CP} \]Which is the desired result.

Let $S$ be the intersection of the three circles. I claim that $S$ is the center of a spiral similarity taking $Y\to B,X\to P, Z\to C$. Clearly this finishes the problem. Note that \[\angle SYZ=\angle SYP=\angle SBP,\]so $SYZ$ and $SBC$ are similar. Now \[\angle YSX=\angle YSR+\angle XSR=\angle YBR+\angle XAR=\angle BYP=\angle BSP,\]as desired. $\blacksquare$

$ $

The spirality returns... Let the intersection of the three circles be $M$. The spiral similarity centered at $M$ that sends $BC$ to $YZ$ shows that $\triangle MBC \sim \triangle MYZ$ (which is also a two-step angle chase, as usual). Meanwhile, \[\angle MXY = \angle MXA = \angle MRA = \angle MPC,\]which means $\triangle MXY \sim \triangle MPB$. $\angle MXZ = \angle MPC$ and $\triangle MXZ \sim \triangle MPC$ immediately follows. We then have that $\tfrac{MX}{XY}=\tfrac{MP}{PB}$ and $\tfrac{MX}{XZ}=\tfrac{MP}{PC}$, and dividing gives $\tfrac{XZ}{XY}=\tfrac{PC}{PB}$, equivalent to the condition. $\blacksquare$

What Miquel point? Let $W = RY \cap QZ$. Now $\triangle WYZ \stackrel{+}{\sim} \triangle ABC$ as $\measuredangle WYZ = \measuredangle RYP = \measuredangle RBP = \measuredangle ABC$ and $\measuredangle WZY = \measuredangle QZP = \measuredangle QCP = \measuredangle ACB$. Indeed, $\measuredangle RAQ = \measuredangle BAC = \measuredangle YWZ = \measuredangle RWQ$, so $W$ lies on $\omega_A$. Now we can check $\measuredangle YWX = \measuredangle RWX = \measuredangle RAX = \measuredangle BAP$, so $X$ and $P$ are in corresponding positions under this similarity, which obviously implies the desired conclusion.