do you mean the intersection of the circumscribed cercles of $ABC$ and $AXY$??

up to this point I have reached that $AXKC$ and $AYKB$ are concyclic . is it useful ?

Let $D$ be the second intersection of $\odot(ABC)$ with $\omega_1$ and $E$ the second intersection of $DC$ with $\omega_1.$ For convenience, denote $\angle AXY=\angle AYX=\theta.$ By Reim's theorem for $\odot(ABC)$ and $\omega_1$ with common chord $BD,$ it follows that $XE \parallel AC$ $\Longrightarrow$ $\angle BXE=\pi -2\theta.$ Since $\angle BDX=\theta,$ then $\angle XDE=\angle XYA=\theta$ $\Longrightarrow$ $X,Y,C,D$ are concyclic. Therefore, $AQ,XY,DC$ concur at the radical center $O$ of $\odot(AXY) ,$ $ \odot(ABC)$ and $\odot(XYC),$ which becomes the exsimilicenter of $\omega_1,\omega_2,$ since $O$ is also center of the positive inversion taking $\omega_1,\omega_2$ into each other. Thus, the tangent of $\odot(XYK) \equiv \omega_3$ at $K$ passes through $O.$ Further, $AB,AC$ are tangent lines of $\omega_3$ at points $X,Y.$ Then $KA$ is the polar of $O$ WRT $\omega_3$ cutting $\omega_3$ again at $R$ such that $OR$ is tangent to $\omega_3$ at $R.$ If $P \equiv AK \cap XY,$ then the cross ratio $(X,Y,P,O)$ is harmonic. Consequently, if $M \equiv PK \cap XY$ (midpoint of XY), it follows that $OR^2=OK^2=OQ \cdot OA=OP \cdot OM.$ Since $K,P,R,A$ are collinear, being K,R double points, $K,$ $M,$ $R,$ $Q,$ $O$ are concyclic, yielding $\angle QKP=\angle QOY.$ But because of $AY^2=AR \cdot AK= AQ \cdot AO,$ we deduce that $\angle QOY=\angle QYA=\angle QXA$ $\Longrightarrow$ $\angle QXA=\angle QKP.$

goodar2006 wrote: up to this point I have reached that $AXKC$ and $AYKB$ are concyclic . is it useful ? This is useful in my solution. After noticing this and that $AX=AY$ we make an inversion with center in $A$. After some angle chasing we actually got following problem: Let tangents in $K, X, Y$ on circumcircle of $KXY$ intersect sides $XY, KY, KX$ in $Q, C, B$ respectively. Show that $Q, C, B$ are collinear (in problem we have that $Q, B, C$ are collinear and need to get that $QK$ is tangent but this is same I don't have nice explanation for this, however it follows from Ceva's theorem.

RaleD wrote: goodar2006 wrote: up to this point I have reached that $AXKC$ and $AYKB$ are concyclic . is it useful ? This is useful in my solution. After noticing this and that $AX=AY$ we make an inversion with center in $A$. After some angle chasing we actually got following problem: Let tangents in $K, X, Y$ on circumcircle of $KXY$ intersect sides $XY, KY, KX$ in $Q, C, B$ respectively. Show that $Q, C, B$ are collinear (in problem we have that $Q, B, C$ are collinear and need to get that $QK$ is tangent but this is same I don't have nice explanation for this, however it follows from Ceva's theorem. beautiful solution! thanks.

frenchy wrote: do you mean the intersection of the circumscribed cercles of $ABC$ and $AXY$?? actualy the point you have reached with the cyclic quadrilaterals folows from the fact that k is the miquel point of the quadrilateral AXPY

As always, Iran have nice problems to propose for contestants.

Here is a solution with spiral similarity. Note that $K$ is the center of spiral similarity that takes $BX$ to $YC$. In particular, $BXK \sim YCK$. But, we know that $\angle KYC = \angle KBX = \angle KXY$ since $XY$ is tangent to $\omega_1$. Similarly, $\angle KXB = \angle KCY = \angle KYX$. Thus, we have that $BXK \sim XYK \sim YCK$, and so there is a spiral similarity at $K$ that takes $BX$ to $XY$ to $YC$. Let $\angle BKX = \angle XKY = \angle YKZ = \theta$. Then, we have $\angle BAC = 180 - \angle AXY - \angle AYX = 180 - \theta - \theta = 180 - 2\theta$. Let $M$, $N$ be the midpoints of $BX$, $CY$, respectively. Then, the spiral similarity at $K$ taking $BX$ to $YC$ takes $M$ to $N$, or $\angle MKN = \angle BKY = 2\theta$. Since $\angle MAN = \angle BAC = 180-2\theta$, we have that $AMKN$ is a cyclic quadrilateral. We know that $Q$ is the center of spiral similarity that takes $BX$ to $CY$, so $\angle MQN = \angle XQY = \angle XAY = 180 - 2\theta$. Thus, the five points $A$, $M$, $K$, $N$, $Q$ are concyclic. Let $PK \cap XY = Z$. Then, by radical-axis, we know that $Z$ is the midpoint of segment $XY$. By the spiral similarity at $K$, we know that $\angle ZKN = \theta$, or $\angle QKP = \theta - \angle QKN$. But, we have $\angle QKN = \angle QAN = \angle QAY = \angle QXY = \angle QAX - \angle QXA = \theta - \angle QXA$. It follows that $\angle QXA = \angle QKP$, as desired.

That is my proof. Let $(O),(O_{4})$ be the circumcircle of $\triangle KXY, \triangle AXY$ And $M$ is the intersection of $QO$ with $XY$ $KP$ meets $XY$ at $N$. Easy to see: $\angle AXY=\angle AYX=\angle XKY$ then $AX,AY$ touch with $(O$ then $O$ lies on $O_{4}$ and $O,A,N$ conllinear. And $KA$ is symmedian of $\triangle XKY$ On the other hand: $\triangle KBX, KXY,KYC$ are similar th­en: $\frac{XM}{MY}=\frac{QX}{QY}=\frac{BX}{BY}=\frac{KX^{2}}{KY^{2}}$ Hence $K,M,A$ is conllinear. Easy to see $M,N,Q,A$ lie on circle.(*) Have $OK^{2}=OX^{2}=ON.OA$ (**) when (*) and (**) have $\angle OQN=\angle KAO=\angle OKN$ th­en $K,O,N,Q$ lie on a circle Hence $\angle QKA=\angle QOA=\angle QXA$ QED

wsjradha wrote: We know that $Q$ is the center of spiral similarity that takes $BX$ to $CY$, so $\angle MQN = \angle XQY = \angle XAY = 180 - 2\theta$. Thus, the five points $A$, $M$, $K$, $N$, $Q$ are concyclic. Sorry to revive, but I'm not following why Q is the center of spiral similarity. Also, how does the fact that $\angle MQN = \angle XQY$ follow from that? Thanks. EDIT: Nevermind.

It's amazing how many geometry problems this paper helps you solve: http://web.mit.edu/yufeiz/www/olympiad/cyclic_quad.pdf Unfortunately I'd forgotten almost everything in there, so it was hard getting this problem to work D:

use of the two centers of rotational homothety and two midpoints of two side...

There is also a nice solution using inversion.Just invert with center $A$ and arbitrary radius.Let $X'$ be the image of $X$ under the inversion.Then $X'P'KB'$ and $KPY'C'$ are cyclic.So $\angle{X'P'Y'}=\angle{X'P'K'}+\angle{Y'PK}=\angle{K'B'A}+\angle{K'C'A}=\angle{BKA}+\angle{CKA}=180-\angle{BAC}$ so $AX'PY'$ is also cyclic.Now some trivial steps yeild you $\angle{AQ'X'}=\angle{AQ'K'}+\angle{AP'K'}$ which means $\angle{QXA}=\angle{QKP}$ as desired.

dang this question took me forever, but it was nice First, WLOG let $Q$ be closer to $Y$ than $X$. Let $\angle BKX = \angle 1, \angle XBY = \angle 2,$ and $\angle XCY = \angle 3$. Since $K$ is the second intersection of the circumcircles of $\triangle PXB$ and $\triangle PYC$, there is a spiral similarity centered at $K$ that takes $\triangle KBX$ to $\triangle KYC$. Notice that this spiral similarity also takes $\triangle KBY$ to $\triangle KXC$, so those triangles are similar as well. Thus, $\angle KYC = \angle 1$ as well. By angle chasing, we find that $\angle CXY = \angle 2$, $\angle BYX = \angle 3$, and $\angle AXY = \angle AYX = \angle 1$. By exterior angles of $\triangle BXY$, we see that $\angle 2 + \angle 3 = \angle 1$. Thus, $\angle XKY = \angle XKP + \angle PKY = \angle 2 + \angle 3 = \angle 1$. Also, we have $\angle KXY = \angle KXC + \angle CXY = \angle KBY + \angle 2 = \angle KBX$, so $\triangle KBX \sim \triangle KXY$, and the spiral similarity that sends $\triangle KBY$ to $\triangle KXC$ also sends $\triangle KBX$ to $\triangle KXY$. It also sends $\triangle KXY$ to $\triangle KYC$ by similar reasoning. Since $Q$ is the second intersection of the circumcircles of $\triangle AXY$ and $\triangle ABC$, $Q$ is the center of spiral similarity that sends $\triangle QXY$ to $\triangle QBC$, which implies that $\triangle QXY \sim \triangle QBC$ are similar. Since quadrilateral $AQYX$ is cyclic, $\angle XQY = \angle XAY = 180 - 2\angle 1$. Let M and N be midpoints of BX and CY, respectively. Lemma 1: Quadrilateral $QMKN$ is cyclic

, spirally similar triangles $QXY$ and $QBC$ are similar to $\triangle QMN$, so $\angle MQN = \angle $XQY$ = 180 - 2\angle 1$. By MGT again, spirally similar triangles $KBY$ and $KXC$ are similar to $\triangle KMN$ are similar, so $\angle MKN = \angle BKY = \angle BKX + \angle XKY = 2\angle 1$. This implies that quad $QMKN$ is cyclic because $\angle MQN + \angle MKN = 180$. Thus by Lemma 1, we have $\angle QKN = \angle QMN = \angle QXY$. By angle chasing, $\angle QKP = \angle PKN - \angle QKN = \angle PKN - \angle QXY$. Now we extend $KP$ to intersect $XY$ at a point $Z$. Lemma 2: $Z$ is the midpoint of $XY$ Proof: As $KP$ is the radical axis of circles $W_1$ and $Q_2$, $Z$ lies on the radical axis. This implies that it has equal power with respect to both circles, so $ZX^2 = ZY^2$, which implies that $ZX = ZY$, so $Z$ is the midpoint of $XY$. Finally we have $\triangle KXZ \sim \triangle KYN$, where Z is midpoint of XY. By Lemma 2, the spiral similarity that takes $\triangle KXY$ to $\triangle KYC$ also takes $Z$ to $N$, as both are corresponding midpoints, so $\triangle KZN \sim \triangle KYC$ by spiral similarity. Thus, $\angle ZKN = \angle YKC = \angle 1$, so $\angle QKP = \angle 1 - \angle QXY = \angle QXA$, as desired. $QED$

My solution: Notice that $\triangle{QXB} \sim \triangle{QYC}$ (same direction) We construct $\triangle{QKL} \sim \triangle{QXB} \sim \triangle{QYC}$ (same direction) $\Rightarrow \exists S_{(Q, \alpha, k)}$ which is a spiral similarity that maps: $X\mapsto{B}, Y\mapsto{C}, K\mapsto{L}$ $\Rightarrow \triangle{BCL} \sim \triangle{XYK}$ $\Rightarrow \angle{BLC} = \angle{XKY} = 180-\angle{XPY} = 180-\angle{BPC}$ $\Rightarrow BPCL$ is cyclic $\Rightarrow \angle{CPL} = \angle{CBL} = \angle{YXK} = \angle{CPK}$ (because $XY$ is tangent to $\omega_1$) $\Rightarrow \overline{P, K, L}$ $\Rightarrow \angle{QKP} = 180-\angle{QKL} = 180-\angle{QXB} = \angle{QXA}$ Q.E.D

My solution: Let $T$ be the second intersection point of $W_2$ and the circumcircle of $\triangle{ABC}$.Also,suppose that $R\equiv TY\cap AB$. We will first show that $\triangle{AXY}$ is isosceles. Indeed,$\angle{AXY}=180^{\circ}-\angle{BXY}=180^{\circ}-\angle{PXY}-\angle{BXP}=180^{\circ}-\angle{PBX}-\angle{BXP}=\angle{BPX}=\angle{CPY}=$ $=180^{\circ}-\angle{PCY}-\angle{PYC}=180^{\circ}-\angle{PYX}-\angle{PYC}=180^{\circ}-\angle{XYC}=\angle{AYX}$ q.e.d. We will go on with showing that $XYTB$ is cyclic.Indeed,$\angle{CTY}=180^{\circ}-\angle{YPC}=180^{\circ}-\angle{AXY}$ and $\angle{CTB}=\angle{CAB}=\angle{YAX}=180^{\circ}-2\angle{AXY}$. Thus,$\angle{YTB}=\angle{CTY}-\angle{CTB}=180^{\circ}-\angle{AXY}-\left(180^{\circ}-2\angle{AXY}\right)=\angle{AXY}$ which is what we wanted. Hence,$RY\cdot RT=RX\cdot RB$ implying that $R$ has equal powers wrt $W_1,W_2$.Thus,it lies on their radical axis which is $PK$. Thus,$P,K,R$ are collinear.We have $\angle{QTR}=\angle{QTY}=\angle{QTC}-\angle{CTY}=$ $=180^{\circ}-\angle{QAY}-\left(180^{\circ}-\angle{YPC}\right)=\angle{YPC}-\angle{QAY}=\angle{AXY}-\angle{QAY}=\angle{QXA}=\angle{QXR}$ whence we get that $QRXT$ is cyclic $(1)$. Also,$\angle{QTK}=\angle{QTY}+\angle{YTK}=\angle{QXA}+\angle{KPB}=$ $=\angle{QXA}+\angle{KXB}=180^{\circ}-\angle{QXK}$ which gives that $QXKT$ is also cyclic $(2)$. From $(1),(2)$ we conclude that $Q,R,X,K,T$ all lie on the same circle. Thus,$\angle{QXA}+\angle{QXR}=\angle{QKR}=\angle{QKP}$ which ends the proof.

By easy angle chasing we get $BKYA$ and $CKXA$ are cyclic. Now let $AQ \cap XY= L $ and $PK \cap XY =R$ and $R$ is the midpoint of $XY$. Notice that by inversion at $A$ and radius $\sqrt{XY}$ point $Q$ goes to $L$ so we have $\angle QXA=\angle QLR$ so we need to prove that $QLKR$ is cyclic so it is enough to prove that $\angle KRY=\angle AQK$. Let $\angle XYP=\angle YKR=x$ and $\angle XYP=\angle XKR=y$. Because $AX$ and $AY$ are tangents to the circumcircle of $KXY$ we have $KA$ is the symmedian in $\triangle KXY$. So we have $\angle AKY=y$. Now let the tangent at $K$ of the circumcircle $KXY$ cut $AY$ at $S$. We get $\angle SKA=\angle YRK$ so we need to prove that $\angle SKA=\angle AQK$ which is equivalent to $SK$ being the tangent to the circle $AKQ$ so we need to prove that circumcircles $KXY$ and $KQA$ are tangent to each other. Now we invert at $A$. Now we let the picture of some point $X$ be $X_1$. Now $K_1$ is the point inside the isosceles triangle $AXY$ such that $\angle AY_1K_1=\angle K_1X_1Y_1$ and the equivalent when we switch $X_1$ and $Y_1$. Now because $BKYA$ and $XKCA$ are cyclic we have that $ B_1= AX_1 \cap K_1Y_1$ and $C_1=AY_1 \cap X_1K_1$ and because $AQBC$ and $AQXY$ are cyclic we have $Q_1= X_1Y_1 \cap B_1C_1$. Now we only need to prove that $Q_1K_1$ is tangent to circumcircle $XKY$. Now because $B$ is the intersection of one side of the triangle and the tangent and the same holds for $C_1$ and we need to prove that the same holds for $Q_1$ it is enough to prove that intersections of tangents at $K,X,Y$ and $XY,KY,KX$ respcetively in any triangle $KXY$ are collinear and it follows from easy Menelaus so we are finished. These angles are like this in my configuration but the idea is the same in any.

RaleD wrote: Let tangents in $K, X, Y$ on circumcircle of $KXY$ intersect sides $XY, KY, KX$ in $Q, C, B$ respectively. Show that $Q, C, B$ are collinear (in problem we have that $Q, B, C$ are collinear and need to get that $QK$ is tangent but this is same I don't have nice explanation for this, however it follows from Ceva's theorem. Actually, this called Pascal's theorem for triangle. Beautiful solution!

First, note that $\angle XKY=\angle XKP+\angle PKY=\angle XBP+\angle YCP=\angle XBP+\angle XYP=\angle AXY$ and similarly $\angle XKY=\angle AYX$, so the circumcircle of $XKY$ is tangent to $AX$ and $AY$. Let $I$ be the incenter of $ABC$. Note that $\angle BPC=180^\circ-\angle XPB=180^\circ-\angle AXY=90^\circ+\angle XAY=\angle BIC$, so $BPIC$ is cyclic. This means that the circumcircle of $BPC$ is symmetric with respect to the angle bisector of $\angle BAC$. Let $M$ be the midpoint of $XY$, $D$ be the harmonic conjugate of $K$ with respect to $X$ and $Y$ (well-known that $D$ is reflection of $P$ across $XY$), $P'$ be the reflection of $P$ over $M$, and $D'$ be the reflection of $D$ over $M$. Note that $DPD'P'$ is a rectangle that is symmetric with respect to $BC$ and the perpendicular bisector of $BC$. Now, perform a $\sqrt{bc}$ inversion with respect to $KXY$. Note that the circumcircle of $KPX$ maps to $AY$ and the circumcircle of $KQY$ maps to $AX$. This means that $A$ maps to $P$ and $B$ maps to $C$. Now, $Q$ maps to the intersection of the circumcircles of $PBC$ and $PXY$. Note that both circles are symmetric with respect to the angle bisector of $\angle BAC$, so $Q$ maps to $D'$. Now, we have $\angle QKP=\angle AKD'$ and $\angle QXA=\angle QXK-\angle AXK=\angle YD'K-\angle YPK=\angle PYD'+\angle PKD'$. Hence, it suffices to show that $\angle AKP=\angle PYD'$. But $\angle AKP=\angle DKP'$, and the circumcircles of $XYK$ and $XYP$ are reflections of each other across $XY$. But since $P'D$ is the reflection of $PD'$ across $XY$, the corresponding arcs and thus angles are equal as desired.

WLOG, let $KX<KY$. Note that $$\angle KXB=\angle KPB=180^{\circ}-\angle KPY=\angle KYX$$from which it follows that $AX$ is tangent to $(KXY)$. Similar argument shows that line $AY$ is tangent to $(KXY)$. Define $R \ne P$ as the other intersection point of circles $(BPC)$ and $(XPY)$. We will show that $XY \parallel PR$. Denote by $T$ the circumcenter of triangle $BPC$ and observe that $$XP\cdot XC=XY^2=YP\cdot YB \Longrightarrow TX=TY \Longrightarrow XY \parallel PR.$$ Apply inversion about $K$ of radius $r^2=KX\cdot KY$ followed by reflection in the bisector of angle $XKY$. Observe that the pairs $\{X,Y\},$ $\{B,C\},$ and $\{P,A\}$ are interchanged under this map. As $(ABC) \rightarrow (PBC)$ and $(AXY) \rightarrow (PXY),$ the points $Q$ and $R$ are also interchanged. Finally, note that $\angle QKP=\angle RKA=\angle RKP+\angle PKA$ and $$\angle QXA=\angle KXQ-\angle KXA=\angle KRY-\angle KPY=\angle RKP+\angle RYP.$$Thus, we have $$\angle QXA=\angle QKP \iff \angle RYP=\angle PKA \iff \angle PXY-\angle PYX=\angle PKA$$which holds as $\angle PXY=\angle PKX$ and $\angle PYX=\angle PKY$ proving the result. $\square$

[asy][asy] size(8cm); defaultpen(fontsize(9pt)); pen pri=deepgreen+linewidth(0.5); pen sec=royalblue+linewidth(0.5); pen tri=rgb(41, 207, 255)+linewidth(0.5); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair X, Y, O1, O2, P, K, B, C, A, Q, L; X=(-1, 0); Y=(1, 0); O1=(-1, -0.9); O2=(1, -1.3); P=intersectionpoints(circle(O1, length(X-O1)), circle(O2, length(Y-O2)))[0]; K=intersectionpoints(circle(O1, length(X-O1)), circle(O2, length(Y-O2)))[1]; B=intersectionpoint(circle(O1, length(X-O1)), (P+(P-Y)*0.01) -- (P+(P-Y)*100)); C=intersectionpoint(circle(O2, length(Y-O2)), (P+(P-X)*0.01) -- (P+(P-X)*100)); A=extension(B, X, C, Y); Q=intersectionpoints(circumcircle(A, B, C), circumcircle(A, X, Y))[1]; L=intersectionpoint(K -- (K+(K-P)*100), circumcircle(B, C, P)); X=(X-B)/(C-B); Y=(Y-B)/(C-B); O1=(O1-B)/(C-B); O2=(O2-B)/(C-B); P=(P-B)/(C-B); K=(K-B)/(C-B); A=(A-B)/(C-B); Q=(Q-B)/(C-B); L=(L-B)/(C-B); B=(0,0); C=(1,0); filldraw(circle(O1, length(X-O1)), fil, pri); filldraw(circle(O2, length(Y-O2)), fil, pri); draw(B -- Y -- X -- C, pri); draw(B -- A -- C -- B, sec); filldraw(circumcircle(A, B, C), sfil, sec); filldraw(circumcircle(A, X, Y), sfil, sec); draw(P -- L, tri); draw(B -- L -- C, tri); filldraw(circumcircle(B, P, C), tfil, tri); dot("$X$", X, dir(100)); dot("$Y$", Y, dir(80)); dot("$P$", P, dir(100)); dot("$K$", K, dir(-30)); dot("$B$", B, dir(210)); dot("$C$", C, dir(-30)); dot("$A$", A, N); dot("$Q$", Q, dir(120)); dot("$L$", L, S); [/asy][/asy] Note that $K$ is the Miquel point of $BXCY$ and $Q$ is the Miquel point of $BXYC$. Let the spiral similarity at $Q$ sending $\overline{XY}$ to $\overline{BC}$ send $K$ to $L$, so $\triangle BLC\sim\triangle XKY$. Note that \begin{align*} \measuredangle BLC&=\measuredangle XKY=\measuredangle XKP+\measuredangle XKP+\measuredangle PKY\\ &=\measuredangle YXP+\measuredangle PYX=\measuredangle YPX=\measuredangle BPC, \end{align*}so $BPCL$ is cyclic. Moreover, \[\measuredangle BPL=\measuredangle BCL=\measuredangle XYK=\measuredangle YCK=\measuredangle BXK=\measuredangle BPK,\]whence $L$ lies on $\overline{PK}$. It follows that $\measuredangle QKP=\measuredangle QKL=\measuredangle QXB=\measuredangle QXA$, as required.

With Aatman Supkar, Alan Chen, Alexandru Girban, Anushka Aggarwal, Daniel Sheremeta, Derek Liu, Dhrubajyoti Ghosh, Gene Yang, Jeff Lin, Paul Hamrick, Raymond Feng, Robin Son, and Sean Li (plus Eric Shen cheerleading looks like we even ended up with the same letters!): Let the spiral similarity at $Q$ mapping $\overline{XY}$ to $\overline{BC}$ take $K$ to $L$. Claim: The point $L$ lies on $(PBC)$ and $\overline{KP}$. Proof. This is only place where the fact $\overline{XY}$ is tangent gets used. Angle chase: \begin{align*} \measuredangle BPK &= \measuredangle YPK = \measuredangle XYK = \measuredangle BCL \\ \measuredangle CPK &= \measuredangle XPK = \measuredangle YXK = \measuredangle CBL. \end{align*}Adding the two gives $BPCL$ cyclic. Then $\measuredangle BPK = \measuredangle BPL$, done. $\blacksquare$ Now, by spiral similarity, $\measuredangle QXA = \measuredangle QXB = \measuredangle QKL = \measuredangle QKP$, as needed. A few behind-the-scenes quotes while we were working on this: ms paint > ggb confirmed if we were in ms paint, "we would [unfortunately] be trying to do spiral similarities at Q and K over and over" evan during the solving: something's wrong i can feel it wow you can turn the random properties we found into a problem now i thought it would be spiral sim the whole way through but then you guys found all of these nice properties

Let $\overline{AK}$ and $\overline{XY}$ intersect at $D$, and let $O$ be the circumcenter of $\triangle KXY$. Let $M$ be the midpoint of $\overline{XY}$, $E$ the second intersection of $\overline{AK}$ and $(KXY)$, and $T$ the intersection of the tangent to $(KXY)$ at $K$ and line $XY$. Denote by $B_1$ and $C_1$ the midpoints of $\overline{BX}$ and $\overline{CY}$ respectively, and $Z$ the image of $Q$ wrt.\ inversion at $K$ with radius $\sqrt{KX\cdot KY}$ followed by a reflection over the angle bisector of $\angle XKY$. By construction, $P$ is an HM-point of $\triangle KXY$. Thus $\overline{KP}$ bisects $\overline{XY}$ (also by radical axes), and $(XPY)$ passes through the orthocenter of $\triangle KXY$.

Note that $K$ is the center of a spiral similarity sending $BX \mapsto YC$ and $Q$ is the center of a spiral similarity sending $XY \mapsto BC$. Also, observe that $PQ \cap XY = M$, the midpoint of $XY$ from a well known lemma. We also have $\measuredangle KYX = \measuredangle KCY$ and $\measuredangle KBX = \measuredangle YXK$ whence $\triangle KXY \sim \triangle KBX \sim \triangle KYC$. Due to this, $\measuredangle BXY = \measuredangle XYC$ whence $\measuredangle AXY = \measuredangle XYA$ so $AXY$ is isosceles. Combined with the angle statements, this implies that $AX$ and $AY$ are tangents to $(KXY)$. Let $AK \cap XY = L$. Observe that $KL$ is a symmedian in $\triangle KXY$. We have $$\frac{QX}{QY} = \frac{BX}{YC} = \frac{BK}{KY} = \frac{KX^2}{KY^2}$$which follows from $\frac{BK}{KX} = \frac{KX}{KY}$. Likewise due to symmedian ratios, $\frac{KX^2}{KY^2} = \frac{XL}{LY}$, so $QL$ bisects $\angle XQY$. Observe that $\measuredangle LQY = \measuredangle MAY$, so $LQ$ and $AM$ intersect on $(AXY)$, say at point $O$. As $\triangle AXY$ is isosceles and $\measuredangle AXY = \measuredangle XYA = \measuredangle XKY$, we have $\measuredangle YAX = 180 - 2 \measuredangle XKY$ so $\measuredangle XOY = 2 \measuredangle XKY$ which, combined with $AM \perp XY$, implies that $O$ is the circumcenter of $\triangle XKY$. Claim: $KOMQ$ is cyclic Proof: Invert about $(KXY)$. Observe that $M \mapsto A$ since they are polar. Now note that as $AO$ is a diameter of $(AXY)$, $\measuredangle OQA = 90$. Also, $AM \perp XY$, so $AQLM$ is cyclic whence $OL \cdot OQ = OM \cdot OA$, so $L$ and $Q$ are swapped after inversion. Therefore, $(MQK)$ is mapped to a line after inversion about $O$, so $KOMQ$ is cyclic as desired. $\blacksquare$ To finish, $\measuredangle QKM = \measuredangle QOA = \measuredangle QXA$, and we are done. $\blacksquare$

[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 = -17.017872221696305, xmax = 27.256889742760723, ymin = -9.900050867795619, ymax = 14.96610115795648; /* image dimensions */ pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqqqcc = rgb(0,0,0.8); pen qqzzqq = rgb(0,0.6,0); pen ccqqqq = rgb(0.8,0,0); draw(circle((-1.8490848403873745,6.896735451806183), 4.883756988528579), linewidth(1.2) + qqzzqq); draw((-3.7239545994373704,11.406274863941478)--(-7.019406119164214,-2.623582796016489)--(14.291278423016704,0.3099251759874293)--cycle, linewidth(2) + ccqqqq); /* draw figures */ draw(circle((6.15395941298003,-0.282390454652267), 8.158847864582619), linewidth(2) + yqqqyq); draw(circle((-4,0), 4), linewidth(1.6) + qqqqcc); draw((-5.53559627430592,3.6935002480489656)--(3.0217853171095594,7.251286198255153), linewidth(2) + linetype("2 2") + blue); draw((-5.53559627430592,3.6935002480489656)--(14.291278423016704,0.3099251759874293), linewidth(2)); draw((-7.019406119164214,-2.623582796016489)--(3.0217853171095594,7.251286198255153), linewidth(2)); draw((-7.019406119164214,-2.623582796016489)--(-3.7239545994373704,11.406274863941478), linewidth(2) + red); draw((-3.7239545994373704,11.406274863941478)--(14.291278423016704,0.3099251759874293), linewidth(2) + red); draw(circle((3.1476653838199984,2.3902501171896704), 11.336130884404211), linewidth(1.6)); draw((-7.019406119164214,-2.623582796016489)--(14.291278423016704,0.3099251759874293), linewidth(2) + red); draw((-6.527616930354384,8.297599401713773)--(-5.53559627430592,3.6935002480489656), linewidth(2)); draw((-6.527616930354384,8.297599401713773)--(-1.4958442565649777,-3.119167198567814), linewidth(2)); draw((-1.4958442565649777,-3.119167198567814)--(-1.3263554851829753,2.9751680638896403), linewidth(2)); draw((-3.7239545994373704,11.406274863941478)--(-7.019406119164214,-2.623582796016489), linewidth(2) + ccqqqq); draw((-7.019406119164214,-2.623582796016489)--(14.291278423016704,0.3099251759874293), linewidth(2) + ccqqqq); draw((14.291278423016704,0.3099251759874293)--(-3.7239545994373704,11.406274863941478), linewidth(2) + ccqqqq); /* dots and labels */ dot((6.15395941298003,-0.282390454652267),dotstyle); label("$O_2$", (6.547143942243271,-0.17767456702917347), NE * labelscalefactor); dot((-4,0),dotstyle); label("$O_1$", (-4.042656266398287,-0.9005278235917346), NE * labelscalefactor); dot((-1.3263554851829753,2.9751680638896403),linewidth(4pt) + dotstyle); label("$P$", (-1.621097856913698,3.5088770414398875), NE * labelscalefactor); dot((-1.4958442565649777,-3.119167198567814),linewidth(4pt) + dotstyle); label("$K$", (-1.837953833882467,-3.9365115011544907), NE * labelscalefactor); dot((-5.738502857552112,-3.6024446996842476),linewidth(4pt) + dotstyle); dot((-5.53559627430592,3.6935002480489656),linewidth(4pt) + dotstyle); dot((-5.53559627430592,3.6935002480489656),linewidth(4pt) + dotstyle); label("$X$", (-6.283501361742234,3.6534476927524), NE * labelscalefactor); dot((3.0217853171095594,7.251286198255153),linewidth(4pt) + dotstyle); label("$Y$", (3.149733636399222,7.55685527819023), NE * labelscalefactor); dot((-7.019406119164214,-2.623582796016489),linewidth(4pt) + dotstyle); label("$B$", (-7.656922549211105,-3.1775155817638017), NE * labelscalefactor); dot((14.291278423016704,0.3099251759874293),linewidth(4pt) + dotstyle); label("$C$", (14.426244438775216,0.5813213523615156), NE * labelscalefactor); dot((-3.7239545994373704,11.406274863941478),linewidth(4pt) + dotstyle); label("$A$", (-3.97037094074203,11.82168949190934), NE * labelscalefactor); dot((-6.527616930354384,8.297599401713773),linewidth(4pt) + dotstyle); label("$Q$", (-7.259353258101695,8.09899522061215), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $O_3$ be circumcircle of $\Delta KXY$ and $O_4$ be the circumcircle of $\Delta AXY$. Let $XY \cap QO_3=R$, $KP \cap XY=S$. Notice, that the angles in circle $\angle XKY=\angle AXY= \angle XKY$, then $O_3$ lies on $O_4$ and hence $O_3, A, N$ are collinear. Hence, by Angle Chasing we can say through AA- similarity that $\Delta KBX, \Delta KXY, \Delta KYC$ are similar triangles. Using the simillarity results we can say that $K, R, A$ are collinear. Hence, points $Q,A, R,S$ form a cyclic quadrilateral with another circle.With the property of power of point $O_3K^2=O_3K\cdot O_3A=O_3X^2$, hence $\angle O_3QS=\angle KAO_3=\angle OKS$, therefore $K, O_3, S, Q$ form a cyclic quadrilateral, hence all the $4$ points lie on a circle. $\therefore \angle QO_3A=\angle QKA=\angle QKA. \blacksquare$

v_Enhance wrote: \begin{align*} \measuredangle BPK &= \measuredangle YPK = \measuredangle XYP = \measuredangle BCL \\ \measuredangle CPK &= \measuredangle XPK = \measuredangle YXP = \measuredangle CBL. \end{align*} I think it should be \begin{align*} \measuredangle BPK &= \measuredangle YPK = \measuredangle XYK = \measuredangle BCL \\ \measuredangle CPK &= \measuredangle XPK = \measuredangle YXK = \measuredangle CBL. \end{align*}

Anshu-Droid's solve again IRAN TST 2010 P5 wrote: Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$ \[\angle QXA=\angle QKP\] Let $D = W_1 \cap \odot ABC, G = W_2 \cap \odot ABC, F = DX \cap \odot ABC, F' = GY \cap \odot ABC, E = DX \cap AC$ and $Z$ be a point on $AF$ such that $A$ lies between $Z$ and $F$. Claim: $\overline{F - Y - G}$ are collinear. Proof: By Reim's on $\{ W_1, \odot ABC \}$ wrt lines $DX$ and $BX$, we get $XX \parallel AF$. Now again by Reim's on $\{ W_2, \odot ABC \}$ wrt lines $GY$ and $CY$, we get $YY \parallel AF'$. But now, $$AF \parallel XX \equiv YY \parallel AF' \implies F' = F.$$ Claim: $DXYC$ is cyclic. Proof: $\measuredangle YCD = \measuredangle ACD = \measuredangle AFD = \measuredangle AFX = \measuredangle YXF = \measuredangle YXD$. Now by Radical Center Theorem on $\{ W_1, W_2, \odot DXYC \}$, we get $E = DX \cap PK \cap CY$. Claim: $AX = AY$. Proof: $\measuredangle AYX = \measuredangle YAF = \measuredangle CAF = \measuredangle CGF = \measuredangle CGY = \measuredangle CPY = \measuredangle XPB = \measuredangle XDB = \measuredangle FDB = \measuredangle FAB = \measuredangle FAX = \measuredangle YXA$. So, from our previous claim, we get that $AF$ is tangent to $\odot AXY$. Claim: $DEYK$ is cyclic. Proof: $\measuredangle YDE = \measuredangle YDX = \measuredangle YCX = \measuredangle YCP = \measuredangle YKP = \measuredangle YKE$. Claim: $QEYD$ is cyclic. Proof: $\measuredangle QYE = \measuredangle QYA = \measuredangle QAZ = \measuredangle QAF = \measuredangle QDF = \measuredangle QDE$. So, from our previous $2$ claims, we come to the conclusion that $QEYKD$ is cyclic. Now, $$\angle QXA = \angle QYA = \angle QYE = \angle QKE = \angle QKP$$and we are done .

wow. We have $\angle AXY=\angle XBY+\angle XYB=\angle XKP+\angle YKP=\angle XKY$, so $\overline{AX}$ is tangent to the circumcircle of $KXY$. Similarly, $\overline{AY}$ is tangent to the circumcircle of $KXY$. Let $M$ be the midpoint of $\overline{BC}$, let $P'$ be the reflection of $P$ over $M$, and let $R$ be the reflection of $P$ over $\overline{BC}$. It's clear that $M$ and $P'$ lie on $\overline{PK}$. We have \[\angle XP'Y=\angle XPY=360^\circ-\angle KPX-\angle KPY=\angle KXY+\angle KYX=180^\circ-\angle XKY,\]so $XKYP'$ is cyclic. Thus, $KR$ is a symmedian of $\triangle KXY$, so $A$, $R$, and $K$ are collinear. Let $\overline{AP'}$ and $\overline{AR}$ intersect the circumcircle of $AXY$ again at $T$ and $T'$, respectively. Notice the following similarities: Since $\angle KBX=\angle KXY$ and $\angle KXB=\angle KYX$, we have $\triangle KBX \sim \triangle KXY$. Similarly, $\triangle KYC \sim \triangle KXY$. Since $Q$ is the Miquel point of $BXYC$, $\triangle QBX \sim \triangle QCY$. We have \[\angle KXT'=\angle KXY-\angle YXT'=180^\circ-\angle AYK-\angle KAY=\angle YKT',\]and similarly, $\angle KYT'=\angle XKT'$, so $\triangle T'XK \sim \triangle T'KY$. Thus, we have \[\frac{T'X}{T'Y}=\frac{T'X}{T'K} \cdot \frac{T'K}{T'Y}=\left(\frac{KX}{KY}\right)^2\]and \[\frac{QX}{QY}=\frac{BX}{CY}=\frac{BX}{XY} \cdot \frac{XY}{YC}=\left(\frac{KX}{KY}\right)^2,\]so $QXT'Y$ is harmonic. Since $\overline{QT'}$ is a symmedian of $QXY$, $\overline{QT}$ is a median. We have $MK \cdot MP'=MX \cdot MY=MQ \cdot MT$, so $KQP'T$ is cyclic, which means $\angle QKP=\angle QTP'=\angle QXA$.

I've come with the chaotic neutral solution. Let $P'$ be the reflection of $P$ over $XY$ and let $M$ be the midpoint of $XY$. Claim: $(P'XKY)$ is cyclic and tangent to $AX, AY$. Proof. \[ \measuredangle XKY = \measuredangle XBP + \measuredangle XCY = \measuredangle YXP + \measuredangle PYX = \measuredangle YPX = \measuredangle XP'Y \]gives that it is cyclic. Then, \[ \measuredangle AXY = \measuredangle BXY = \measuredangle BXP + \measuredangle PXY = \measuredangle BXP + \measuredangle PBX = \measuredangle BPX = \measuredangle YPX \]as desired, symmetry suffices. $\blacksquare$ Claim: The center $G$ of that $(P'XKY)$ lies on $(AXY)$ and the angle bisector of $\angle BAC$. Proof. $\measuredangle AXG = \measuredangle AYG = 90^\circ$, then SAS gives the angle condition. $\blacksquare$ Claim: $A$ lies on $P'K$. Proof. Since $AB$ and Quadrilateral $P'XKY$ is harmonic. As such, since $AX$ and $AY$ are tangents, the result follows. $\blacksquare$ Thus, $XY$, and the tangents from $P'$ and $K$ concur at some point $E$. Claim: $(EP'MGK)$ is cyclic. Proof. This follows as \[ \measuredangle EQG = \measuredangle EMG = \measuredangle EP'G = \measuredangle EK'G = 90^\circ \]$\blacksquare$ Claim: $Q$, $D$, $G$ are collinear. Proof. Length bash. $\blacksquare$ Claim: $E$ lies on $AQ$. Proof. Radical axis on $(P'MGK)$, $(AQXFGY)$, and $(PX'YK)$. $\blacksquare$ As such, $\measuredangle EQG = \measuredangle AQG$ holds as well, and thus $Q$ lies on $(EP'MGK)$. As such, it follows that \[ \measuredangle QXA = \measuredangle QGA = \measuredangle QGM = \measuredangle QKM = \measuredangle QKP \]as desired.

Cute Spiral

Let $K$ go to $L$ under the spiral at $Q$ Claim: $BPCL$ is cyclic

Claim 2: $P-K-L$

Now $\measuredangle QKP = \measuredangle QKL = \measuredangle QXA$. Done. $\blacksquare$