Proposed by the Daniel, Nordic Country

We claim the fixed points are the points forming equilateral triangles with $B$ and $C$. Let $O_1$ be the center of $\Omega_1$. Since $O_1$ lies on the external bisector of $XBP$ and the perpendicular bisector of $XP$, it lies on the circumcircle of $BXP$. Let $Z=XY\cap BC$, and let $\Omega_1\cap BC=Z_1,Z_2$. Then, $ZB\cdot ZO_1=ZX\cdot ZP=ZZ_1\cdot ZZ_2$, so $(Z_1,Z_2;B,P)=-1$. This implies $\Omega_1$ is an Apollonian circle of $B$ and $Z$ passing through $X$. Let $JBC$ be an equilateral triangle. We need to show \begin{align*} \frac{BJ}{JZ}&=\frac{BX}{XZ}\\ \frac{BC}{BX}&=\frac{JZ}{XZ}\\ \frac bc&=\frac{JZ}{XZ}. \end{align*} Now, we present two ways to compute $\frac{JZ}{XZ}$.

Clearly, $J$ is fixed as $A$ varies, so the two circles intersect at the two possible points $J$

Claim 1: $XY \cap BC = Z$ is the exsimilicenter of $\Omega_1, \Omega_2$. Proof: If $T$ is the intersection of tangents to $(ABC)$ at $B$ and $C$, then $BX \cap CY$ is the reflection $T'$ of $T$ over $BC$. Furthermore, $BX \parallel CQ, BP \parallel CY$, and if $O_1$ and $O_2$ are the centers of $\Omega_1$ and $\Omega_2$ then $BC$ bisects $\angle{PBX}$ and $\angle{QCY}$, so $PBO_1X$ and $QCO_2Y$ are cyclic. $\triangle{PO_1X} \sim \triangle{YO_2Q}$, and $Z$ is their center of homothety, so $Z$ is the exsimilicenter of $\Omega_1$ and $\Omega_2$. Claim 2: The radical axis of $\Omega_1, \Omega_2$ is the perpendicular bisector of $BC$. Proof: We claim that $T$ lies on the radical axis. This is true because if $X', Y'$ are the reflections of $X$ and $Y$ over $BC$, then $X' \in \Omega_1, Y' \in \Omega_2$ $T = BX' \cap CY'$, and $\triangle{TX'Y'} \cong \triangle{T'XY} \sim \triangle{TQP}$, so $X'Y'PQ$ is cyclic and $TX' \cdot TP = TY' \cdot TQ$. Thus, if $V$ is one of the intersection points of $\Omega_1$ and $\Omega_2$ then $ZV^2 = ZP \cdot ZQ = ZX \cdot ZY$. If $M$ is the midpoint of $BC$, then it suffices to show that $ZX \cdot ZY^2 - ZM^2$ is fixed as $A$ varies. (Since this equals the square of the distance from one intersection point to $BC$.) Claim 3: $M$ has fixed power with respect to $(AXY)$; this is in fact $\frac{3}{4}BC^2$. Proof: Note that $(AXB)$ and $(AYC)$ are tangent to $BC$, so $AM$ is their radical axis, and let their second intersection point be $W \in AM$. Also let $K = (AXY) \cap AM$. Then by Spiral Similarity on $(AXY)$ and $(AYC)$, $\triangle{YKW} \sim \triangle{YXC} \implies KW = YW \cdot \frac{XC}{YC}$. We also know by the Law of Sines on $\triangle{AYW}$ that $YW = YA \cdot \frac{\sin \angle{BAM}}{\sin \angle{ACY}}$ which by the Law of Sines on $\triangle{ACY}$ equals $YC \cdot \frac{\sin \angle{BAM}}{\sin \angle{A}}$. Also, $\triangle{CBX} \sim \triangle{CAB} \implies XC = \frac{BC^2}{AC}$, so putting this all together we get $KW = \frac{BC^2}{2 \cdot AM}$. Note that $MC^2 = AM \cdot MW \implies WM = \frac{BC^2}{4 \cdot AM}$, meaning that $MK = 3 \cdot MW$. Thus, $Pow_{(AXY)}(M) = MK \cdot MA = 3 \cdot MW \cdot MA = \frac{3}{4}BC^2$. *DottedCaculator's alternate proof of this claim is that by Linearity of PoP, $Pow_{(AXY)}(B) - Pow_{M}(B) = Pow_{(AXY)}(M) - Pow_{M}(M) = Pow_{(AXY)}(C) - Pow_{M}(C) = BC^2 - \frac{BC^2}{4}$. Claim 4: $(AXY)$ is centered on the perpendicular bisector of $BC$. Proof: $\triangle{BYC} \sim \triangle{BCA}$ and $\triangle{CBX} \sim \triangle{CAB}$, so $\frac{YB}{XC} = \frac{AC}{AB} \implies$ the Miquel point of $ABCXYZ$ is the reflection $A'$ of $A$ over the perpendicular bisector of $BC$. Thus, $AA'XY$ is cyclic, and its center lies on the perpendicular bisector of $AA'$. *DottedCaculator's alternate proof of this claim is that $Pow_{(AXY)}(B) = Pow_{(AXY)}(C)$. If $O$ is the center of $(AXY)$, then by Claim 3 we know that $MO^2 - OA^2$ is fixed. Note that $ZX \cdot ZY - ZM^2 = (ZO^2 - OA^2) - (ZO^2 - MO^2) = MO^2 - OA^2$, so this is fixed, and we are done. $\square$

We claim that the fixed points are $U$ and $V$ such that $\triangle BUC$ and $\triangle BVC$ are equilateral. We show that $U$ (so that $A$ and $U$ lie on opposite sides of $BC$) lies on $\Omega_{1}$, which suffices. Let the perpendicular bisector of $PX$ meet $BC$ at $O_1$, the center of $\Omega_{1}$. It suffices to show that $O_1U=O_1X$. Claim. $BXPO_1$ is cyclic. Proof. Simply note that $BO_1$ is the external angle bisector of $\angle XBP$ and $O_1X=O_1P.$ Claim. $CXYO_1$ is cyclic. Proof. We have $$\angle PXO_1=\angle PBO_1=\angle BAC=\angle YCO_1,$$which proves our claim. Claim. $O_1X$ is tangent to $\left(AXY\right)$. Proof. Simply note that $$\angle YXO_1=\angle YCB=\angle XAY.$$ Claim. $BC$ is the radical axis of point circle $U$, which we label $\omega_{0}$, and $\left(AXY\right)$. Proof. Note that $$CU^2=BC^2=CX\cdot CA,$$so $C$ lies on the radical axis. Similarly, $B$ lies on the radical axis, proving our claim. We are now ready to finish. Since $O_1$ lies on $BC$, we have $$\text{Pow}\left(O_1, \omega_{0}\right)=\text{Pow}\left(O_1, \left(AXY\right)\right)\Leftrightarrow O_1U^2=O_1X^2\Leftrightarrow O_1U=O_1X,$$as desired.

Complex bash with $b=0$, $c=1$. We claim that $\omega=e^{i\pi/3}$ and its conjugate lie on $\Omega_1$. By symmetry, they lie on $\Omega_2$ as well. Note that $CBX\sim CAB\sim YCB$, so $\frac{x-b}{b-c}=\frac{b-c}{c-y}$, implying that $y=1-\frac1x$. We have $\measuredangle PBC=\measuredangle CBA=\measuredangle CBX$, so $p=r\overline x$, $r\in\mathbb R$. From $PXY$ collinear, we have $\frac{p-x}{x-y}\in\mathbb R$ $\frac{r\overline x-x}{x-1+\frac1x}=\frac{rx-\overline x}{\overline x-1+\frac1{\overline x}}$ $r(\overline x^2-\overline x+1-x^2+x-1)=-x\overline x+\overline x-\frac{\overline x}x+x\overline x-x+\frac x{\overline x}$ $r(\overline x-x)(x+\overline x-1)=(\overline x-x)(1-\frac{x+\overline x}{x\overline x})$ $r=\frac{x\overline x-x-\overline x}{x\overline x(x+\overline x-1)}$ Now it suffices to show that $p,x,\omega,\overline\omega$ are concyclic; their circumcenter obviously lies on $BC$, the perpendicular bisector of $\omega$ and $\overline\omega$. Indeed, $(p-\omega)(x-\overline\omega)=(\frac{x\overline x-x-\overline x}{x(x+\overline x-1)}-\omega)(x-\overline\omega)$ $=\frac{x^2\overline x-x^2-x\overline x+\overline\omega(-x\overline x+x+\overline x)+\omega(x^3+x^2\overline x-x^2)+x^2+x\overline x-x}{x(x+\overline x-1)}$ $=\frac{x^2\overline x-x+\overline\omega(-x\overline x+x+\overline x)+\omega(-x^3-x^2\overline x+x^2)}{x(x+\overline x-1)}$ $=\frac{\overline\omega(-x\overline x+x+\overline x+x^2\overline x-x)+\omega(-x^3-x^2\overline x+x^2+x^2\overline x-x)}{x(x+\overline x-1)}$ $=\frac{\overline\omega(-x\overline x+\overline x+x^2\overline x)+\omega(-x^3+x^2-x)}{x(x+\overline x-1)}$ $=\frac{(x^2-x+1)(\overline{\omega x}-\omega x)}{x(x+\overline x-1)}$ The above quantity divided by its own conjugate is $\frac{(p-\omega)(x-\overline\omega)}{\overline{(p-\omega)(x-\overline\omega)}}=-\frac{\overline x(x^2-x+1)}{x(\overline x^2-\overline x+1)}$ $=-\frac{x-y}{\overline{x-y}}$ $=\frac{(p-x)(\overline\omega-\omega)}{\overline{(p-x)(\overline \omega-\omega)}}$, proving that the desired cross ratio is real.

Here is a different hybrid solution that mostly doesn't use the circumcenters.

My solution from the contest. The answer is points $U$ and $V$ such that $\triangle BUC$ and $\triangle BVC$ are equilateral. We first prove the radical axis of $\Omega_1$ and $\Omega_2$ is the bisector of $BC$. Let $S=BP\cap CQ$, and $T$ be its reflection in $BC$, ie. $BX\cap CY$. Also let $X'$ and $Y'$ be second intersections of $XB$ and $YC$ with $\Omega_1$ and $\Omega_2$ respectively, which are coincidentally reflections of $P$ and $Q$ in $BC$. Now $\triangle SPQ \sim \triangle TYX$ yields: \[TX\cdot TX'=TX\cdot SP=TX\cdot TY\cdot\frac{PQ}{XY}=TY\cdot SQ=TY\cdot TY'\] So the radical axis is the line through $T$ perpendicular to $BC$ better known as its bisector. Now let $W= BQ\cap CP$. I claim it is enough to show that $AUVW$ is cyclic. To show this, we invert through $B$, fixing $U$, $V$, and $C$. Clearly, $Y$ and $A$ are swapped, and I claim that $W$ and $Q$ are also swapped. This would mean that the mentioned concyclicity is equivalent to $YUQV$ and analogously $XUPV$, finishing the problem. Now we are trying to show $\triangle BWC \sim \triangle BCQ$, or $\angle BWC= 180^{\circ}-\angle BAC$. This is due to $BX\cdot CY=BC^2$. Now let $W'$ be the reflection of $W$ in $BC$. It lies on the circumcircle of $\triangle ABC$. So we've proved everything we've talked about so far due to the fact that $X'P \parallel Y'Q$ since $BC$ is their common bisector. We will prove $AUVW$ concyclic by first proving $UV \cap AW'= T$, and then calculating the powers from T. Let $PQ \cap BC=F$ and let $F'$ be its harmonic conjugate wrt. $BC$. From quads $BYXC$ and $BPQC$, we get that lines $AT$ and $SW$ pass through $F'$, so they're reflections of each other over $BC$, meaning $W'$ is on $AT$. Finally, we're left to prove that after drawing a circle $\gamma$ though $BC$ and defining $S$ and $T$ using the tangents and reflecting, $TU\cdot TV$ is the power of $T$ wrt. $\gamma$, which can be proved in two lines using Pythagoras' theorem or complex numbers. (Interestingly, I didn't have enough time to carry this calculation out on the contest... I hope I won't get a huge deduction )

Let $O_1$ be the centre of $\Omega_1$ and let $E \neq F$ be the points such that $\triangle BEC$ and $\triangle BFC$ are equilateral. A quick angle chase shows $BC$ bisects $\angle XBP$. This combined with $O_1$ lying on the perpendicular bisector of $PX$ is enough to show $PXO_1B$ is cyclic. Therefore: $$\measuredangle YXO_1 = \measuredangle PXO_1=\measuredangle PBO_1=\measuredangle PBC= \measuredangle BAC=\measuredangle YAX$$which gives that $O_1X$ tangent to $\odot AXY$. The angle conditions also give $BC$ tangent to $\odot BXA$ and $\odot CYA$ which gives: $$CE^2=BE^2=BC^2=BA \cdot BY=CA \cdot CX$$. This shows $BC$ is the radical axis of $\odot AXY$ and the degenerate circle at $E$. Applying radical axis theorem to the circle $\odot AXY$ and the degenerate circles at $X,E$ we get $BC$, the tangent to $\odot AXY$ at $X$ (which from above is $O_1X$) and the perpendicular bisector of $XE$ concur. As $O_1 \in BC$, this shows that $O_1X=O_1E$ and hence $E \in \Omega_1$. A similar argument gives $F \in \Omega_1$ and thus $E,F$ lies on $\Omega_1$. Similarly they lie on $\Omega_2$ which completes the proof.

Let $\omega$ be the circumcircle of triangle $AXY$, $\Omega$ the circumcircle of $ABC$, and $A'$ the reflection of $A$ in the perpendicular bisector of $BC$. The angle conditions imply that line $BC$ is tangent to the circumcircles of triangles $AXB$ and $AYC$, so from PoP, \[ BY \cdot BA = BC^2 = CX \cdot CA. \]Thus $B$ and $C$ have equal powers with respect to $\omega$, so they are equidistant from its centre. The line connecting the centres of $\omega$ and $\Omega$ is therefore the perpendicular bisector of $BC$, and this implies that $AA'$ is the radical axis. $A'$ lies on $\Omega$, so it also lies on $\omega$. We conclude that $A'$ is the miquel point of $BCXY$. Let $D,E$ be the centres of $\Omega_1, \Omega_2$ and let $U,V$ be the intersections of $\Omega_1$ and $\Omega_2$. Let further $T$ be the second intersection of $PB$ and $\Omega_1$. From $\angle CBX = \angle BAC = \angle TBC$, $T$ is the reflection of $X$ in $BC$, so $\angle BPX = \frac{1}{2}\angle TDX = \angle BDX$ and we have that $P,X,B,D$ are cyclic. Now $\angle YXD = \angle PBD = \angle TBC = \angle YCD$ gives that also $D,Y,X,C$ are cyclic. Note that we also get the cyclic quadrilaterals $QYCE$ and $EYXB$ in the same way. From $\angle DYX = \angle DCX = \angle BCA = \angle A'BC = \angle A'AX = \angle A'YX$ we get that $D,Y,A'$ are collinear. Similarly $E,X,A'$ are collinear. Now $\angle DA'X = \angle YA'X = \angle YAX = \angle BAC = \angle CBX = \angle DBX$ so $A'$ lies on the circle $PXBD$. Similarly $A'$ lies on $QYCE$. Let $Z$ be the intersection of $XY$ and $BC$. Noticing also the circles $PQBE$ and $PQCD$ (this is clear from say $BP \parallel CY$), we get that there is an inversion at $Z$ swapping the pairs $(X,Y)$, $(P,Q)$, $(C,D)$, $(B,E)$. Thus, the circles $A'PDBX$ and $A'QCEY$ swap under this inversion, so $A'$ is fixed. Also $\Omega_1$ and $\Omega_2$ swap, so if $U,V$ are their intersections, these points are also fixed. So the circle through $A',U,V$ has centre $Z$, call this circle $k$. Let $R$ be the second intersection of $k$ and $\omega$ and let $M$ be the midpoint of $BC$. Note that $\omega$ inverts to itself under this inversion at $Z$, so $ZA$ and $ZR$ are tangents to $\omega$. Now quadrilateral $A'RXY$ is harmonic, and projecting this from $A$ onto $BC$ gives $(\infty,AR\cap BC, C,B) = -1$ so $AR \cap BC = M$ and we conclude that $A,R,M$ are collinear. Before the last step, we show that the perpendicular bisector of $BC$ is indeed the radical axis of $\Omega_1$ and $\Omega_2$: Let $T$ and $S$ be the second intersections of $PB$ with $\Omega_1$ and $QC$ with $\Omega_2$, then $(S,T)$ also swap under the inversion, so $PT \cap QS$ lies on the radical axis of $\Omega_1$ and $\Omega_2$ ($PQST$ cyclic), but this point lies on the perpendicular bisector of $BC$, so this line is the radical axis. Now for the last step. From $\angle MRA' = \angle ARA' = \angle AYA' = \angle BYA' = \angle BZA' = \angle MZA'$ we get $MRA'Z$ cyclic. Note that we used that $A'$ was the miquel point again. But this circle can also be described as the locus of points $W$ with $Pow_{\omega}(W)/Pow_k(W) = -1$ since this locus is a circle and the points $R,A'$ and $Z$ all satisfy the condition. So we get that $M$ does as well, which gives $Pow_k(M) = -Pow_{\omega}(M)$. However $Pow_{\omega}(M)$ can easily be calculated (by linearity of PoP) to be $\frac{3}{4}BC^2$, and in particular it does not depend on $A$. So $MU^2 = MV^2 = Pow_k(M)$ does not depend on $A$ and $M,U,V$ are collinear so therefore $U,V$ are fixed. very good problem daniel.

what Okay first degenerate cases to claim that the desired points form equilateral triangles with $B$ and $C$. Assuming you have all the necessary angle stuff (I'll implicitly use but I sorta forgot what I proved) then let's write $S$ is an intersection of $\Omega_1$ and $\Omega_2$. Let $O_1$ and $O_2$ be the circles. Note that $XPBO_1$ cyclic and $YQCO_2$ cyclic. Claim: $\triangle XBS\sim \triangle SCY$. Proof: Let $K$ be the obvious center of homothety sending $\Omega_1\to \Omega_2$. Write $KS\cap \Omega_1=T$; since $\triangle SCY$ is homothetic to $\triangle TBP$ it suffices to show that $B$ is the center of spiral similarity sending $TP\to XS$. This is true by well-known properties of Miquel Point (provable with angle chase) as $B=(XPO_1)\cap KO_1$. Claim: $\triangle BSC$ is equilateral. Proof: Notice by length chase that $XB\cdot YC=BC^2$ and this is also equal to $SB\cdot SC$. But also since $XB$ and $YC$ form an angle which can be bisected by a line parallel to $BC$ this essentially proves $S$ is on the perpendicular bisector of $BC$ hence $SB=SC=BC$ done.

(Solved with islander7) A nice synth sol that does not use the centers (because we're blind and washed or something idk). Define $S$ to be the intersection of $CQ$ and $BP$, $S'$ to be the reflection of $S$ over $BC$ (thus $S'$ is also $BX \cap CY$). Further, define $T$ to be $XY \cap BC$, and let $U$ and $V$ be the points such that $BCU$ and $BCV$ are equilateral, with $U$ on the same side of $BC$ as $X$, $Y$, $P$, and $Q$. We claim the desired intersection points are exactly $U$ and $V$ (it suffices to show that one intersection point is $U$). Claim: $BX \parallel CQ$, $BP \parallel CX$ Proof: Simple angle chase (too lazy to write). Claim: The perpendicular bisector of $BC$ (also $SS'$) is the radical axis of the two circles. Proof: Let $M$ be the $SS' \cap XY$. Then, from parallel lines, we get $\frac{MX}{MQ}=\frac{MY}{MP}=\frac{MS'}{MS}$, so $MX\cdot MP=MY\cdot MQ$, and thus $M$ is on the radical axis. Moreover, the radical axis is perpendicular to $BC$, since both the centers are on $BC$, so the claim is shown. Claim: It suffices to show $TU^2=TX\cdot TY$, or equivalently, $TU$ is tangent to $(UXY)$. Proof: The parallel lines give a homothety centered at $T$ taking the two circles to each other. Thus, if $U'$ is an intersection between the two circles, it is straightforward to show that $TU'^2$ is the geometric mean of the power of $T$ to both circles. Moreover, the parallel lines also give $\frac{TQ}{TX}=\frac{TY}{TP}=\frac{TC}{TB}$, so we have $TP\cdot TQ = TX \cdot TY$. Since $TP\cdot TQ \cdot TX\cdot TY$ is the product of the power of $T$ to both circles, we have that $TU'^2=\sqrt{TP\cdot TQ\cdot TX\cdot TY}=TX\cdot TY$. Moreover, $U'$ is on the perp. bisector of $BC$ from the previous claim, so $U'$ is the intersection of $TU'^2=TX\cdot TY$ and the perp. bisector of $BC$. Since $U$ is on the perp. bisector already, it only remains to show that $TU^2=TX\cdot TY$. From here, we delete all the points in the diagram except for $B$, $C$, $X$, $Y$, $T$, and $U$. The only leftover condition from the original problem is that $\triangle BXC \sim \triangle CBY$. This condition gives us $\frac{BC}{XB}=\frac{YC}{BC}$, so $BX\cdot YC=BC^2=BU\cdot CU$, or $\frac{YC}{CU}=\frac{BU}{BX}$. Now, we also have that $\angle XBC = \angle YCB$ and $\angle UBC = \angle UCB$, so $\angle UBX = \angle YCU$. Thus, we have that $\triangle UXB \sim \triangle YUC$. This similarity gives us $\frac{UX}{UY}=\frac{BU}{CY}=\frac{BX}{CU}=\sqrt{\frac{BX}{CY}}$. Now, we have from LoS that $\frac{BX}{TX}=\frac{\sin T}{\sin A}=\frac{CY}{TY}$ ($T$ is the angle between $XY$ and $BC$, and $A$ is the original angle at $A$), so $\frac{BX}{CY}=\frac{TX}{TY}$. Hence, we have $\frac{UX}{UY}=\sqrt{\frac{TX}{TY}}$, which is equivalent to the condition $TU$ being tangent to $(UXY)$, so we're done.

Nice problem! We will use the following notation: triangles $BCU, BCV$ are equilaterals, the intersection point of $XY$ and $UV$ is $D$ (then $DB=DC$), and $L$ is the point on $BD$ for which $CL=CB$. Let $CY$ and $BD$ intersect at the point $K$. Finally, let the circles $(ABC)$ and $(AXY)$ intersect a second time at point $T$, and $BX$ intersect a second time $(AXY)$ at point $Y'$. Let's prove that the points $U$ and $V$ are the desired ones. It is enough to show that the $PUXV$ quadrilateral is inscribed, then, due to the symmetry of the points $U, V$ onto $BC$, it will follow that this circle is .$\Omega_1$, similar with $\Omega_2$. The condition for angles means that the lines $BP$ and $CY$ are parallel. According to the degree of the point $D$, the condition of $PUXV$ is inscribed is equivalent to $DX*PD=^?UD*DV=BD*DL$. Then it is enough to check the fit of $PBXL$. From the parallel of $BP$ and $CY$, it is enough to prove that $YKXL$ is inscribed. Let's prove that the angles $CYX$ and $BLX$ are equal, from where the desired fit will follow. By counting angles, for example, it is easy to check that the lines $AT, YY', BC$ are parallel. Then, from symmetry with respect to the perpendicular bisector to $BC$, the angles $CYD, BY'D$ are equal. Then the condition of equality of the angles $CYX$ and $BLX$ is now equivalent to the $DXY'L$ is inscribed (?) or the equality of $BD*BL=^?BX*BY'=BY*BA$. But note that both parts of the checked equality $BD*BL=BY*BA$ are equal to $BC^2$. For the left part, this is true from the equality of the angles $BCY$ and $BAC$, and for the right part it is true from the equalities $BD=DC$ and $CB=CL$, that is, the following angles are equal: $BLC = LBC = BCD$. Then we solved the problem.

Points moved to 7 The fixed points are the points $R_1,R_2$ such that $BCR_1$ and $BCR_2$ are equilateral triangles. It suffices to fix $A$, define $X,Y,P,Q$ as before, and show that $XPR_1R_2$ is cyclic ($YQR_1R_2$ cyclic follows similarly), whence $\Omega_1=(XPR_1R_2)$ since $\overline{BC}$ is the perpendicular bisector of $\overline{R_1R_2}$. Construct points $B',C' \in (ABC)$ such that $BC=BC' $ and $CB=CB' $; the angle condition implies that $B,X,B'$ and $C,Y,C'$ are collinear. Hence we can instead view the problem as fixing a choice of $(BCB'C')$ and moving $A$ on its circumference, defining $X=\overline{AC} \cap \overline{BB'}$ and $Y$ similarly; clearly as $A$ varies $X$ hits every point on line $\overline{BB'}$. Let $T=\overline{BB'} \cap \overline{CC'}$ and note that $TB=TC$ by symmetry. Let $\ell_B$ and $\ell_C$ denote the tangents to $(BCB'C')$ at $B$ and $C$ respectively. We have $\measuredangle BCC'=\measuredangle CC'B=\measuredangle (\overline{CB}, \ell_B)$, so $\ell_B$ (resp. $\ell_C$) is parallel to $\overline{CC'}$ (resp. $\overline{BB'}$). We now use moving points with the following setup, essentially aiming to prove a converse of the claim. Delete $A$ and instead move $X$ linearly on $\overline{BB'}$. Let $P_1$ be the point on $\overline{CC'}$ such that $XP_1R_1R_2$ is cyclic, and let $P'$ be its reflection over $\overline{BC}$, so $XP'R_1R_2$ is cyclic and $P'$ lies on $\ell_C$.. Let $Y'=\overline{XP'} \cap \overline{CC'}$. It suffices to show that $\measuredangle (\overline{BY'},\overline{CX})=\measuredangle BB'C$, whence $A':=\overline{BY'} \cap \overline{CX}$ will lie on $(BCB'C')$. If true, then $A'$ can clearly hit every point on $(BCB'C')$, which implies the forward result. By power of a point, $TP_1\cdot TX=TR_1\cdot TR_2$, which is fixed, so $P_1$ moves with degree $1$ and hence $P'$ does as well since $\overline{BC}$ is fixed. Thus $\overline{XP'}$ has degree at most $2$ and so does $Y'$. But note that when $X=T$ we have $P_1=\infty_{\overline{BB'}}$ and thus $P'=\infty_{\overline{CC'}}$, so $\overline{XP'}=\overline{T\infty_{\overline{CC'}}}=\overline{CC'}$. Thus by strong Zack's lemma we actually have $\deg Y' \leq 1$. Thus both $\overline{BY'}$ and $\overline{CX}$ have degree (at most) $1$. To show the desired angle equality, we will show that $\tan\measuredangle (\overline{BY'},\overline{CX})=\tan \measuredangle BB'C$; note the latter is constant. $\tan \measuredangle(\overline{BC},\overline{BY'})$ and $\tan(\overline{BC},\overline{CX})$ are quotients of linear polynomials (think slope), so by tangent subtraction $\tan\measuredangle (\overline{BY'},\overline{CX})$ is a quotient of quadratic polynomials, and $\tan\measuredangle (\overline{BY'},\overline{CX})=\tan \measuredangle BB'C$ is (upon cross-multiplication) a polynomial identity of degree at most $2$. Hence it suffices to establish that $\measuredangle (\overline{BY'},\overline{CX})=\measuredangle BB'C$ for three cases of $X$: When $X=B$, we have $\overline{XP'}=\ell_B$, so $Y'=\infty_{\ell_B}$ and $\measuredangle(\overline{BY'},\overline{CX})=\measuredangle(\ell_B, \overline{CB})=\measuredangle BB'C$. When $X=\infty_{\overline{BB'}}=\infty_{\ell_C}$, we have $P_1=T$, so $P'=\ell_B \cap \ell_C$ and $Y'=\ell_C \cap \overline{CC'}=C$. Thus $\measuredangle(\overline{BY'},\overline{CX})=\measuredangle(\overline{BC},\ell_C)=\measuredangle BB'C$. When $X=B'$, because $CB'=CB=CR_1=CR_2$ we have $P_1=B \implies P'=B$ and $Y'=T$, so $\measuredangle(\overline{BY'},\overline{CX})=\measuredangle(\overline{BT},\overline{CB'})=\measuredangle BB'C$. This finishes the problem. $\blacksquare$

IAmTheHazard wrote: Points moved to 7 The fixed points are the points $R_1,R_2$ such that $BCR_1$ and $BCR_2$ are equilateral triangles. It suffices to fix $A$, define $X,Y,P,Q$ as before, and show that $XPR_1R_2$ is cyclic ($YQR_1R_2$ cyclic follows similarly), whence $\Omega_1=(XPR_1R_2)$ since $\overline{BC}$ is the perpendicular bisector of $\overline{R_1R_2}$. Construct points $B',C' \in (ABC)$ such that $BC=BC' $ and $CB=CB' $; the angle condition implies that $B,X,B'$ and $C,Y,C'$ are collinear. Hence we can instead view the problem as fixing a choice of $(BCB'C')$ and moving $A$ on its circumference, defining $X=\overline{AC} \cap \overline{BB'}$ and $Y$ similarly; clearly as $A$ varies $X$ hits every point on line $\overline{BB'}$. Let $T=\overline{BB'} \cap \overline{CC'}$ and note that $TB=TC$ by symmetry. Let $\ell_B$ and $\ell_C$ denote the tangents to $(BCB'C')$ at $B$ and $C$ respectively. We have $\measuredangle BCC'=\measuredangle CC'B=\measuredangle (\overline{CB}, \ell_B)$, so $\ell_B$ (resp. $\ell_C$) is parallel to $\overline{CC'}$ (resp. $\overline{BB'}$). We now use moving points with the following setup, essentially aiming to prove a converse of the claim. Delete $A$ and instead move $X$ linearly on $\overline{BB'}$. Let $P_1$ be the point on $\overline{CC'}$ such that $XP_1R_1R_2$ is cyclic, and let $P'$ be its reflection over $\overline{BC}$, so $XP'R_1R_2$ is cyclic and $P'$ lies on $\ell_C$.. Let $Y'=\overline{XP'} \cap \overline{CC'}$. It suffices to show that $\measuredangle (\overline{BY'},\overline{CX})=\measuredangle BB'C$, whence $A':=\overline{BY'} \cap \overline{CX}$ will lie on $(BCB'C')$. If true, then $A'$ can clearly hit every point on $(BCB'C')$, which implies the forward result. By power of a point, $TP_1\cdot TX=TR_1\cdot TR_2$, which is fixed, so $P_1$ moves with degree $1$ and hence $P'$ does as well since $\overline{BC}$ is fixed. Thus $\overline{XP'}$ has degree at most $2$ and so does $Y'$. But note that when $X=T$ we have $P_1=\infty_{\overline{BB'}}$ and thus $P'=\infty_{\overline{CC'}}$, so $\overline{XP'}=\overline{T\infty_{\overline{CC'}}}=\overline{CC'}$. Thus by strong Zack's lemma we actually have $\deg Y' \leq 1$. Thus both $\overline{BY'}$ and $\overline{CX}$ have degree (at most) $1$. To show the desired angle equality, we will show that $\tan\measuredangle (\overline{BY'},\overline{CX})=\tan \measuredangle BB'C$; note the latter is constant. $\tan \measuredangle(\overline{BC},\overline{BY'})$ and $\tan(\overline{BC},\overline{CX})$ are quotients of linear polynomials (think slope), so by tangent subtraction $\tan\measuredangle (\overline{BY'},\overline{CX})$ is a quotient of quadratic polynomials, and $\tan\measuredangle (\overline{BY'},\overline{CX})=\tan \measuredangle BB'C$ is (upon cross-multiplication) a polynomial identity of degree at most $2$. Hence it suffices to establish that $\measuredangle (\overline{BY'},\overline{CX})=\measuredangle BB'C$ for three cases of $X$: When $X=B$, we have $\overline{XP'}=\ell_B$, so $Y'=\infty_{\ell_B}$ and $\measuredangle(\overline{BY'},\overline{CX})=\measuredangle(\ell_B, \overline{CB})=\measuredangle BB'C$. When $X=\infty_{\overline{BB'}}=\infty_{\ell_C}$, we have $P_1=T$, so $P'=\ell_B \cap \ell_C$ and $Y'=\ell_C \cap \overline{CC'}=C$. Thus $\measuredangle(\overline{BY'},\overline{CX})=\measuredangle(\overline{BC},\ell_C)=\measuredangle BB'C$. When $X=B'$, because $CB'=CB=CR_1=CR_2$ we have $P_1=B \implies P'=B$ and $Y'=T$, so $\measuredangle(\overline{BY'},\overline{CX})=\measuredangle(\overline{BT},\overline{CB'})=\measuredangle BB'C$. This finishes the problem. $\blacksquare$ I'd like to point out you can actually simplify the moving points approach a lot. Fix $(ABC)$ and move $A$ on it. As $BA,CA$ both have deg 1, $X,Y$ have deg $1$ as well. Now by the inversion argument, $P_1,$ and hence $P',$ has deg 1. Now we want to show $X-Y-P$ which has deg at most $3$ and then it suffies to check four cases; then $A=B,C$ and the arc midpoints of $BC$ suffies( the arc midpoints case is just an simple angle chasing).

HAHAHA Won't post full sol since im too lazy but here's a sketch. We add the miquel point $G$ of $BYXC$. We then angle chase with this to get $GPBOX$ is cyclic where $O=GY\cap BC$. This gives that $O$ is one of the circumcenters we want. This allows us to "delete P" and simplify to just $X,Y$. Now, we use radical axis on $(AGXY)$ at the point circle at $A'$ with $A'BC$ equilateral which $B,C$ lie on from the angle conditions and thus $O$ lies on it but $OX$ is tangent to $(AGXY)$ done. Remark. I was stuck after deleting $P,Q$ but then saw the words point circles when scrolling the thread and solved it a minute later :sobbb:

[asy][asy] //24rmm5 //setup; size(10cm); 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(234,218,255);// blu1 lighter //defn pair U,B,V,C,O; U=(0,0.4); B=(-1,0); V=-U; C=-B; O=2*circumcenter(B,V,C)-V; real r=distance(O,B); pair A=O+r*dir(143.4); /*actually works well*/ pair A1,X,Y,P,Q,R,Mb,Mc; A1=2*foot(A,U,V)-A; X=extension(B,U,A,C); Y=extension(C,U,A,B); P=extension(B,V,X,Y); Q=extension(C,V,X,Y); R=extension(X,Y,B,C); Mb=extension(A1,Y,B,C); Mc=extension(A1,X,B,C); //draw filldraw(A--B--U--C--cycle,blu1+opacity(.3),blu); filldraw(B--P--X--cycle,lightpurple+opacity(.3),purple); filldraw(C--Q--Y--cycle,lightpurple+opacity(.3),purple); draw(Q--R,purple); draw(Mb--R,blu); draw(Mb--A1--Mc,magenta); draw(circle(O,r),blu); draw(circumcircle(B,X,P)^^circumcircle(C,Y,Q),purple); //label void pt(string s,pair P,pair v,pen a){filldraw(circle(P,.04),a,linewidth(.3)); label(s,P,v);} pair points[]={A,B,C,A1,X,Y,P,Q,R,Mb,Mc}; string labels[]={"$A$","$B$", "$C$", "$A'$", "$X$", "$Y$", "$P$", "$Q$", "$R$", "$M_b$", "$M_c$"}; real dirs[]={130,-90,-80,50,100,80,60,100,-90,-90,-90}; pen colors[]={blu,blu,blu,blu, purple,purple,purple,purple,purple, magenta,magenta}; for (int i=0; i< 11; ++i) { pt(labels[i], points[i], dir(dirs[i]), colors[i]); } [/asy][/asy] Define: $M_b=(BXP)\cap\overline{BC}$ and $M_c$ similarly. These are the centers of the desired circles because $M_b$ (as $\overline{BC}$ bisects $\angle XBP$) is the midpoint of one of the arcs $PX$, and similarly for $M_c$. $\gamma_b$ be the circle at $M_b$ through $X$ and similarly for $\gamma_c$. $M=(B+C)/2$. The wording of the problem statement implies that $M$ should have a fixed power wrt. each of these $\gamma$'s, which is stated in earlier posts to be $\boxed{\frac34 a^2}$. (Here, we use $a=BC$, etc. for ease of computation.) $R=\overline{XYPQ}\cap\overline{BC}$; $A'$ be the reflection of $A$ in the perpendicular bisector of $\overline{BC}$. Claim 1: $A'$ is the Miquel point of $BCXY$, so $AA'XY$ cyclic. Proof: Check that that $A'B/BY=a^2/bc=A'C/CX$, triangle similarity follows using equal angles.$\qquad\square$ Claim 2: Define $A'$ as the reflection of $A$ in the perpendicular bisector of $\overline{BC}$. Then $A'\in\overline{M_bY},\overline{M_cX}$. Proof: We assert that $CXYM_b$ is cyclic, so that the claim will follow by Reim converse: \[\measuredangle RCQ=\measuredangle M_bBX=\measuredangle M_bPX.\qquad\qquad\square\]It immediately follows (by angles, say) that $A'\in(BXP),(CXQ)$. As a result, using power of a point and the so-called `shooting lemma' we may obtain \[M_bX^2=M_bY\cdot M_bA'=M_bM_c\cdot M_bC.\]The rest of the problem is computation, but we make use of directed lengths as well as $BM_bY\overset+\sim AA'Y$ and the similar $CM_cX\overset+\sim AA'X$: \begin{align*} & M_bM^2-M_bM_c\cdot M_bC = -3a^2/4\\ &\iff\frac12(M_bB^2+M_bC^2)-MB^2-M_bM_c\cdot M_bC =-3a^2/4\\ &\iff M_bB^2+(M_bB+BC)^2-(M_bB+BC+CM_c)(M_bB+BC)+BC^2 =0\\ &\iff \frac{AA'}{M_bB}+\frac{AA'}{BC}+\frac{AA'}{CM_c} =0\\ &\iff \frac{AA'}{BC} =\frac{AY}{YB}-\frac{AX}{XC}=\frac{AB}{YB}-\frac{AC}{XC}=\frac{b^2-c^2}a \end{align*}which is evident.