This is my solution during the test. Let $N$ be the midpoint of arc $\widehat{ABC}$, $P^*$ be the isogonal conjugate of $P$ wrt $\triangle ABC$, $P'$ is on $\overline{XY}$ and satisfying $\overline{XP}=\overline{YP'}$. We have $\overline {JX}=\overline{JY}$, which implies $AX, AY$ is isogonal wrt $\angle A$ and $CX,CY$ is isogonal wrt $\angle C$, hence $X,Y$ is isogonal conjugate. It is well-known that $BP,BQ$ is isogonal wrt $\angle B$, so the rest is to prove $Q=P^*$. We have $BN$ pass through $J$, so $BJ$ bisect $\overline{XY}$, implies $\triangle BXY$ is isosceles triangle. $\textbf{Lemma}$ The Miquel point of $\mathcal{Q}\{PX,XP^*,P^*Y,YP\}$ is on $(ABC)$ Then consider Newton line of $\mathcal{Q}$, take $\mathcal{H}(P,2)$, we will get newton line is parallel to $P'P^*=BQ//AC$, so Miquel point should be its isogonal conjugate, which is $B$. Hence$\triangle BXP\sim\triangle BP^*Y$, implies that $\angle BXY=\angle BXP=\angle BP*Y$, is equivalent to $P^*\in(BXY)$. Further $BP^*=BQ$, so $P^*=Q$.

One can easily observe that $X$ and $Y$ is a pair of isogonal conjugate wrt $\triangle ABC$; $BP$, $BQ$ are isogonal wrt $\angle ABC$. So we are required to show that $P$ and $Q$ is a pair of isogonal conjugate. It is equivalent to show that $ABCXYQ$ lie on the same conic. Let $U = XQ \cap CY$, $V = AX \cap BY$. Apply Pascal’s theorem on $BQXACY$, it suffices to show that $UV \parallel AC$. On the other hand, by Reim’s theorem, we only need to show that $UVXY$ are concyclic. The final step is just some angle chasing, we show that $YV + XU - VX - YU \equiv 0$. Notice that $BY + CQ \equiv BQ + CY$ and $AX + CY \equiv AC + XY$. We have $$ YC + AX - XQ - YB \equiv YC + AX - YP - BQ \equiv CA - AX + AX - BQ \equiv 0. $$Thus we complete the proof. Remark. (Convention invented by Li4 and Untro368)The equivalent relationship here is defined as $$ \sum_{i=1}^n \ell_i \equiv \sum_{i=1}^n \ell_i’ \iff \sum_{i=1}^n \measuredangle (\ell_i, \ell_i’) = 0^\circ. $$

Cindy.tw wrote: It is equivalent to show that $ABCXYQ$ lie on the same conic. A simpler way to prove this is to apply Pascal's theorem twice: As above, let $U = XQ\cap CY$, $V = AX\cap BY$. It suffices to show that $UV\parallel AC$. Let $D$, $E$ be the second intersection of $XA$, $YC$ and $\odot(BQXY)$. Then Reim's theorem tells us that $DE\parallel AC$. Apply Pascal’s theorem on $XDEYBQ$, we see that $U$, $V$, and $BQ\cap DE = \infty_{AC}$ are collinear, which implies $UV\parallel AC$.

Inversion Solution The proposition after inverting at $A$ is the following: Let $J$ be the incenter of $\triangle{ABC}$. $X$, $Y$ are two points on $CJ$ such that $\measuredangle BAX=\measuredangle YAC$. Let the circle passing $A$, $B$ and tangent to $BC$ be $\Gamma_1$, the circle passing $A$, $B$ and tangent to $AC$ be $\Gamma_2$. Then suppose $\odot (AXY) \cap \Gamma_1=\{A, P\}$ and $\odot (BXY)\cap \Gamma_2=\{B, Q\}$. Prove that $\measuredangle BAP =\measuredangle QAC$. Let $\odot (AXY)$ intersects $AC$ again at $E$, $\odot(BXY)$ intersects $BC$ again at $D$. Since $CE\cdot CA=CX\cdot CY=CD\cdot CB$, $ABDE$ are concyclic. Suppose $AQ$ intersects $\odot(BXY)$ again at $B'$, $BP$ intersects $\odot(AXY)$ again at $A'$. Then $\measuredangle CDE=\measuredangle BAC=\measuredangle BQA=\measuredangle BQB'= \measuredangle BDB'$ $\implies B'$ is on $DE$. Similarly, $A'$ is also on $DE$. Claim. The perpendicular bisector of $\overline{XY}$ passes $S$. Proof. In fact, we can prove $SX$, $SY$ are tangent to $\odot (ABX)$ and $\odot (ABY)$, respectively. (Recall 2021 IMO P3 ) Since we know $X$, $Y$ are isogonal conjugates in $(AB, BC, CA, DE)$, we have $\measuredangle BXS=\measuredangle CXE$. $\because \measuredangle SAX=\measuredangle BAX=\measuredangle YAC=\measuredangle YAE=\measuredangle CXE=\measuredangle BXS$. $\therefore SX$ is tangent to $\odot (ABX)$. Similarly, $SY$ is tangent to $\odot(ABY)$. Hence $SX^2=SA\cdot SB=SY^2$. It is well-known that the inner angle bisector of ($AB$, $DE$) (called it $\ell$) is perpendicular to $XY$. So $DE$ is the reflection of $AB$ with respect to $\ell$. Then $AA'B'B$ is an isosceles trapezoid. Therefore, $\measuredangle BAQ = \measuredangle BAB' = \measuredangle BA'B'= \measuredangle PA'E= \measuredangle PAC$.

$\textbf{Lemma}$ The Miquel point of $\mathcal{Q}\{PX,XP^*,P^*Y,YP\}$ is on $(ABC)$ i dont understand this. can u help me?

LoloVN wrote: $\textbf{Lemma}$ The Miquel point of $\mathcal{Q}\{PX,XP^*,P^*Y,YP\}$ is on $(ABC)$ i dont understand this. can u help me? Means the miquel point of the quadrilateral formed by the lines $PX,XP^*,P^*Y,YP$.

LoloVN wrote: $\textbf{Lemma}$ The Miquel point of $\mathcal{Q}\{PX,XP^*,P^*Y,YP\}$ is on $(ABC)$ i dont understand this. can u help me? To prove this, a good insight is by perfect hexagon. A perfect hexagon $ABCDEF$ on complex plane is defined as $\dfrac{a-b}{b-c}\cdot\dfrac{c-d}{d-e}\dfrac{e-f}{f-a}=-1$. Notice the definition give us if we have five points $B,C,D,E,F$， there must be an only $A$ such that these six points form a perfect hexgon. $\textbf{Corollary}$ A complete quadrangle is a perfect hexagon. $\textbf{Lemma}$ If $ABCDEF$ forms a perfect hexagon, then $ACBDFE, AECDBF, ABFDEC$ also form perfect hexagons. The proof of above lemma is trivial by easy calculate. Hence we can view a perfect hexagon as $(AD)(BE)(CF)$. $\textbf{Theorem}$ Given a perfect hexagon, $(BE)(CF),(CF)(AD), (AD)(BE) $ share same Miquel point. $\textbf{Proof}$ Let the Miquel point of $(BE)(CF)$ be $M$. If $M$ is a point at infinity, then $(BE)(CF)$ form a parallelogram. We have $b+e=c+f$. Define $A'$ as the point of coordinate $-d$. It can easily verify $A'BCDEF$ form a perfect hexagon, hence $a+d=0$. If $M$ isn't a point at infinity, we can perform a Möbius transform such that $m=0, be=cf=1$. Define $A'$ as the point of coordinate $d^{-1}$. It can easily verify $A'BCDEF$ form a perfect hexagon, hence $ad=1$. All in all, a perfect hexagon can be regarded as three pairs of ponts $(AD)(BE)(CF)$ and exist a involution to match the three pairs of points. Analogously, if $n$ pairs of points share the same involution, they form a "Perfect $2n$-tuple" and share the same Miquel points (can regard as involution center) . $\textbf{Theorem}$ Let $Q$ be the isogonal conjugate of $P$ wrt $(AD)(BE)$, then $(AD)(BE)(PQ)$ form a perfect hexagon. $\textbf{Proof}$ Let $Q'$ be the point such that $(AD)(BE)(PQ')$ form a perfect hexagon. Let $C=BD\cap EA, F=AB\cap DE$. Hence $(AD)(BE)(CF)(PQ')$ form a perfect 8-tuple. Since $P$ have isogonal point, we have $\measuredangle{APC}+\measuredangle{DPF}=0^{\circ}$. Further, we have $AQ'CDPF$ is a perfect hexagon, we have $\measuredangle{APC}+\measuredangle{AQ'C}=\measuredangle{FPD}+\measuredangle{PDF}+\measuredangle{AFP}=\measuredangle{ABC}$. Likewise, $\measuredangle{BPC}+\measuredangle{BQ'C}=\measuredangle{BAC}$. Hence $P, Q'$ are a pair of isogonal conjugate wrt $\triangle{ABC}$. Hence $Q=Q'$. $\textbf{Corollary}$ If $(P,P^*)(Q,Q^*)$ be two pairs of isogonal conjugates wrt $(AD)(BE)$, then Miquel point of $(PP^*)(QQ^*)$ is the Miquel point of $(AD)(BE)$. $\textbf{Proof}$ Since $(AD)(BE)(PP^*)$ and $(AD)(BE)(QQ^*)$ form perfect hexagons, hence $(AD)(BE)(PP^*)(QQ^*)$ share the same Miquel point. Back to the problem, regard $C$ as $BD\cap EA$, then since Miquel point of $(AD)(BE)$ on $(ABC)$, hence the Miquel point of $(PP^*)(QQ^*)$ on $(ABC)$

But if you only want to prove the last lemma, it can be done in fundemental way. Here introduce formal sum: $\measuredangle{ABC}=BC-AB$. Choose $X$ such that $\triangle{MAX}\overset+\sim\triangle{MPQ}\overset+\sim\triangle{MQ^*P^*}$, hence $\triangle{APQ^*}\overset+\sim\triangle{XQP^*}$. $XP^*-XQ=AQ^*-AP=(AB+AC-AQ)-(AB+AC-AP^*)=AP^*-AQ\implies A,Q,P^*,X$ are concyclic. Hence $$AX=AQ+P^*X-P^*Q$$Analogously, if we define $Y$ such that $\triangle{MBY}\overset+\sim\triangle{MPQ}\overset+\sim\triangle{MQ^*P^*}$$$BY=BQ+P^*Y-P^*Q$$Moreover, $\triangle{ABQ^*}\overset+\sim\triangle{XYP^*}$ $$BY-AX=BQ+P^*Y-AQ-P^*X=BQ-AQ+BQ^*-AQ^*=BA+BC-AB-AC=BC-AC$$Hence $AX\cap BY$ on $(ABC)$. However, $\triangle{MAX}\overset+\sim\triangle{MBY}$. Hence $M$ in $(ABC)$

Lemma (Projective Reim's). Let $A,B,C,D,E,F$ be points such that $(A,B,C,D)$ and $(C,D,E,F)$ concyclic and $AB \parallel EF$. Then $A,B,C,D,E,F$ lie on a common conic. Proof of Lemma. Let $\ell = AB$ intersect $(C,D,E,F)$ at $A',B'$, intersect $CD$ at $O$, and intersect $EF$ at $\infty$. Consider the involution $f$ (by DIT) on $\ell$ where $f(P)$ is the second intersection of $\ell$ and conic $CDEFP$. Then $f$ swaps $O \leftrightarrow \infty$ and $A' \leftrightarrow B'$, hence $f$ is in fact the inversion at $O$ with inversion power $OA \cdot OB$. Hence $f$ swaps $A \leftrightarrow B$ and so $A,B,C,D,E,F$ lie on a common conic. $\square$ In the context of the problem, this implies that $A,B,C,X,Y,Q$ lie on a common conic $\omega$ (as $AC \parallel BQ$, $(A,C,X,Y)$ concyclic, and $(X,Y,B,Q)$ concyclic). Let $A'$ and $P'$ be the reflections of $A$ and $P$ across $BJ$ - it is easy to show that $A'$ lies on $(ACJ)$ and $BC$, while $P'$ lies on $BQ$. Also let $T=BC \cap XY$. Now \[ A'(Y,P';X,C)=(Y,P';X,T) \stackrel B= (Y,Q;X,C)_\omega \stackrel{A}= A(Y,Q;X,C) \]implies that $A'P' \cap AQ$ lies on $(ACJ)$, so \[ \angle BAP = \angle P'A'B = \angle QAC \]as desired.