Henceforth assume $\angle A \neq 60^{\circ}$; we prove the concurrence. Let $L$ denote the center of $\omega$, which is the midpoint of minor arc $BC$. Claim: Let $K$ be the point on $\omega$ such that $\overline{KL} \parallel \overline{AB}$ and $\overline{KC} \parallel \overline{AL}$. Then $\overline{KA}$ is tangent to $\gamma$, and we may put \[ x = KA = LB = LC = LX = LY = KX = KY. \] Proof. By construction, $KA = LB = LC$. Also, $\overline{MO}$ is the perpendicular bisector of $\overline{KL}$ (since the chords $\overline{KL}$, $\overline{AB}$ of $\omega$ are parallel) and so $KXLY$ is a rhombus as well. Moreover, $\overline{KA}$ is tangent to $\gamma$ as well since \[ \measuredangle KAD = \measuredangle KAL = \measuredangle KAC + \measuredangle CAL = \measuredangle KBC + \measuredangle ABK = \measuredangle ABC. \]$\blacksquare$

Up to now we have not used the existence of $Q$; we henceforth do so. Note that $Q \neq O$, since $\angle A \neq 60^{\circ} \implies O \notin \omega$. Moreover, we have $\angle AOM = \angle ACB$ too. Since $O$ and $Q$ both lie inside $\triangle ABC$, this implies that $A$, $M$, $O$, $Q$ are concyclic. As $Q \neq O$ we conclude $\angle CQA = 90^{\circ}$. The main claim is now: Claim: Assuming $Q$ exists, the rhombus $LXKY$ is a square. In particular, $\overline{KX}$ and $\overline{KY}$ are tangent to $\omega$.

We finish by proving that \[ KD = KA \]and hence line $\overline{KD}$ is tangent to $\gamma$. Let $E = \overline{BC} \cap \overline{KL}$. Then \[ LE \cdot LK = LC^2 = LX^2 = \frac{1}{2} LK^2 \]and so $E$ is the midpoint of $\overline{LK}$. Thus $\overline{MXOY}$, $\overline{BC}$, $\overline{KL}$ are concurrent at $E$. As $\overline{DL} \parallel \overline{KC}$, we find that $DLCK$ is a parallelogram, so $KD = CL = KA$ as well. Thus $\overline{KD}$ and $\overline{KA}$ are tangent to $\gamma$.

Here is my solution on contest. I like it a lot even though it has been considered "overkill." First note that since $\angle BAC\neq 60^\circ$, point $Q$ is distinct from point $O$. We will assume this for the rest of the proof. Let $L$ be the midpoint of minor arc $BC$, and set $S=\overline{OM}\cap \overline{BC}$, $P=\overline{OM}\cap \overline{AD}$, $R=\overline{BP}\cap \overline{AS}$. By the Incenter-Excenter Lemma, $L$ is the center of $(BIC)$. Observe that $\angle AQM=\angle C=\angle AOM$ and $\angle ASC=2\angle B=\angle AOC$, so $AOQM$ and $AOSC$ are cyclic. Furthermore, $$\angle SAC = \angle SOC=\angle QOM=\angle QAB,$$so $\overline{AS}$, $\overline{AQ}$ are isogonal. We also know that $$\angle QBC=180^\circ - \angle QCB-\angle BQC = \angle A/2=\angle ABP,$$so $Q$ and $R$ are isogonal conjugates. Lines $AP$ and $BP$ are reflections across $\overline{OM}$, hence $R$ and $D$ are symmetric. This shows that $ARDB$ is cyclic and an isosceles trapezoid. In addition, we can angle chase to get \begin{align*} \angle BRC &= 180^\circ - \angle PBC-\angle RCB = 180^\circ - (\angle B - \angle A/2)-(90^\circ -\angle B) \\ &=90^\circ+\angle A/2=\angle BQC, \end{align*}so $BQRC$ cyclic. [asy][asy] size(375); defaultpen(fontsize(10pt)); pair A, B, C, D, M, O, L, P, Q, X, Y, R, S, T, I; B = dir(195); C = dir(345); O = (0,0); L = dir(270); Q = IP(Line(C, O, 20), CP(L, B), 1); A = IP(Line(Q, rotate(90, Q)*C, 20), CP(O, B), 1); I = incenter(A, B, C); M = midpoint(A--B); D = IP(A--L, B--C); P = extension(O, M, A, D); S = extension(O, M, B, C); R = extension(B, P, A, S); T = 2*S-L; Y = IP(Line(O, M, 20), CP(L, B), 0); X = IP(Line(O, M, 20), CP(L, B), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--Q--C, heavygreen); draw(CP(L, B), blue); draw(A--S, heavygreen); draw(M--Y, purple); draw(B--T, dashed+magenta+linewidth(0.9)); draw(circumcircle(B, A, D), heavycyan); draw(A--L, heavyred); draw(L--T, purple); draw(circumcircle(A, M, O), royalblue+dotted); draw(circumcircle(A, O, C), deepcyan+dotted); draw(D--R--C, deepgreen); draw(A--T--D^^X--T--Y, magenta+linewidth(0.8)); draw(CP(T, A), heavymagenta+dashed); clip((-1.5, 1.2)--(1.5, 1.2)--(1.5,-1.1)--(-1.5,-1.1)--cycle); dot("$A$", A, dir(90)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$D$", D, dir(255)); dot("$L$", L, dir(270)); dot("$I$", I, dir(45)); dot("$M$", M, dir(130)); dot("$S$", S, dir(270)); dot("$R$", R, dir(90)); dot("$T$", T, dir(T)); dot("$X$", X, dir(60)); dot("$Y$", Y, dir(240)); dot("$P$", P, dir(65)); dot("$Q$", Q, dir(270)); dot("$O$", O, dir(270)); draw(rightanglemark(A, Q, O, 2)); draw(rightanglemark(L, S, O, 2)); draw(rightanglemark(D, R, C, 2)); [/asy][/asy] Since $\overline{CR}$, $\overline{CO}$ isogonal, we have that $\angle RCA=90^\circ - \angle A$, so $\overline{CR}$ is an altitude. Thus $\triangle DRC$ is right with circumcenter $S$. Then both $L$ and $S$ lie on the perpendicular bisector of $\overline{RC}$, which proves that $\overline{LS}\parallel \overline{AB}$. Now let $T$ be the reflection of $L$ across $S$; by symmetry across $\overline{OM}$, $T$ lies on $(ABC)$ and line $\overline{BPR}$. We will show $T$ is the desired point of concurrency. Project $$-1=(A, B; M, \infty)\stackrel{S}{=} (R, B; P, T),$$so $P$ lies on the polar of $T$ with respect to both $(ABD)$ and $(BIC)$. Then $\overline{TL}\perp \overline{XPY}$ implies that $\overline{XY}$ is the polar of $T$, so $\overline{TX}$ and $\overline{TY}$ are tangent to $(BIC)$. To finish, Power of a Point at $P$ gives $$PX\cdot PY=PR\cdot PB=PA\cdot PD,$$so $AXDY$ cyclic. Then observe that $$TY=TX=LX=LB=TA,$$so $T$ is the circumcenter of $AXDY$. But $T$ also lies on the radical axis of $(BAD)$ and $(BIC)$! Hence the power of $T$ with respect to $(BAD)$ is $TA^2=TD^2$, which proves that $TA$, $TD$ are tangent to $(BAD)$, as desired.

It is clear that the hypothesis is equivalent to having the foot from $A$ to $CO$ lie on $(BIC)$. Let $J$ be the midpoint of minor arc $BC$ on $\Gamma$, and let $K\neq J$ be the point on $\Gamma$ for which $JK \parallel AB$. Furthermore, let $JK \cap BC = L$ and let $\star$ be the condition that $L$ is the midpoint of $JK$. Lemma. Redefine $Q$ as the foot from $A$ to $CO$. Then $Q \in (BIC) \iff \star$. Proof. Let $L'$ be the intersection of $MO$ and $BC$. Then it is not hard to see that, under $\sqrt{bc}$ inversion, $Q$ gets sent to the reflection $A^*$ of $A$ in $L'$. Then noting that $MO \parallel BA^*$, \begin{align*} Q \in (BIC) &\iff A^* \in (BIC) \iff JB = JA^*\\ &\iff BA^* \perp JL' \iff MO \perp JL'\\ &\iff \star, \end{align*}where the last equivalence follows from the fact that $K$ is the reflection of $J$ about $MO$ since $ABJK$ is an isosceles trapezoid. $\blacksquare$ Lemma. The tangents at $X$ and $Y$ to $(BIC)$ intersect at $K \iff \star$. Proof. It is clear that the tangents from $X$ and $Y$ intersect at $K$ iff $MO$ is the polar of $K$ w.r.t. $(BIC)$. Note that $L$ is the image of $K$ under inversion w.r.t. $(BIC)$. Since $MO$ is perpendicular to $JK$, this is enough to imply that \[ MO \text{\;is\;the\;polar\;of\;} K \text{\;w.r.t.\;} (BIC) \iff MO \text{\;passes\;through\;} L,\]which is clearly equivalent to $\star$. $\blacksquare$ Lemma. The tangents at $A$ and $D$ to $(ABD)$ intersect at $K \iff \star$. Proof. We aim to show that \[AD \text{\;is\;the\;polar\;of\;} K \text{\;w.r.t.\;} (ABD) \iff \star.\]Since $BJ = CJ$, $AKCJ$ is an isosceles trapezoid as well. Thus \[ \measuredangle DAK = \measuredangle JAK = \measuredangle CJA = \measuredangle CBA = \measuredangle DBA, \]implying that $AK$ is tangent to $(ABD)$. Thus by La Hire's Theorem, it suffices to show that \[K \text{\;lies\;on\;the\;polar\;of\;} J \text{\;w.r.t.\;} (ABD) \iff \star.\]Note that $(ABD)$ is orthogonal to $(BIC)$, implying that the polar of $J$ w.r.t. $(ABD)$ is the radical axis of $(BIC)$ and $(ABD)$. Thus by Power of a Point, $K$ lies on the polar of $J$ iff \[ AK^2 = JK^2 - JB^2 \iff JK^2 = 2\cdot JB^2 \iff \frac{JB^2}{JK} = \frac{JK}{2}. \]But by the Shooting Lemma, \[ \frac{JB^2}{JK} = JL, \]so \[ \frac{JB^2}{JK} = \frac{JK}{2} \iff JL = \frac{JK}{2}, \]which is clearly equivalent to $\star$. $\blacksquare$ Having shown the desired equivalences with $\star$, we may conclude the proof.

Here is a proof with no computation. The angle condition implies that $\overline{AQ} \perp \overline{CO}$; in particular; this means that if $E$ is the midpoint of minor arc $BC$ in $\odot(BIC)$, then $\overline{AQ} \cap \overline{CE} \in \odot(BIC)$. Hence, $\sqrt{bc}$ inversion in $A$ tells us that $\overline{AT}, \overline{OM}, \overline{BC}$ are concurrent at a point $L$, where $\{B, T\} \equiv \odot(ADB) \cap \odot(BIC)$, and that $ATDB$ is a harmonic quadrilateral. Moreover, $ATDB$ is an isosceles trapezoid by symmetry; since $E \equiv \overline{AD} \cap \overline{BB}$, we know that $\overline{EL}$ is the polar of $\overline{AD} \cap \overline{BT}$ and is hence parallel to $\overline{AB}$. Hence, if $K$ is the intersection of the tangents to $\odot(ADB)$ at $A$ and $D$, then $\angle AKB = \angle AEB$ by symmetry, and so $K \in \odot(ABC)$. Finally, $\overline{EL} \parallel \overline{AB} \perp \overline{XY}$, and since $L$ is the inverse of $K$ in $\odot(BIC)$, $\overline{XY}$ is the polar of $K$ in $\odot(BIC)$ with midpoint $L$, which solves the problem.

Here's another proof. Define $P$ and $S$ as the intersections of $MO$ and $AC$, $BC$ respectively. Let $K$ be the midpoint of minor arc $BC$, $N$ as the midpoint of $AC$, and $O'$ as the center of $(ABD)$. $T$ is the intersection of $(ABD)$ and $(BIC)$, and $BT$ and $BQ$ meet $\Gamma$ at $Z(\neq B),E(\neq B)$. Notice that since $\angle BAC\neq 60^{\circ}$, $Q$ and $O$ are distinct points, and also $\angle AQM = \angle ACB$ implies $\angle AQC=90^{\circ}$. Claim 1. $\angle O'BK=90^{\circ}$ Proof. $\angle O'BK=\angle ABK-\angle ABO'=\angle ADC-(90^{\circ}-\angle ADB)=90^{\circ}$ $\blacksquare$ Claim 2. The tangents to $\gamma$ at $A$ and $D$ meet on $BT$. Proof. $\omega$ and $\gamma$ are orthogonal circles, so using La Hire's Theorem, we are done. $\blacksquare$ Claim 3. $KN$ and $BQ$ meet on $\Gamma$ Proof. $\angle EBC=\angle QBC=\angle CKN$, so $E$,$N$,$K$ are collinear. $\blacksquare$ Claim 4. $KS\perp XY$ Proof. Note that $\angle OPC=90^\circ -\angle BAC=\angle OCS=\angle OKN$ so $O,K,P,N$ are cyclic, and $OK^2=OC^2=OS\times OP$. Hence $\angle OKP= 180^\circ-\angle ONP =90^\circ$ and $\angle OSK=90^\circ\blacksquare$ Claim 5. The tangents to $\omega$ at $X$ and $Y$ meet at $Z$. Proof. Invert at $K$ with power $KC$. $(XK)$ and $(YK)$ meet at $S$, so the tangents to $\omega$ at $X$ and $Y$ meet at $Z$. $\blacksquare$ Claim 6. The tangents to $\gamma$ at $A$ passes $Z$. Proof. Note that $\triangle ABD$ and $\triangle KSD$ are homothetic $\Longrightarrow$ $(ABD)$ and $(KSD)$ are tangent. Again inverting at $K$ with power $KC$ finishes Claim 6. $\blacksquare$ Claim 7. The tangents to $\gamma$ at $D$ passes $Z$. Proof. This is trivial by Claim 2. and Claim 6. $\blacksquare$ This shows that the four tangent lines meet at $Z$$(\in\Gamma)$, so we are done.

Firstly, note that if $N$ is the midpoint of $AC$, then $\angle AQM=\angle ANM=B$, so $AMNQ$ is cylic. But $(AMN)$ has diameter $AO$, so $Q$ is the foot from $A$ to $CO$. Also, if $L$ is the arc midpoint of $BC$ on $(ABC)$, then $(BIC)$ has center $L$. These two observations allow us to draw an accurate diagram, as well as get rid of $I$. [asy][asy] unitsize(0.25inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -19.346146341463424, xmax = 8.98556097560978, ymin = -21.452146341463425, ymax = 13.136146341463416; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-7.446829268292673,3.267268292682925), 7.2758664615433), linewidth(2) + wrwrwr); draw(circle((-8.252995512378936,-3.9637986239695873), 8.963419359442142), linewidth(2) + wrwrwr); draw((-14.658885134475598,2.305767929049566)--(-0.6236097560975494,0.7410243902438998), linewidth(2) + wrwrwr); draw((-0.6236097560975494,0.7410243902438998)--(-8.81761937476474,10.412838056544231), linewidth(2) + wrwrwr); draw((-8.81761937476474,10.412838056544231)--(-14.658885134475598,2.305767929049566), linewidth(2) + wrwrwr); draw((-10.979020637046352,4.575034647204184)--(-7.446829268292673,3.267268292682925), linewidth(2) + wrwrwr); draw((-7.446829268292673,3.267268292682925)--(-0.6236097560975494,0.7410243902438998), linewidth(2) + wrwrwr); draw(circle((-8.132224321528708,6.840053174613579), 3.6379332307716514), linewidth(2) + wrwrwr); draw((-8.252995512378936,-3.9637986239695873)--(-0.8427502894756254,6.320852024588341), linewidth(2) + wrwrwr); draw((-11.738252254620171,6.3593029927968905)--(-7.446829268292673,3.267268292682925), linewidth(2) + wrwrwr); draw((-7.446829268292673,3.267268292682925)--(-4.547872900927279,1.178526700309378), linewidth(2) + wrwrwr); draw(circle((-11.124452998553817,5.917051412212093), 5.053076183791322), linewidth(2) + wrwrwr); /* dots and labels */ dot((-7.446829268292673,3.267268292682925),dotstyle); label("$O$", (-7.848195121951217,2.653414634146339), NE * labelscalefactor); dot((-0.6236097560975494,0.7410243902438998),dotstyle); label("$C$", (-0.17502439024389035,0.4813170731707289), NE * labelscalefactor); dot((-8.252995512378936,-3.9637986239695873),dotstyle); label("$L$", (-8.296780487804876,-4.429512195121956), NE * labelscalefactor); dot((-14.658885134475598,2.305767929049566),linewidth(4pt) + dotstyle); label("$B$", (-15.21443902439025,2.2520487804878027), NE * labelscalefactor); dot((-10.979020637046352,4.575034647204184),linewidth(4pt) + dotstyle); label("$Q$", (-10.893853658536585,4.7546829268292665), NE * labelscalefactor); dot((-8.81761937476474,10.412838056544231),linewidth(4pt) + dotstyle); label("$A$", (-8.721756097560974,10.60990243902439), NE * labelscalefactor); dot((-11.738252254620171,6.3593029927968905),linewidth(4pt) + dotstyle); label("$M$", (-11.649365853658539,6.549024390243901), NE * labelscalefactor); dot((-0.8427502894756254,6.320852024588341),linewidth(4pt) + dotstyle); label("$P$", (-0.7416585365853544,6.501804878048779), NE * labelscalefactor); dot((-4.547872900927279,1.178526700309378),linewidth(4pt) + dotstyle); label("$F$", (-4.448390243902432,1.3784878048780462), NE * labelscalefactor); dot((-8.472136045757011,1.6160290103748562),linewidth(4pt) + dotstyle); label("$D$", (-8.367609756097558,1.803463414634144), NE * labelscalefactor); dot((-4.720614565431147,5.576931223394066),linewidth(4pt) + dotstyle); label("$N$", (-4.63726829268292,5.769902439024389), NE * labelscalefactor); dot((-6.205037222166256,4.762526688254988),linewidth(4pt) + dotstyle); label("$E$", (-6.101073170731702,4.9435609756097545), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] How to construct diagram: Let $O$ be any point, let $C$ be any point, and let $L$ be on the circle with center $O$ and radius $OC$. Then, let $B$ be the second intersection of $\omega$ (center $L$ passes through $C$) with $(O)$. Finally, let $A$ be the intersection of the perpendicular to $CO$ at $O$ with $(O)$. From this construction alone, we will be able to prove an important lemma. Lemma: $AC^2=2LC^2$ Proof of Lemma: We use complex numbers with $O=0$, $C=1$, and $L=z$. Note then that $\bar{z}=z$. We see that $q$ is real and satisfies $|z-q|=|z-1|$, so we have \begin{align*} (z-q)(1/z-q)=(z-1)(1/z-1)&\implies 1-qz-q/z+q^2=1-z-1/z+1\\&\implies (1/z+z)(1-q)=(1-q)(1+q)\\&\implies q=z+1/z-1. \end{align*}We have that $AQ\perp CO$, so $a-q$ is pure imaginary. Thus, $a-q=-(1/a-q)$, or $a+1/a=2q$. Therefore, \begin{align*} AC^2&=(a-1)(1/a-1)\\&=2-(a+1/a)\\&=2(1-q)\\&=2(2-z-1/z)\\&=2(z-1)(1/z-1)\\&=2LC^2, \end{align*}as desired. $\blacksquare$ We now proceed with the rest of the solution. Let $P$ be on $(ABC)$ such that $PL\parallel AB$ (this is motivated by the fact that we want the pole of $MO$ wrt $\omega$ to lie on $(ABC)$, and the fact that $MO\perp AB$). We claim that $P$ is the desired concurrence. Let $F$ be the midpoint of $LP$. By the above mentioned motivation, we have that $M,O,F$ collinear. Note that $\angle OLP$ is the angle between $OL$ (the perpendicular bisector of $BC$) and $AB$ (since $LP\parallel AB$). This is simply $\pi/2-B$, so since $OL=OP$, we have that $\angle LOP=2B=\angle AOC$. Thus, $LP=AC$, so by the lemma, we have that $LF\cdot LP=LC^2$. Thus, the inverse of $F$ about $\omega$ is $P$, and since the inverse of $(ABC)=(LBC)$ is $BC$, we then have that $F\in BC$. We want to show that the tangents at $X,Y$ to $\omega$ concur at $P$, or that the polar of $P$ wrt $\omega$ is $MO$. But the inverse of $P$ is $F$ in $\omega$, so the polar passes through $F$. By the motivation of the definition of $P$, we have that the polar is parallel to $MO$. Therefore, since $M,O,F$ collinear, the polar of $P$ wrt $\omega$ is $MO$, as desired. We now want to show that the polar of $P$ with respect to $\gamma=(ABD)$ is $AD$, or $AL$ (now we basically got rid of $D$, we'll still use it to make the angle chases cleaner). We see that \[\angle PAD=\angle PAL=\frac{1}{2}\widehat{PL}=\frac{1}{2}\widehat{AC}=\angle ABD,\]so $PA$ is tangent to $\gamma$. Therefore, it suffices to show that the polar of $P$ wrt $\gamma$ passes through $L$. By Lahire, it suffices to show that $P$ is on the polar of $L$. We see that \[\angle LBD=\angle LBC=\angle LAC=\angle BAD,\]so $LB$ is tangent to $\gamma$. Let $E=\gamma\cap\omega$. Since $LB=LE$, we must have that $LE$ is the other tangent, so the polar of $L$ is $BE$, or the radical axis of $\gamma$ and $\omega$. However, we have that the power of $P$ with respect to $\gamma$ is $PA^2$, and the power of $P$ with respect to $\omega$ is $PL^2-LC^2=2LC^2-LC^2=LC^2$. Since $LP=AC$, there is a rotation at $O$ that sends $L\to C$ and $P\to A$, so $LC=PA$, so $P$ has equal power with respect to $\omega$ and $\gamma$, so $P$ is on the radical axis of $\gamma$ and $\omega$, which is the polar of $L$. Therefore, as we noted before, by Lahire, the polar of $P$ wrt $\gamma$ is $AD$. This completes the proof. $\blacksquare$

Solution. It isn't hard to see that $\angle ACB=\angle AQM$ implies that $Q$ lies on $(AOM)$, and thus $\angle AQO=AQC=90^\circ$. Let $S=\overline{AQ}\cap\Gamma$, $A\neq S$; $P=\overline{AQ}\cap (BIC)$, $P\neq Q$; $E=\overline{AD}\cap \Gamma$, $E\neq A$. We start proving the following. Claim 1. $SD$ is tangent to $(ABD)$. Proof. Since $\angle PQC=90^\circ$ we have that $E$ is the midpoint of $\overline{PC}$. Clearly, $AC=SC$, so $\angle PAC=\angle SAC=\angle ASC=\angle B$, hence $$\angle SPE=180^\circ-\angle B-\angle ACE=\dfrac{\angle A}{2}=\angle DCE$$and also $\angle PSE=\angle ACE=\angle EDC$, leading to infer that $\bigtriangleup SPE\cong \bigtriangleup DCE$ and by symmetry $SD\parallel PC$, then $\angle BDS=\angle BCE=\angle BAD$ and the claim follows. $\blacksquare$ Now, let $R=\overline{SD}\cap \Gamma,\ R\neq S$. We have that $SECR$ is an isosceles trapezoid with $SR\parallel EC$, and being $SE=SD$, $DRCE$ must be a parallelogram and $ARCE$ another isosceles trapezoid with $RC\parallel AE$, so $AR=CE=RD$; therefore, $AR$ must be tangent to $(ABD)$ as well. Finally, let $X',\ Y'$ the common points of $(BIC)$ and $O(R,\ RA)$. These circles are congruent because $AR=EC$. Moreover, since $ED\cdot EA=EC^2$, they are orthogonal, too; thus, we conclude that $RX'$ and $RY'$ are tangent to $(BIC)$ and that $RX'EY'$ is a square; hereby, $X'Y'$ is the perpendicular bisector of $\overline{RE}$. Since $BA\parallel RE$ due to $AR=BE$, $OM$ must be the perpendicular bisector of $\overline{RE}$, which forces to have $X=X'$ and $Y=Y'$. We are done. $\blacksquare$

oops realized first 70% was basically evan's solution

Another solution :

Complex bash solution: Settings. Set $\odot(ABC)$ be unit circle with center at origin. Let $N$ be midpoint of $\widehat{BC}$ (arc not containing $A$). Set $n=1$. Then, $c = \overline{b} = 1/b$. Let $T$ be intersection of tangents to $\omega$ at $X$ and $Y$. Step 1. $q = \frac{b^2 - b +1}{b^2}$

Step 2. $\angle AQM=\angle ACB \iff (1-ab)^2 = 2a(1-b)^2$

Step 3. $t=ab$, so $T\in\Gamma$

$t=ab \Longrightarrow \widehat{NT} = \widehat{CA} \Longrightarrow \angle TAD=\angle ABC \Longrightarrow$ $TA$ is tangent to $\gamma$ Step 4. $TA=TD$, so $TD$ is tangent to $\gamma$.

Thankfully trig is here when synthetic fails you. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.35)); dotfactor *= 1.5; pair B = dir(190), C = dir(350), Ma = dir(270), C1 = extension(B,B+dir(270),C,Ma), O = origin, Q = foot(C1,C,O), A = intersectionpoint(Q--Q+dir(C1--Q)*100,unitcircle), D = extension(A,Ma,B,C), P = intersectionpoint(C+dir(D--A)*0.001--C+dir(D--A)*100,unitcircle), M = (A+B)/2, I = incenter(A,B,C); draw(A--B--C--A); draw(unitcircle, heavyblue); draw(arc(Ma,abs(Ma-B),0,180), purple); draw(A--Ma^^C--P, heavyred); draw(A--Q--C^^O--A^^B--Q); draw(Ma--P^^M--O, heavygreen); draw(circumcircle(A,Q,O), magenta); draw(circumcircle(A,B,D), orange); draw(A--P--D, orange+dashed); dot("$A$", A, dir(130)); dot("$B$", B, dir(170)); dot("$C$", C, dir(10)); dot("$N$", Ma, dir(270)); dot("$O$", O, dir(270)); dot("$Q$", Q, dir(260)); dot("$D$", D, dir(230)); dot("$P$", P, dir(45)); dot("$M$", M, dir(170)); dot("$I$", I, dir(50)); [/asy][/asy] Let $N$ be the midpoint of arc $BC$ not containing $A$ and $P$ be the unique point on $\Gamma$ such that $\overline{CP} \parallel \overline{AN}$. For brevity let $r' = NB = NI = NC = \tfrac{a}{2} \sec \tfrac{A}{2}$ be the radius of $\omega$. I claim that $P$ is the desired point of concurrence. We proceed in 3 steps. Step 1: Interpreting the given condition. Note that $$\angle AQM = \angle ACB = \angle AOM,$$so $AOQM$ is cyclic and thus $\angle AQC = 90^\circ$. We compute the length of $\overline{CQ}$ in two ways. Elementary angle chase gives $$\angle QBC = 180^\circ - \angle BQC - \angle QCB = 180^\circ - \angle BIC - \angle OCB = \frac{1}{2} \angle A$$and $$\angle CAQ = 90^\circ - \angle ACQ = 90^\circ - (\angle C - 90^\circ + \angle A) = \angle B.$$By Law of Sines in $\triangle BCQ$, $$CQ = 2 r' \sin \angle QBC = 2r' \sin \frac{A}{2}.$$Also $$CQ = AC \sin \angle CAQ = b\sin B.$$Setting them equal gives $$2r' \sin \frac{A}{2} = a \tan \frac{A}{2} b \sin B \implies \frac{\sin B \cos \frac{A}{2}}{\sin \frac{A}{2}} = \frac{a}{b} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin B} \implies 2 \sin^2 \frac{A}{2} = \sin^2 B.$$By Law of Sines in $\triangle ACN$ we have $$\frac{b}{r'} = \frac{AC}{CN} = \frac{\sin \angle ANC}{\sin \angle CAN} = \frac{\sin B}{\sin \frac{A}{2}} = \sqrt{2}.$$To sum things up, we have reduced the given condition to $\boxed{b = \sqrt{2}r'}$. Step 2: Proving $\overline{PA}$, $\overline{PD}$ tangent to $\gamma$. First note that $$\angle PAD = \angle CNA = \angle CBA,$$so $\overline{PA}$ is a tangent. Now by Ptolemy's in $ANCP$ and Shooting Lemma we have $$CP = \frac{AC \cdot NP - AP \cdot CN}{AN} = \frac{b^2 - (r')^2}{AN} = \frac{(r')^2}{AN} = DN,$$so $CNDP$ is a parallelogram and $PD = CN = PA$ as desired. $\square$ Step 3: Proving $\overline{MO}$ is the polar of $P$ with respect to $\omega$. Observe that $AP = CN = BN$, so $\overline{AB} \parallel \overline{NP}$; whence $\overline{MO}$ is the perpendicular bisector of $\overline{NP}$. We now check that $$(\tfrac{1}{2}NP)(NP) = \tfrac{1}{2}b^2 = (r')^2,$$and we are done by the definition of an inverse. $\square$ Combining the results from steps 2 and 3 finished the problem. $\blacksquare$

New solution with radical axis theorem... please correct me if there is an error. Redefine $P$ as the intersection point of the tangent to $\gamma$ at $A$ and $\Gamma,$ let $L=\overline{AT}\cap\overline{BC},$ and let $N$ be the midpoint of $\overline{AC}.$ Additionally, let $\Omega_1$ and $\Omega_2$ denote the circles with diameters $\overline{PL}$ and $\overline{AC}$ respectively. $\textbf{Claim: }$ $Q$ lies on $\Omega_2$ $\emph{Proof: }$ Note that $\angle AQM=\angle ACB=\angle ANM,$ so quadrilateral $AMQN$ is cyclic. But since $\angle AMO,\angle ANO$ are right, $O$ lies on this circle as well. Therefore, $\angle AQO=90^\circ,$ so $Q$ lies on $\Omega_2.$ $\blacksquare$ $\textbf{Claim: }$ $O$ lies on the radical axis of $\Omega_1$ and $\Omega_2$ $\emph{Proof: }$ Remark that $$\angle PAL=\angle PAD=\angle ABD=\angle ABC=\angle ALC,$$so quadrilateral $APCL$ is an isosceles trapezoid. Since $O$ is the center of $(APCL),$ the claim follows. $\blacksquare$ $\textbf{Claim: }$ $\overline{PX},\overline{PY}$ are tangent to $\omega.$ $\emph{Proof: }$ Since $O$ lies on the radical axis of $\omega$ and $\Omega_2$ (which is $\overline{QC}$) and the radical axis of $\Omega_1$ and $\Omega_2,$ it also lies on the radical axis of $\omega$ and $\Omega_1.$ Thus, it suffices to show that the line through the centers of $\omega,\Omega_1$ is parallel to $\overline{AB}.$ Note that $L$ is the center of $\omega$ and the midpoint of $\overline{PL}$ is the center of $\Omega_1,$ so it suffices to show $\overline{PL}\parallel\overline{AB}.$ But observe that $\angle PLA=\angle CAL=\angle BAL,$ so we are done. $\blacksquare$ $\textbf{Claim: }$ $\overline{PD}$ is tangent to $\gamma$ $\emph{Proof: }$ Let $T=\overline{PL}\cap\overline{BC}.$ By radical axis theorem on $\Gamma,\omega,\Omega_1,$ we know $T$ lies on the radical axis of $\omega,\Omega_1.$ Therefore, $\overline{OT}\perp\overline{PL},$ so $T$ is the midpoint of $\overline{PL}.$ This implies that $\triangle PCT\cong\triangle LDT,$ so $\overline{PD}\parallel\overline{CL}.$ Hence, $\angle PDA=\angle CLA=\angle ABC=\angle ABD,$ as desired. $\blacksquare$

Let $U=(AQM) \cap AC \implies \angle AUM= \angle AQM= \angle ACB \implies UM \parallel BC \implies U$ is the midpoint of $AC \implies Q \in (AMU) \implies$ since $AO$ is a diameter of $(AMU) \implies \angle AQO= 90º$. Thus, $$\angle QCB= \angle OCB= 90º- \angle BAC \implies \angle QCA= \angle ACB - \angle QCB= \angle BAC+ \angle ACB - 90º= 90º- \angle ABC$$so $\angle QAC= \angle ABC \implies \angle AQD= \angle QAC - \angle DAC= \angle ABC- \frac{\angle BAC}{2}$. $(\star)$ On the other hand, $\angle QBC= 180º- \angle BQC - \angle QCB= 180º- \angle BIC - 90º + \angle BAC= 180º- 90º- \frac{\angle BAC}{2}-90º+ \angle BAC= \frac{\angle BAC}{2} \implies \angle QBA= \angle ABC- \angle QBC= \angle ABC- \frac{\angle BAC}{2} \implies$ from $(\star)$, $\angle QBA= \angle QAD$. $(\heartsuit)$ Furthermore, it's well known that $LQ^2=LB^2=LD.LA$, where $L$ is the midpoint of $\widehat{BC}$, and $L$ is the center of $\omega$. Hence, $LQD \sim LAQ \implies \angle QDA= 180º- \angle LDQ= 180º- \angle LQA= \angle QLA+ \angle QAD \implies$ from $(\heartsuit)$, $\angle QLA+ \angle QAD= \angle QLA + \angle QBA= \angle QLI + \angle QBA= 2 \angle QCI+ \angle QBA= 2(90º- \angle ABC- \frac{\angle ACB}{2})+ \angle QBA= 180º- 2 \angle ABC- \angle ACB + \angle QBA= \angle BAC - \angle ABC + \angle QBA \underbrace{=}_{(\heartsuit)} \frac{\angle BAC}{2}= \angle QBC= \angle QBD \implies$ from $(\heartsuit)$, $\angle QBA= \angle QAD$ and $\angle QDA= \angle QBD \implies Q$ is the $B$-humpty point WRT $ABD$. Now, let $N$ be the midpoint of $AD$, $S= \omega \cap \gamma, S \neq B$ and $T= BS \cap AD$ and let $P$ be the intersection of the two tangents to $\gamma$ at $A$ and $D$. Observe that $(A,D;I,I_A)=-1$, where $I_A$ is the $A$-excenter of $ABC$. Therefore, since $\omega$ has diameter $II_A \implies \omega$ is the Apollonius circle WRT $AD,I \implies$ since $B,S \in \gamma \implies \frac{BD}{BA}= \frac{SD}{SA} \implies (A,D;B,S)=-1$, so $B,S,P$ are collinear and $(B,S;T,P)=-1$. Now, note that since $\angle QBC= \frac{\angle BAC}{2}= \angle PDC \implies PD \parallel QB \implies$ let $R= DP \cap AQ \implies$ since $DR \parallel QN \implies Q$ is the midpoint of $AR \implies \angle ARP= \angle AQN= \angle QBA+ \angle QAB \underbrace{=}_{(\heartsuit)} \angle QAN + \angle QAB= \angle BAN= \frac{\angle BAC}{2}= \angle QBD= \angle ABP \implies P \in (ABR)$. Furthermore, since $Q$ is the midpoint of $AR$ and $CQ \perp AR \implies CA=CR \implies \angle CRA= \angle CAR= \angle CAQ = \angle ABC \implies R \in \Gamma \implies P \in \Gamma$. Then, notice that $\angle STBA= \angle ABP= \frac{\angle BAC}{2}= \angle TAB \implies TA=TB$, so $T \in XY \implies$ since $T$ lies on the radical axis of $\omega, \gamma$, $TX.TY=TA.TD \implies AXDY$ is cyclic. Moreover, observe that since $\angle ABP= \frac{\angle BAC}{2}= \angle LAB \implies AB \parallel LP \implies$ since $XY \perp AB \implies XY \perp PL$, and since $LX=LY$, then $PX=PY$. Therefore, since $P$ already lies on the perpendicular bisector of $AD$, and also lies on the perpendicular bisector of $XY$, $P$ is the circumcenter of $(AXDY) \implies PX^2=PY^2=PA^2=PD^2=PB.PS \implies PX,PY$ are both tangent to $\omega$, so we are done. $\blacksquare$

here's a sketch of my solution: first let $S,T=AQ \cap BC,(ABC)$ $P=(BIC) \cap (CAD)$ $S',T' = AQ \cap BC,(ABC)$ $M,N$ be the midpoints of $AD$ and arc $BC$ $C'$ is the antipode of $C$ in $(ABC)$ some observation: $\angle AQO=90$ $C'-S-N$ are collinear $AP-AQ$ are isogonal $C'-D-T'$ are collinear $C'A=C'D$ now let $R \in (ABC )$ such that $CR \parallel AI$ we claim that $R$ is the desired point we have $\angle AC'M=\angle AC'R$ so $CN=RA=RD=$ so $RDNC$ is parallelogram since $S'$ is the midpoint od $CD$ then $R-S'-N$ are collinear and $CN \parallel RD$ so $\angle RDA = \angle CAN=\angle RBA $ so $RA,RD$ are tangent to $(ADB)$ now we have $OS' \perp RN$ so $S'$ is the midpoint of $XY$ so $RXNY$ is parallelogram and then $RY,RX$ are tangent to $(BIC)$

here's my trig bash solution Step1: Tangents to $\gamma$ at $A$ and $D$ meet on $\Gamma$ let $N=BQ\cap AD , L$ the midpoint of $AC$ we claim that $N$ is the midpoint of $AD$ $\angle{C}=\angle{AQM}=\angle{AOM}$ so $A,M,Q,O,L$ are concyclic and $AQ\perp QC$ and so $\angle{QAC}=\angle{B}$ easy to get $\angle{CBQ}=\frac{\angle{A}}{2}$ so $DN.DA=DB^2$ so it suffices to prove that $\frac{AD}{BD}=\sqrt{2}$ we have $\frac{BCsin\frac{\angle{A}}{2}}{cos\frac{\angle{A}}{2}}=CQ=ACsin\angle{B}$ so $\frac{sin\angle{B}}{sin\frac{\angle{A}}{2}}=\frac{BC}{AC.cos\frac{\angle{A}}{2}}=\frac{sin\angle{A}}{sin\angle{B}cos\frac{\angle{A}}{2}}=\frac{2sin\frac{\angle{A}}{2}}{sin\angle{B}}$ $\Longrightarrow \frac{AD}{BC}=\frac{sin\angle{B}}{sin\frac{\angle{A}}{2}}=\sqrt{2}$ as desired Let $F$ be the intersection of tangents to $\gamma$ at $A,D$ $\angle{CAF}=\angle{B}-\frac{\angle{A}}{2}=\angle{ABQ}=\angle{ABN}=\angle{CBF}$ So $F\in \Gamma$ $\blacksquare$ Step2: Tangents to $\omega$ at $X$ and $Y$ meet at $F$ Let $G$ be the centre of $\omega$ which is the midpoint of arc $BC$ Main claim $FG$ and $XY$ meet at the midpoint of $CD$ Proof. $\angle{FDC}=\angle{ADC}-\angle{B}=\frac{\angle{A}}{2}=\angle{DCG}$ $\angle{FCA}=\angle{FBA}=\angle{CBN}=\frac{\angle{A}}{2}=\angle{CAG}$ So $CFDG$ is a parallelogram , in particular $FG$ bisects $CD$ so it suffices to prove that $XY$ bisects $CD$ Let $K$ be the midpoint of $CD , K'=XY\cap CD , Q'$ the reflection of $A$ on $Q$ which lies on $\Gamma$ $\angle{CFD}=\angle{AFC}-\angle{AFD}=\angle{B}=\angle{CAQ}$ So $F, D, Q'$ are collinear So $\angle{AQ'D}=\angle{AQ'F}=\frac{\angle{A}}{2}$ and from $\angle{Q'AD}=\angle{B}-\frac{\angle{A}}{2}$ we get $\angle{ADQ}=180-\angle{B}$ angle chasing gives that $\triangle BNK\sim \triangle ADC$ $\Longrightarrow \frac{BK}{b}=\frac{NK}{CD}=\frac{\frac12 b}{CD}$ so $BK=\frac{\frac12 b^2}{CD}$ on other hand $BK'=\frac{c}{2cos\angle{B}}$ So $BK'=BK \Longleftrightarrow b^2cos\angle{B}=c.CD$ $\Longleftrightarrow \frac{bc.cos\angle{B}}{BD}=c$ $\Longleftrightarrow bcos\angle{B}=BD$ $\Longleftrightarrow AQ=BD$ $\Longleftrightarrow AD=\sqrt{2}AQ$ $\Longleftrightarrow AQ'=\sqrt{2}AD$ $\Longleftrightarrow\frac{sin\angle{B}}{sin\frac{\angle{A}}{2}}=\sqrt{2}$ which we proved in step1 $\blacksquare$ $\angle{FGA}=\angle{FBA}=\angle{CBN}=\frac{\angle{A}}{2}=\angle{BAG}$ So $FG\perp XY$ radical axis theorem on $\Gamma , \omega , (FXY)$ gives that $F,X,G,Y$ are concyclic so $F$ is the pole of $XY$ wrt $\omega$ . done!

Let $N$ be the midpoint of arc $BC$, and let $T$ be the point on $(ABC)$ such that $TC\parallel AN$. Let $R$ be the radius of $(BIC)$. We will show that $T$ is the desired concurrency point. Since \[\measuredangle AQM=\measuredangle ACB=\measuredangle AOM,\]it follows that $A,M,Q,O$ are concyclic, so $\angle AQO=90^\circ$. Also, $\measuredangle TAD=\measuredangle ANC =\measuredangle ABD$, so the tangent to $\gamma$ at $A$ passes through $T$. Now we claim that $AC^2=2R^2$. As $\measuredangle OQN=-\measuredangle OCN=\measuredangle ONC$, we have $CO\cdot CQ = R^2$. Therefore, \[CA^2 = CQ^2-OQ^2+OC^2 = 2OQ\cdot OC+OC^2+OC^2= 2OC\cdot CQ = 2R^2.\]This means that \[\text{Pow}(T,\omega) = TN^2-R^2= AC^2-R^2 = R^2 = TA^2 = \text{Pow}(T,\gamma)\]and hence $T$ lies on the radical axis of the two circles. Let the second intersection of those circles be $D'$. Then since $\omega$ is orthogonal to $\gamma$, it follows that $ND'$ is tangent to $\gamma$. Reflecting over $\overline{OM}$ shows that $TD$ is also tangent to $\gamma$, as desired. Now to show the second part, it suffices to show that $O$ lies on the polar of $T$ with respect to $\omega$, as $XY \perp NT$. By La Hire's theorem, this is equivalent to showing that $T$ lies on the polar of $O$. Let $Q$ be the foot of perpendicular from $T$ to $ON$, and let $N'$ be the $N$-antipode. Then by power of a point, \[NO\cdot NQ= \frac{NN'\cdot NQ}{2}=\frac{NT^2}{2}=\frac{AC^2}{2}=R^2\]as desired.

Very nice! Solved with Psyduck909, Pujnk Let $Q'$ be the isogonal conjugate of $Q$ and $K$ be midpoint of arc $BC$ not containing $A$. Since $\angle BQC + \angle BQ'C = 180 + \angle BAC$ and $\angle BQC = 90 + \frac{A}{2}$, we have that $\angle BQ'C = \angle BQC$ and so $Q' \in (BIC)$. Note that this means $Q'$ is just the reflection of $Q$ across $AI$. See that $\angle QBC = 180 - \angle BQC - \angle QCB = 180 - \angle BIC - \angle OCB = \frac{\angle BAC}{2}$ and $\angle QAD = 90 - \angle QCA - \angle DAC = \angle ABC - \frac{\angle BAC}{2}$, so we obtain $\angle QBA = \angle QAD$ So, we have $\angle Q'BD = \angle QBA = \angle QAD = \angle Q'AD$ , which means $Q'$ lies on $(ABD)$. This means $\angle QBD = \angle Q'BA = \angle Q'DA = \angle QDA$, so this means $Q$ is the $B$ humpty point in $\triangle ABD$. Now, just angle chase to see that $DQ' || AB$ and so $AQ', MO$ meet at a point $S$ on $BC$ because $AQ'DB$ is an isosceles trapezoid. Now, let ray $OM$ meet $(ANK)$ at $Z$. By APMO 2017/2, we know that $ZK$ is diameter in $(ANK)$. Notice that $KS || AB$ and so there is a homothety at $A$ sending $KS$ to $DQ'$, which means $S$ lies on $(ANK)$ as well. So this means $\angle ZSK = 90$, which means $S$ is the midpoint of $XY$ Define $O_1$ to be the center of $\odot (ABD)$. Note that $\measuredangle KBD = \measuredangle KAC= \measuredangle BAD$ which implies that the circles $\odot (ABD)$ and $\odot (BIC)$ are orthogonal. Define $V$ to be the intersection point of the tangents to $\omega$ and $X,Y$, and we will show that $V$ is the desired intersection point.\ Consider the inversion about $(BIC)$ with radius $KB$. Due to the orthogonality, we see that $(ABD)$ remains fixed with $A,D$ swapped; while $V$ moves to $S \in BC$, which gives $V \in \odot (ABC)$. Since $AB \perp XY, KV \perp XY$ we deduce $AB \parallel KV$, which is enough to imply that $\odot(ABD)$ is tangent to $\odot (KSD)$ at $D$. Thus, to finish off, it suffices to show that $\odot (ASK), \odot (ABD)$ are tangent. Let $B'$ be the intersection of $\odot (BIC), \odot (ABD)$ other than $B$. Since the polar $XY$ of $V$ passes through $O_1$, by La Hire's the polar of $O_1$ i.e. $BB'$ passes through $V$. Due to this we have $\measuredangle BB'D=\measuredangle BAK=\measuredangle BVK \Rightarrow DB' \parallel KV \parallel AB$ which gives, due to the concyclicities, that $O-M-O_1$ is also the perpendicular bisector of $DB'$ along with $AB,KV$ i.e. $AB'$ is the reflection of $BD$ about $O-M-O_1$ which is enough to imply that $AB',BD, OM \equiv XY$ concur, or that $A,B',S$ are collinear. Thus, we have $DB'\parallel KS\equiv KV$, and so by homothety the circles $\odot (ABD), \odot (ASK)$ are tangent at $A$, as needed.

[asy][asy] size(400); pair NN = dir(270); pair B = dir(196); pair C = dir(344), O = origin; pair X1 = extension(C, O, (-50, -50), (-50, 50)); pair X2 [] = intersectionpoints(O--X1, circle(NN, abs(B-NN))); pair Q = X2[0]; pair X3 = dir(degrees(C-Q)+90) + Q; pair X4 [] = intersectionpoints(Q--X3, unitcircle); pair A = X4[0]; pair M = (A+B)/2; pair I = incenter(A, B, C), D = extension(A, I, B, C), J = (C+D)/2; pair CC = 2*NN - C; pair R = extension(B, CC, Q, C); pair Z = extension(D, CC, A, J); pair P = extension(NN, J, B, Z); pair X6 = extension(M, O, (50, -50), (50, 50)); pair X5 [] = intersectionpoints(X6--M, circumcircle(B, Z, C)); pair X = X5[1], Y = X5[0]; pair M1 = (A+C)/2; draw(A--B--C--cycle, red); draw(circumcircle(A, B, C), green); draw(circumcircle(A, M, Q), orange+dashed); draw(circumcircle(B, I, C), purple); draw(circumcircle(A, B, D), blue); draw(circumcircle(A, X, D), yellow); draw(circumcircle(R, Q, D), gray+dashed); draw(A--NN, magenta); draw(B--NN--C, cyan); draw(CC--R--C--cycle, brown+linewidth(1)); draw(P--X--NN--Y--cycle, orange+linewidth(1)); draw(M--Y, springgreen); draw(NN--P, pink); draw(A--P--NN--B, lightolive); draw(A--Q, dashed); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$D$", D, SE); dot("$N$", NN, S); dot("$Y$", Y, SE); dot("$J$", J, S); dot("$Z$", Z, SE); dot("$O$", O, N); dot("$Q$", Q, S); dot("$X$", X, N); dot("$I$", I, NE); dot("$M$", M, NW); dot("$M_1$", M1, NE); dot("$R$", R, NW); dot("$P$", P, NE); dot("$C'$", CC, SW); [/asy][/asy] Let $P$ be a point on $(ABC)$ such that $\overline{PN} \parallel \overline{AB}$, where $N$ is the midpoint of minor arc $BC$, and let $M_1$ be the midpoint of $\overline{AC}$. It is well-known that $(BIC)$ has center $N$. We will show that $PX$ and $PY$ are tangents to $\omega$, whilst $PA$ and $PD$ are tangents to $\gamma$. Claim. $PXNY$ is a rhombus. Proof. First, observe that $P$ and $N$ are symmetric around line $\overline{XY}$, because $\overline{XY}$ is the perpendicular bisector of $\overline{AB}$. From this, it follows that $\overline{PN} \perp \overline{XY}$. But since $X$ and $Y$ are two points in a circle centered at $N$, it follows that $\overline{PN}$ bisects $\overline{XY}$. The claim follows. $\blacksquare$ Claim. $\angle AQC = 90^{\circ}$. Proof. Observe that by definition, $$\measuredangle AQM = \measuredangle ACB = \measuredangle AOM,$$which implies that $AMQO$ is cyclic. The claim follows as a corollary. $\blacksquare$ Claim. The triangles $NOC$ and $QNC$ are similar. Proof. This follows because $ON=OC$ and $QN = NC$ by definition. $\blacksquare$ From the second claim we obtain $CO \cdot CQ = CN^2$. But then, $$CN^2 = CO \cdot CQ = CM_1 \cdot CA = \frac 12 CA^2 = \frac 12 PN^2,$$by power of a point $C$ with respect to the circumcircle of cyclic $(AMM_1QO)$. ($CA = PN$ follows because $PANC$ is an isosceles trapezoid, by construction.) But $PN^2 = XY^2$, so it follows that $XY = \sqrt 2 CN = \sqrt 2 YN$. From here, we deduce that $PXNY$ is actually a square, because $XNY$ is isosceles. Hence, it follows that $\overline{PX}$ and $\overline{PY}$ are tangents to $\omega$, finishing the first part. $\square$ Next, we show that $\overline{PA}$ is tangent to $\gamma$. This is simply because $$\measuredangle BAP = \measuredangle BNP = \measuredangle ANC = \measuredangle ABD,$$by the definition of arc midpoint $N$ and the well-known similarity $\triangle ABD \sim \triangle ANC$. $\blacksquare$ The final part of the problem lies in the following claims: Claim. $P$ is the center of $(AXY)$. Because $N$ is the center of $(ABC)$, and $(BIC)$ and $(AXY)$ are images of each other under reflection about $\overline{XY}$, the center of $(AXY)$ is the image of $N$ under reflection about $\overline{XY}$. This point is precisely $P$ by definition, proving the claim. $\blacksquare$ Claim. Let $Z$ be the intersection of $(ABD)$ and $(BIC)$ that is not $B$. Then $Z$ is the reflection of $D$ over line $\overline{OM}$. Proof. Observe that the images of $N$ and $B$ under reflection about $\overline{OM}$ are $P$ and $A$, respectively, by definition. Therefore, since $D$ lies on the line $\overline{AN}$, $D'$ must lie on the line $\overline{BP}$. Moreover, since $ABDD'$ is an isosceles trapezoid, $D'$ must also lie on $(ABD)$. Now we show that $D'=Z$. Because $PA=PX$, and $\overline{PA}$ and $\overline{PX}$ are tangents to $(ABD)$ and $(BIC)$, respectively, $BP$ is the radical axis of $(ABD)$ and $(AIC)$. Then, $D'$ is the intersection of the radical axis of $(ABD)$ and $(AIC)$ and $(ABD)$. However, this point is precisely $Z$, so indeed $D'=Z$, proving the claim. $\blacksquare$ To finish, notice that because $D'$ lies on $(BIC)$, the reflection of $D'$ over $\overline{XY}$, $D$, will lie on $(AXY)$, because $(AXY)$ and $(BIC)$ are images under reflection about $\overline{XY}$. Then, since the center of $(AXY)$ is $P$, we indeed have $$PA=PD,$$so $\overline{PD}$ must be a tangent to $\gamma$. This completes the proof. $\square$ Remarks. There are many more interesting observations that come from this configuration, that are not particularly useful to solving the problem, but are nontrivial corollaries of the problem. Some of them include Line $AQ$ meets $(BIC)$ again at the $C$-antipode $C'$; If we let $R = \overline{BC'} \cap \overline{QC}$, $\triangle RC'C$ is isosceles. Lines $\overline{RN}, \overline{QC'}, \overline{BC}$ concur at the orthocenter of $\triangle RCC'$. The lines $\overline{PN}, \overline{XY}, \overline{BC}, \overline{AL}$ concur, where $L$ is the intersection point of $(AXY)$ and $\omega$ that is not $X$. The lines $\overline{AD}, \overline{BP}, \overline{XY}$ concur. The lines $\overline{AF}$ and $\overline{PD}$ meet on $\Gamma$; The lines $\overline{AJ}$ and $\overline{PC}$ meet on $\omega$. $\overline{AF}$ and $\overline{AJ}$ are isogonal in $\triangle ABC$. and so on...

A very fast proof: Our notations and diagram will be similar to Evan's one. Let $L$ denote the center of $\omega$, which is the midpoint of minor arc $BC$ ; $E = \overline{OM} \cap \overline{BC}$, $K = \overline{LE} \cap \Gamma$ and $E',Q'$ be the reflection of $A$ in points $E,Q$, respectively. Since $\angle AQM = \angle ACB = \angle AOM$, so points $A,M,Q,O$ lie on the circle with diameter $\overline{AO}$. It follows that $Q' \in \Gamma$. [asy][asy] size(9cm); pair L = dir(270); pair B = dir(196); pair C = L*L/B; real r = 2**0.5 * abs(C-L); pair A = IP(unitcircle, CR(C, r)); draw(unitcircle, lightblue); draw(A--B--C--cycle, lightblue); pair M = midpoint(A--B); pair O = origin,Q=foot(A,O,C); pair E = extension(M, O, B, C); pair K = 2*E-L; pair X = IP(M--O, CP(L, B)); pair Y = 2*E-X,Qp=2*Q-A,Ep=2*E-A; draw(CP(L,B,-40,160),heavygreen); pair D = extension(A, L, B, C); draw(A--K, red); draw(B--L, heavycyan); draw(K--D, red); draw(C--L, heavycyan);draw(Y--K--X,red+dashed); draw(A--L, heavycyan); draw(K--C, heavycyan); draw(A--Q--C, orange); draw(K--L, purple); dot("$L$", L, dir(L)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$A$", A, dir(A)); dot("$M$", M, dir(M)); dot("$O$", O, dir(45)); dot("$E$", E, dir(210)); dot("$K$", K, dir(K)); dot("$X$", X, dir(100));dot("$Y$",Y,dir(Y)); dot("$D$", D, dir(245)); dot("$Q$", Q, dir(Q));dot("$E'$",Ep,dir(Ep));dot("$Q'$",Qp,dir(Qp));draw(A--Qp,orange);draw(A--Ep,purple);draw(Y--M,purple); draw(circumcircle(A,B,D),heavygreen); [/asy][/asy] Claim:(Key Claim) $\angle LEO = 90^\circ$. proof: Let $\Phi$ be the inversion at $A$ with squared radius $AB \cdot AC$ composed with reflection in angle bisector of $\angle BAC$. Note that $\overline{AQ},\overline{AE}$ are isogonal wrt $\angle BAC$ (as $\angle QAC = \angle EAB = \angle ABC$) so $\Phi(Q') = E$ (as $Q' \in \Gamma$). Consequently, $\Phi(Q) = E'$ and since $\Phi$ fixes $\omega$, so $\boxed{E' \in \omega}$. Then since $EB=EA=EE'$, so $\overline{LE'}$ is the perpendicular bisector of segment $BE'$ and thus $\overline{LE} \perp \overline{BE'}$ and since $\overline{BE'} \parallel \overline{ME}$, so we are done. $\square$ As $K$ is image of $E$ under inversion wrt $\omega$, so it follows that $K$ is the polar to line $OM$ wrt $\omega$ which implies $\overline{KX},\overline{KY}$ are tangent to $\omega$. Now by shooting lemma $LD \cdot LA = LE \cdot LK$ so quad $ADEK$ is cyclic. So $\angle KAD = \angle XED = \angle ABD$ so $\overline{KA}$ is tangent to $\gamma$. Also, as $\angle AED = 180^\circ - 2 \angle B, \angle LED = \angle B$, so $\overline{LE} \equiv \overline{KE}$ is the external angle bisector of $\angle AED$ which implies $KA = KD$. Hence both of $\overline{KA},\overline{KD}$ are tangent to $\gamma$, which completes the proof of the problem. $\blacksquare$

I skip over a lot of the angle chasing but hopefully this makes sense. Let $N$ be the midpoint of $AC$. $A,Q,M,O$ are cyclic, so $\angle OMA=90$, so $\angle CMA=90$ and $\overline{CO}$ is the radical axis of $(BIC)$, $(AC)$. Let $L$ be the midpoint of $(BIC)$. $\overline{LN}\perp\overline{CO}$, so $\angle CLN=\angle LAC$. Let $P$ be the $L$-dumpty point of $\triangle LAC$. $\angle PCL=\angle PLA=\angle CLN=\angle LAC$. $A,P,O,C$ are cyclic, so this angle condition means that $P\in \overline{BC}$, $P-O-N$, $\angle OPL=90$. $\angle APL=180-\angle CLA$ and $\angle DPL=\angle CLA$. Let $P'$ be the inversion of $P$ in $(BIC)$. $LC^2=LP\cdot LP'$, so $P'$ is on $\Gamma$. $D$ goes to $A$ under the inversion, so by angle chasing $P'$ is the intersection of the $4$ desired lines.

Let $L$ be the center of $\odot(BIC)$. The angle condition implies that $A,Q,O,M$ are concyclic, so as $\angle AQC = \angle AMO = 90^\circ$. Combining with $\angle BQC = 90^\circ + \tfrac{\angle A}{2}$, we have $\angle BQA = 180^\circ - \tfrac{\angle A}{2}$, so $\odot(BQA)$ is tangent to $AD$. Inverting this around $\odot(BIC)$ gives $\odot(BQD)$ is tangent to $AD$. Thus, $Q$ is the Humpty point of $\triangle ABD$, implying that $BQ$ bisects $AD$. Now, let the tangent to $\odot(ABD)$ at $A$ meet $\odot(ABC)$ again at $T$. Since $\angle TAM = \angle B = \angle AMC$, we have that $ATCM$ is an isosceles trapezoid, or $\angle ABT = \angle MBC = \tfrac{\angle A}{2} = \angle QBC$, so $BT$ is $B$-symmedian of $\triangle ABD$, implying that $BT$ is tangent. Finally, we show that $TX$ and $TY$ are tangent. In doing so, let $S$ be the center of $\odot(ABD)$, noticing that $SB$ is tangent to $\odot(BIC)$. Moreover, since $\odot(BIC)$ is the $B$-Apollonius circle of $\triangle BAD$, it must intersect $\odot(ABD)$ again at point $K\in BT$. Then, $S$ is the polar of $BK$, which passes through $T$, so by La Hire's theorem, the polar of $T$ w.r.t. $\odot(BIC)$ passes through $S$. Finally, we note that since $AT=MB$, we have $MT\parallel AB \perp XY$, done.

Such a nice problem ! Let $S$ be the midpoint of arc $BC$ of $(ABC)$ not containing $A$, and let $I_A$ be the excenter opposite $A$. First note that the problem condition gives $\angle AQM=\angle ACB=\angle AOM$, hence $A,Q,M$ and $O$ are concyclic and therefore we have $\angle AQO=\angle AMO=90$. Now let line $(CO)$ cut $(ABC)$ again at point $P$. We have the following claim : Claim : Point $P$ lies on the perpendicular bisector of $AD$. Proof : Note that $\triangle PAC$ has a right angle at $A$ and $Q$ is the foot of the $A$-altitude, hence we have $PA^2=PQ\cdot PC$. It follows that $P$ lies on the radical axis of circle $(BIC)$ and the circle-point $A$. Now let $N$ be the midpoint of $AD$. Since $A,D,I$ and $I_A$ form a harmonic bundle, we have by Mac Laurin that $NA^2=NI\cdot NI_A$, i.e. $N$ also lies on the radical axis. Our $2$ circles are therefore centered at $A$ and $S$ respectively and have line $PN$ as their radical axis, thus $PN\perp \overline{A-D-I}$, and the claim follows. $\square$ We then naturally introduce point $R$, defined as the common point of $(ABC)$ with the perpendicular bisector of $AD$ other than $P$. We claim $R$ is actually the point of interest. Claim : Lines $RA$ and $RD$ are tangent to circle $(BAD)$. Proof : By simple angle chasing we get that line $RA$ is tangent to $(BAD)$. The other tangency follows from the fact that $RA=RD$. $\square$ Lastly, we introduce point $Z$ : the common point of line $MO$ and line $AI$. We then prove the final claim : Claim : Points $Z$ and $N$ are inverse w.r.t $(BIC)$. In particular, $R$ lies on line $PR$ which is exactly the polar of $Z$. Proof : It obviously suffices to prove that $SB^2=SN\cdot SZ$, or in other words that line $SB$ tangent to circle $BZN$. Again this is simple angle chasing : one can easily compute angle $\angle SBZ$ in terms of elements of the triangle, and we can compute angle $\angle SNB=\angle DNB$ by simply noting that $P,B,D$ and $N$ are concyclic because of the two right angles. $\square$ To complete the proof, note that $BARS$ is an isosceles trapezoid by angle chasing, so $R$ and $S$ are symmetric with respect to the perpendicular bisector of $AB$, i.e. line $XY$. It follows that $SR\perp XY$, so $R$ lies on the perpendicular bisector of segment $XY$. This finishes because the common point of the tangents to $(BIC)$ at $X$ and $Y$ must lie on line $PR$ (la hire combined with the last claim), and must also lie on the perpendicular bisector of $XY$. One single point in the plane satisfies these two properties, and this point $R$. $\blacksquare$

Great problem. Let $N$ be the arc midpoint of $BC$ (i.e. on line $AI$) and let $P$ be the midpoint of $AC$. Let $T$ be the point on $\gamma$ such that $NT \parallel AB$. Consequently by a simple angle chase $CT \parallel AN$. Let $MO$ intersect $BC$ at $E$. By the condition, $A$ is the center of the spiral similarity sending $MQ$ to $EC$. $\textbf{Claim.}$ $E$ is the midpoint of $CD$. $\textit{Proof.}$ Constructing parallelogram $AQBQ'$ if the spiral similarity sends $Q'$ to $D$, then we're done. In particular, it remains to prove that $\angle AQB = 180^{\circ} - \frac 12 \angle A$. Indeed, note that $AOQM$ is cyclic, whence $\angle AQB = 270^{\circ} - \angle BIC$ from which it is clear. $\textbf{Claim.}$ $PN \perp CQ$. Remark: I would consider N to be the key point in this problem as without it so many of the properties are just locked away. It took so long to notice though, I had a proof that relied on the fact that AQ=BD (extremely infeasible without bash) $\textit{Proof.}$ Since $AQC$ is a right angle, $NP$ is perpendicular to the radical axis of $(AQC)$ and $(BIC)$, which is $CQ$. A quick angle chase reveals that $\angle NPC = \frac 12 \angle A$, so $\triangle NPC \sim \triangle PAC$. In particular, \[ PC^2 = CN \cdot CA = \frac 12 CA \implies \sqrt{2} PC = CA. \]Now because $\triangle APC \sim \triangle CPD \sim \triangle ABD$, we must have that \[ \frac{PC}{AC} = \frac{PD}{CD} = \frac{BD}{AD} = \sqrt{2}. \]Moreover, \[ \frac{PD}{ED} = \sqrt{2}, \]as $E$ is the midpoint of $CD$ (established earlier.) Therefore, $\triangle ADB \sim \triangle PDE$, so $\angle PEB = \angle ABC$. In particular it is easily computed that $\angle MEP = 90^{\circ}$. In particular, $P$, $E$, and $T$ are collinear. Define $R$ as the intersection of $AE$ with $BIC$. As angle $AEB$ is bisected by $EM$, it can be concluded that $ER=EC=ED$, and hence $ARDB$ is an isosceles trapezoid, i.e. $R$ lies on $(ADB)$. Now it's easy to prove that $PB$ is tangent to $(ABD)$, implying that the circles $(BIC)$ and $(ABD)$ are orthogonal; so, by reflecting across $OM$, we find that $TA$ and $D$ are tangent to $(ABD)$. It remains to tackle $X$ and $Y$. Compute $\frac{PX}{PT} = \frac{PC}{AC} = \frac{1}{\sqrt{2}}$. Since $P$ and $T$ are reflections of each other across $XY$ and $PX=PY$, $PXTY$ is a rhombus; from here the ratio $PX$ to $PT$ forces it to be a square, hence $TX$ and $TY$ are the tangents from $T$. We're done.

Who cares about config issues?? Define $F$ to be the second intersection of $(BIC)$ and $(BAD)$. Note that by orthogonal circles that we have $AFDB$ harmonic Claim: Let $BQ\cap AC=H.$ Then $H$ lies on $(ABD)$. Proof. Notice that we have \[\angle HBD=180^{\circ} -\angle BQ-\angle CQB=180 ^{\circ} - (90^{\circ} +a/2)-(90^{\circ}-1)=a/2=\angle HAD\]proving the lemma. $\blacksquare$ Claim: Line $AN$ is the perpendicular bisector of $FQ$ where $N$ is the arc midpoint. Proof. We first prove that $ACSD$ is a parallelogram. This is since \[\measuredangle DSA=\measuredangle QSF=\measuredangle QBF=\measuredangle HBF=\measuredangle HAF=\measuredangle HAS\]and \[\measuredangle ASC=\measuredangle FSC=\measuredangle FBC=\measuredangle FBD=\measuredangle FAD=\measuredangle SAD.\]Now, this means that under $\sqrt{bc}$ inversion, as $\omega$ inverts to itself we have $Q$ is sent to $Q'$ on $\omega$ with $\angle AQ'M'=\angle BCA$ but this is just $S$ by homothety. Thus $Q$ is sent to $S$ and $\angle BAQ=\angle JAC$ so $\angle QAD=\measuredangle FAD$ and pop gives $AF=AQ$. Also, trivially $N$ is on it as well. $\blacksquare$ Now, note that this implies $Q$ is the reflection of $F$ over $AD$ so it's the humpty point. In particular, we have $\angle BAD=\angle QBD=\angle ABF$ so $AFDB$ is an isosceles trapezoid. Now, by symmetry $G$, the intersection of the tangents at $D,A$ lies on $\Gamma$ and also by pascal $NG\parallel AB,DF$. To finish from here one could note, say that $NG, AS, BC$ concur at $J$ by symmetry and by Reim $FJNB$ cyclic and inverting that around $\omega$ gives $J,G$ are inverses and then as $G$ is the pole of the line through $J$ perpendicular to $GN$ which is $ON$. This is line $XY$ as well, finishing.

Define $R$ as the point on $(ABC)$ such that $RT\parallel AB$. Also, $P=MO\cap BC$. Claim: $AMQO$ and $APOC$ are cyclic Proof: For $AMQO$: \[\angle AQM=\angle ACB=\angle AOM.\]For $APOC$: \[\angle APC=180-\angle APB=180-2\angle MPB=2\angle B=\angle AOC\]$\square$ Therefore: \[\angle BAQ=\angle MOQ=\angle COP=\angle CAP,\]so $AQ$ is isogonal to $AP$. Claim: $\angle TPM=90$ Proof: Consider the transformation $*$ that is an inversion around $A$ with radius $\sqrt bc$, and then reflects across $AT$. This results in $B$ and $C$ being swapped, and $\omega$ being fixed. Therefore: \[Q*=AP\cap \omega.\] Note that as an inversion preserves angles, $Q*$ is the reflection of $A$ about $P$. \[\therefore BQ*\parallel MP,\]by homothety. Also: \[BQ*\perp TP,\]as $PA=PB=PQ*$. \[\therefore \angle TPM=90\]$\square$ Claim: $TXRY$ is a square, implying that $RX$ and $RY$ are tangent to $\omega$ Proof: As $\angle TPM=90$, and $MO$ bisects $RT$ ($RTBA$ is an isosceles trapezoid), $RXTY$ is a rhombus, and $T,P,R$ are collinear. Now, we take an inversion around $(BIC)$. This sends $BC$ to $(ABC)$, implying that $P$ goes to $R$. Therefore: \[\angle TPX=\angle TXR=90,\]implying that $TXRY$ is a square $\square$ Claim: $RA$ is tangent to $\gamma$ Proof: \begin{align} \angle RAD & =\angle RAT\\ & =\angle RAC+\angle TAC\\ & =\angle RBC+\angle TAB\\ & =\angle RBC+\angle RAB\\ &=\angle ABD\\ \end{align}$\square$ Claim: $RA=RD$ Proof: Note that $PT=PR$ and $RATC$ is an isosceles trapezoid. \[\therefore \triangle DTP\cong \triangle CPR,\]\[\therefore DP=PC,\]\[\therefore DRTC\text{ is a parallelogram},\]\[\therefore CT=DR,CT=AR,\]\[\therefore DR=AR\]$\square$ We now finish by noting that $RD$ tangent to $(ABD)$, as desired $\blacksquare$

Let $J$ be the midpoint of minor arc $BC$ with respect to $\Gamma$ and let $K$ be the point on $\Gamma$ such that $\overline{AB} \parallel \overline{JK}$ and $\overline{AJ} \parallel \overline{KC}$. Claim 1: $JXKY$ is a rhombus, and $\overline{KA}$ is tangent to $\gamma$. Proof: Notice that $\overline{MO}$ is the perpendicular bisector of segment $JK$ since $\overline{AB} \parallel \overline{JK}$. Hence, the diagonals of $JXKY$ are perpendicular. Furthermore, $\omega$ is centered at $J$ from Fact 5, so $X$ and $Y$ are symmetric around $\overline{JK}$. This proves that $JXKY$ is a rhombus. As for the second part, note that $AK=JB=JC$ from the defintion of point $K$, so the measure of arc $AK$ equals the measure of arc $JC$. Then, we simply have \[\angle KAD = \angle KAJ = \angle KAC+\angle CAJ = \angle KBC+\angle ABK = \angle ABD. \ \square\] We have \[\angle AQM = \angle ACB = \angle AOM,\] meaning that points $A$, $O$, $Q$, and $M$ are concyclic. Hence, $\angle AQC = 90^\circ$. Claim 2: $JXKY$ is a square as well. Proof: Denote the midpoint of side $AC$ as $N$, and let the radius of $\omega$ be $r$. Notice that $\triangle QJC \sim \triangle JOC$, that $AC=KL$ (due to equal arc lengths), and that $AQON$ is cyclic. Hence, \[JX^2+XK^2 = 2CJ^2 = 2 (CO \cdot CQ) = 2 (CN \cdot CA) = CA^2=JK^2,\] proving our claim. $\square$ Claim 2 directly implies that $\overline{KX}$ and $\overline{KY}$ are tangents to $\omega$. Let $L = \overline{BC} \cap \overline{JK}$. Observe that $\triangle JLC \sim \triangle JCK$, so \[JL \cdot JK = JC^2 = JX^2 = \frac{1}{2}JK^2,\] implying that $L$ is the midpoint of $\overline{JK}$. This also means that $\overline{XY}$ passes through $L$. Thus, $DJCK$ is a parallelogram, meaning $KA = CL = KD$, which is the last desired tangency. $\square$

Let $M$, $N$ be the midpoints of $AB$, $AC$, and $E = AI \cap (ABC)$. Define $P$ as the reflection of $C$ over the perpendicular bisector of $AE$. We claim $P$ is the desired concurrence. First note that $ABEP$ and $APCE$ are isosceles trapezoids, inducing numerous equal lengths along with Incenter-Excenter Lemma. We also have $\angle AQM = \angle AOM = \angle C$, so $AMNOQ$ is cyclic. Then $EXPY$ is a rhombus, as $XY$ is the perpendicular bisector of $PE$. Further, it is a square as \[PE^2 = AC^2 = 2 CN \cdot CA = 2 CO \cdot CQ = 2 CE^2 = 2 YE^2.\] This shows $P$ is the intersection of the tangents at $X$, $Y$. For circle $\gamma$, we have $PA$ is tangent to $\gamma$, as $\angle PAD = \angle AEC = \angle ABD$. It suffices to prove $PD = PA = CE$, or $CEDP$ is a parallelogram, which holds as \[DE = \frac{EC^2}{AE} = \frac{PE \cdot AC - AP \cdot EC}{AE} = PC. \quad \blacksquare\]

Let $N$ be the midpoint of $AC$, let $E = XY \cap BC$, and let $AI$ meet $(ABC)$ again at $M_A$. Then $\measuredangle AQM = \measuredangle ACB = \measuredangle AOM$, so $AMQON$ is cyclic. We first prove that the tangents to $(BIC)$ at $X$ and $Y$ concur on $(ABC)$. Let $A'$ be the reflection of $A$ over $E$. Note that $AQ$ and $AE$ are isogonal in $\angle BAC$, since $\measuredangle BAE = \measuredangle QAC = \measuredangle CBA.$ Then a $\sqrt{bc}$ inversion maps $Q$ to $A'$, so $A'$ lies on $(BIC)$ since $(BIC)$ is fixed under this transformation. Then $BE = EA = EA'$, so $BA' \perp EM_A$, and since $BA' \parallel ME$, we find that $XY \perp EM_A$, so $E$ is the midpoint of $XY$. Let the tangents to $(BIC)$ at $X$ and $Y$ meet at point $P$. Then an inversion about $(BIC)$ swaps $E$ and $P$, and since line $BC$ is sent to $(ABC)$, we find that $P$ lies on $(ABC)$, as desired. We now prove that $PA$ and $PD$ are tangent to $(ABD)$. Since $AB$ and $PE$ are both perpendicular to $ME$, we have $PE \parallel AB$. From Shooting Lemma, $DM_A \cdot AM_A = EM_A \cdot PM_A = CM_A^2$, so $ADEP$ is cyclic. Then $\measuredangle ABD = \measuredangle PEC = \measuredangle PAD$ and $\measuredangle ABD = \measuredangle EAB = \measuredangle PEA = \measuredangle PDA,$ so $PA$ and $PD$ are tangent to $(ABD)$, as desired.

USA TSTST 2018 p3 Nice problem. Here is a sketch of the solution. 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dot((11.907444277290818,8.704802863819081),linewidth(4.pt) + dotstyle); label("$E$", (12.05256018288888,9.019209784701548), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $K$ be a point such that $AL \| KC$ and $KL \| AB$. Then we show the following claim. Claim 1: $KA=LC=LB=LX=LY=KY=KX.$ Proof: we first show that $KXLY$ is a rhombus, then by some angle chase we show that $KA$ is tangent to $\gamma$. $\blacksquare$ Claim 2: $KLXY$ is a square. Proof: By PoP. $\blacksquare$ Claim 3: $KD=KA.$ Proof: By PoP. $\blacksquare$.

$L$ is center of $(BIC),$ $K$ is where $\overline{KL}\| \overline{AB}, \overline{KC} \| \overline{AL}.$ Then note that $KXLY$ is rhombus by lengths and angle chasing. Note taht $LC=KX$ as well. Thus we want to prove $KXLY$ is square, which is simply by showing $KL= \sqrt{2} LC$ by using the fact that $\triangle LQC, \triangle LOC$ are isosceles$.\blacksquare$

Because $\angle A\neq 60$, $Q\neq O$. Let $E$ be the center of $\omega$ (equivalently, the arc midpoint of $BC$). Let $N$ be the midpoint of $AC$. Let $K$ be the point on $\Gamma$ such that $EK\parallel AB$. Let $R = KE\cap BC$. We have \[\angle MQA = \angle BCA = \angle MOA.\]Therefore, $MQOA$ is cyclic. So, $\angle AQC = \angle AQO = \angle AMO = 90^\circ$. Therefore, $Q$ is on $(AC)$. So, $QC$ is the radical axis of $\omega$ and $(AC)$. So, $O$ is on the radical axis of $\omega$ and $(AC)$. Therefore, \[\text{Pow}_\omega (O) = \text{Pow}_{(AC)} (O)\]\[ON^2 - CN^2 = OC^2 - EC^2\]\[OC^2 - 2CN^2 = OC^2 - EC^2\]\[\frac{AC^2}2 = EC^2\]\[AC = \sqrt{2}EC.\] Claim: $R$ is on $MO$ Proof: Because $KA = BE = CE$, $KC\parallel AE$. We have \[\angle RCE = \angle BAE = \angle EKC.\]Therefore, $\triangle RCE \sim \triangle CKE$. We have \[ER = \frac{EC^2}{EK} = \frac{EC^2}{AC} = \frac12AC = \frac12EK.\]So, $R$ is the midpoint of $EK$. So, $R$ is on the perpendicular bisector of $KE$, which is the perpendicular bisector of $AB$, which is $MO$. Claim: $K$ is the pole of $XY$ with respect to $\omega$. Proof: Since $ER \parallel AB \perp XY$, $R$ is the midpoint of $XY$. So, it suffices to show that inversion about $\omega$ swaps $R$ and $K$. Because $R$ is the intersection of $BC$ and $ER$, the inverse of $R$ is the intersection of $\Gamma$ and $ER$, which is $K$. Claim: $KA$ and $KD$ are tangent to $\gamma$ Proof: Because $CD$ bisects $KE$ and $CK \parallel ED$, $CKDE$ is a parallelogram. So, \[\angle KDA = \angle CEA = \angle CBA = \angle DBA.\]So, $KD$ is tangent to $\gamma$. We have \[\angle KAD = \angle KAE = \angle AEC = \angle ABC = \angle ABD.\]So, $KA$ is tangent to $\gamma$. Therefore, the four given tangents concur at $K$, which is on $\Gamma$.

Fun problem! Claim 1. $(ABD)$ is the $B$-Apollonian circle of $\triangle BXY$. (This is true without the angle condition) Proof: Let $MO$ intersect $(ABD)$ at $E$ and $F$ and let $G$ be the center of this circle. It suffices to prove that $GB$ is tangent to $(BIC)$. Notice that $\angle BGD=A\Rightarrow \angle GBI=\angle GBC-\angle IBC=C/2$. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = 9.574376232695768, xmax = 35.39148632139925, ymin = -10.858555769218876, ymax = 2.7438891793940954; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); draw((24.,0.)--(18.,-7.)--(27.,-7.)--cycle, linewidth(0.8) + qqzzcc); /* draw figures */ draw((24.,0.)--(18.,-7.), linewidth(0.8) + qqzzcc); draw((18.,-7.)--(27.,-7.), linewidth(0.8) + qqzzcc); draw((27.,-7.)--(24.,0.), linewidth(0.8) + qqzzcc); draw(circle((22.5,-4.785714285714286), 5.015282766154845), linewidth(0.8)); draw(circle((22.5,-9.800997051869128), 5.300526811985722), linewidth(0.8)); draw((24.,0.)--(22.5,-9.800997051869132), linewidth(0.8)); draw(circle((20.464340212305366,-3.0408630391188822), 4.663447025216459), linewidth(0.8)); draw((22.178594294998966,-4.510223681427682)--(18.,-7.), linewidth(0.8)); draw((18.,-7.)--(27.77768514449523,-9.309444409567337), linewidth(0.8)); draw((16.923587268628975,-0.0059319445391192716)--(27.77768514449523,-9.309444409567337), linewidth(0.8)); draw((20.464340212305366,-3.0408630391188822)--(18.,-7.), linewidth(0.8)); /* dots and labels */ dot((24.,0.),dotstyle); label("$A$", (24.046937093001386,0.11151319338721262), NE * labelscalefactor); dot((18.,-7.),dotstyle); label("$B$", (17.346857327328012,-7.179837864087919), NE * labelscalefactor); dot((27.,-7.),dotstyle); label("$C$", (27.14277921933558,-6.993470602011856), NE * labelscalefactor); dot((22.5,-4.785714285714286),linewidth(4.pt) + dotstyle); label("$O$", (22.24901602983429,-5.301659634583531), NE * labelscalefactor); dot((23.30188567571449,-4.561477622793264),linewidth(4.pt) + dotstyle); label("$I$", (23.342033063275696,-4.4703629998297885), NE * labelscalefactor); dot((22.928680424610725,-7.),linewidth(4.pt) + dotstyle); label("$D$", (22.967552797483922,-7.415498852940784), NE * labelscalefactor); dot((22.5,-9.800997051869132),linewidth(4.pt) + dotstyle); label("$N$", (22.24901602983429,-10.213086720914626), NE * labelscalefactor); dot((21.,-3.5),linewidth(4.pt) + dotstyle); label("$M$", (21.24008084120273,-3.6130069552412494), NE * labelscalefactor); dot((22.178594294998966,-4.510223681427682),linewidth(4.pt) + dotstyle); label("$X$", (22.218592265900373,-4.426306497971932), NE * labelscalefactor); dot((27.77768514449523,-9.309444409567337),linewidth(4.pt) + dotstyle); label("$Y$", (27.82478212731252,-9.217451075013749), NE * labelscalefactor); dot((24.005093155981754,-6.075794133698647),linewidth(4.pt) + dotstyle); label("$E$", (24.346937093001386,-6.190312313925813), NE * labelscalefactor); dot((16.923587268628975,-0.0059319445391192716),linewidth(4.pt) + dotstyle); label("$F$", (16.364854419351074,0.17847081699382077), NE * labelscalefactor); dot((20.464340212305366,-3.0408630391188822),linewidth(4.pt) + dotstyle); label("$G$", (20.51140281890846,-2.9504136857337637), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim 2. $\angle AQC=90^\circ$ Proof: Let $\angle MQB=x$. From ratio lemma, we have: $$\frac{\sin C}{\sin x}=\frac{BQ}{AQ}$$From LOS in $\triangle AQC$ and $\triangle BQC$ we get: $$\frac{BQ}{AQ}=\frac{\sin A\cos A\cos(B+\frac{A}{2}-x)}{\sin B\cos B\cos \frac{A}{2}}$$ Therefore, $$\sin C\sin B\cos B\cos \frac{A}{2}=\sin A\cos A\cos(B+\frac{A}{2}-x)\sin x$$ Now, $\cos(B+\frac{A}{2}-x)\sin x$ is increasing hence the equation has at most $1$ solution so it suffices to check it for $x=B+\frac{A}{2}$, this is left as an exercise. $\square$ Claim 3. $S\in XY\cap BC$ is the midpoint of $XY$. Proof: Let $P$ be the isogonal conjugate of $Q$ wrt $\triangle ABC$. Note that from Claim 2, we know every angle involving point $Q$ so a quick angle chase gives $\angle DBP=\angle QBA=B-\frac{A}{2}$ and $\angle BAP=\angle QAC=B$ so $\angle DAP=B-\frac{A}{2}$, hence $P$ lies on $(BDA)$. Now $\angle ACP=\angle BCQ=90^\circ-B$ so $CP\perp AB$ and since $\angle PBC=B-\frac{A}{2}$, we get that $P$ lies on $(BIC)$. Also, since $\triangle ASB$ is isosceles and $\angle BAS=B$, we have $\overline{A-P-S}$. Now from Claim 1, we know that $BP$ is the $B$-symmedian in $\triangle BXY$. We also have that $\angle BYS=\angle BYX=\angle BPX$ and $\measuredangle BSY= \measuredangle XSB=\measuredangle PCB=\measuredangle BXP$, hence $\triangle BXP\sim \triangle BSY$ so $BP$ and $BS$ are isogonal in $\angle XBY$, hence $S$ is the midpoint of $XY$. $\square$ We can finally finish the problem. Let $\{N,T\}=NS\cap (ABC)$ and notice that $T$ is the symmetric of $N$ in $XY$. PoP gives $XS=SN$, hence $NXTY$ is a square so $TX$ and $TY$ are tangent to $(BXY)$. We also have $\angle NBC=\angle BAN$ so $NB$ and $NP$ are tangent to $(ABD)$. Therefore, reflecting across $MO$ gives that $TA$ and $TD$ are tangent to $(ABD)$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = 7.746808274616942, xmax = 40.886825822447946, ymin = -15.809589035765393, ymax = 1.6511284009118148; /* image dimensions */ pen qqzzcc = rgb(0.,0.6,0.8); draw((23.844013239452686,-3.5824750216139964)--(18.,-11.)--(28.,-11.)--cycle, linewidth(0.8) + qqzzcc); /* draw figures */ draw((23.844013239452686,-3.5824750216139964)--(18.,-11.), linewidth(0.8) + qqzzcc); draw((18.,-11.)--(28.,-11.), linewidth(0.8) + qqzzcc); draw((28.,-11.)--(23.844013239452686,-3.5824750216139964), linewidth(0.8) + qqzzcc); draw(circle((23.,-8.928417144337091), 5.412158121108115), linewidth(0.8)); draw(circle((23.,-14.340575265445192), 6.013272245965939), linewidth(0.8)); draw((23.844013239452686,-3.5824750216139964)--(23.,-14.34057526544519), linewidth(0.8)); draw(circle((20.631040317877716,-7.061995152312377), 4.736059051015864), linewidth(0.8)); draw((20.922006619726343,-7.291237510806998)--(28.971517563892615,-13.633170792582476), linewidth(0.8)); draw((23.844013239452686,-3.5824750216139964)--(25.629360500578684,-11.), linewidth(0.8)); draw((20.922006619726343,-7.291237510806998)--(21.827646586272994,-8.442691697806142), linewidth(0.8)); draw((21.827646586272994,-8.442691697806142)--(23.844013239452686,-3.5824750216139964), linewidth(0.8)); draw((18.,-11.)--(21.827646586272994,-8.442691697806142), linewidth(0.8)); draw((21.827646586272994,-8.442691697806142)--(28.,-11.), linewidth(0.8)); draw((18.,-11.)--(28.971517563892615,-13.633170792582476), linewidth(0.8)); draw((28.,-11.)--(25.07777102245801,-8.697676127383454), linewidth(0.8)); /* dots and labels */ dot((23.844013239452686,-3.5824750216139964),dotstyle); label("$A$", (23.906808299007235,-3.466911851782233), NE * labelscalefactor); dot((18.,-11.),dotstyle); label("$B$", (17.457709426933298,-11.216728136143065), NE * labelscalefactor); dot((28.,-11.),dotstyle); label("$C$", (28.190692819908595,-11.129833998081056), NE * labelscalefactor); dot((23.262080635755428,-11.),linewidth(4.pt) + dotstyle); label("$D$", (23.355417222059533,-11.485354558245793), NE * labelscalefactor); dot((23.470318537144518,-8.345723842955852),linewidth(4.pt) + dotstyle); label("$I$", (23.60990541142001,-8.231496285920063), NE * labelscalefactor); dot((23.,-14.34057526544519),linewidth(4.pt) + dotstyle); label("$N$", (23.05851433447231,-14.226106968633532), NE * labelscalefactor); dot((20.922006619726343,-7.291237510806998),linewidth(4.pt) + dotstyle); label("$M$", (20.582258862970266,-7.114575899282411), NE * labelscalefactor); dot((23.,-8.928417144337091),linewidth(4.pt) + dotstyle); label("$O$", (23.05851433447231,-8.81116382835226), NE * labelscalefactor); dot((22.290367498066242,-8.369322062280599),linewidth(4.pt) + dotstyle); label("$X$", (22.29504976639088,-8.24563451866231), NE * labelscalefactor); dot((28.971517563892615,-13.633170792582476),linewidth(4.pt) + dotstyle); label("$Y$", (28.99966919862955,-13.872651150077314), NE * labelscalefactor); dot((21.827646586272994,-8.442691697806142),linewidth(4.pt) + dotstyle); label("$Q$", (21.61538222395868,-8.954610897383266), NE * labelscalefactor); dot((25.07777102245801,-8.697676127383454),linewidth(4.pt) + dotstyle); label("$P$", (24.52682579805766,-8.995993224805753), NE * labelscalefactor); dot((25.629360500578684,-11.),linewidth(4.pt) + dotstyle); label("$S$", (25.47615213339684,-11.541907489214787), NE * labelscalefactor); dot((28.25872100115737,-7.659424734554809),linewidth(4.pt) + dotstyle); label("$T$", (28.317936914588834,-7.552861114292122), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

v_Enhance wrote: Let $ABC$ be an acute triangle with incenter $I$, circumcenter $O$, and circumcircle $\Gamma$. Let $M$ be the midpoint of $\overline{AB}$. Ray $AI$ meets $\overline{BC}$ at $D$. Denote by $\omega$ and $\gamma$ the circumcircles of $\triangle BIC$ and $\triangle BAD$, respectively. Line $MO$ meets $\omega$ at $X$ and $Y$, while line $CO$ meets $\omega$ at $C$ and $Q$. Assume that $Q$ lies inside $\triangle ABC$ and $\angle AQM = \angle ACB$. Consider the tangents to $\omega$ at $X$ and $Y$ and the tangents to $\gamma$ at $A$ and $D$. Given that $\angle BAC \neq 60^{\circ}$, prove that these four lines are concurrent on $\Gamma$. Evan Chen and Yannick Yao Claim 1 : Let $P$ be the second intersection of line $AQ$ and $\Gamma$. Then $PD$ is tangent with $\gamma$ $\textit{Prove : }$ $\, \,$Let $E$ be the second intersection of line $AD$ and $\Gamma$; $L$ be the second intersection of line $AQ$ and $\omega$ $\, \,$Cuz $\angle APB = \angle ACB = \angle AQM$ so $MQ$ $\|$ $BP$. Which means $Q$ is the midpoint of $AP$ $\, \,$Concurently, cuz $\angle CQL = \angle OQP = 90^\circ$ so $CL$ is diameter of $\omega$ $\, \,$Cuz $OC \perp AP$ so $BC$ bisects $\angle ABP$. And $\angle LBC = 90^\circ$ so $BL$ is the extorior bisector of $\angle ABP$ $\, \,$Therefore, we have : $\hspace{1cm}$ $\dfrac{BP}{AP} = \dfrac{LP}{LA} = \dfrac{LP}{LC} .\dfrac{AE}{LA} .\dfrac{LC}{AE}$ $\hspace{1.8cm}$ $= \dfrac{EP}{AC} .\dfrac{CE}{AC} .\dfrac{LC}{AE} $ (Cuz $\triangle CAE \sim \triangle CLA$ follow shooting lemma, $\triangle LPE \sim \triangle LCA$ ) $\hspace{1.8cm}$ $= \dfrac{EP}{AE}$ (Cuz $CA^2 = CE. CL$ follow shooting lemma ) (*) $\hspace{1.8cm}$ $= \dfrac{EP}{EQ} .\dfrac{EC}{AE} = \dfrac{\sin \angle EQP}{\sin \angle EPQ}. \dfrac{EC}{AE} $ (Use Sin theorem for $\triangle EPQ$ ) $\hspace{1.8cm}$ $= \dfrac{\sin \angle EAC}{\sin \angle ECA} .\dfrac{EC}{AE} $ (Cuz $\angle EQP = 90^\circ - \angle EQC = 90^\circ - \angle OCB - \angle ECB = \angle BAC - \angle ECB = \angle EAC$ ) $\hspace{1.8cm}$ $= \left( \dfrac{EC}{AE} \right)^2 = \left( \dfrac{BD}{AB} \right)^2 $ (Cuz $\triangle ABD \sim \triangle AEC$) $\, \,$So we have that $\dfrac{BP}{AB} = \left( \dfrac{BD}{AB} \right)^2 \Leftrightarrow \dfrac{BP}{BD} = \dfrac{BD}{BA} $ $\, \,$Combine with $\angle PBD = \angle DBA$, we see that $\triangle BPD \sim \triangle BDA$ $\, \,$Lead to $\angle BAD = \angle BDP$, so PD is tangent with $\gamma$ Claim 2 : Let $S$ be the second intersection of line $PD$ and $\Gamma$; $R$ be the second intersection of line $SC$ and $\omega$ $\hspace{1.75cm}$ $W$ be the midpoint of segment $DC$ Then lines $AR$, $ES$, $XY$ concur at $W$ $\textit{Prove : }$ $\, \,$Cuz $\angle DEP = \angle ACP = 180^\circ - \angle ABP = 180^\circ - 2\angle ABD = 180^\circ - 2\angle PDE$ so $\triangle DEP$ is a isosceles triangle at $E$ $\, \,$Lead to $\angle SAE = \angle SPE = \angle EDP = \angle ABC = \angle AEC$, so ASCE is a isosceles trapezoid $\, \,$Combine with $EC$ $\|$ $DP$ and $EC = ER$ we see that $ASRE$, $DSCE$ are parallelograms $\, \,$Therefore $ACRD$ is a paralellogram. Include, we proved that $AR$, $ES$ passes through $W$ $\, \,$Next, let $W'$ be the intersection of lines $XY$, $BC$ and $J$ be the circumcenter of $\triangle ABD$ $\, \,$Cuz $XY$ is the perpendicular bisector of segment $AB$ so $\triangle AW'B$ is a isosceles triangle at $W'$ $\, \,$Lead to $\angle W'AB = \angle W'BA = \angle PBC = \angle PAC$. So $AW'$ is the reflection of $AO$ through the internal bisector of $\angle BAC$ $\, \,$Cuz $\dfrac{BP}{BA} = \dfrac{EP}{EA}$ from (*) so $ABPE$ is a harmonic quadrillateral $\, \,$Reflect through internal bisector of $\angle BAC$, we have $A(dW'CD) = A(APBE) = -1 $ (With $d$ is the lie passes through $A$ and parallel $BC$) $\, \,$Therefore $W'$ is the midpoint of segment $DC$. Which means $XY$ passes through $W$ From these claims, I will prove that $SA$, $SD$ is tangent with $\gamma$ and $SX$, $SY$ is tangent with $\omega$ : $\, \,$Let $K$ be the intersection of lines $AR$, $BS$ $\, \,$Follow Reim theorem for $AD$, $BK$, $CR$, we see that $K$ lie of $\gamma$ $\, \,$Cuz $EB^2 = ED.EA$ follow shooting lemma so $EB$ tangent with $\gamma$, which means $JB$, $JK$ are tangent with $\omega$ $\, \,$ but $J$, $X$, $Y$ all lie on perpendicular bisector of segment $AB$ $\, \,$So $BXKY$ is a harmonic quadrilateral. $\, \,$Lead to $(XYCR) = P(XYCR) = P(YXBK)$, which means $CXRY$ is a harmonic quadrillateral too! $\, \,$Combine with $S$ is the intersection of lines $CR$, $BK$, we see that $SX$, $SY$ tangent with $\omega$ $\, \,$At the end, we ez to see that $SD$, $SA$ are tangent with $\gamma$. Done