Any other solutions?

If $O_1, O_2$ are the centers of the 2 circumcircles, prove easily that $O_1POO_2$ is a parallelogram; since $Q$ is the reflection of $P$ in $O_1O_2$, clearly $O_1O_2OQ$ is an isosceles trapezoid, done. Best regards, sunken rock

By angle chasing it is easy to prove $ PROS $ is a parelleogram, where $ R, S $ be the centers of triangles. now $ RS $ bisects $ PO $ and $ RS $ bisects $ PQ $ perpendicularly . So, we have, $ RS $ is parellel to $ OQ $ , which means $ \angle PQO=90^0 $

By angle chasing it is easy to prove $ PROS $ is a parelleogram, can someone tell me how to angle chasing?

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Dear Mathlinkers, we have also that circle (ADQ) goes through Q. (This was the original problem...) Sincerely Jean-Louis

Lemma: $COQB$ is cyclic. We would instead prove that $\angle COB=\angle CQB$. $\angle COB=\overarc{CD}$ (I am referring to the minor arc $CD$). $\angle CQB=360-\angle CQP-\angle BQP=360-(180-\angle CDP)-(180-\angle BAP)=\angle CDP+\angle BAP=2\angle CDP=\overarc{CD}$. (because of cyclic quads $DPQC$, $APQB$, and $ABCD$). Hence $\angle COB=\angle CQB$ and $COQB$ is cyclic. $\angle OQP=360-\angle PQB-\angle OQB=\angle OCB+\angle PAB=\angle OCB+\angle CAB=(90-\frac{\angle COB}{2})+\angle CAB=(90-\angle CAB)+\angle CAB=90$. Q.E.D. Is this correct?

Does anyone have a solution involving the use of spiral symmetry This problem was an exercise in Yufie Zhao's notes under the topic of spiral symmetry. So someone please help !

Dear Mathlinkers, in my archive this problem comes from Russia in 1999 and IMO Shortlist... Can anyone precises my references? For a proof, you can see http://jl.ayme.pagesperso-orange.fr/ vol. 16, Triangles adjacents p. 67-72 with a generalization after... Sincerely Jean-Louis

utkarshgupta wrote: Does anyone have a solution involving the use of spiral symmetry This problem was an exercise in Yufie Zhao's notes under the topic of spiral symmetry. So someone please help ! Yeah there is one and relatively easier than others My Solution: Let $AB\cap CD\equiv R$.Let $M_1,M_2$ denote midpoints of $AB$ and $CD$. Note that $Q$ is the unique spiral centre that carries $A$ to $B$ and $C$ to $D$.Thus it also carries $A$ to $M_1$ and $C$ to $M_2$.Thus by spiral similarity lemma $\odot (RAC) \cap \odot (R M_1 M_2) \equiv Q$ and $R$.Thus $R,M_1,M_2,Q$are concyclic .It is easy to see that $O,M_1,M_2,R$ are concyclic with $OR$ as diametre. Thus $M_1,Q,O,M_2,R$ all lie on a circle with $OR$ as diametre.Thus $\angle OQP=\angle OQR =90^{\circ}$ ($P,Q,R$ are colliner due to radical axis theorem) Q.E.D

Let $AB \cap CD=E$ and $AD \cap BC=F$. Prove that quadrilaterals $ADOQ$ and similarly $BCOQ$ are cyclic using simple angle chasing. Radical Axis Theorem: 1. $AB, CD, PQ$ are concurrent i.e. $P, Q, E$ collinear. 2. $AD, BC, OQ$ are concurrent i.e. $O, Q, F$ collinear. Brokard's Theorem: 1. $OF \perp PE$ i.e. $OQ \perp PQ$ i.e. $\angle OQP=90^{\circ}$, as required.

Suppose line PQ meets BC in S and <PAB=<PDC=α. Since <BQS=<PAB, <CQS=<PDC, <BQC=2α and <BOC=2<BDC=2<PDC=2α.Hence B,Q,O,C are concylic, <OQC=<OBC=90˚-α. Hence <OQP=<OQC+<CQS=90˚-α+α=90˚.

Let $R$ be the intersection of the lines $AB,QP$ and $DC$ (they are concurrent by radical axis). $X(Y)$ are the midpoint of $AB(CD)$. We know that $OX(OY) \perp AB(CD)$. So, $O,R,X,Y$ are concyclic. Or, $OR$ is the diameter. And the spiral similarity center at $Q$ carries $AB \rightarrow DC$ must also carries $XB \rightarrow YD$. Which implies $\angle QXB=\angle QYD \Rightarrow \angle QXO =\angle QYO$. So $O$ lies on $(XQY)$ or $Q,R,X,Y,O$ are concyclic and $QR$ is a diameter so $\angle OQP=90^o$.

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thunderz28 wrote: We know that $OX(OY) \perp AB(CD)$. So, $Q,R,X,Y$ are concyclic. This time, $O,R,X,Y$ con-cyclic. thunderz28 wrote: $QR$ is a diameter so $\angle OQP=90^o$. $OR$ is diameter, not $QR$ Just small typo.

Let $AB\cap CD\equiv E$, $AD\cap BC\equiv F$. From Brocard, $\triangle EPF$ is autopolar and O is its orthocenter. We also know that E-P-Q are collinear (E is the radical center of (PQCD), (ABQP) and (ABCD)). => $FO \perp EQ$, $QO \perp PQ$, $\angle OQP=90^{\circ}$ and we are done.

Q is the center of spiral similarity that takes CD to AB hence :triangle QCA is similar to triangle QDB. and QDB' is similar to QCA' where A' (B') is the midpoint of AC (BD). QA'PB' is Cyclic then OQPB' L cyclic hence OQP=90.

Firstly, noting that $Q$ is spiral center sending $AB$ to $CD$, and by angle chasing, its easy to show that $ADOQ$ and $BCOQ$ are cyclic. Let $X=AB \cap CD$ and $Y=AD \cap BC$. As $X$ is radical center of $(ABCD)$, $(ABP)$ and $(CDP)$, $P,Q,X$ are collinear. Similarly $O,Q,Y$ are collinear. From Brocard's theorem, $OY \perp PX$, and the result follows. $\square$

Seriously, no inversion?! Inversion trivializes it! Consider an inversion about $\odot(ABCD)$. It sends $P \mapsto \odot(AOC) \cap \odot(OBD)=M$, where $M$ is the miquel point of $ABCD$. Also, $Q \mapsto \odot(AMD) \cap \odot(BMC) \ne M$, and so $Q \mapsto R=AB \cap CD$. Hence, $\angle OQP=\angle OMR=90^o$, as desired. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -300.7946837651921, xmax = 919.8259515006099, ymin = -300.0925604705601, ymax = 400.6581024171497; /* image dimensions */pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen ccwwff = rgb(0.8,0.4,1); draw((-52.809239371112085,121.75362521700968)--(-25.201516745702648,125.24128271840691)--(-28.68917424709987,152.84900534381634)--(-56.29689687250931,149.36134784241912)--cycle, linewidth(0.5)); /* draw figures */draw(circle((-24.440312181446032,-102.81013218146497), 144.7438719258341), linewidth(0.5) + dotted + wrwrwr); draw((-116.50567949948035,8.88031772665245)--(117.97219812409817,-76.93610333871355), linewidth(0.5) + red); draw((-114.8108985564882,-215.87624371942007)--(-9.233564035210499,41.132715056246816), linewidth(0.5) + red); draw((-116.50567949948035,8.88031772665245)--(-9.233564035210499,41.132715056246816), linewidth(0.5) + blue); draw((-9.233564035210499,41.132715056246816)--(117.97219812409817,-76.93610333871355), linewidth(0.5) + blue); draw((117.97219812409817,-76.93610333871355)--(-114.8108985564882,-215.87624371942007), linewidth(0.5) + blue); draw((-114.8108985564882,-215.87624371942007)--(-116.50567949948035,8.88031772665245), linewidth(0.5) + blue); draw(circle((-111.28144309810753,-103.46495930997465), 112.46667914726531), linewidth(0.5) + linetype("4 4") + ccwwff); draw(circle((52.070078152762676,-20.3788633005303), 86.84359974950932), linewidth(0.5) + linetype("4 4") + ccwwff); draw(circle((24.714874905976057,31.497542263144066), 143.02022176013625), linewidth(0.5) + dotted + red); draw(circle((-262.9300206565725,-4.840365479525634), 257.8282688848979), linewidth(0.5) + dotted + red); draw(circle((331.81784999133004,-100.1237610243733), 461.38473780507456), linewidth(0.5) + dotted + ccwwff); draw(circle((289.43688817128776,235.3569557768216), 356.268290344114), linewidth(0.5) + dotted + ccwwff); draw((-9.233564035210499,41.132715056246816)--(645.6950503440642,238.0433269339125), linewidth(0.5) + wrwrwr); draw((645.6950503440642,238.0433269339125)--(117.97219812409817,-76.93610333871355), linewidth(0.5) + wrwrwr); draw((-24.440312181446032,-102.81013218146497)--(-56.29689687250931,149.36134784241912), linewidth(0.5) + wrwrwr); draw((-56.29689687250931,149.36134784241912)--(645.6950503440642,238.0433269339125), linewidth(1) + wrwrwr); /* dots and labels */dot((-24.440312181446032,-102.81013218146497),dotstyle); label("$O$", (-25.970517215370467,-138.9451581797426), NE * labelscalefactor); dot((-9.233564035210499,41.132715056246816),dotstyle); label("$B$", (-3.6701843885026935,64.38140582993256), NE * labelscalefactor); dot((117.97219812409817,-76.93610333871355),dotstyle); label("$C$", (144.56143969597133,-102.21519822960775), NE * labelscalefactor); dot((-114.8108985564882,-215.87624371942007),dotstyle); label("$D$", (-136.16039706577592,-255.69395944981417), NE * labelscalefactor); dot((-116.50567949948035,8.88031772665245),dotstyle); label("$A$", (-155.8371613247769,10.59825018866364), NE * labelscalefactor); dot((-34.77105276389883,-21.033690429039996),linewidth(4pt) + dotstyle); label("$P$", (-65.76512431370543,3.662897336863476), NE * labelscalefactor); dot((0.39746601887208044,-90.17680003732414),linewidth(4pt) + dotstyle); label("$Q$", (5.512305599031095,-79.91486540274015), NE * labelscalefactor); dot((-56.29689687250931,149.36134784241912),linewidth(4pt) + dotstyle); label("$M$", (-112.54827995497476,147.023815717736), NE * labelscalefactor); dot((645.6950503440642,238.0433269339125),linewidth(4pt) + dotstyle); label("$R$", (650.9101732942631,248.0312055806069), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

REDACTED

I think inversion and spiral sim are overkill when we can finish only by straightforward angle chasing. Solution. Let $X=BC\cap PQ$, now we have $\angle BAC=\angle BDC=\angle SQX=\angle BQX\implies\angle CQB=2\angle CAB$, but $\angle CCB=2\angle CAB\implies COQB$ is cyclic. $$\angle OQX=\angle OCQ+\angle CQX=\angle OBC+ \angle CAB=90^{\circ}-\angle CAB+\angle CAB=90^{\circ}.\blacksquare$$

Let w be the circumcircle of quadrilateral ABCD and T be the circumcircle of (APQB) and AQ intersects w at Z.Then by Reim's theorem, PQ is parallel to ZD. Let OQ intersects DZ at M.So it suffices to prove that Angle OMD =90°. By angle chasing,Angle AQP=ABP=ACD=PQD.So angle AQD=2Angle ACD=AOD. So, AQOD is cyclic. Hence, Angle DOM+ODM=DAQ+90°-DAQ=90°, so Angle OMD=90°... And we r done

Let $AB \cap CD=M ; AD \cap BC=N ; PQ \cap AD=L$ -Note that: $\angle DCA=\angle DBA=\alpha$ but $DCPQ,AQPB$ are concyclics. $\implies \angle DQL=\angle LQA=\alpha \implies \angle DQA=2 \alpha$ -$DOA=2 \alpha=\angle DQA \implies AOQD$ is concyclic. similarly $CQOB$ are concyclics. By axis radical in: $P-Q-M$ and $O-Q-N$ collinears. By polars: $\triangle MNP$ is autopolar $\implies ON \perp PM$ $\implies \angle OQP=90$ $\blacksquare$.

Notice that $Q$ is the Miquel point of $ABDC.$ Then, by 10.12 of EGMO, $\overline{OQ}\perp\overline{PQ}.$ $\square$

It's clear that $Q$ is the Miquel Point of cyclic quadrilateral $ABDC$. Now, Master Miquel finishes. $\blacksquare$

Observe that by radical axis, $\overline{AB},\overline{CD},\overline{PQ}$ are concurrent, say at $K$. Note that $AQOD$ and $BQOC$ is cyclic by easy angle chase, hence by radical axis again, we obtain that $\overline{BC},\overline{AD},\overline{OQ}$ are concurrent, say at $L$. Now, by Brokard, $\overline{PK}\perp\overline{OL}$, thus $\measuredangle OQP=90^\circ$. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -13.881545592384889, xmax = 6.885390591378926, ymin = -8.01706105179004, ymax = 3.922118287307623; /* image dimensions */ /* draw figures */ draw(circle((-4.312112455441459,-1.2153467616374318), 4.927367255649258), linewidth(2)); draw((-1.0709231392391878,-4.926632238204918)--(-7.723470822102511,2.340153507840286), linewidth(2)); draw((-9.105873856745824,-0.07569782472356396)--(-0.9023078635506594,2.341643639975273), linewidth(2)); draw((-7.723470822102511,2.340153507840286)--(-0.9023078635506594,2.341643639975273), linewidth(2)); draw((-0.9023078635506594,2.341643639975273)--(-1.0709231392391878,-4.926632238204918), linewidth(2)); draw((-1.0709231392391878,-4.926632238204918)--(-9.105873856745824,-0.07569782472356396), linewidth(2)); draw((-9.105873856745824,-0.07569782472356396)--(-7.723470822102511,2.340153507840286), linewidth(2)); draw(circle((-6.274241728573986,-4.465362257040207), 5.223724182768531), linewidth(2)); draw(circle((-4.3132536155531,4.008377296483783), 3.796386703806889), linewidth(2)); draw((-4.312112455441459,-1.2153467616374318)--(-3.9783741256843217,0.22678930569116218), linewidth(2)); draw((-4.312112455441459,-1.2153467616374318)--(-6.275382888685626,0.7583618010810067), linewidth(2)); draw((-6.275382888685626,0.7583618010810067)--(-3.9783741256843217,0.22678930569116218), linewidth(2)); draw((-3.9783741256843217,0.22678930569116218)--(-0.9023078635506594,2.341643639975273), linewidth(2)); draw((-3.9783741256843217,0.22678930569116218)--(-1.0709231392391878,-4.926632238204918), linewidth(2)); draw((-4.312112455441459,-1.2153467616374318)--(-1.0709231392391878,-4.926632238204918), linewidth(2)); draw((-4.312112455441459,-1.2153467616374318)--(-0.9023078635506594,2.341643639975273), linewidth(2)); draw((-4.312112455441459,-1.2153467616374318)--(-0.9911458022765348,-1.4877759928063137), linewidth(2)); /* dots and labels */ dot((-4.312112455441459,-1.2153467616374318),dotstyle); label("$O$", (-4.23975379278026,-1.0344501049844375), NE * labelscalefactor); dot((-9.105873856745824,-0.07569782472356396),dotstyle); label("$B$", (-9.033515194084625,0.10519883192943043), NE * labelscalefactor); dot((-1.0709231392391878,-4.926632238204918),dotstyle); label("$A$", (-1.0017036386916518,-4.742831566370834), NE * labelscalefactor); dot((-7.723470822102511,2.340153507840286),dotstyle); label("$C$", (-7.658700603521864,2.529214031079562), NE * labelscalefactor); dot((-0.9023078635506594,2.341643639975273),dotstyle); label("$D$", (-0.8388966477039563,2.529214031079562), NE * labelscalefactor); dot((-6.275382888685626,0.7583618010810067),dotstyle); label("$P$", (-6.211527350297905,0.937323452533207), NE * labelscalefactor); dot((-3.9783741256843217,0.22678930569116218),dotstyle); label("$Q$", (-3.91413981080487,0.4127231482395218), NE * labelscalefactor); dot((-0.9911458022765348,-1.4877759928063137),dotstyle); label("$E$", (-0.9112553103651544,-1.30579508996393), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] in quadrilateral $DCPQ$ we get $\angle{DPQ}=180^{\circ}-\angle{DCP}$ in quadrilateral $PQAB$ we get $\angle{PQB}=180^{\circ}-\angle{PBA}$ and since $\angle{ACD}=\angle{ABD}$ we get $\angle{AQD}=(180^{\circ}-\angle{DCP})+(180^{\circ}-\angle{PBA})=2\angle{ABD}=\angle{AOD}$ so from this we get points $A,O,Q,D$ are concyclic. now $\angle{PQO}=\angle{PQA}-\angle{OQA}$ $=180^{\circ}-\frac{AOD}{2}-\left(90^{\circ}-\frac{\angle{AOD}}{2}\right)$ $=90^{\circ}$ as desired $\blacksquare$