Let $X$ be the intersection between $AB$ and $\ell_Q$, $Y$ the intersection between $AB$ and $\ell_P$, $Z$ the intersection between $\ell_P$ and $\ell_Q$, $R$ the intersection between $AB$ and $MP$, $S$ the second intersection point betwen the ray $PM$ and $\Omega$, and, finnaly, let $W$ be the second intersection point between $YS$ and $\Omega$. First, consider a circle $\omega_{P}$ internally tangent to $\omega$ at $P$ and tangent to $AB$ in $R'$. Since $M$ is midpoint of the arc $AB$, the tangent to $\omega$ at $M$ is parallel to $AB$. Thus, the homotety with center $P$ that takes $\omega_{P}$ to $\omega$ takes $R'$ into $M$, so $P, R', M$ are colinear. Because of that, we conclude that $R=R'$, and so $YP$ and $YR$ are both tangents to $\omega_{P}$ and we get $YP=YR$. Now, observe that, by power of a point, $YP^2=YB \times YA=YW \times YS=YR^2$. From $YP^2=YW \times YS$ we get $\angle YPW=\angle PSW$ and from $YR^2 =YW \times YS$, we get $\angle YRW=\angle RSW= \angle PSY$. Then, $\angle YPW=\angle YRW$, so $YPRW$ is cyclic. Also, $ZQ$ is tangent to $\Omega$ at $Q$, so $\angle XQW = \angle ZQW = \angle QSW = \angle RSW = \angle YRW = \angle XRW$. So, $\angle XQW = \angle XRW$, and then $XQRW$ is cyclic. Because $YPRW$ is cyclic, we have $\angle YPW = \angle YRW$ and because $XQRW$ is cyclic, we have $\angle XRW = \angle XQW$. So, $\angle ZPW = \angle ZQW$, which give us that $ZPQW$ is cyclic. From this, we get $\angle XZW = \angle QPW$ and because $\angle QPW = \angle XYW$ ($YPRW$ is cyclic), we get $\angle XZW = \angle XYW$. This proves that $XWZY$ is cyclic. So, the circuncircle of the triangle formed by $\ell_P$, $\ell_Q$ and $AB$ touches $\Omega$ at one point $W$. It remains to prove that $W$ is the only intersection point. To do this, we'll prove that the tangent to $\Omega$ at $W$, say $\ell$, is tangent to $(XYZ)$ as well. Let $T$ be the intersection point between $\ell$ and $AB$. Then, $\angle QWT = \angle QSW$, and from $180^{\circ}=\angle QSW + \angle QWS + \angle SQW = \angle QWT +\angle QWS + \angle TWY$, we obtain $\angle TWY = \angle WQS$. But $\angle WQS= \angle WZY$ ($ZPQW$ is cyclic), and finally we gain $\angle TWY = \angle WZY$. Therefore, $\ell$ is tangent to both $(XYZ)$ and $\Omega$ at $W$, which implies $(XYZ)$ tangent to $\Omega$, as desired.

I'll use Davi's diagram. Let $K$ be the Miquel point of lines $MP$, $AB$, $\ell_P$ and $\ell_Q$. Point $K$ is actually the desired tangency point. (How I conjectured that? By drawing a good diagram.) By the way, if you want to follow the diagram, picture $K$ in place of $W$, okay? I did not name it $W$ because, even though they coincide, they have different definitions. First, as Davi proved before, $YR = YP$. Oh, look, $Y$ is on the radical axis of $\omega$ and $\Omega$! So it seems to be a good idea to invert about $Y$ and radius equal to the square root of the common power of $Y$ wrt $\omega$ and $\Omega$. It fixes both circles and maps $K$ to the intersection $K'$ of lines $YK$ and $PR = P'R'$ ($K$ lies in circle $YPR$). I'm using the standard notation that $C'$ is the inverse of $C$. Now consider circle $KK'Q$. Because of the inversion, $YR^2 = YK\cdot YK'$, so $\angle XRK = \angle YRK = \angle YK'R = \angle KK'R$. Also, as $KXQR$ is cyclic, $\angle XRK = \angle XQK$, so $\angle XQK = \angle KK'R$ and $\ell_Q$ is tangent to this circle. But $Q'$ also lies in this circle (because $YQ\cdot YQ' = YK\cdot YK'$), so circle $KK'Q$ is actually $\Omega$ and $K$ lies on it. Well, it obviously also lies on circle $XYZ$ (it's a Miquel point). (Yes, $K'$ coincides with $S$ in Davi's diagram.) It remains to prove that it is the only intersection point of circles $\Omega$ and $XYZ$. Use the inversion again: circle $XYZ(K)$ is mapped to line $X'K'$ and $\Omega$ is fixed. Now use the cyclic quad $XX'K'K$: $\angle(X'K',KK') = \angle KXX' = \angle KXR = \angle KQR = \angle KQK'$, so $X'K'$ is tangent to $\Omega$, finishing the problem. EDIT: I felt guilty not writing the angle chasing down, so I wrote it.

Well, I am sure you being skilled would have remembered about IMO 2011, am I right? hahaha First thing I did, was actually let the second intersection of, if $O_A, O_B$ are circumcentres of $w, \Gamma$, $O_AO_B$ with $w$ be $T$ and $AB \cap MP = X, MP \cap \Gamma = Y$. Let $PP, QQ \cap AB = D, C$ and $PP \cap QQ = E$. Notice $\angle PXD = \angle PTM = \angle DPM \implies DP=DX$. It follows circles $w, \Gamma$ stay fixed under inversion with centre $D, $radius $DP$. Therefore, $DY \cap \Gamma = Y'$ is the image of $Y$ under this inversion. $\angle DXY' = \angle DYX = \angle (QQ, QY)$ so $Y' \in \odot CQX$ and we know $Y' \in DPX$ and so it is the miquel point of $EDPQXC$, or $Y' \in CDE$. Tangency follows since $\angle YQY' = \angle Y'QX = \angle Y'ED$.

Let $\ell_Q \cap AB=X,\ell_P \cap AB=Y,\ell_P \cap \ell_Q=Z$. $MQ \cap AB=S,MQ \cap \Omega=R,RY \cap \Omega=W$. Let $O$ be the centre of $\omega$. $YP^2=YA.YB=YW.YR \Rightarrow \angle YPW=\angle YRP=\angle WRQ=\angle ZQW$. $\Rightarrow ZWQP$ is cyclic $\Rightarrow \angle XZW=\angle QZW=\angle QPW$. Now $\angle YPS=\frac 12 \angle MOP=90^{\circ}-\angle PMO=\angle YSP \Rightarrow YP=YS$ $\Rightarrow YS^2=YW.YR \Rightarrow \angle YSW=\angle WRS=\angle YPW$ $\Rightarrow YWSP$ is cyclic $\Rightarrow \angle XYW=\angle SYW=\angle SPW=\angle QPW$. So we have $\angle XZW=\angle XYW \Rightarrow W \in \odot XYZ$. Draw tangent $WE$ at $W$ to $\Omega$. Since $\angle XSW=\angle XQW$, we have $WSQX$ is cyclic. Now $\angle XWE=\angle XWQ-\angle EWQ=\angle QSY-\angle YRS=\angle WYX \Rightarrow WE$ is tangent to $\odot XYZ$ meaning $\Omega,\odot XYZ$ are tangent.

Let $S$ be intersection of $PM and \Omega$,$T$ intersection of $PM$ and $AB$,$X$ intersection of $\ell_Q$ and $AB$,$Y$ intersection of $\ell_P$ and $\ell_Q$, $Z$ intersection of $AB$ and $\ell_P$, $N$ intersection of $ZS$ and $\Omega$. We want to prove that circle around $XYZ$ is tangent to $\Omega$ at $N$. First note that angle chase gives us that $ZP=ZT$ hence $ZNQ$ and $ZQS$ are similar so $\angle XQN=\angle QSN=\angle XTN$ hence $XNTQ$cyclic. Now $\angle YXN=\angle XTQ + \angle XTN = \angle ZPS + \angle ZSP = \angle YZN$ hence $XZYN$ cyclic. So now we have $ \angle XNQ= \angle XTQ =\angle XZN + \angle QSN$. Let $t$ be tanget to $\Omega$ at $N$. $\angle (t,NQ)=\angle XNQ - \angle (t,XN)$ hence $\angle (t,XN) = \angle XZN$ hence $t$ is tangent to circle $XZYN$ at $N$,and we are done.

Let $X=l_P\cap AB$, $Y=l_Q\cap AB$, $Z=l_P\cap l_Q$, and $N=MP\cap AB$. By Casey's theorem for the quadruple ($\Omega, Y,X,Z$) it is enough to prove that $XP\cdot YZ=QZ\cdot XY+QY\cdot XZ$ (1). Now, if $O_1$ is the center of $\omega$ then $\triangle{MO_1P}$ is isosceles, which implies that $XP=XN$ (2). Finally, by Menelaus' Theorem for $\triangle{XYZ}$ and the transversal $P-Q-N$ we have: \[\frac{(XP+XZ)}{XP}\cdot \frac{XN}{XN-XY}\cdot \frac{QY}{QZ}=1,\] which together with (2) yield (1).

[asy][asy] /* APMO 2014 Problem 5, free script by liberator, 25 October 2014 */ /* Manually converted from a geogebra file, see original geogebratube.org/student/mNmSZAvbW */ unitsize(3.2cm); pointpen=black; /* Initialize objects */ pair A = (0.97, 0.79); pair B = (0.97, -0.79); pair C = (1.63, -1.06); pair M = (1.25, 0); pair N = (2.53, 0.63); pair P = (-0.75, -1); pair Q = (0.65, -0.3); pair R = (0.97, -0.14); pair S = (2.53, -0.64); pair X = (0.97, -1.37); pair Y = (0.97, -2.28); pair Z = (1.31, -2.54); pair G = (0.96,0); /* Draw objects */ draw(X--Y--Z--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(A--X, rgb(0.4,0.6,0.8)); draw(P--Y, rgb(0.4,0.6,0.8)); draw(Q--X, rgb(0.4,0.6,0.8)); draw(P--N, rgb(0.4,0.6,0.8)+linewidth(1)); draw(N--Y, rgb(0.4,0.6,0.8)); draw(X--S, rgb(0.4,0.6,0.8)); draw(P--C, rgb(0.4,0.6,0.8)+dashed); draw(Q--C, rgb(0.4,0.6,0.8)+dashed); draw(R--C, rgb(0.4,0.6,0.8)+dashed); draw(N--S, rgb(0.4,0.6,0.8)); draw(circumcircle(P,A,B), rgb(0.9,0,0)); draw(circumcircle(A,B,C), rgb(0.9,0,0)); draw(circumcircle(Q,R,C), rgb(0.7,0,0.8)); draw(circumcircle(P,R,C), rgb(0.7,0,0.8)); draw(circumcircle(X,Y,Z), rgb(0,0.6,0)); draw(circumcircle(P,R,G), rgb(0.7,0.7,0.7)); draw(circumcircle(R,C,N), rgb(0.7,0.7,0.7)); /* Place dots on and label each point */ Drawing("A", A, dir(90)); Drawing("B", B, dir(-75)); Drawing("C", C, dir(-50)); Drawing("M", M, dir(-40)); Drawing("N", N, dir(70)); Drawing("P", P, dir(210)); Drawing("Q", Q, dir(140)); Drawing("R", R, dir(60)); Drawing("S", S, dir(-20)); Drawing("X", X, 2*dir(-150)); Drawing("Y", Y, dir(-90)); Drawing("Z", Z, dir(-90)); [/asy][/asy] Define $X = \ell_Q \cap AB, Y = \ell_P \cap AB, Z = \ell_P \cap \ell_Q$. Let $R = PM \cap AB, N = PM \cap \Omega, C = YN \cap \Omega, S = XC \cap \Omega$. Consider the circle tangent to $\omega$ at $P$ and to $AB$ at $R'$. The tangent to $\omega$ at $M$ is parallel to $AB$, so by homothety, it follows that $P,R,M$ are collinear; i.e. $R=R'$. Hence $YP = YR$. $YP^2 = YC \cdot YN$, so $\angle YCP = \angle YPN = \angle YRP$, and $YCRP$ is cyclic. Also, $YR^2 = YC \cdot YN$, hence $YR$ is tangent to $(RNS)$. Therefore $\angle XQC = \angle QNC = \angle XRC$, so that $XQRC$ is cyclic. It follows that $C$ is the Miquel point of $\ell_P, \ell_Q, \overline{AB}, \overline{PR}$: hence $CXYZ$ is cyclic. By Reim's theorem on $(XQR)$ and $\Omega$, we have $RX \parallel NS$; by a converse of Reim's theorem on $(XYZ)$ and $\Omega$, these two circles are tangent at $C$, as required.

My solution: Let $ D=\ell_Q \cap AB, E=\ell_P \cap AB, F=\ell_Q \cap \ell_P, T=PQ \cap \Omega, S=PQ \cap AB, W=TE \cap \Omega $ Since $ EP^2=EA \cdot EB=EW \cdot ET $ , so we get $ \triangle EPW $ ~ $ \triangle ETP $ and $ \angle EPW=\angle ETP=\angle WQF $ , hence $ P, Q, W, F $ are concyclic . Since $ \angle EPS=\angle ESP $ (notice that arc $ AM $ = arc $ MB $ ) , so we get $ \angle DET=\angle ESP-\angle ETS=\angle EPS-\angle WPE=\angle QPW=\angle QFW $ . i.e. $ D, E, F, W $ are concyclic Since $ \angle DWQ=\angle DWP+\angle PWQ=\angle DWP+\angle PFQ $ $ =\angle DEP+\angle EWD=\angle EWP=\angle WTP+\angle TPW=\angle WTP+\angle DFW $ , so we get $ \odot (DEF) $ is tangent to $ \Omega $ at $ W $ . Q.E.D

Let $X = PM \cap AB$ and $(C, D, E) = (AB \cap QQ, PP \cap QQ, AB \cap PP)$. Notice that $\angle MPE = 90^\circ - \angle PMO_1 = \angle PXE$ so $EP = EX$. So that means $EX^2 = EP^2 = EB \cdot EA$. Let $Y = PM \cap \Omega$ and $T = EY \cap \Omega$. Therefore, $EX^2 = ET \cdot EY$ so $\angle TYX = \angle EXT = \angle TQE$ so $TCQX$ is cyclic. Similarly, we have $EP^2 = ET\cdot EQ$ so $\angle EQT = \angle EYP = \angle EPT$ so $EPXT$ is cyclic. Therefore, $T$ is Miquel point of $ECQP$. Therefore, $T$ lies on circumcircle of $DEC$. We now have $\angle EDT + \angle TYQ = \angle TCX + \angle TYQ = \angle TQX + \angle TYQ = \angle ETQ$. This directly implies the desired statement, so we are done.

Let $T$ be the Miquel point of $\ell_P$, $\ell_Q$, $AB$, and $PQ$. Let $\ell_P$ and $\ell_Q$ intersect $AB$ at $X$ and $Y$, let $\ell_P$ and $\ell_Q$ intersect at $Z$, and let $PQ$ intersect $AB$ at $C$ and $XT$ at $R$. Note that $\angle XPC=\angle XPA+\angle APC=\angle PBA+\angle ABM=\angle PBM=\angle PQC$. Hence, $X$ is the midpoint of arc $PC$ on the circumcircle of $XPC$. Since $T$ lies on this circle, $TX$ bisects $\angle PTC$. Hence, $\angle XTP=\angle CTR$, and because $\angle TXP=\angle TCR$ we have that $TXP$ is similar to $TCR$. Now, since $T$ is the center of spiral similarity taking $YQ$ to $CR$, the circumcircle of $QTR$ is tangent to $YQ$. Because $\angle QBT=\angle QRT=\angle YQT=\angle YCT=180^{\circ}-\angle TCB$, the circumcircle of $TCB$ is tangent to $QB$. Hence, $T$ is the center of spiral similarity taking $AQ$ to $CB$, so $\angle TAC=\angle TQB$. Thus, $TAQB$ is cyclic. It remains to show that the circumcircles of $TAQB$ and $TXYZ$ are tangent. But this just follows from Miquel's pivot theorem on triangle $XPR$ with points $Q$, $T$, and $Z$.

The hardest part of the solution is to identify the point of tangency. The subsequent arguments are fairly straightforward. Let the lines $\ell_P,\ell_Q,$ and $AB$ form the triangle $XYZ$. Let $PQ$ and $AB$ meet at $D$ and $PQ$ intersect $\Omega$ at $V \not=Q$. Let $T$ be the Miquel Point for the quadrilateral $PQYX$. The main step is to prove that $T$ is the desired tangency point. We show that $T$ lies on $\Omega$. Let $S$ be a point on $AB$ such that the circumcircle of triangle $VSD$ is tangent to $\Omega$ at $V$ and let $VS$ meet $\Omega$ at $L \not= V$. Let $ML$ meet $AB$ at $R$. Note that lines $VS$ and $VD$ are isogonal in $\angle AVB$, thus, $LQ \parallel AB$. As $M$ lies on the perpendicular bisector of $AB$, it follows that $MR=MD$. Apply an inversion at $D$ of power $\sqrt{DA\cdot DB}$ followed by reflection in $D$. The map fixes $\omega$ and $\Omega$ as $D$ lies on their radical axis $AB$. This maps $P$ to $M$ and $X$ to $R$ since $(DMR)$ is tangent to $\omega$. Thus, the circle $(DPX)$ is mapped to the line $MR$. Point $Q$ is mapped to $V$ and $Y$ is mapped to $S$ since $(VSD)$ is tangent to $\Omega$ and $QY$ is also tangent to $\Omega$. Thus, the circle $(DQY)$ is mapped to the line $VS$. As the lines $VS$ and $MR$ meet at $L$ on $\Omega$, it follows that $T=(DQY) \cap (DPX)$ lies on $\Omega$ establishing the claim. Let $W$ be the second intersection point of line $YT$ with $\Omega$. As spiral similarity occurs in pairs, we note that $T$ lies on $(XYZ)$. We have the following angle relations: $$\angle VTW=\angle XTY=\angle XZY=\angle YQD-\angle PXD=\angle QAV-\angle PDX=\angle BAV-\angle AQV,$$yielding that $AV=BW$ and hence, $VW \parallel AB$. It follows that $(XYZ)$ is tangent to $\Omega$.

Reposting. This solution looks shorter, neater, and different from the previous one, so I see no reason to not post it. v_Enhance wrote: Circles $\omega$ and $\Omega$ meet at points $A$ and $B$. Let $M$ be the midpoint of the arc $AB$ of circle $\omega$ ($M$ lies inside $\Omega$). A chord $MP$ of circle $\omega$ intersects $\Omega$ at $Q$ ($Q$ lies inside $\omega$). Let $\ell_P$ be the tangent line to $\omega$ at $P$, and let $\ell_Q$ be the tangent line to $\Omega$ at $Q$. Prove that the circumcircle of the triangle formed by the lines $\ell_P$, $\ell_Q$ and $AB$ is tangent to $\Omega$. Ilya Bogdanov, Russia and Medeubek Kungozhin, Kazakhstan Let $X=\ell_P \cap \overline{AB}$, $Y=\ell_P \cap \ell_Q$, $Z=\ell_Q \cap \overline{AB}$ and $\Gamma$ denote the circle $(XYZ)$. Let $\overline{PQ}$ meet $\Omega$ again at $S$, parallel from $S$ to $\overline{AB}$ meet $\Omega$ again at $C$. Let $\overline{ZC}$ meet $\Omega$ again at $T$. Claim: $T$ is the desired point of tangency. (Proof) First, we show $T$ lies on $(QRZ)$; indeed, $\angle QTZ=\angle QAC=\angle QRT$, so that is true. Next, we show $T$ lies on $\overline{XS}$. Let $X'=\overline{TS} \cap \overline{AB}$, then $\angle TRZ=\angle TQZ=\angle TSR,$ so $$X'R^2=X'S \cdot X'T=X'A \cdot X'B.$$Consequently, $X$ and $X'$ coincide. Finally, $\angle XTZ=\angle CTS=\angle XYZ$ so $T$ lies on $\Gamma$. It follows that $T$ is the center of a homothety $\Omega \mapsto \Gamma$, hence $\Gamma$ and $\Omega$ are tangent at $T$, as desired. $\blacksquare$

A solution for those of us who are unable to think of Miquel points. Let's define $X = \ell_P \cap AB$, $Y = \ell_Q \cap AB$, and $Z = \ell_P \cap \ell_Q$. Our goal is to show that $(XYZ)$ is tangent to $\Omega$. We can assume without loss of generality that $PA \le PB$. Now we can invert about $A$. Let's let $K^*$ denote the image of figure $K$. We see that $\omega^*$ and $\Omega^*$ are lines that intersect at $B^*$. $P*$ lies on $\omega^*$ such that the segment $P^*A$ does not intersect $\Omega^*$ and $Q*$ lies on $\Omega^*$ such that the segment $Q^*A$ does intersect $\omega^*$. Now $AM = MB$, so after inversion $AB^* = B^*M^*$. Furthermore, $B^*$ is between points $P^*$ and $M^*$ on line $\omega^*$. Finally, points $A, P^*, Q^*, M^*$ lie on a circle. We will let $\theta = \measuredangle B^*AM^* = \measuredangle AM^*B^*$. Now let's try to find $X^*, Y^*, Z^*$. We know that $\ell_P^*$ is a circle passing through $A$ and tangent to $\omega^*$ at $P^*$. Then $X^*$ is the second intersection of line $AB^*$ with $\ell_P^*$. It's clear that $A$ lies between $X^*$ and $B^*$ on line $AB^*$. Similarly, $\ell_Q^*$ is a circle passing through $A$ and tangent to $\Omega^*$ at $Q^*$. Then $Y^*$ is the second intersection of line $AB^*$ with $\ell_Q^*$. It's clear that $A$ lies between $Y^*$ and $B^*$ on line $AB^*$. Now $Z^*$ is the intersection of $\ell_P^*$ with $\ell_Q^*$. Let's let $a = B^*P^*$ and $b = B^*Q^*$. Let $x = B^*A = B^*M^*$. It's clear that $B^*X^* = \frac{a^2}{x}$ and $B^*Y^* = \frac{b^2}{x}$. Now define $R$ such that $B^*$ lies between $Q^*$ and $R$ on line $\Omega^*$ and such that $B^*R = \frac{ab}{x}$. It follows that $(RX^*Y^*)$ is tangent to $\Omega^*$ at $R$. If we can show that $R,X^*, Y^*, Z^*$ are concyclic, then we deduce that $(X^*Y^*Z^*)$ is tangent to $\Omega^*$, yielding that $(XYZ)$ is tangent to $\Omega$, as desired. Hence we only need to show that $R, X^*, Y^*, Z^*$ are concyclic. We need to show that $\measuredangle X^*RY^* = \measuredangle X^*Z^*Y^*$. We do this in two steps: First we note that $\measuredangle X^*RY^* = \measuredangle X^*RB^* + \measuredangle B^*RY^* = \measuredangle X^*RB^* + \measuredangle RX^*B^*$. Now we define $S$ to be the second intersection of $\Gamma$ with $\Omega^*$. We see that $B^*S = \frac{ax}{b}$ and thus $A, X^*, R, S$ are concyclic. Thus $\measuredangle X^*RB^* + \measuredangle RX^*B^* = \measuredangle B^*AS + \measuredangle B^*SA = \theta + \measuredangle M^*AS + \measuredangle Q^*SA = \theta + \measuredangle M^*AS + \measuredangle Q^*M^*A $. We conclude that $\measuredangle X^*RY^* = \theta + \measuredangle M^*AS + \measuredangle Q^*M^*A$. Now $\measuredangle X^*Z^*Y^* = \measuredangle X^*Z^*A + \measuredangle AZ^*Y^* = \measuredangle X^*P^*A + \measuredangle AQ^*Y^*$. We note that $\measuredangle X^*P^*A = \measuredangle X^*P^*B^* + \measuredangle B^*P^*A = \measuredangle B^*AP^* + \measuredangle B^*P^*A$. Similarly, $\measuredangle AQ^*Y^* = \measuredangle AQ^*B^* + \measuredangle B^*Q^*Y^* = \measuredangle AQ^*B^* + \measuredangle Q^*AB^*$. Thus $\measuredangle X^*Z^*Y^* = (\measuredangle B^*AP^* + \measuredangle B^*P^*A) + (\measuredangle AQ^*B^* + \measuredangle Q^*AB^*) $ $= \measuredangle B^*P^*A + \measuredangle AQ^*B^* + \measuredangle Q^*AP^* = \measuredangle M^*P^*A + \measuredangle AQ^*S + \measuredangle Q^*M^*P^*$. $ = \measuredangle M^*Q^*A + \measuredangle AQ^*S + \measuredangle Q^*M^*A + \measuredangle AM^*P^*$ $ = \measuredangle M^*Q^*S + \measuredangle Q^*M^*A + \theta$ $ = \measuredangle M^*AS^* + \measuredangle Q^*M^*A + \theta$. We conclude that $\measuredangle X^*RY^* = \measuredangle X^*Z^*Y^*$, as desired. $\blacksquare$

My solution from 2 years ago. (I personally find this problem to be somewhat easier for its place on the exam) Define the following points : $MP \cap AB = T, MP \cap \Omega = U, AB \cap \ell_P = X, AB \cap \ell_Q = Y, \ell_P \cap \ell_Q = Z$. $X$ is on the radical axis of $\omega, \Omega$, thus the circle with radius $XP$ centered at $X$ keeps both of them fixed. Also by the Shooting lemma, the circle tangent to $\omega$ at $P$ and to $AB$ is tangent to $AB$ at $T$. So we have $XP=XT$. Let $U = MP \cap \Omega$. $K = XU \cap \Omega$. So under this inversion $U$ is mapped to $K$. We have $XT^2=XP^2=XA \cdot XB = XK \cdot XU$, so $\angle XTK = \angle XUT = \angle YQK$. So $YQTK$ is cyclic. $180^\circ - \angle KXP = \angle XPU + \angle XUP = \angle XTQ + \angle KTA = \angle KTQ$, so $KTPX$ is cyclic. So $K$ is the Miquel point of $\{ Y, Q, Z \}$ wrt $\triangle XPT$, so $K \in (XYZ)$. Now to finish, observe that $\angle YKQ = \angle YTQ = \angle TXU + \angle TUX = \angle YXK + \angle KUQ$, which leads to the required tangency.

A more easy and intuitive synthetic solution: Let $l_P$ $\cap$ $AB = R$, $l_Q$ $\cap$ $AB = S$, $l_P$ $\cap$ $l_Q$ = $T$, $PQ$ $\cap$ $AB = G$. Also, Let $(RST)$ meet $\Omega$ at some point $C$. Invert the figure about $A$, and denote the inverse of an object $Z$ by $Z'$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = red; /* point style */ real xmin = -38.368087309032624, xmax = 81.61356975341747, ymin = -19.21786149825803, ymax = 36.10192542213895; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((xmin, 1.5326890347032844*xmin-6.358405360890629)--(xmax, 1.5326890347032844*xmax-6.358405360890629), linewidth(2) + wrwrwr); /* line */ draw(circle((9.970196908102775,20.056993645508648), 9.050855001698539), linewidth(2) + wrwrwr); draw(circle((42.89021553368776,15.158395549757062), 25.20897737828278), linewidth(2) + wrwrwr); draw(circle((-0.20881934577113626,35.277224051083245), 26.80315024067216), linewidth(2) + wrwrwr); draw(circle((15.461079922841508,2.2703159269949396), 20.338210699148654), linewidth(2) + wrwrwr); draw((1.447574781434927,17.010210366736906)--(13.968159719736342,-18.01302708420825), linewidth(2) + wrwrwr); draw((-1.5966697153831308,-8.805603525701168)--(26.579288786182353,34.379379111903035), linewidth(2) + wrwrwr); draw((-10.725834647419006,10.623598517118714)--(32.998721494901744,-8.028904554813845), linewidth(2) + wrwrwr); draw((-10.725834647419006,10.623598517118714)--(1.447574781434927,17.010210366736906), linewidth(2) + wrwrwr); draw((1.447574781434927,17.010210366736906)--(11.404035533489628,11.12043485265554), linewidth(2) + wrwrwr); draw((11.404035533489628,11.12043485265554)--(-10.725834647419006,10.623598517118714), linewidth(2) + wrwrwr); draw((-1.5966697153831308,-8.805603525701168)--(32.998721494901744,-8.028904554813845), linewidth(2) + wrwrwr); draw((32.998721494901744,-8.028904554813845)--(13.968159719736342,-18.01302708420825), linewidth(2) + wrwrwr); draw((13.968159719736342,-18.01302708420825)--(-1.5966697153831308,-8.805603525701168), linewidth(2) + wrwrwr); /* dots and labels */ dot((18.72719483590407,22.34456081485151),dotstyle); label("$A$", (16.49375435540066,23.279458255159515), NE * labelscalefactor); dot((1.447574781434927,17.010210366736906),dotstyle); label("$P'$", (-1.3661106271777863,17.326169927633348), NE * labelscalefactor); dot((32.998721494901744,-8.028904554813845),dotstyle); label("$Q'$", (34.71997554274995,-7.402873894398413), NE * labelscalefactor); dot((13.968159719736342,-18.01302708420825),dotstyle); label("$M'$", (13.929260922004783,-16.378600911284014), NE * labelscalefactor); dot((6.3321519958585935,3.3468145692363507),dotstyle); label("$B'$", (7.518027338515083,3.4046341463413943), NE * labelscalefactor); dot((-10.725834647419006,10.623598517118714),dotstyle); label("$C'$", (-12.814742026266535,8.075675757169616), NE * labelscalefactor); dot((26.579288786182353,34.379379111903035),dotstyle); label("$S'$", (27.30126239614044,33.07948673277951), NE * labelscalefactor); dot((11.404035533489628,11.12043485265554),dotstyle); label("$R'$", (11.639534642187032,8.899977217904008), NE * labelscalefactor); dot((17.68174884647451,15.31885656817893),dotstyle); label("$T'$", (18.60030253283299,14.12055313588849), NE * labelscalefactor); dot((-1.5966697153831308,-8.805603525701168),dotstyle); label("$G'$", (-5.304439828464316,-9.509422071830748), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Now, As $AM = BM$, so $B'A = B'M' \Rightarrow B'P' = B'G'$. And, By Reim's theorem, $R'P' \parallel G'M'$. Also, $\triangle C'P'R'$ is perspective to $\triangle Q'M'G'$ $\Rightarrow$ By Desargues' Theorem of two triangles, $R'C' \parallel G'Q'$ $\Rightarrow \triangle C'B'R' \stackrel{+}{\sim} \triangle Q'B'G' \Rightarrow \frac{{B'}{G'}}{{B'}{Q'}} = \frac{{B'}{R'}}{{B'}{C'}}$ Now, ${B'}{P'}^2 = {B'}{R'} \cdot {B'}{A}$ and ${B'}{Q'}^2 = {B'}{S'} \cdot {B'}{A}$ $\Rightarrow$ $\frac{{B'}{R'}}{{B'}{S'}}$ = $\left(\frac{{B'}{P'}}{{B'}{Q'}}\right)^2$ = $\left(\frac{{B'}{G'}}{{B'}{Q'}}\right)^2$ = $\left(\frac{{B'}{R'}}{{B'}{C'}}\right)^2$ $\Rightarrow$ ${B'}{C'}^2 = {B'}{R'} \cdot {B'}{S'} \Rightarrow B'C'$ is tangent to $(C'R'S'T') \Rightarrow (RST)$ is tangent to $\Omega$ Q.E.D.

Converse of Casey's theorem kills this the only thing you need after applying the theorem is menelause and some computation.

Let $X=l_p \cap AB$, $Y=l_q \cap l_p$ ,$Z=l_q \cap AB$, $U=MP \cap \Omega$ with $U\neq M$ and $V$ the intersection of the line through $U$ parallel to $XY$ with $\Omega$, and $W$ the intersection of the line through $U$ parallel to $XZ$ with $\Omega$ we will show that $XYZ$ and $UVW$ are homothetic and the center of homothety is a point on $\Omega$ so we will have that the two circles are homotheticso they are tangent. We clearly have $\angle VUW = \angle YXZ$ Let $O_1$ and $O_2$ be the centers of $\omega$ and $\Omega$ respectively, We have $AB$ is perpendicular to $O_1O_2$ so must be $UW$ and hence $\angle WVU=\frac{\angle WO_2U}2=\pi-\angle MO_2U=\angle O_2UQ+\angle O_2MU=\angle O_2UQ+\angle PMO_1=\frac{\pi}2-(\pi-\angle QAU)+\frac{\pi}2-\angle PAM=\angle QAU-\angle PAM$ and we have $\angle ZYX=\angle PYQ=\pi -\angle YPM-\angle YQP=-\angle PAM+\angle YQU=-\angle PAM+\angle QAU$ so the two triangle are similar and we can easily see that we must also have $WV$ and $YZ$ parallel, so by desargues $YV$, $XU$ and $WZ$ intersect and by thales that point of intersection is the center of homothety that sends $XYZ$ to $UVW$ Let $J=XU \cap \Omega$ with $J\neq U$ We have $XJ\times XU=XB\times XA=XP^2$ so $XP$ is tangent to $(PJU)$ Hence $\angle YPJ=\angle XPJ=\angle JUQ=\angle YQJ$ so $Y,Q,J,P$ are concyclic so $\angle YJQ=\pi-\angle YPQ=\pi-\angle YPU=\pi-\angle PUV=\pi-\angle QJV$ So $J$ lies on $YV$ and hence $J$ is the center of homothety that takes $XYZ$ to $UVW$ but $J$ lies on $(UVW)$ so it must lie on $(XYZ)$ and if the two circles have another point of intersection, that point must be fixed by that homothety which is impossible and hence the two circles are tangent

[asy][asy] size(8cm); 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 = -23.35875138992646, xmax = 13.353764604630443, ymin = -22.891318352238255, ymax = 8.294582331310082; /* image dimensions */pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen zzttff = rgb(0.6,0.2,1); /* draw figures */draw(circle((3.169240834286882,-1.8885049704659111), 4.92514158595303), linewidth(0.8) + fuqqzz); draw(circle((-3.6581942639594653,-1.0765937695933727), 5.201235230798455), linewidth(0.8) + linetype("4 4") + blue); draw(circle((-6.660544337272981,-11.72675515483047), 5.864027689454336), linewidth(0.8) + zzttff); draw(circle((1.125517764194021,-8.607727788885091), 6.689794777823583), linewidth(0.8) + zzttff); draw((0.39601670251866206,2.1816605736689128)--(-1.6128847895576797,-14.711374700609413), linewidth(0.8)); draw((-3.090335130925679,-16.3786823861616)--(1.2959522692152765,-2.660664492109178), linewidth(0.8)); draw((-3.090335130925679,-16.3786823861616)--(6.855384045403722,-5.1549048034351035), linewidth(0.8)); draw((-1.6128847895576797,-14.711374700609413)--(-8.135332855898149,1.5706836912468014), linewidth(0.8)); draw((-8.135332855898149,1.5706836912468014)--(6.855384045403722,-5.1549048034351035), linewidth(0.8)); /* dots and labels */dot((0.39601670251866206,2.1816605736689128),dotstyle); label("$A$", (-1.0549110276418898,2.1100366261338297), NE * labelscalefactor); dot((-0.4811813317002967,-5.194777441354147),dotstyle); label("$B$", (-1.5483588232676546,-5.42326638708618), NE * labelscalefactor); dot((6.855384045403722,-5.1549048034351035),dotstyle); label("$P$", (7.366598017704497,-4.929818591460416), NE * labelscalefactor); dot((-1.721440855113482,-1.306910391185868),linewidth(4pt) + dotstyle); label("$M$", (-2.8642196116030276,-1.9033387782890576), NE * labelscalefactor); dot((1.2959522692152765,-2.660664492109178),dotstyle); label("$Q$", (1.2478453519450128,-3.646854322833427), NE * labelscalefactor); dot((-1.0537021954048034,-10.00915743159659),linewidth(4pt) + dotstyle); label("$X$", (-2.2062892174353412,-9.83140002800968), NE * labelscalefactor); dot((-3.090335130925679,-16.3786823861616),linewidth(4pt) + dotstyle); label("$Z$", (-4.47614907731386,-16.37780744997816), NE * labelscalefactor); dot((-1.6128847895576797,-14.711374700609413),linewidth(4pt) + dotstyle); label("$Y$", (-1.4167727444341174,-15.686980536102089), NE * labelscalefactor); dot((-0.10507818223774487,-2.032091866328144),linewidth(4pt) + dotstyle); label("$K$", (0.42543235923540473,-1.5743735812052144), NE * labelscalefactor); dot((-5.069451193781412,-6.082710208923217),linewidth(4pt) + dotstyle); label("$N$", (-6.219664621858229,-7.035195852797012), NE * labelscalefactor); dot((-8.135332855898149,1.5706836912468014),linewidth(4pt) + dotstyle); label("$U$", (-9.410627033571508,1.846864468466755), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Label the intersection of the three lines as $X,Y,Z$ as labelled in the diagram. Suppose $(KPY)$ intersect $(XYZ)$ again at $N$ and suppose $NY\cap MP=U$. By easy angle chasing we have $YK=YP$. Then $N$ is the miquel point of $XQPY$. Now from $$\angle UNK=\angle KPY=\angle YKP=\angle YNP, \angle NKU=\angle NYP$$we obtain $\triangle UNQ\sim\triangle PNY$. Therefore, $\angle NUQ=\angle NPY=\angle NQX=\angle NKX$. Hence $QX$ is tangent to $(UNQ)$, and $\triangle YNK\sim\triangle YKU$ so $$YN\times YU=YK^2=YP^2=YA\times YB$$Hence $N,U,A,B$ are concyclic. Now we claim that $(NUAB),(NUQ),(ABQ)$ are the same circle, indeed, suppose they are distinct then $AB, UN, QX$ are their respective radical axis hence they are concurrent by radical axis theorem, which is clearly impossible. Now we show that the two circles are tangent at $N$, which is easy since $$\angle BNY+\angle ANX=180^{\circ}-\angle NYA-\angle NAY+\angle NBX$$and $$\angle NBX=\angle NBA-\angle NXB=\angle NBA-\angle NQU=\angle UBA=\angle UNA=\angle NYA+\angle NAY$$hence those two circles are tangent at $N$ as desired.

Nice problem needing only angle chasing and PoP! Let $\ell_Q \cap \ell_P = Y, \ell_Q \cap AB = X, \ell_P \cap AB = Z$. Let $PQ \cap \Omega = C, PQ \cap AB = R$ and finally let $YC \cap \Omega = K$. I claim $K$ is the desired point of tangency. Observe that in $\triangle PAB$, $R$ is the foot of angle bisector and $Y$ is the point where tangent at $P$ meets $AB$. So its well known (Just angle chasing) that $YP = YR$. Now, $YP^2 = YA.YB = YK.YC$ which means $\angle ZPK = \angle YPK = \angle QCK = \angle ZQK \implies ZKQP$ is cyclic. Also, since $YR^2 = YP^2 = YK.YC$, we also have that $\angle XRK = \angle YRK = \angle KCQ = \angle XQK$ and so $XKRQ$ is also cyclic. So this means that $K$ is the Miquel point of the quadrilateral $PYXQ$ and so $K \in (XYZ)$ Let $T$ be a point on the tangent to $(XYZ)$ at $K$. Then $\angle TKQ = \angle XKQ - \angle XKT = \angle XRQ - \angle XYK = \angle YRQ - \angle RYC = \angle KCQ$ which means $T$ lies on the tangent to $\Omega$ at $K$ as well. So, we are done. $\blacksquare$

Let the tangent distance from $X$ to $\Omega$ is $t$. By the converse of Casey theorem on $\Omega$ and degenerate circles $X, Y, Z$ we have to prove $$XZ\cdot YQ+QZ\cdot XY=YZ\cdot t$$[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.cm); 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 = 0., xmax = 12., ymin = -5., ymax = 8.; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); /* draw figures */ draw(circle((3.84,-0.8), 3.120256399721023), linewidth(0.8)); draw(circle((7.50250544956587,-0.5537961114859398), 3.900461511154802), linewidth(0.8)); draw((1.2464276596812818,0.9347571920974094)--(5.1229360812554505,-3.6443057169391806), linewidth(0.8)); draw((1.2464276596812818,0.9347571920974094)--(4.730709168465898,2.190424247027633), linewidth(0.8)); draw((5.096806909680528,6.691319071781163)--(4.489211202498332,5.7829248438034355), linewidth(0.8)); draw((4.585144514076699,4.355830094929869)--(5.1229360812554505,-3.6443057169391806), linewidth(0.8)); draw((5.096806909680528,6.691319071781163)--(4.585144514076699,4.355830094929869), linewidth(0.8)); draw((1.2464276596812818,0.9347571920974094)--(4.489211202498332,5.7829248438034355), linewidth(0.8)); draw((4.585144514076699,4.355830094929869)--(3.6924083853136516,0.28092561518506287), linewidth(0.8)); draw((1.2464276596812818,0.9347571920974094)--(3.6924083853136516,0.28092561518506287), linewidth(0.8)); draw((4.489211202498332,5.7829248438034355)--(4.585144514076699,4.355830094929869), linewidth(0.8)); draw((3.6924083853136516,0.28092561518506287)--(4.880418652840075,-0.03663969449416248), linewidth(0.8)); draw(circle((6.3560594333040425,5.191647790239139), 1.9582468871248302), linewidth(0.8) + aqaqaq); draw((4.880418652840075,-0.03663969449416248)--(6.953230132195835,-0.5907199388668184), linewidth(0.8)); /* dots and labels */ dot((4.730709168465898,2.190424247027633),dotstyle); label("$A$", (5.010046026056555,2.113372576984483), NE * labelscalefactor); dot((5.1229360812554505,-3.6443057169391806),dotstyle); label("$B$", (4.9721869033492165,-4.303748721909448), NE * labelscalefactor); dot((6.953230132195835,-0.5907199388668184),dotstyle); label("$M$", (7.035509090899176,-0.404259083053549), NE * labelscalefactor); dot((1.2464276596812818,0.9347571920974094),dotstyle); label("$P$", (0.48588086252957907,0.8261624049349628), NE * labelscalefactor); dot((3.6924083853136516,0.28092561518506287),dotstyle); label("$Q$", (3.079230767982281,-0.29068171493153255), NE * labelscalefactor); dot((5.096806909680528,6.691319071781163),dotstyle); label("$Z$", (4.820750412519861,7.129706335706878), NE * labelscalefactor); dot((4.585144514076699,4.355830094929869),dotstyle); label("$Y$", (4.082497519726757,4.403849500778482), NE * labelscalefactor); dot((4.489211202498332,5.7829248438034355),dotstyle); label("$X$", (3.779624538068047,5.76677791824268), NE * labelscalefactor); dot((4.880418652840075,-0.03663969449416248),dotstyle); label("$R$", (4.536806992214821,-0.536766012529235), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Clearly $t=XP$. Applying Menelaus' theorem on $\triangle XYZ$ and line $\overline{PQR}$ we have, \begin{eqnarray*} \frac{XR}{RY}\cdot \frac{YQ}{QZ}\cdot \frac{ZP}{PX}&=&1\\ \implies \frac{YQ}{XP-XY}\cdot \frac{PX+XZ}{QZ}&=&1\\ \implies YQ(PX+ZX)&=&QZ(XP-XY)\\ \implies XZ\cdot YQ+QZ\cdot YQ\cdot PX&=&QZ\cdot XP-QZ\cdot XY\\ \implies XZ\cdot YQ+QZ\cdot XY&=&YZ\cdot XP \end{eqnarray*}as desired.

Let PQ meet $\Omega$ at S, AB meet $\ell_P$ at Z, AB meet $\ell_Q$ at X, $\ell_P$ and $\ell_Q$ meet at Y, AB meet PQ at F and SZ meets $\Omega$ at K. Claim1 : PQKY is cyclic. Proof : ∠YQK = ∠QSK and ZP^2 = ZB.ZA = ZK.ZS so ∠YPK = ∠PSK so ∠YQK = YPK. Claim2 : ZXKY is cyclic. Proof : ∠ZPF = ∠ZFP and ∠ZPK = ∠FSZ so ∠SZF = ∠KPF so ∠KZX = ∠KPQ = ∠KYQ = ∠KYX. Claim3 : K is the Tangency point. Let tangent at W to XYZ meet $\Omega$ at N. ∠NQS = ∠NKZ = ∠KYZ = ∠KQS so N is K. we're Done.

Let $\ell_Q ,\ell_P$ intersect at $X$ and intersect $AB$ at $Y,Z$ respectively, $PQ\cap AB=R.$ Let $PQ$ meet $\Omega$ again at $S, ZS$ meet $\Omega$ again at $T.$ Since $PM$ bisects angle $APB$ it follows that $|ZR|=|ZP|.$ $$|ZP|^2=|ZA|\cdot |ZB|=|ZT|\cdot |ZS|\implies \measuredangle ZTP=\measuredangle SPZ=\measuredangle ZRP\implies T\in \odot (ZRP)$$$$|ZR|^2=|ZT|\cdot |ZS|\implies \measuredangle YRT= \measuredangle ZRT=\measuredangle QST=\measuredangle YQT\implies T\in \odot (YRQ).$$Thus $T$ is the Miquel point of $PZYQ,$ in particular $T\in \odot (XYZ),$ and this is the desired tangency point by $$\measuredangle TXY+\measuredangle QST=\measuredangle TZY+\measuredangle QST=\measuredangle SZR+\measuredangle RSZ=\measuredangle SRZ=\measuredangle QTY.$$

Let $D,E$ be the intersection of $AB$ with $l_P,l_Q$ ; $F=l_P \cap l_Q$ ; $R=PM \cap \Omega$ ; $L=DR \cap \Omega$ ; $I = AB \cap PQ$ $\bullet$ $DL.DR=DA.DB=DP^2 \Rightarrow \measuredangle DPL = \measuredangle LRP = \measuredangle FQL \Rightarrow PQLF$ is cyclic Main claim:We'll prove $(DEF)$ tangent to $\Omega$ at $L$. Proof: This is equivalent to $\measuredangle ELQ = \measuredangle LDE + \measuredangle LRQ = \measuredangle EIQ \Leftrightarrow ELIQ \textrm{ is cyclic}\Leftrightarrow \measuredangle DIL= \measuredangle FQL \Leftrightarrow \measuredangle DIL = \measuredangle LRI \Leftrightarrow DI^2=DL.DR \Leftrightarrow DI=DP $ $\Leftrightarrow \measuredangle DIP = \measuredangle DPI \Leftrightarrow \measuredangle MIB = \measuredangle PBM \Leftrightarrow \measuredangle MBA = \measuredangle MPB \Leftrightarrow arcAM=arcBM$, which is exactly true. Diagram:

[asy][asy] //14APMO5 //setup; size(9cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(237,186,255); // blu1 lighter //defn real ao1,ao2; ao1= 26.3843; ao2=62.7204;// <(AO1,O1O2) and <(AO2,O1O2) resp pair O1,O2,A,B; O1=(0,0); O2=(2,0); A=extension(O1,dir(ao1),O2,O2+dir(ao2)); B=extension(O1,dir(-ao1),O2,O2+dir(-ao2)); real r1,r2; r1=distance(O1,A); r2=distance(O2,A); path w,W; w=circle(O1,r1); W=circle(O2,r2); pair M,P,D,Q1,Q2; M=(r1,0); P=r1*dir(263); D=extension(M,P,A,B); Q1=intersectionpoints(P--D,W)[0]; Q2=intersectionpoints(2*D-P--D,W)[0]; pair D1,X,K,Y1,Y2,Z1,Z2; D1=extension(-M,P,A,B); X=(D+D1)/2; K=2*circumcenter(O2,Q1,Q2)-O2; Y1=extension(K,Q1,A,B); Y2=extension(K,Q2,A,B); Z1=extension(K,Q1,P,X); Z2=extension(K,Q2,P,X); //draw fill(w,blu1); fill(W,blu1); fill(A--arc(O1,r1,ao1,-ao1)--arc(O2,r2,360-ao2,ao2)--cycle,blu2); draw(w,blu); draw(W,blu); draw(2*A-B--3.2*B-2.2*A,blu); draw(circle(X,distance(X,D)),purple+dashed); draw(1.2*P-.2*Z1 -- 1.1*Z1-.1*P ^^ 1.2*P-.2*D -- 2.5*D-1.5*P,purple); draw(1.3*Y2-.3*Q2 -- 3.3*Z2-2.3*K ^^ 1.7*Q1-.7*K -- 1.2*Z1-.2*K,magenta); draw(1.2*D1-.2*K -- 2.7*K-1.7*D1,red); draw(circumcircle(X,Y1,Z1),red); //label void pt(string s,pair P,pair v,pen a){filldraw(circle(P,.06),a,linewidth(.3)); label(s,P,v);} pt("$A$",A,-dir(0),blu); pt("$B$",B,-dir(0),blu); pt("$M$",M,dir(120),purple); pt("$P$",P,dir(-70),purple); pt("$D$",D,dir(150),purple); pt("$Q_1$",Q1,-dir(0),purple); pt("$Q_2$",Q2,dir(90),purple); pt("$D'$",D1,dir(-45),blu); pt("$X$",X,-dir(40),purple); pt("$Y_1$",Y1,-dir(30),red); pt("$Y_2$",Y2,-dir(0),red); pt("$K$",K,dir(20),magenta); pt("$Z_1$",Z1,dir(-80),magenta); pt("$Z_2$",Z2,-dir(70),magenta); pt("",extension(P,D,D1,K),(0,0),red); pt("",extension(Q2,X,O2,circumcenter(X,Y1,Z1)),(0,0),red); label("$\omega$",(-2.5,2.6),blu); label("$\Omega$",(.9,1.5),blu); label("$d$",(6.5,4),red); label("$\ell_p$",(-2.5,-3),purple); label("$\ell_{q1}$",(8.8,-4.7),magenta); label("$\ell_{q2}$",(5.2,-6.5),magenta); [/asy][/asy] We'll consider both $Q$'s at once, the one inside and outside. Call them $Q_1,Q_2$ in any order. Define (here $k=1,2$): $X=\ell_p\cap\overline{AB}$, $Y_k=\ell_{qk}\cap\overline{AB}$, $Z_k=\ell_{qk}\cap\ell_p$; $D$ and $D'=2X-D$ as the intersections of the internal and external bisectors of $\angle APB$ with $\overline{AB}$, respectively, so that $XP=XD=XD'$; $K=\ell_{q1}\cap\ell_{q2}$ as the pole of $\overline{Q_1Q_2}$ wrt $\Omega$, so that $KQ_1=KQ_2$. Claim 1: $Y_1Y_2Z_1Z_2$ is cyclic. Proof. Note that triangles $PXD,KQ_1Q_2$ are both isosceles. Then \[\measuredangle(\ell_p,\ell_{q1})=\measuredangle XPD+\measuredangle PQ_1K \overset{\text{isosceles}}=-\measuredangle XDP-\measuredangle PQ_2K=-\measuredangle(\overline{AB},\ell_{q2}),\]whence the quadrilateral formed by $\ell_p$, $\ell_{q1}$, $\overline{AB}$, $\ell_{q2}$ (in order) is cyclic. $\qquad\qquad\square$ Let $i$ denote inversion at $X$ with power $XP^2=XD^2=XA\cdot XB$ (last equality by midpoints of harmonic bundles lemma). Claim 2: $i$ swaps $Y_1,Y_2$ as well.} Proof. Consider the polar $\overline{KD'}$ of $D$ wrt $\Omega$, which we call $d$. Then \[(Y_1Y_2; DD') \overset K=(Q_1,Q_2; D,d\cap\overline{Q_1DQ_2})=-1,\]the last harmonic bundle holding by definition of polar. The claim follows by another application of midpoints of harmonics bundles lemma. $\qquad\qquad\square$ By the previous two claims and power of a point at $X$, $i$ also swaps $(Z_1,Z_2)$. Applying $i$ to the given ``$\overline{Y_2Z_2}$ touches $\Omega$'' yields $(XY_1Z_1)$ also tangent to $\Omega$, concluding the proof.

$X$ = $AB$ $\cap$ $l_Q$; $Y$ = $AB$ $\cap$ $l_P$; $Z$ = $l_Q$ $\cap$ $l_P$; $R$ = $AB$ $\cap$ $MP$; $S$ = $PM$ $\cap$ $\Omega$; $W$ = $YS$ $\cap$ $\Omega$ Goal is to show that $\Omega$ and $(XYZ)$ share a tangent. Claim 1: $YP$ = $YR$ From Shooting Lemma We know that $P$ and $R$ lie on a circle that's tangent to $\omega$ and $AB$ as well as $M, P, R$ being collinear, Now since $YP$ is tangent to $\omega$ it is tangent to the inscribed circle and with $AB$ being tangent to it, $YR$ is also a tangent, thus $YP$ = $YR$. Claim 2: $YPRW$ and $XQRW$ are cyclic From Power Of Point we get $YR^2$ = $YP^2$ = $YB \cdot YA$ = $YW \cdot YS$ From which we get $\angle YPW=\angle YRW$ so $YPRW$ is cyclic. $ZQ$ is also a tangent to $\Omega$ from which we get $\angle XQW = \angle XRW$ $\Longrightarrow$ $XQRW$ is cyclic $\newline$ Claim 3: $XWZY$ is cyclic $\angle XRW = \angle XQW$ and $\angle YPW = \angle YRW$ $\Longrightarrow$ $\angle ZPW = \angle ZQW$ $\Longrightarrow$ $ZPQW$ is cyclic. $\angle XZW = \angle QPW$ and $\angle ZPW = \angle ZQW$ $\Longrightarrow$ $\angle XZW = \angle XYW$ $\Longrightarrow$ $XWZY$ is cyclic. $W$ = $(XYZ)$ $\cap$ $\Omega$. let's draw a tangent of $\Omega$ in $W$ (call it $l_W$) and finish it off with more angle-chasing. $K$ = $AB$ $\cap$ $l_W$ Claim 4: $\angle KWY = \angle WZY$ $$ \angle QWK = \angle QSW$$$$180^{\circ}=\angle QSW + \angle QWS + \angle SQW = \angle QWK +\angle QWS + \angle KWY \Longrightarrow \angle KWY = \angle WQS $$$$ \angle KWY = \angle WQS \hspace{5mm} \angle WQS = \angle WZY \Longrightarrow \angle KWY = \angle WZY $$ after this its clear that $l_W$ is tangent to both circles, thus $(XYZ)$ is tangent to $\Omega$

whenever you have zero claims, just invert the diagram

A picture is worth a thousand words. Unsaid thumbrule: for two circles being tangent, the problem is usually taking a similar triangle to XYZ in circle 2 and take homothety center as the tangency point and show homothety. A similar idea is used in ISL 2018 G5.

Remark: This is inspired by 2020 G6, another tangency problem where the answer is miquel Let $AB$ intersect the $P$ and $Q$ tangents at $X$ and $Y$, and the two tangents meet at $Z$. Let $MP$ hit $AB$ at $K$ and $\Omega$ again at $T$. Claim: $XP^2=XK^2=XA\cdot XB$ Proof. This is just an angle chase: \[\measuredangle XKP=\measuredangle BKM=\measuredangle MBP=\measuredangle MAP=\measuredangle MPX=\measuredangle KPX\]as desired, where we used $\triangle MKB \sim \triangle MBP$. $\blacksquare$ Now, let $T'=XT\cap \Omega$ be the image of $T$ under inversion at $X$ fixing $P,K$. $XT'KP$ is cyclic since it inverts to line $TPQ$. $T'KQY$ is cyclic as \[\measuredangle YQT'=\measuredangle QTT'=\measuredangle KTX=\measuredangle XKT'=\measuredangle YKT'\]from $\triangle XT'K \sim \triangle XKT$. $YXZT'$ is cyclic since the first two bullets imply $T$ is the miquel point of $PQYX$. Now, to show $(XYZ)$ and $\Omega$ are tangent at $T$, it suffices to show $\measuredangle T'YX=\measuredangle T'QT$. This is true as \[\measuredangle T'YX=\measuredangle T'YK=\measuredangle T'QK=\measuredangle T'QT.\]

Solved with gepassup. First we restate the problem, set the problem on $\triangle PBA$. New Problem Statement: $ABC$ is a triangle, $M$ is the midpoint of the arc $BC$ not containing $A$. The tnagent at $A$ to $(ABC)$ intersect $BC$ at $T$. Take an arbitrary point $Q$ on $AM$. Tangent at $Q$ to $(QBC)$ intersect $AT,BC$ at $X,Y$ respectively. Prove that $(TXY)$ and $(QBC)$ are tangent to each other. Take $\sqrt{bc}$ inversion and reflect the diagram over $AM$. $T^*$ is the point on $(ABC)$ which satisfies $AT^*\parallel BC$. $X^*,Y^*$ are the intersections of the circle passes through $A,Q^*$ and tangent to $(QBC)$ with $AT^*,(ABC)$ respectively. We will show that $(T^*X^*Y^*)$ and $(Q^*BC)$ are tangent. New Problem Statement After $\sqrt{bc}$ inversion: $ABC$ is a triangle and $T$ is the intersection of the parallel line passing through $A$ to $BC$ with $(ABC)$. $M$ is the midpoint of the arc $BC$ not containing $A$. Take an arbitrary point $Q$ on $AM$. The circle passes through $A,Q$ which is tangent to $(QBC)$ intersects $(ABC),AT$ at $X,Y$ respectively. Prove that $(TXY)$ and $(QBC)$ are tangent. Proof: Let $MT$ intersect $(QBC)$ at $N$. Claim: $Q,N,F$ are collinear. Proof: Let $K=AM\cap (QBC)$. Note that $MK=MN$ and $MA=MT$ hence $KN\parallel AT$. By homothety centered at $Q$ which sends $(QBC)$ to $(QAF)$ takes $K$ to $A$. If $N'$ is the point $N$ goes, then $N'A\parallel KN$ and $N'\in (QAF)$ thus $N'=Q$. This yields $Q,N,F$ are collinear.$\square$ Claim: $T,X,Y,N$ are concyclic. Proof: \[\measuredangle YTN=\measuredangle YTM=\measuredangle YAM=\measuredangle YAQ=\measuredangle YXQ=\measuredangle YXN\]Which gives the result.$\square$ Claim: $(TXYN)$ and $(QBCN)$ are tangent. Proof: \[\measuredangle KNX=\measuredangle NKQ+\measuredangle KQN=\frac{\measuredangle A}{2}+\measuredangle C+\measuredangle KQN=\measuredangle NTX+\measuredangle KQN\]Thus, $(TXYN)$ and $(QNBC)$ are tangent to each other as desired.$\blacksquare$

I solved this problem in the contest back in 2014. It's been a while, but since no one seems to have solved it like I did, I guess I will post my solution. I will use inversion on the unit circle centered at $Q$. Also assume WLOG $QA<QB$. I will use $'$ to denote the points mapped by inversion, for example $P'$ is the point mapped by inversion from $P$, etc. Since $\angle APQ=\angle APM=\angle BPM=\angle BPQ$, we have $\angle P'A'Q=\angle APQ=\angle BPQ=\angle P'B'Q$. Also, $QA'>QB'$. $\omega$ is mapped to the circumcircle of $\triangle P'A'B'$, so $\ell_P$ is mapped to the circle that passes through $P'$, $Q$ and is tangent to the circumcircle of $\triangle P'A'B'$. Call this circle $\Gamma$. $\Omega$ is mapped to the line $A'B'$, so $\ell_Q$ is mapped to the line passing through $Q$ and is parellel to line $A'B'$. Call this line $\ell$. Finally, line $AB$ is mapped to the circumcircle of $\triangle QA'B'$. We need to prove that the circle passing through the three intersecting points of $\Gamma$, $\ell$, and the circumcircle of $\triangle QA'B'$ is tangent to the line $A'B'$. From now on, we will not revisit the diagram before inversion. So, for the sake of conveniency, from now on we will rename the points so that we rename $A'$ as simply $A$, $B'$ as simply $B$, etc. Define $X$ as the intersecting point of $\Gamma$ and $\ell$, $Y$ as the intersecting point of $\ell$ and the circumcircle of $\triangle QAB$, and $Z$ as the intersecting point of $\Gamma$ and the circumcircle of $\triangle QAB$. The following facts can be checked easily: (1) $\square QYAB$ is an isosceles trapezoid. This is quite obvious, because the four points are on a circle and $YQ//AB$. (2) $\angle APX=\angle BPQ$. This is simple angle chasing. Consider any point $S$ on the line passing through $P$ and tangent to both $\Gamma$ and the circumcircle of $\triangle PAB$, such that $S$ is closer to $B$ than $A$. Then $\angle APX=\angle PTB-\angle PAB=\angle PXQ-\angle PAB=\angle QPS-\angle BPS=\angle BPQ$. Define $T$ as the intersecting point of line $PX$ and $AB$. We will show that the circumcircle of $\triangle XYZ$ is tangent to $AB$ and the point of tangency is $T$, which is sufficient to solve the problem. We will prove this in two steps. 1. We will prove $TX=TY$. Consider a point $U$ on segment $AB$ such that $AT=BU$. Also define $V$ as the intersecting point of line $PQ$ and $AB$. First we will prove $\angle BQU =\angle AQV$. We see that $$\frac {QBsinBQU}{QAsinAQU}=\frac {BU}{AU}=\frac {AT}{BT}=\frac {APsinAPT}{BPsinBPT}=\frac {APsinBPQ}{BPsinAPQ}=\frac {APsinPBQ\cdot\frac {QB}{PQ}}{BPsinPAQ\cdot\frac {QA}{PQ}}=\frac {AP\cdot QB}{BP\cdot QA}$$so $\frac {sinBQU}{sinAQU}=\frac {AP}{BP}$. Also, $$\frac {QAsinAQV}{QBsinBQV}=\frac {AV}{BV}=\frac {APsinAPQ}{BPsinBPQ}=\frac {APsinPAQ\cdot\frac {QA}{PQ}}{BPsinPBQ\cdot\frac {QB}{PQ}}=\frac {AP\cdot QA}{BP\cdot QB}$$so $\frac {sinAQV}{sinBQV}=\frac {AP}{BP}$. Therefore $\frac {sinBQU}{sinAQU}=\frac {sinAQV}{sinBQV}$, from which it follows $\angle BQU =\angle AQV$. Now, proving $TX=TY$ is simple angle chasing. Define $\angle PAQ=\angle PBQ=\alpha$, $\angle APX=\angle BPQ=\beta$, and also I will denote the angles $\angle PAB, \angle ABP, \angle APB$ as simply $\angle A, \angle B, \angle P$. We see that $$\angle XYT=\angle YTA=\angle QUB(\because \triangle YTA\equiv\triangle QUB)=180-\angle QBA-\angle BQU=180-\angle QBA-\angle AQV=180-(\angle B-\alpha)-(\angle APQ+\angle PAQ)$$$$=180-(\angle B-\alpha)-(\angle P-\beta+\alpha)=180-\angle B-\angle P+\beta=\angle A+\beta=\angle A+\angle APT=\angle PTB=\angle YXT$$so $TX=TY$. 2. We will prove $X,Y,Z,T$ are on a circle, which will solve the problem, since $\angle XYT=\angle YXT=\angle XTB$, so $AB$ is tangent to the circumcircle of $\triangle XYT$. Again this is just angle chasing. We see that $$\angle XZY=\angle QZY-\angle QZX=180-\angle QAY-\angle QPX=180-(\angle YAB-\angle QAB)-\angle QPX=180-(\angle QBA-\angle QAB)-\angle QPX$$$$=180-((\angle B-\alpha)-(\angle A-\alpha))-(\angle P-2\beta)=180-\angle B-\angle P+\angle A+2\beta=2\angle A+2\beta=2\angle PTB=2\angle TXY=180-\angle XTY$$which implies that $X,Y,Z,T$ are on a circle. The above steps solve the problem, and we are done.

Pure angle chasing! Let $C = \ell_P \cap \ell_Q$, $D = \ell_P \cap \overline{AB}$, and $E = \ell_Q \cap \overline{AB}$. I claim that the desired tangency point is the Miquel point of quadrilateral $PQED$. Define $T = (DPX) \cap \Omega$. Notice that $DP=DX$ because $\angle DXP = \angle MBP$. Claim: $T$ lies on $(CPQ)$. Proof: Let $Y = \overline{DT}\cap \overline{PM}$. Then $\measuredangle DTQ = \measuredangle DPX = \measuredangle YXD$, so $DT \cdot DY = DX^2 = DP^2$ by similarity, and it follows that $X$ also lies on $\Omega$ as $D$ lies on the radical axis $\overline{AB}$. Then \[\measuredangle TQC = \measuredangle TYQ = \measuredangle TXD = \measuredangle TPC\]so $T$ lies on $(CPQ)$ too. $\blacksquare$ This characterizes $T$ as our Miquel point. Finally, \[\measuredangle EDT + \measuredangle TYQ = \measuredangle XPT + \measuredangle TYQ = \measuredangle DTP = \measuredangle DXP = \measuredangle ETQ\]so it follows that $(CDE)$ and $\Omega$ are tangent at $T$, as needed.