first T3MO

We divide our solution into two main steps:

Step 1: Let $O$ be the midpoint of $AB$ and $R$ the midpoint of $PQ$, then $C$, $R$, $O$ are collinear.

Proof. We will denote the complex number of each point by its lowercase letter.
Let $AP \cap BQ = V$, $AQ \cap BP = U$. Let $\angle PAC = \alpha$, $\angle PBC = \beta$, then $\angle PVQ = 90^\circ - \alpha - \beta$ and $\angle PUQ = 90^\circ + \alpha + \beta$, therefore quadrilateral $PUQV$ is cyclic.
Let $AC \cap BP = W$, $AQ \cap BC = X$, $AC \cap BQ = Y$, and $AP \cap BC = Z$. Since $C$ is on the angle bisector of both $\angle PAQ$ and $\angle PBQ$, and since $\angle ACB = 90^\circ$, $WC = CY$ and $XC = CZ$. Thus $\square WXYZ$ is a rhombus, hence $c = \frac{w+x+y+z}{4}$.

Since $PUQV$ is cyclic, $\frac{AU}{AP} = \frac{AV}{AQ}$. By the angle bisector theorem, $\frac{UW}{PW} = \frac{AU}{AP}$ and $\frac{AV}{AQ} = \frac{VY}{YQ}$. Therefore $\frac{UW}{UP} = \frac{VY}{VQ} = d$ for some $d \in \mathbb{R}$. It follows that $w = dp + (1-d)u$ and $y = dq + (1-d)v$. Similarly there exists $e \in \mathbb{R}$ such that $x = eq + (1-e)u$ and $z = ep + (1-e)v$. Therefore
$$c = \frac{(d+e)(p+q) + (2-d-e)(u+v)}{4}.$$

Let $T$ be the midpoint of $UV$. Clearly $t = \frac{u+v}{2}$, and $r = \frac{p+q}{2}$. Therefore
$$c = \left(\frac{d+e}{2}\right)t + \left(1 - \frac{d+e}{2}\right)r$$
with $\frac{d+e}{2} + \left(1 - \frac{d+e}{2}\right) = 1$, so $C$, $R$, $T$ are collinear.

Using the well-known fact that in a complete quadrilateral, the midpoints of three diagonals are collinear, we have $R$, $T$, $O$ are collinear. Hence we have $C$, $R$, $O$ collinear. $\Box$

Step 2: Showing that $M$, $S$, $N$ are collinear.

Proof. First, since $\angle PAN = \angle NAQ$, $N$ is the midpoint of arc $PQ$, so $\angle NRQ = 90^\circ$, hence quadrilateral $NCRS$ is cyclic.

Let $\angle OBC = \theta$. We have $\angle MNQ = \angle MAQ = 90^\circ - \theta - \alpha$. Since $\angle SNQ = 90^\circ - \angle RNS - \angle PQN = 90^\circ - \angle OCB - \angle PAN = 90^\circ - \theta - \alpha$, so $\angle MNQ = \angle SNQ$, therefore $M$, $S$, $N$ are collinear. $\square$

(by Kritkorn Karntikoon)
Let $D = AB \cap PQ$, $\angle PAC = x$, $\angle PBC = y$, $\angle QAB = z$, and $\angle PQA = w$, and $S' = PQ \cap AC$.

First we will show that $DP \cdot DQ = DS' \cdot DS$, or, equivalently, $\frac{DP}{PS'} = \frac{DS}{SQ}$.
From the triangles $ADP$ and $AS'P$ we have
$$\frac{DP}{\sin(2x + z)} = \frac{AP}{\sin \angle ADP} \quad \text{and} \quad \frac{PS'}{\sin x} = \frac{AP}{\sin \angle PS'A}$$

Hence,
$$\frac{DP}{DS'} = \frac{\sin(2x + z) \sin \angle AS'P}{\sin \angle ADP \sin x}$$

A similar argument with triangles $BDS$ and $BQS$ yields
$$\frac{DS}{SQ} = \frac{\cos(x + z) \sin \angle BQS}{\sin \angle BDS \sin y}$$

Therefore we have
$$\begin{align*} \frac{DP}{PS'} = \frac{DS}{SQ} &\iff \frac{\sin(2x+z) \sin \angle AS'P}{\sin x} = \frac{\cos(x+z) \sin \angle BQS}{\sin y} \\ &\iff \frac{\sin(2x+z) \sin(x+w)}{\sin x} = \frac{\cos(x+z) \cos(x+w-y)}{\sin y} \\ &\iff \sin(2x+z) (2 \sin(x+w) \sin y) = \cos(x+w-y) (2 \sin x \cos(x+z)) \\ &\iff \sin(2x+z) (\cos(x+w-y) - \cos(x+y+w)) \\ &\phantom{\iff \sin(2x+z) (\cos(x+w-y) -} = \cos(x+w-y) (\sin(2x+z) - \sin z) \\ &\iff \sin(2x+z) \cos(x+y+w) = \sin z \cos(x+w-y) \\ &\iff \sin \angle PAB \sin \angle BPQ = \sin \angle QAB \sin \angle BQP \\ &\iff \frac{\sin \angle PAB}{\sin \angle QAB} = \frac{\sin \angle BQP}{\sin \angle BPQ}. \end{align*}$$

Observe that $\angle APB + \angle AQB = (90^\circ - x + y) + (90^\circ + x - y) = 180^\circ$, so $\sin \angle APB = \sin \angle AQB$. Therefore
$$\frac{\sin \angle PAB}{\sin \angle QAB} = \frac{BP \sin \angle APB}{AB} \cdot \frac{AB}{BQ \sin \angle AQB} = \frac{BP}{BQ} = \frac{\sin \angle BQP}{\sin \angle BPQ}.$$

Hence we have proved that $DS \cdot DS' = DP \cdot DQ$. Moreover, since $DP \cdot DQ = DA \cdot DM$, so $DS \cdot DS' = DA \cdot DM$. Therefore, $AS'SM$ is cyclic.
Let $NS$ intersect the circumcircle of $APQ$ again at $M'$. Since $N$ is the midpoint of arc $PQ$, by a well-known lemma, we have $NS' \cdot NA = NP^2 = NS \cdot NM'$, so $AS'SM'$ is cyclic. Therefore $M$ and $M'$ are both the second intersection point of the circumcircles of triangles $AS'S$ and $APQ$, so $M = M'$.