Let $Q$ be intersection of circumcircles $\omega_1, \omega_2$ of $\triangle ABP, \triangle CDP$ other than $P$ $\Longrightarrow$ $\angle QCA \equiv \angle QCP = \angle QDP \equiv \angle QDB$ and $\angle QAC \equiv \angle QAP = \angle QBP \equiv \angle QBD$. In addition, $|AC| = |BD|$ $\Longrightarrow$ $\triangle ACQ \cong \triangle BDQ$ are congruent by ASA $\Longrightarrow$ their Q-altitudes are congruent $\Longrightarrow$ $PQ$ internally bisects $\angle BPC = \angle DPA$. $BM, CN$ internally bisect $\angle PBC \equiv \angle PBS, \angle PCB \equiv \angle PCT,$ resp. $\Longrightarrow$ $I \equiv BM \cap CN$ is incenter of $\triangle BCP$ $\Longrightarrow$ $I \in PQ$. Let $PI$ cut circumcircle $\omega_3$ of $\triangle BCP$ again at $X$, the circumcenter of $\triangle BCI$. Since $I \in PQ$, the radical axis of $\omega_1, \omega_2$ $\Longrightarrow$ $\overline {IC} \cdot \overline {IN} = \overline {IB} \cdot \overline {IM}$ $\Longrightarrow$ $BCMN$ is cyclic $\Longrightarrow$ $MN$ is antiparallel of $BC$ WRT $\angle BIC$ $\Longrightarrow$ $MN \perp (IX \equiv PQ)$ $\Longrightarrow$ $MN \parallel O_1O_2$.

yetti wrote: Let $Q$ be intersection of circumcircles $\omega_1, \omega_2$ of $\triangle ABP, \triangle CDP$ other than $P$ $\Longrightarrow$ $\angle QCA \equiv \angle QCP = \angle QDP \equiv \angle QDB$ and $\angle QAC \equiv \angle QAP = \angle QBP \equiv \angle QBD$. In addition, $|AC| = |BD|$ $\Longrightarrow$ $\triangle ACQ \cong \triangle BDQ$ are congruent so $QD=QC$ and $QB=QA$ and $\angle DQC=\angle DPC=\angle APB=\angle AQB$ so $\triangle QDC\sim \triangle QAB$ and $QB/QC=O_{1}M/O_{2}N$ if $K$ is the intersection of $O_{1}M$ and $O_{2}N$ than $\angle QBC=\angle QBS=\angle QO_{1}S=\angle O_{2}O_{1}K$ because $O_{2}O_{1}$ bisects arc $PQ$ and similarly$\angle QCB=\angle O_{1}O_{2}K$ so $\triangle O_{1}O_{2}K\sim \triangle QCB$ so $O_{1}K/O_{2}K=QB/QC=O_{1}M/O_{2}N$ so $O_{1}O_{2}\parallel MN$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(450); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.500000000000002, xmax = 5.500000000000006, ymin = -1.200000000000001, ymax = 3.500000000000004; /* image dimensions */ pen xfqqff = rgb(0.4980392156862751,0.000000000000000,1.000000000000000); /* draw figures */ draw((-0.3955027041764394,2.769703318376122)--(0.000000000000000,0.000000000000000)); draw((0.000000000000000,0.000000000000000)--(3.251247363475243,-0.6510004022174213)); draw((3.251247363475243,-0.6510004022174213)--(4.868735292508128,1.138163719104525)); draw((-0.3955027041764394,2.769703318376122)--(4.868735292508128,1.138163719104525)); draw((-0.3955027041764394,2.769703318376122)--(3.251247363475243,-0.6510004022174213)); draw((0.000000000000000,0.000000000000000)--(4.868735292508128,1.138163719104525)); draw(circle((0.7249288513237442,1.516606751193169), 1.680957428742995)); draw(circle((3.463961311185914,0.7824206313670016), 1.449118036280306)); draw((xmin, -0.2680457901046969*xmin + 1.710920877915932)--(xmax, -0.2680457901046969*xmax + 1.710920877915932), xfqqff); /* line */ draw((xmin, -0.2680457901047006*xmin + 0.4255385698408272)--(xmax, -0.2680457901047006*xmax + 0.4255385698408272), xfqqff); /* line */ /* dots and labels */ dot((0.000000000000000,0.000000000000000),dotstyle); label("$B$", (-0.09487963689043608,-0.2176445980104207), NE * labelscalefactor); dot((4.868735292508128,1.138163719104525),dotstyle); label("$D$", (4.939487282933190,1.152467918144111), NE * labelscalefactor); dot((3.251247363475243,-0.6510004022174213),dotstyle); label("$C$", (3.206135634588853,-0.8740240824937546), NE * labelscalefactor); dot((-0.3955027041764394,2.769703318376122),dotstyle); label("$A$", (-0.5537080143933489,2.834838635654792), NE * labelscalefactor); dot((2.047063050201595,0.4785417063121247),dotstyle); label("$P$", (1.944357596455843,0.2284385467840780), NE * labelscalefactor); dot((0.8100380300812474,-0.1621946977392190),dotstyle); label("$S$", (0.7463057218649041,-0.3769600068655989), NE * labelscalefactor); dot((3.108372887091018,-0.6223925077097222),dotstyle); label("$T$", (3.034074993025261,-0.8421610007227189), NE * labelscalefactor); dot((3.463961311185913,0.7824206313670030),dotstyle); label("$O_2$", (3.320842728964581,0.8975632639758261), NE * labelscalefactor); dot((0.7249288513237442,1.516606751193169),dotstyle); label("$O_1$", (0.6507164765517973,1.243646408770325), NE * labelscalefactor); dot((2.420677410180655,-0.2233138191596462),dotstyle); label("$N$", (2.447794288438206,-0.1857815162393851), NE * labelscalefactor); dot((1.498051911284965,0.02399206166259225),dotstyle); label("$M$", (1.447293520827688,-0.1794088998851780), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use barycentric coordinates with respect to $\triangle PBC$. Let $P = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$. Let $BC = a, CP = b, BP = c,$ and let $AC = BD = d$. Since $AP = d-b$ and $DP = d-c$, we see that $A = \left(1+\frac{d-b}{b}, 0, -\frac{d-b}{b}\right) = (d:0:b-d)$ and $D = \left(1+\frac{d-c}{c}, -\frac{d-c}{c}, 0\right) = (d:c-d:0)$. It is well-known that the general equation of a circle is $a^2yz+b^2zx+c^2xy - (x+y+z)(ux+vy+wz) = 0$. Substituting $A, B, P, C, D$ yields the equations of the circumcircles of $\triangle ABP$ and $\triangle DCP$: \begin{align*} \odot ABP &: a^2yz+b^2zx+c^2xy-(x+y+z)(bdz) = 0 \\ \odot DCP &: a^2yz+b^2zx+c^2xy-(x+y+z)(cdy) = 0\end{align*} Let $P'$ be the intersection other than $P$ of the two circumcircles. Since $PP'$ is their radical axis, it follows that $O_1O_2 \perp PP'$. The equation of the radical axis of the circumcircles can be obtained by subtracting their equations. Thus $PP': cy - bz = 0$, which implies that $PP'$ is the angle bisector of $\angle BPC$. Let $I$ be the incenter of $\triangle BPC$. Since $\angle PBM = \angle PAM = \angle MAS = \angle MBS$, $I$ lies on $BM$. Similarly, $I$ also lies on $CN$. By PoP, we have $IM \cdot IB = IP \cdot IP' = IN \cdot IC$, so $BMNC$ is cyclic. Therefore $MN \perp IP$, so $MN \perp PP'$, implying that $MN \parallel O_1O_2$.

robinpark wrote: Let $I$ be the incenter of $\triangle BPC$. Since $\angle PBM = \angle PAM = \angle MAS = \angle MBS$, $I$ lies on $BM$. Similarly, $I$ also lies on $CN$. By PoP, we have $IM \cdot IB = IP \cdot IP' = IN \cdot IC$, so $BMNC$ is cyclic. Therefore $MN \perp IP$, so $MN \perp PP'$, implying that $MN \parallel O_1O_2$. Darn I barybashed this part too on the test. It was not fun.

Another solution:Let $O3$ be the circumcenter of $PTS$.Now,from PoP and using that $AC=BD$ we get that $BT/BP=CS/CP$ and since $O2,N,O3$ are collinear and $O1,M,O3$ are collinear,so we have to prove $O2N/O2O3=O1M/O1O3$,or $O2P/O2O3=O1P/O1O3$.Now it is easy to see that $PO3T$ is similar to $PBO1$(simple angle chase) so we have $PO1O3$ is similar to $PTB$,so we have $O1O3/PO1=BT/BP$ and in the same way($PO3S$ is similar to $PO2C$) we get that $O2O3/O2P=CS/CP$,so from this we are finished.

Let $Q$ be the second point of intersection of $\odot{CPD},\odot{APB}$.Then note that $\angle{QCA}=\angle{QCP}=\angle{QDP}=\angle{QDB}$ and similarly $\angle{QAC}=\angle{QBD}$.Combining this with $AC=BD$ we see that $\triangle{AQC} \cong \triangle{BQD}$.Thus $Q$ is the midpoint of arcs $CPD$ and $BPA$.So $\triangle{CQD} \sim \triangle{BQA} \implies \frac{BQ}{CQ}=\frac{R_1}{R_2}$.Let $X=O_1O_2 \cap PQ$.Then since $N$ is the midpoint of minor arc $PT$ we get $\angle{O_1O_2N}=\angle{XO_2N}=\angle{PO_2N}-\angle{PO_2X}=\angle{PCT}-\angle{PCQ}=\angle{QCT}$.Analogously we get $\angle{O_2O_1M}=\angle{MBS}$.Let $Y=O_2N \cap O_1M$.We have $\triangle{O_1O_2Y} \sim \triangle{BCQ}$.Thus $\frac{YO_1}{YO_2}=\frac{QB}{QC}=\frac{R_1}{R_2}=\frac{MO_1}{NO_2} \implies \frac{MO_1}{YO_1}=\frac{NO_2}{YO_2}$ or $MN \parallel O_1O_2$ as desired //

Somewhat similar to the above approaches, but still... [asy][asy]size(12cm); pair B=(0,0), C=(10,0), A=(3,3),D,P,Q,O1,O2,Ss,T,Nn,M,I; D=abs(A-C)*dir(35);P=extension(A,C,B,D); O1=circumcenter(A,B,P);O2=circumcenter(C,D,P); Q=extension((A+B)/2,bisectorpoint(A,B),(C+D)/2,bisectorpoint(C,D)); T=C*abs(P)*abs(D)/(abs(C)^2); Ss=C+(B-C)*abs(P-C)*abs(A-C)/(abs(B-C)^2); M=extension(B,bisectorpoint(P,B,C),O1,bisectorpoint(P,Ss)); Nn=extension(C,bisectorpoint(P,C,B),O2,bisectorpoint(P,T)); I=extension(B,M,C,Nn); draw(circumcircle(A,B,P)^^circumcircle(C,D,P)^^arc(circumcenter(B,M,Nn),abs(B-circumcenter(B,M,Nn)),45,135),cyan+green); D(MP("A",A,N)--MP("B",B,S)--MP("S",Ss,S)--MP("T",T,S)--MP("C",C,S)--MP("D",D,N)--A); D(B--MP("P",P,N)--D,magenta);D(C--MP("N",Nn,W)--MP("M",M,N)--B,magenta);D(T--P--Ss,magenta);D(P--MP("I",I,SE)--MP("Q",Q,S),magenta);D(C--A--Q--B);D(D--Q--C);D(MP("O_1",O1,S)--MP("O_2",O2,SE),magenta); dot(A^^B^^C^^D^^Ss^^T^^O1^^O2^^P^^Q^^M^^Nn^^I); [/asy][/asy] Let $Q$ be the second intersection of $\odot (ABP)$ and $\odot(CDP)$; clearly $Q$ is the center of the spiral similarity taking $AC\mapsto BD$, but since $AC=BD$, this is actually a rotation. So $QAB$ and $QCD$ are similar isosceles triangles. Clearly $O_1O_2\perp PQ$, so it suffices to show that $MN\perp PQ$. Note that $$\angle BPQ=\angle BAQ=\angle CDQ=\angle CPD,$$so $PQ$ is the bisector of $\angle BPA$. Also, $BM$ bisects $\angle PBC$ and $CN$ bisects $\angle PCB$, so $PQ,BM,CN$ concur at the incenter $I$ of $\triangle PBC$. Now by power of point, $NI\cdot IC=PI\cdot IQ=MI\cdot IB$, so $B,M,N,C$ are concyclic. Shoving configuration issues under the carpet, this implies $\angle MNI=\angle MBC=\tfrac12\angle PBC$. Also, by an easy angle chase, $$\angle PIN=\angle 90^\circ-\tfrac12 \angle PBC,$$so $MN\perp PI\equiv PQ$, as required. $\blacksquare$

A very intuitive approach by trig: Let $R_1, R_2$ be the radius's of $w_1, w_2$ and $Q$ be the 2nd intersection of $w_1, w_2$. If we want $O_1O_2||MN$, we want the projections from $O_1$ and $O_2$ onto $MN$ to be the same length, in other words we want $R_1*sin(MO_1O_2)=R_2*sin(NO_2O_1)$ or $\frac{R_1}{R_2}=\frac{sin(NO_2O_1)}{sin(MO_1O_2)}$. Also since $O_1O_2$ perpendicular $PQ$ and $O_1M$ perpendicular $QS$ and $NO_2$ perpendicular $PT$, we have $\frac{sin(NO_2O_1)}{sin(MO_1O_2)}=\frac{sin(QPT)}{sin(QPS)}$. Also by sine law on $O_1PO_2$, we get $\frac{R_1}{R_2}=\frac{sin(PO_2O_1)}{sin(PO_1O_2)}=\frac{sin(QTP)}{sin(QSP)}$. So the problems reduces to showing $\frac{sin(QPT)}{sin(QTP)}=\frac{sin(QPS)}{sin(QSP)}$ or $QT=QS$ which follows from easy angle chase From AC=BD, spiral, we get AQ=QB, QC=QD, so $AQB$, $DQC$ similar so $QSB=QAB=QDC=QTC$ so $QT=QS$.

Could someone please tell me how to immediately conclude after showing $BMNC$ cyclic? I had to do a lot more angle chasing to show $MN\perp PQI$ from that.

Generic_Username wrote: Could someone please tell me how to immediately conclude after showing $BMNC$ cyclic? I had to do a lot more angle chasing to show $MN\perp PQI$ from that. The easiest way IMO to think about it is as follows: letting $I_B$, $I_C$ be the $B$-, $C$-excenters of $\triangle PBC$, we know that $BCMN$ and $BCI_BI_C$ are cyclic, so by Reim we obtain $\overline{MN} \parallel \overline{I_BI_C}$, which is perpendicular to $\overline{PI}$.

Let $X$ be the second intersection of $\omega_1, \omega_2$. The length condition gives us a spiral congruence $\triangle XAC \cong \triangle XBD$, which is equivalent to $\triangle XAB$ and $\triangle XCD$ being similar isosceles triangles. Thus $\angle APX = \angle DPX$, i.e. $XP$ is the internal angle bisector of $\angle BPC$, and so $BM, CN, XP$ concur at the incenter $I$ of $\triangle BPC$. By power of a point, this implies $BMNC$ cyclic. Generic_Username wrote: Could someone please tell me how to immediately conclude after showing $BMNC$ cyclic? I had to do a lot more angle chasing to show $MN\perp PQI$ from that. I think this is reasonably fast: consider $(BIC)$. Note that $II \parallel MN$, as both are antiparallel to $BC$ wrt $\angle BIC$. But by Fact 5, $PI$ passes through the center of $(BIC)$, hence $PI \perp II$. Finally, we're done because $XPI \perp MN \parallel O_1O_2$. $\Box$

Let $X$ be the second intersection of $(ABP)$ and $(CDP)$. Observe that $\triangle ACX \sim \triangle BDX$, so $\triangle ABX$ and $\triangle CDX$ are similar and isosceles. Now observe that $$\angle BPX = 180 - \angle BAX = 180 - \angle XDC = \angle XPC, $$so $XP$ is the angle bisector of $\angle BPC$. Note that by definition, $BM$ is the angle bisector of $\angle CBP$, and $CN$ is the angle bisector of $\angle PCB$, so $BM, CN, PX$ concur at the incenter $I$ of $\triangle PBC$. In addition, by radical axes $BCNM$ is cyclic. Now easy angle chasing shows that $PI \perp MN$, and since $P,I,X$ are collinear, $MN \perp PX$. Finally $O_1O_2 \perp PX$, so $MN \parallel O_1O_2$ as desired.

Let $X$ be the second intersection of $\omega_1$ and $\omega_2$. By spiral similarity lemma, $X$ is the center of spiral similarity sending $\overline{AC}$ to $\overline {BD}$. Moreover since $BD=AC$ we have $$\triangle XBD\cong\triangle XAC$$Hence $XA=XB$ and $XC=XD$. Hence $XP$ is the external angle bisector of $\angle APB$. Let $l$ be the internal angle bisector of $\angle APB$, since $O_1O_2\perp XP$ we have $O_1O_2\|l$. Hence it suffices to prove $MN\|l$. Let $BM$ and $NC$ meet $l$ at $Q$ and $R$ respectively. Since $M$ and $N$ are the arc-midpoints we conclude that $Q$ and $R$ are the $C$- and $B$- excenters of $\triangle PBC$ respectively. Now $QB,XP,RC$ interesects at the $P-excenter$ of $\triangle PBC$. Let it be $Y$, then $$YB\times YM=YP\times YX=YC\times YN$$Hence $M,B,C,N$ are concyclic. Meanwhile, $\angle YBC=90^{\circ}-\frac{\angle PBC}{2}=\angle PRC$. Hence $QBCR$ is cyclic. Applying Reim's theorem to $(BCQR)$ and $(BCNM)$ we have $$MN\|QR$$as desired.

Solution with hint from mathlogician. Let $\omega_1,\omega_2$ meet again at $E$. Note that $E$ is the Miquel point of $ABDC$, so there is a spiral similarity about $E$ taking $AC$ to $BD$, and $\triangle EAC\sim\triangle EBD$. Since $BD=AC$, this implies $EB=EA,EC=ED$, so $E$ is an arc midpoint of $CD$ with respect to $\omega_2$. Now, note that $BM$ is the interior angle bisector of $\angle SBP$ by the definition of $M$, and $CN$ is the interior angle bisector of $\angle TCP$. Since $\angle EPC=\frac{1}{2}\angle EO_2C=\frac{1}{4}\angle DO_2C=90^\circ-\frac{1}{2}\angle DPC=\frac{1}{2}\angle BPC$, $PE$ is an interior angle bisector of $\angle BPC$, hence $PE,BM,CN$ concur at the incenter of $\triangle BPC$. Now, we delete as many points as is possible; rewrite the problem in terms of $\triangle BPC$ as follows, noting that $EP$, the radical axis of $\omega_1$ and $\omega_2$, must be perpendicular to $O_1O_2$. Claim: [Reformulated Problem] Given $\triangle BPC$, let $E$ be a point on the interior angle bisector of $\angle BPC$, then define $M$ to be the second intersection of $(BEP)$ and the interior angle bisector of $\angle PBC$, and $N$ to be the second intersection of $(BEC)$ and the interior angle bisector of $\angle PCB$. Prove that $MN$ is perpendicular to line $EP$. Let $I$ denote the incenter of $\triangle BPC$. We demonstrate that $I$ is the orthocenter of $\triangle MEN$, which suffices. Indeed, observe that \[\measuredangle IEM=\measuredangle PEM=\measuredangle PBM=90^\circ-\measuredangle BCN-\measuredangle EPB=90^\circ-\measuredangle BMN-\measuredangle EMB=90^\circ -\measuredangle EMN,\]so $IE\perp MN$ and similar results hold. Remark: I enjoyed this problem and especially the symmetric version (although it died to angle chase), although I disliked how the statement ``E is the arc midpoint of CD on $\omega_2$ and the arc midpoint of AB on $\omega_1$'' was disguised with the length condition ``$AC=BD$,'' which is essentially begging to be used in a clever way because it is mostly useless as is. @this thread the parts after noting $BCMN$ concyclic are not trivial, it took me a while to work out the necessary angle chase after I realized that it was feasible.

Let $Q$ be the spiral center sending $AC \to BD$; then by spiral similarity/rotation we get that $\Delta QAB$ and $\Delta QCD$ are similar and isosceles. Add in the incenter $I$ of $\Delta BPC$; since $\angle BPQ = \angle BAQ = \angle CDQ = \angle CPQ$ we have that $I$ is the concurrence of $PQ, BM, CN$. In particular since $BM$ and $CN$ intersect on the radical axis we get $BCMN$ cyclic; from here (letting $PQ \cap MN = E$) it is easy to compute $$\angle MIE = \angle IPB + \angle IBP = 90^\circ - \angle NCB = 90^\circ - \angle IME$$as desired.

Let $Q$ be the second intersection of $(ABP)$ and $(CDP)$. First note it is obvious that $PQ\perp O_1O_2$. By Spiral, we have $\triangle QCA\cong \triangle QDB$ which implies $\triangle QAB$ and $\triangle QCD$ is isosceles and they are similar. Hence \[\measuredangle QPB=\measuredangle QAB=\measuredangle CDQ=\measuredangle CPQ\]so $PQ$ bisects $\measuredangle BPC$. Also we know that $BM$ and $CN$ bisects $\measuredangle PBC$ and $\measuredangle BCP$; therefore, $I=PQ\cap BM\cap CN$ is the incenter of $\triangle PBC$. Thus by the converse of Raxis Theorem we have that $BMNC$ is cyclic. Therefore, letting $R=PQ\cap MN$, we have \[\measuredangle RIM=\measuredangle PIB=90^\circ+\measuredangle NCB=90^\circ+\measuredangle NMB=90^\circ - \measuredangle IMR,\]so \[\measuredangle PRM=\measuredangle IRM=\measuredangle RIM+\measuredangle IMR=90^\circ. \]Hence, $MN||O_1O_2$ as desired.

Solved with pbj2006 and tigershark22. We let $Q=\omega_1\cap\omega_2$ and $R=BM\cap CN$. Note that $Q$ is the Miquel point of $ACBD$, hence we have the spiral congruence of \[\triangle AQC \cong \triangle BQC\]This spiral congruence sends $AB$ to $CD$, hence \[\angle APQ=\angle ABQ=\angle DCQ=\angle DPQ\]so $PQ$ bisects $\angle BPC$. It follows from this that $BM$, $CN$, and $PQ$ concur at $R$, since it is the incenter of $\triangle BPC$. We now claim that $R$ is the orthocenter of $\triangle QMN$, and this finishes the problem, as $PQ\perp O_1O_2$. This is equivalent to showing that $\angle QRN=90-\angle RNM$. But this follows from \[\angle RNM=180-\angle MNC=\angle MBC=\frac12\angle PBC\]and \[\angle QRN=180-\angle PQN-\angle RNQ=\angle QNC-\angle PQN=90-\frac12\angle PBC\]as desired.

Let $Q$ be the intersection of $\omega_1$ and $\omega_2$ other than $P$. It suffices to show that $PQ \perp MN$ since $O_1O_2 \perp PQ$ by radical axis. Note that $\triangle QBD \cong \triangle QAC$, so $QC=QD$ and $QB=QA$ and $\triangle QAB \sim \triangle QCD$. This means that $\triangle QO_2C \sim \triangle BO_1Q$, so $\angle QPC=\frac{1}{2} \angle QO_2C=\frac{1}{2} \angle BO_1Q=\angle QPB$. Hence $PQ$ is the angle bisector of $\angle BPC$. Lines $PQ, BM, CN$ concur at the point $I$, so we can deduce by radical axis that $BMNC$ is cyclic. Now, if $\angle NIQ=x$, then $\angle CBP=2x$ and $\angle CIP=90+x$. Hence $MN \perp PQ$.

Let the circles intersect again at $Q$. It suffices to show $\overline{PQ} \perp \overline{MN}$ by radical axis. Claim. $\overline{PQ}, \overline{BM}, \overline{CN}$ concur. We will show that they concur at the incenter $I$ of $\triangle PBC$. First, observe $Q$ is the Miquel Point of $ACDB$, and thus sends $\overline{AC} \to \overline{BD}$. Yet $AC=BD$, implying that $QA=QB$ and $QC=QD$. As a result, Fact 5 implies that $\overline{PQ}$ bisects the exterior $\angle APB$, and thus bisects $\angle BPC$. Next, $$\measuredangle MBC = \measuredangle MBS = \measuredangle MPS = \measuredangle PBM,$$so $\overline{BM}$ bisects $\angle PBC$. Similarly, $\overline{CN}$ bisects $\angle PCB$, implying the result. $\blacksquare$ Thus $BMNC$ is cyclic by radical axis. Let $X = \overline{PI} \cap \overline{MN}$; then we have $$\angle INM + \angle XIN = \frac 12 \angle BPC + \frac 12 \angle BCP + \frac 12 \angle CBP = 90^{\circ}$$by our cyclic quadrilateral, giving the desired perpendicularity.

Let $Q = \omega_1 \cup \omega_1$. By spiral similarity lemma $\triangle QBD\cong\triangle QAC$, therefore $QA=QB$ and $QC=QD$. Claim. $PQ$ bisects $\angle CPB$.

Claim. $BM$ and $CN$ bisect $\angle PBC$ and $\angle BCP$, respectively.

By previous claims we conclude, that $PQ$, $BM$, $CN$ are concurrent, say this point $I$. Obviously, $I$ is the incenter of $\triangle PBC$. Note that $$IM \cdot IB = IQ \cdot IP = IN \cdot IC$$so $M$, $B$, $C$, $N$ are concyclic. It's well-known, that $PQ \perp O_1O_2$, so it's sufficient to show $PQ \perp MN$.

gives us $\angle MRP = 90^\circ$. Thus, $PQ \perp MN$. $\blacksquare$

I just wanted to point out that after proving $\triangle QST$ is isosceles (where $Q = \omega_1 \cap \omega_2$) the problem can be generalised to the following problem Generalised Problem wrote: Let $\omega_1$ and $\omega_2$ be two circles intersecting at $P$ and $Q$ and $S$ and $T$ be two points on $\omega_1$ and $\omega_2$ such that $QS=QT$. Let $M$ and $N$ be the midpoints of minor arcs $\widehat {SP}$ (not including $B$) and $\widehat {TP}$ (not including $C$). Prove that $MN \perp PQ$ Note that the TSTST problem immediately follows from this generalisation The proof is by considering the two circles $\gamma_1$ and $\gamma_2$ centred at $M$ and $N$ with radius $MP$ and $MQ$ respectively and let $\Gamma$ denote the circle centred at $Q$ with radius $QS=QT$ Then consider the point $X=PQ \cap \Gamma$, it can be shown by easy angle chasing that $2 \cdot \angle PXS = 180 - \angle SMP$ which is enough to imply that $X$ lies on $\gamma_1$ and similarly on $\gamma_2$ Now radical axis on $\gamma_1$ and $\gamma_2$ gives the desired conclusion $\blacksquare$

Let $Q$ be the second intersection between $\omega_1$ and $\omega_2$ and $R=\overline{MN}\cap\overline{PQ}.$ Note $Q$ is the Miquel point of $ABDC,$ so there is a spiral similarity centered at $Q$ from $\overline{AC}$ to $\overline{BD}.$ Since we also know $AC=BD,$ we conclude that $\triangle QAC\cong\triangle QBD.$ Also, $\triangle AQB\sim\triangle DQC$ with $AQ=BQ,$ so $$\measuredangle QPC=\measuredangle QDC=\measuredangle BAQ=\measuredangle BPQ.$$But $\measuredangle NCT=\measuredangle PCN$ and $\measuredangle SBM=\measuredangle MBP$ so $\overline{PQ},\overline{BM},$ and $\overline{CN}$ concur at incenter $I$ of $\triangle BPC.$ It follows that $BMNC$ is cyclic from Radical Axis. Since $\overline{PQ}$ is the radical axis of $\omega_1,\omega_2,$ $\overline{O_1O_2}\perp\overline{PQ}.$ Moreover,\begin{align*}180-\angle IRM&=\angle IMR+\angle RIM\\&=180-\angle BMR+\angle NIM-\angle RIN\\&=\angle BCI+90+\tfrac{1}{2}\angle CPB-\angle RIN\\&=\angle ICP+90+\angle IPC-(180-\angle CIP)\\&=90\end{align*}and $\overline{MN}\perp\overline{PQ}.$ $\square$

Had to use GGB for this one. :cry: [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11, xmax = 11, ymin = -9, ymax = 8; /* image dimensions */ pen wwccqq = rgb(0.4,0.8,0); /* draw figures */ draw((-5.036817252067716,4.370696189752862)--(-6.357855203599197,-3.7041912997544517), linewidth(0.5)); draw((-6.357855203599197,-3.7041912997544517)--(5.219022656010006,-3.759209079015626), linewidth(0.5)); draw((5.219022656010006,-3.759209079015626)--(-5.036817252067716,4.370696189752862), linewidth(0.5) + blue); draw((-6.357855203599197,-3.7041912997544517)--(0.9053141489813683,7.182680742067465), linewidth(0.5) + blue); draw((5.219022656010006,-3.759209079015626)--(0.9053141489813683,7.182680742067465), linewidth(0.5)); draw((0.9053141489813683,7.182680742067465)--(-5.036817252067716,4.370696189752862), linewidth(0.5)); 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label("$Q$", (-1.9032464281198098,-0.20893535982089118), NE * labelscalefactor); dot((-2.4521659403727973,-1.6266905202116078),linewidth(3pt) + dotstyle); label("$M$", (-3.3636445853086763,-1.449327197054375), NE * labelscalefactor); dot((-1.3511865669195364,-1.4534293782562635),linewidth(3pt) + dotstyle); label("$N$", (-1.38308210927996,-1.289276637411345), NE * labelscalefactor); dot((-1.8182679346440949,-1.289509676499491),linewidth(3pt) + dotstyle); label("$X$", (-2.5634549866473115,-1.1492323977236936), NE * labelscalefactor); dot((-6.028919833012769,0.3874989626276793),linewidth(3pt) + dotstyle); label("$O_1$", (-6.724769537366109,0.6713427182157748), NE * labelscalefactor); dot((3.5114813222192196,1.8888720480133765),linewidth(3pt) + dotstyle); label("$O_2$", (3.358415720144824,2.1318040749584255), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $Q=\odot(\omega_1)\cap\odot(\omega_2)\not=P$. This implies that $Q$ is the center of the spiral similarity mapping $\overline{BA}\mapsto\overline{DC}$, and moreover due to our given length condition, we get that the spiral similarity is just a rotation about $Q$. And thus as $B\mapsto A$, we get that $QB=QA$ and similarly that $QD=QC$. Now note that due to Fact 5, we get that $BQ$ is the angle-bisector of $\angle SBP$ and similarly, we also get that $CN$ is the angle-bisector of $\angle TCP$. Now note that due to our spiral similarity, we have that $\measuredangle QPB=\measuredangle QAB=\measuredangle QBA=\measuredangle QDC=\measuredangle QPC\implies PQ$ is also the angle-bisector of $\angle BPC$. Now letting $X=$ the incenter of $\triangle BPC$, we get that the lines $\left\{PQ,BM,CN\right\}$ concur at $X$. Then we have $XM\cdot XB=XQ\cdot XP=XN\cdot XC\implies BMNC$ is cyclic. Now finally we have, \begin{align*} \measuredangle (PX,MN)&=\measuredangle NMP+\measuredangle MPX\\ &=\measuredangle XMP+\measuredangle NMX+\measuredangle MPQ\\ &=(2\measuredangle BPM)+\measuredangle MPQ+\measuredangle NCB\\ &=\measuredangle BPM+\measuredangle BPQ+\measuredangle NCB\\ &=\measuredangle MBP+\measuredangle BPQ+\measuredangle NCB\\ &=\measuredangle BXP+\measuredangle NCB\\ &=90^\circ+\measuredangle BCN+\measuredangle NCB\\ &=90^\circ-\measuredangle NCB+\measuredangle NCB\\ &=90^\circ .\end{align*}This proves that $PQ\perp MN$, but as $PQ$ is the radical axis of $\left\{\omega_1,\omega_2\right\}$, we also have that $PQ\perp O_1O_2$ which finally implies $O_1O_2\parallel MN$ and we are done.

Solved with everythingpi3141592 Let $X$ be the second intersection of $(APB)$ and $(BPC)$. Note that $X$ becomes the center of spiral similarity from $AB$ to $CD$ thus making it the center of spiral similarity from $AC$ to $BD$. However since $AC = BD$, we have $\triangle AXC \equiv \triangle BXD$, which means $X$ is the midpoint of arcs $AB$ and $CD$ on $(APB)$ and $(BPC)$ respectively. Next let $K$ be the incenter of $\triangle BPC$ and $G = PK \cap MN$. We have: $$\measuredangle BPX = \measuredangle BAX = \measuredangle XDC =\measuredangle XPC$$Also note that $CN$ and $BM$ are the angle bisectors of $\angle CBP$ and $\angle BCP$ respectively, since $M$ and $N$ are the midpoints of arcs $PT$ and $PS$ respectively. Thus we obtain $PX$, $CN$, $MB$ are concurrent, which by radical axis implies $MBNC$ is a cyclic quadrilateral. Now note that $\angle BKP = 90^{\circ} + \frac{1}{2} \cdot \angle BCP$, implying $\angle MKG = \angle BKG = 90 - \frac{1}{2} \cdot \angle BCP$. Also note that $$\angle KMG = \angle BMG = \angle BMN = \angle BCN = \angle BCK = \angle BCP / 2$$Thus we obtain $PX \perp MN$, and since $PX \perp O_1O_2$ by radical axis theorem, we have $O_1O_2 \parallel MN$. 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/* end of picture */ [/asy][/asy] Not one word about the asymptote, not one!

Let Q be the second intersection point of the circles, and note that Q sending AC to BD is a rotation, whence AQB and CQD are isosceles, and also BPQ=BAQ=CDQ=CPQ, so PQ is the angle bisector of BPC. Since BM,CN are angle bisectors of CBP,BCP, BM,CN,PQ concur at the incenter of BPC I, and IN\cdot IC=IP\cdot IQ=IM\cdot IB means BCMN is cyclic. The problem is now evident, because NMI+MIP=(NCB)+(180-BIP)=(NCT)+(180-(90+PCN))=90 (since NCT=PCN), from which it follows that MN perp. PQ perp. O_1O_2 means MN//O_1O_2.

InterLoop wrote:

just a small remark, the solution is really cool especially the asymptote diagram.

nice problem

Let $Q = (ABP) \cap (CDP)$, so that $Q$ is the center of spiral similarity taking $\overline{AB} \mapsto \overline{CD}$ and also $\overline{AC} \mapsto \overline{BD}$. Then since $QA = QB$ and $QC = QD$, we have $$\measuredangle QPB = \measuredangle QAB = \measuredangle CDQ = \measuredangle CPQ,$$so $Q$ lies on the bisector of $\angle CPB$. But we also have $$\measuredangle PBM = \measuredangle PSM = \measuredangle MPS = \measuredangle MBS = \measuredangle MBC,$$so $BM$ bisects $\angle PBC$, and similarly $CN$ bisects $\angle BCP$. Thus, $PI$, $BM$, and $CN$ concur at some point $I$ which is the incenter of $\triangle PBC$. Moreover, if $R = \overline{PI} \cap \overline{MN}$, then $$\angle IMR + \angle RIM = 180^{\circ} - \angle BMN + 180^{\circ} - \angle PIB = \angle BCN + 180^{\circ} - \angle PIB = \tfrac{1}{2} \angle C + 180^{\circ} - (90^{\circ} + \tfrac{1}{2} \angle C) = 90^{\circ},$$so $\overline{IR} \perp \overline{MN}$. Since $\overline{PIR} \perp \overline{O_1O_2}$ by radical axis, we are done.

Let $Q$ be the other point of intersection. We need $NQ^2-NP^2$ to be constant, or \[\sin^2\left(\frac{\angle PCB}{2}+\angle QCP\right)-\sin^2\left(\frac{\angle PCB}{2}\right)\]and since $\sin^2(a)-\sin^2(b)\propto \sin(a+b)\sin(a-b)$ we simplify down and we're done by LoS.

Claim 1: $QS = QT$. Spiral similarity at $Q$ along with $AC=BD$ tells us \[\triangle QAC \cong \triangle QBD \implies QA=QB, \quad QC=QD.\] Thus we also have similar isosceles triangles $\triangle QAB \sim \triangle QCD$, and \[\measuredangle TSQ = \measuredangle BAQ = \measuredangle QDC = \measuredangle QTS. \quad {\color{blue} \Box}\] Claim 2: The distances to $O_1O_2$ from $M$ and $N$ are the same, implying the desired. Note that we have $SP \perp MO_1$, $TP \perp NO_2$, and $PQ \perp O_1O_2$, so $\measuredangle MO_1O_2 = \measuredangle SPQ$. As a result, \begin{align*} MO_1 \sin \angle MO_1O_2 &= r_1 \sin \angle SPQ = \frac{QS}{2}, \\ NO_2 \sin \angle NO_2O_1 &= r_2 \sin \angle TPQ = \frac{QT}{2}, \end{align*} and hence the distances are equal. $\blacksquare$

Let $Q = (ABP) \cap (CPD)$, where $Q \neq P$. We take directed angles $\measuredangle$ modulo $\pi$. Observe the spiral similarity at $Q$ that sends $AB \stackrel{Q}{\mapsto} CD$. Now as $AC = BD$, we have that $\triangle QAC \cong \triangle QBD$, and also $\triangle QAB \sim \triangle QCD$ are isosceles. We now make a following claim: Claim: $PQ, BM, CN$ concur at the incenter $I$ of triangle $\triangle BPC$. Proof: Observe that $BM$ bisects $\angle SBP$: \[\measuredangle SBM = \measuredangle SPM; \text{ and } \measuredangle MBP = \measuredangle MSP. \]It similarly follows that $CN$ bisects $\angle PCT$. Now we also have $PQ$ is the external bisector of $\angle BPC$: \[\measuredangle PQB = \measuredangle QAB = \measuredangle CDQ = \measuredangle CPQ. \square\] As $I$ now lies on the radical axis of $\omega_1$ and $\omega_2$, we have that $BMNC$ is cyclic as $IB \cdot IM = IP \cdot IQ = IN \cdot IC$. Now since $I$ is the incenter, $MN$ is parallel to the tangent of $(BIC)$ as $\angle IMN = \angle NCB$. As we also have that the tangent at $I$ of $(BIC)$ is perpendicular to $PI$ (use the incenter/excenter lemma to find out why...), we find that $PI \perp MN$, and similarly $PI = PQ \perp O_1O_2$, so we indeed have $MN \parallel O_1O_2$, as desired. $\blacksquare$

Let $P'$ be the spiral center taking $CA$ to $DB$, it is the arc midpoint of $AB$ on $(PAB)$ as well as the arc midpoint of $CD$ on $(PCD)$ since $AC=BD$. Claim: This rotation at $P'$ sending $AC$ to $BD$ also sends $S$ to $T$. In particular, $P'S=P'T$. It suffices to show that $SCA$ and $TDB$ are congruent. We have $CA=DB$ given, and $\angle SAC=\angle TBD$ as $PSAB$ is cyclic. Furthermore, $\angle SCA=\angle PCT=\angle TDB,$ which shows the claim. Now, draw $\omega_1$ and $\omega_2$ intersecting at $P$ and $P'$ such that $O_1O_2$ is horizontal, and let $S$ be on $\omega_1$ and $T$ be on $\omega_2$ such that $P'S=P'T$. Let $\theta_1=\angle P'O_1O_2$ and $\theta_2=\angle P'O_2O_1$. On $\omega_1$, $P'$ is an angle of $\theta_1$ above the horizontal, while $T$ is $\theta_1+\angle TO_1P'$ below. Thus, the argument of the arc midpoint $M$ is $-\angle TO_1P'/2$, possibly plus 180 degrees depending on which arc midpoint. Similarly, the argument of $N$ is $-\angle SO_2P'/2$ possibly plus 180. The height of $M$ is then $\pm R_1\sin -\angle TO_1P'/2=\pm TP'/2$, and similarly the height of $N$ is $\pm SP'/2$, so as long as we choose the right one they are the same height, done.

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