Edit: @below We do not need $G$. By the similarity $\angle QJN = \angle PSE = \angle MSF = \angle MJF$. Marvellous solution

Let $V = AS \cap (ABC) , L = (ADE) \cap (ABC) , F = DE \cap BC , J = (LSFV) \cap BC \neq F , H = QJ \cap (ABC) , G = VN \cap (ABC) , K = AP \cap DE$ and $M = OP \cap (LSFV)$. $\angle LQP = \angle LQA = \angle LBA = \angle LFK$, thus $LKQF$ are concyclic, and since $QK \perp FK$, we have $90^\circ = \angle QLF$. So, lines $FL$ and $OQ$ meet on $(ABC)$, call that point $I$. Also, $\angle VQN = \angle VQI = \angle VLF = \angle VJF$, thus $VJNQ$ are concyclic. So, $\angle QJN = \angle QVN = \angle QVG = \angle QHG$, so $HG \parallel BC$. Now, using the fact that $\triangle DSP \sim \triangle BJQ$, we have $\angle QJN = \angle QVN = \angle CAQ + \angle CAG = \angle PDE + \angle BQJ = \angle PDE + \angle DPS = \angle PSE = \angle MSF = \angle MJF$, so $M,J,Q$ are collinear. So, $\angle SON = 180^\circ - \angle QMO - \angle OQM = 180^\circ - \angle JFS - \angle NQJ = 180^\circ - \angle JVS - \angle NVJ = 180^\circ - \angle SVN$, so $SONV$ are concyclic. Finally, since $W$ is the orthocenter of $SAO$, $\angle SWO = \angle SAO = \angle OVS$, implying that $W \in (SONV)$, we are done. $\square$

The Last Problem I Solved in Taiwan TST. First, notice that $\measuredangle SWO= -\measuredangle SAO$, so it is equivalent to showing $\measuredangle SAO+\measuredangle SNO=0^\circ$. Consider the locus of $S$ when we fix the point $Q$ and move $D$, $E$ on $AB$, $AC$, respectively. For any four points $D_1$, $D_2$, $D_3$, $D_4$ on $AB$, we construct the corresponding points $E_i\in AC$, $P_i\in AQ$, and $S_i$ such that $D_iE_i\perp AQ$, $(AD_iE_iP_i)$ is concyclic, and $S_i=OP_i\cap D_iE_i$. It is easy to see \[(D_1, D_2; D_3, D_4)=(E_1, E_2; E_3, E_4)=(P_1, P_2; P_3, P_4).\]It implies $\infty_{DE}(S_1, S_2; S_3, S_4)=(D_1, D_2; D_3, D_4)=(P_1, P_2; P_3, P_4)=O(S_1, S_2; S_3, S_4)$. Hence the locus of $S$ is a conic passing $O$ and $\infty_{DE}$. (Denote it by $\mathcal{H}$.) Claim. $\mathcal{H}$ is a rectangular hyperbola passing $A$, $N$, $O$, $\infty_{DE}$, and $\infty_{AQ}$. Proof. We have already proved $\mathcal{H}$ passes $O$ and $\infty_{DE}$. When we move $P$ to $A$ and $\infty_{AQ}$, it follows that $S=A$ and $\infty_{AQ}$ respectively, which implies that $\mathcal{H}$ passes $A$ and $\infty_{AQ}$. So it remains to show $N$ is on $\mathcal{H}.$ We choose $D$ and $E$ such that $DE$ passes $N$, then it is equivalent to showing $P=Q\iff (ADEQ)$ is concyclic. In fact, we have \[\measuredangle BDN=\measuredangle ADN=90^\circ-\measuredangle QAB= \measuredangle BQO=\measuredangle BQN\implies (BDQN)\text{ is concyclic}.\]Similarly, $(CEQN)$ is concyclic. By Miquel's theorem, it concludes $(ADQE)$ is concyclic. $\Box$ To solve the original problem, we claim that $A$ and $N$ are antigonal conjugate on $\mathcal{H}$, which means for any two points $X$, $Y\in \mathcal{H}$, we have \[\measuredangle XAY+\measuredangle XNY=0^\circ.\]We know the angle bisectors of $\angle AON$ are parallel to $DE$ and $PQ$. Therefore, we have \[\measuredangle \infty_{DE}AO+\measuredangle \infty_{DE}NO=0^\circ. \]On the other hand, suppose $A'$ is the antigonal conjugate of $A$ on $\mathcal{H}$, then we have \[\measuredangle \infty_{DE}AO+\measuredangle \infty_{DE}A'O=0^\circ, \]which implies $O$, $A'$, $N$ are collinear. But a line only can intersect a conic at two points, hence $N=A'$. In particular, if we choose $X=S$, $Y=O$ in the property of antigonal conjugate, then \[\measuredangle SAO+\measuredangle SNO=0^\circ.\]It completes the proof.

Let $AS$ intersect $\Omega$ second time at $A_1$. Let midpoint of $QN$ be $M$. Let $OQ$ intersect $\Omega$ at $Q'$. Since $AA_1 \perp OW$ , $A_1$ is the reflection of $A$ across $OW$ hence it is on circle $(WSO)$ Let $H$ be the orthocenter of $ADE$ and $AH$ intersect $DE$ at point $X$. Perpendicular from $Q$ to $BC$ intersect $BC$ and $\Omega$ at $X'$ and $A'$. $\textcolor{red}{Claim:}$ $AHXP$ is similar to $ONMQ$. For this, it is enough to show $\frac{AX}{XP}=\frac{OM}{MQ}$ Notice there is a spiral similarity that sends $A$ to $A'$ $P$ to $Q$ $D$ to $B$ $E$ to $C$ $X$ to $X'$ Notice $X'M = MN = MQ$ and $OQ=OA'$ Which implies $QMX'$ and $QOA'$ are similar and $MX'\parallel OA'$ $\frac{AX}{XP}=\frac{A'X'}{X'Q}=\frac{OM}{MQ}$, our claim is proven. Let $L$ be the circumcenter of $A_1NQ$. $2 \angle SAX = \angle A_1OQ = 2\angle LOQ$ and $\angle SXA =90 = \angle LMO$ We can conclude $AHXPS$ is similar to $ONMQL$ $(*)$ $\textcolor{red}{Claim:}$ $SONLA_1$ are cyclic. $\textcolor{red}{Proof:}$ $\angle A_1ON + \angle NLA_1 = 2\angle A_1AQ + 2\angle OQA = 2\angle A_1AQ + 2\angle Q'QA = 180$ $ONLA_1$ are cyclic. On the other hand, $\angle A_1SP = 180-\angle ASP=180-\angle OLQ=180-\angle OLA_1$ $SOLA_1$ are cyclic. At the end we showed $SNOWA_1L$ are cyclic. Q.E.D.

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Really nice First note that $\angle SWO = \angle 180 - \angle SAO$ so we need to prove $\angle SAO = \angle SNO$. Let $AS$ meet $\Omega$ at $T$. we need to prove $\angle SNO = \angle STO$. we'll prove $SONT$ is cyclic. Let $ADE$ meet $\Omega$ at $K$. Let $DE$ meet $BC$ at $R$. Let $AP$ meet $DE$ at $L$. Claim $: KDPE$ and $KBQC$ are similar.

Proof $:$ Click to reveal hidden text

Now we have for any point $W$ on $DE$, if $AW$ meets $\Omega$ at $W'$ then $RKWW'$ is cyclic so $RKST$ and $ RKLQ$ are cyclic. Since $RKLQ$ is cyclic we have $\angle RKQ = \angle 90$ so $RK$ and $QO$ meet on $\Omega$ at $Q'$. Let $RKST$ meet $BC$ at $X$. Claim $: TXNQ$ is cyclic.

Proof $:$ Click to reveal hidden text

Note that $\angle STN = \angle STX + \angle XTN = \angle SRX + \angle OQN$ and we want to prove $\angle SOQ = \angle 180 - \angle STN$ so we just need to prove $OS$ and $QX$ meet on $RKST$. Let $SO$ meet $RKST$ at $Y$. Claim $: Q,X,Y$ are collinear.

Proof $:$ Click to reveal hidden text

Now note that $\angle NTS = \angle NTX + \angle XRS = \angle OQX + \angle SYX = \angle 180 - \angle SON$

After hours of failing to complex bash, I realized that things I wanted to show are equal are linear, so i can just use moving points. Let $N'$ be the reflection of $N$ on line $OP.$ Then, clearly condition is equivalent to $A,O,S,N'$ are concyclic. We first prove the following claim (which comes from doing a good diagram and observing that apparently $(SOW)$ cuts $(ABC)$ at a point in line $AS$. Claim 1: Let $AS$ cut $(ABC)$ again at $K$ and let $K'$ be the reflection of $K$ over line $OP,$ then $K \in (AOS).$ Proof: Let also $A'$ be the reflection of $A$ over line $OP$ then $A',K',S$ are colinear by definiton and $ \measuredangle AK'S = \measuredangle AK'A' .$ But by definition, $OP$ bisects $ \measuredangle AOA'$ and so \[ \measuredangle AK'S = \measuredangle AK'A' = \frac{1}{2} \measuredangle AOA' = \measuredangle AOP = \measuredangle AOS \]and so Claim is proved. $\square.$ We will use tethered moving points. Fix $ABC$ and $Q$ and move $P$; then $D$ is on $ AB$ such that $ \measuredangle (AP, PD ) = \frac{\pi}{2} - \measuredangle QAC$ and so $D$ moves projectively. Analogously, so does point $E.$ Now here goes the main trick, although point $S$ does not moves projectively, line $AS$ does. So Claim 2: Line $AS$ moves projectively. Proof: Let $D',E'$ be the corresponding $D,E$ when $P=Q.$ Then note that there exists a homothety centered at $A$ that sends $(P,D,E)$ to $(Q,D',E');$ let this homothety send $O$ to $O'$ and $S$ to $S'.$ Note that $O'$ is the intersection of line $AO$ with the line passing through $P$ and the point of infinity of $OP$ (which clearly moves projectively). Then $O'$ moves projectively; and then so does $S' = D'E' \cap QO'$ and so line $ AS = AS'$ moves projectively as desired. $\square$ Now we are able to finish. Note that by the second paragraph observation and Claim 2, our problem is equivalent to showing $N$ lies on $(A'OK)$ which by inversion at $(ABC)$ is equivalent to show that the inverse of $N$ (denote this point by $M$) is on line $A'K$. Now let $K^*$ be the second intersection of $MA'$ and $(ABC).$ Clearly $K^*$ moves projectively as $M$ is fixed, and so does point $K$ by Claim 2. So to finish we only need to verify three cases. The following works : We first check that the point of infinity of $AQ$ and midpoint of $AQ$ work; which are simple ones: $\bullet $ When $P$ is the midpoint of $AQ;$ then $OP \perp AQ$ and so $S$ is the point of infinity with direction perpendicular to line $AQ$ and $A'=Q.$ Therefore, $K$ is the antipode of $Q$ with respect to $(ABC)$ and so clearly $QK$ passes through $M$ in this case; $\bullet$ When $P$ is the point of infinity of $AQ$, then $S$ is the point of infinity of $AQ$ (so $K=Q$) and $A$ is the reflection of $A$ across line through $O$ parallel to $AQ,$ which is the antipode of $Q$ in $(ABC)$ and so this case works again. To conclude, we check the following case ( which is obtained by reverse engineering the case $K= B.$) $\bullet$ The final case is when $O,P$ and $D$ are colinear; i.e., the case where $P$ is the intersection of $AQ$ and the line through $O$ which forms angle $\frac{\pi}{2} - \measuredangle CAQ$ with line $AQ.$ In this case, $K=B;$ now we need to check that $A'$ is the second intersection of $(ABC)$ and $BM$; we claim this point is $C"$, the reflection of $C$ across $OQ.$ Indeed, let $R$ be the antipode of $Q,$ so $CQRC"$ is harmonic, and projection through $B$ into line $QR$ implies $BC"$ passes through $M.$ Finally, we need to chceck $K = C".$ Indeed, we have \[ \measuredangle KBA = \measuredangle POA = \measuredangle DOA - \measuredangle OAP = \]\[ = \frac{\pi}{2} - \measuredangle CAQ - ( \measuredangle CAQ - \frac{\pi}{2} - (\frac{\pi}{2} - \measuredangle ABC)) = \measuredangle ABC - 2 \measuredangle CAQ \]on the other hand, \[ \measuredangle C"BA = \measuredangle C"BR - \measuredangle ABR = \measuredangle RQC - \measuredangle ABR = \]\[ =\frac{\pi}{2} - CAQ - \frac{\pi}{2} - \measuredangle QCA = \frac{\pi}{2} - CAQ - \frac{\pi}{2} -( - \measuredangle CAQ - \measuredangle ABC)\]so $K = C"$ and we are done.

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Let $Z$ be the reflection of $A$ across $DE$, $F$ be the midpoint of $AZ$, $G = NO \cap DE$, $H = AO \cap DE$, $I = AN \cap DE$, $J$ be the reflection of $Q$ across $O$. Note that $AJ \parallel DE$. We first show that $PI \parallel ZH \parallel QG$. It is easy to see that $ZH \parallel QG$, as $\measuredangle AZH = -\measuredangle ZAH = -\measuredangle QAO = \measuredangle AQO$. Also, angle chasing gives $\Delta ADP \stackrel{-}\sim \Delta JNB$ and $\Delta ADF \stackrel{-}\sim \Delta JQB$, we have \[ \frac{AP}{AQ} = \frac{AP}{AD} \cdot \frac{AD}{AF} \cdot \frac{AF}{AQ} = \frac{JB}{JN} \cdot \frac{JQ}{JB} \cdot \frac{JG}{JQ} = \frac{JG}{JN} = \frac{AI}{AN} \]so $PI \parallel QN$ as claimed. Now $(F,P;Z,Q)=(F,I;H,G)=(Q,N;O,G)=(G,O;N,Q)$ implies $S,Z,N$ are collinear. Finally, \[ \measuredangle SWO = - \measuredangle SAO = \measuredangle SZH = \measuredangle SNO, \]so $S,N,O,W$ are concyclic, as desired.