What is $O$?

@above $O$ is the center of circle $\omega$. I’ll tell him to fix the statement

Let $Y$ be the point on $\omega$ that $AX$ and $AY$ are isogonal, and $S, L$ be the intersection of $XM, YM$ and $\omega$, respectively. We can see that $SL \parallel BC$. All polars taken in the solution is a polar wrt the $A$ mixtilinear incircle.

Using Well Known Fact

, it suffices to prove that $PM$ and $AQ$ are parallel. Now, as $-1 = (B,C;M, \infty_{BC}) = (YB,YC;YM,Y\infty_{BC}) = (B,C;L,X)$, we know that $LX$ passes through the intersection of tangents from $B,C$ to $\omega$. This implies that $O,M,X,L$ are concyclic. Since $\angle YAX = \angle YLX = \angle MLX$, it suffices to prove that $P,M,X,L$ are concyclic. We will prove that $P,O,M,X$ are concyclic by proving that $\angle XPO = \angle XMO$. Since $P$ lies on the $A$ polar, $A$ lies on the polar of $P$. Since $-1 = (B,C;L,X) = (AD, AE; AL, AP)$, we can easily see that $AL$ is $P$ polar. This implies that $AL \perp PO$. Hence, $\angle XPO = \angle (XP, OP) = \angle (XA, LA) + \angle (LA, OP)$. From $\angle (LA, OP) = \angle (CM, MO)$ and $\angle (XA, LA) = -\angle LAX = -\angle LSX = -\angle CMX = \angle XMC$, we obtain that $\angle XPO = \angle (XA, LA) + \angle (LA, OP) = \angle XMC + \angle CMO = \angle XMO$, as desired.

i swear the motivation is so hard to find so i rate 40 mohs but @above is so orz or maybe i didn't do enough taiwan-style config geo Let $N$ be the midpoint of major arc $BC$, and $F$ midpoint of minor arc $BC$. Let $O'$ be the centre of the mixtilinear incircle. Let $I$ be the incentre of $\triangle ABC$, and $M$ the midpoint of $BC$. We reconstruct $Q$ as the intersection of the reflection of $AP$ over $AI$ with $MX$. It suffices to show that $QC$ and $CP$ are isogonal in $\angle BCA$ to prove $P$ and $Q$ are isogonal conjugates. The following results are well-known or trivial to see: $DE\perp AI$, and $D,I,E$ collinear. $O,O', T$ collinear. $T,I,N$ collinear. $DBTI$ and $ECTI$ cyclic. $N,O,M,F$ collinear. $A,I,O',F$ collinear. Our solution depends on the following claim: Claim: $XMOP$ is cyclic. Proof: We will show that $\triangle TIX\sim\triangle FMX$. Clearly, $\angle XFN=\angle XTN$ by Bowtie, hence we just need \[\frac{IT}{MF}=\frac{TX}{FX}\]To conclude by $SAS$ similarity. Let us proceed by trigonometric bash. First evaluate the right hand side, \[\frac{TX}{FX}=\frac{\sin\angle XFT}{\sin\angle FTX}=\frac{\sin\angle TAX}{\sin\angle FAX}\]However, by Ratio Lemma in triangle $\triangle ATO'$, we have \[\frac{\sin\angle TAX}{\sin\angle FAX}=\frac{TP}{PO'}\times \frac{AO'}{TA}\]Now, by Ratio Lemma in triangle $\triangle TIO'$ now, we have \[\frac{TP}{PO'}=\frac{IT}{IO'}\times\frac{\sin\angle DIT}{\sin 90^\circ}\]Hence we have \[\frac{TX}{FX}=\frac{AO'}{TA}\times \frac{ IT\sin\angle DIT}{IO'}\]Now, let's move to the left hand side of our initial equation. First of all, we prove an important similarity: $\triangle IMF\sim\triangle AIT$. Indeed, by Shooting Lemma and Fact 5 we have $FB^2=FI^2=FM\times FN$ hence $(NIM)$ is tangent to $FI$. Thus, $\measuredangle FIM=\measuredangle INF=\measuredangle TAF$, and by Bowtie we have $\angle MFI=\angle ITA$. This is sufficient due to $AA$ similarity. Now, this implies that $\frac{IT}{MF}=\frac{AT}{IF}$. Thus, it simply suffices to show that \[\frac{AT}{IF}=\frac{AO'}{TA}\times\frac{IT\sin\angle DIT}{IO'}\]Now, by Ratio Lemma in $\triangle ATI$, we have \[\frac{AO'}{IO'}=\frac{AT}{TI}\times\frac{\sin\angle ATO'}{\sin\angle ITO'}\]Hence it simply suffices to show that \[\frac{AT}{IF}=\frac{\sin \angle DIT\sin\angle ATO'}{\sin\angle ITO'}\]Now, $BDIT$ is cyclic as mentioned before, hence \[\sin{\angle DIT}=\sin\angle TBA=\sin\angle ANT=\frac{AI}{NI}\]Because $\triangle ANI$ is right. However, we also have $\triangle AIT\sim\triangle NIF$ because of Bowties and $AA$ similarity. Hence, $\frac{AI}{NI}=\frac{AT}{NF}$. Hence it suffices to prove \[\frac{NF}{IF}=\frac{\sin\angle ATO'}{\sin\angle ITO'}\]However, from $TO=ON$ by circumcentre definition, $\angle ITO'=\angle ONI$. Further, we have \[\measuredangle ATO'=\frac{180^\circ-\measuredangle TOA}{2}=90^\circ-\measuredangle TCA=90^\circ-\measuredangle TNA=\measuredangle AIN\]due to angle at centre theorem. This implies that \[\frac{\sin\angle ATO'}{\sin\angle ITO'}=\frac{\sin \angle AIN}{\sin\angle FNI}\]And the right hand side is $\frac{NF}{IF}$ due to sine law in $\triangle INF$. This concludes our trigonometric bash. Hence, $\triangle TIX\sim\triangle FMX$. Note that Reim's Theorem on $(ABC)$ and parallel lines $AN$ and $DE$ (Both are perpendicular to $AI$) implies that $IPXT$ is cyclic. Using our similarity, \[\measuredangle FMX=\measuredangle TIX=\measuredangle TPX=\measuredangle OPX\]Which finally gives $OPMX$ cyclic. This concludes our claim. $\square$ Lemma: $PM\parallel AQ$ Proof: We will angle chase: \[\measuredangle MPX=\measuredangle FOX=2\measuredangle FNX=2\measuredangle FAX=\measuredangle QAX\]Where we use $XMOP$ cyclic, angle at centre theorem, and the fact that $AI$ bisects $\angle QAP$. This suffices by corresponding angles in parallel lines.$\square$. Now, in order to show that $CP$ and $CQ$ are isogonal, we will employ DDIT at $C$ on quadrilateral $AQMP$. Note that $CA$ and $CM$ are clearly isogonal. If we show that $AP\cap QM=X$ and $PM\cap AQ=P_\infty$ are isogonal, this would characterize our involution, then imply $CP$ and $CQ$ also being isogonal. Indeed, it suffices to show that $\measuredangle(AC,CX)=\measuredangle(NM,BC)$ to prove isogonality. By Pitot Theorem on $\triangle NOT$, $NIMX$ is cyclic. In particular, it is tangent to $AI$ from our earlier result. Now we claim that $I$ is the incentre of $\triangle AQX$ as well. Indeed, $AI$ bisects $\angle QAP$. Further, \[\measuredangle AQX=\measuredangle PMX=\measuredangle POX=180^\circ-2\measuredangle INX=180^\circ-2\measuredangle FIX=2\measuredangle AIX-180^\circ\]Which suffices by well-known incentre angle formula, where we used angle at center theorem and $XMOP$ cyclic. Now, let us angle chase: \begin{align*}\measuredangle ACX&=\measuredangle ANX\\ &=\measuredangle INX+\measuredangle ANI\\ &=\measuredangle FIX+\measuredangle AXT\\ &=\measuredangle FIX+\measuredangle IXF\\ &=180^\circ-\measuredangle XFA\\ &=180^\circ-\frac{1}{2}\measuredangle XOA\\ &=90^\circ+\measuredangle PXO\\ &=90^\circ+\measuredangle PMO\\ &=\measuredangle PMB \end{align*}As desired, where we used $(NIMX)$ tangent to $AI$, $MOPX$ cyclic, and multiple Bowties, as well as because from $XI$ bisecting $\measuredangle AXM$, \[\measuredangle AXI=\measuredangle IXM=\measuredangle TXF,\]implying $\measuredangle AXT=\measuredangle IXF$. (Similar switch gives $\triangle IMX\sim\triangle TFX$, and our angle equality.) This concludes the problem, as $PC$ and $QC$ are now isogonal, and hence $Q,P$ are isogonal conjugates, but $Q$ lies on $MX$, so we are done.