Solved with Alan Bu, Alex Zhao, Ankit Bisain, David Dong, Edward Yu, Eric Shen, Isaac Zhu, Kevin Wu, Luke Robitaille, Let $S$ be the intersection of $PQ\cap \Gamma$. Then, we have \[\angle AQT = \angle AQS = \overarc{AP} = \overarc{AR} = 180 - \angle AQR\]This proves that $T,Q,R$ are collinear. Now, I claim $M, S, T$ are collinear. Observe that $\angle AMP = \angle ASP = 90$, so $(AMSP)$ is cyclic. This means \[\angle MSA = 180 - \angle MPA = \angle RQA = 180 - \angle AQT = 180 - \angle AST\]Therefore, $M,S,T$ are collinear. This means that $N$ is the inverse of $Q$ with respect to $\Gamma$. Taking an inversion centered at $A$ with circle $\Gamma$ sends $\omega$ to a line. Since $N$ goes to $Q$, and now the locus of $Q$ is a line, this means that $N$ lies on a line. Q.E.D

Solved with Alan Bu, Alex Zhao, Ankit Bisain, David Dong, Edward Yu, Eric Shen, Jeffery Chen, Kevin Wu, Luke Robitaille: Remove all the subscripts and we show $N$ is always on the radical axis of $\omega$ and $\Gamma$. Note $QRT$ are collinear, since $QA$ bisects both $\angle SQT$ and $\angle PQR$. Claim: $SMT$ are collinear. Proof: Since $AMSP$ and $ASQT$ are cyclic with diameters $AP$ and $AQ$ respectively, $$\measuredangle MSA = \measuredangle MPA = \measuredangle RPA = \measuredangle RQA = \measuredangle TQA = \measuredangle TSA.$$ Now since $\angle QTA = 90^\circ$ and $\angle TCQ = 90^\circ$, we have $NQ \cdot NA = NT^2$ and hence $N$ has equal power to $\omega$ and $\Gamma$ (so it moves with locus the radical axis).

Actual Solution Solved with Alan Bu, Alex Zhao, Ankit Bisain, David Dong, Edward Yu, Eric Shen, Jeffery Chen, Kevin Wu, Luke Robitaille.

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