All right, lessgo! One of my favourite ones again cuz I complex bashed part (b) . I provide two ways to solve the part (a) and one complex bash for part (b) . Firstly, $\textbf{part (a) SOLUTION 1}$. Let $D,E$ be the midpoints of $AB,BC$ respectively, and $F = OR \cap AC$. Firstly, as $QC = QB$ and $PB = PA$, then $P, Q$ lie on the perpendicular bisectors of $AB$ and $BC$ respectively $\implies Q \in OE$ and $P \in OD$. As $O$ is circumcenter, $\measuredangle OEB = \measuredangle ODB = 90^{\circ} \implies ODBE$ is cyclic $\implies \measuredangle POQ = \measuredangle DOE = \measuredangle DBE = \measuredangle ABC$. Now, \begin{align*} \measuredangle PRQ &= \measuredangle PQR + \measuredangle RPQ\\ &= \measuredangle PQM + \measuredangle NPQ\\ &= \measuredangle PBM + \measuredangle NBQ\\ &= \measuredangle PBA + \measuredangle CBQ\\ &\stackrel{PB = PA \text{ and } QB = QC}{=} \measuredangle BAP + \measuredangle QCB\\ &= \measuredangle BAC + \measuredangle ACB\\ &= \measuredangle ABC\\ &= \measuredangle POQ \implies POQR \text{ is cyclic.} \end{align*} Now, $$\measuredangle FOE = \measuredangle ROQ = \measuredangle RPQ= \measuredangle NPQ = \measuredangle NBQ = \measuredangle CBQ =\measuredangle QCB = \measuredangle FCE \implies OFEC \text{ is cyclic.}$$ Thus, $\implies \measuredangle OFC = \measuredangle OEC = 90^{\circ} \implies OF \perp AC.$ Thus, $OF$ is the perpendicular bisector of $AC$ $\implies R \in $ perpendicular bisector of $AC \implies RA = RC$ and we are done. Now $\textbf{part (a) SOLUTION 2}$ which I like more cuz Orthologic Triangles . Let $H$ be the foot of $B-$altitude of $\triangle ABC$ and let $T=BH\cap\{ABC\}$. Now $X=NP\cap AT,Y=MQ\cap CT$. Now $$\measuredangle QYC=\measuredangle QCY+\measuredangle YQC=\measuredangle ACT+\measuredangle MQP=\measuredangle ABT+\measuredangle MBP$$$$=\measuredangle ABH+\measuredangle ABP\stackrel{AP=PB}{=}\measuredangle ABH+\measuredangle PAB=\measuredangle ABH+\measuredangle HAB=\measuredangle AHB=90^{\circ}\implies QY\perp CT$$ Similarly $$\measuredangle PXA=\measuredangle XPA+\measuredangle PAX=\measuredangle NPQ+\measuredangle CAT=\measuredangle NBQ+\measuredangle CBT$$$$=\measuredangle CBQ+\measuredangle CBH\stackrel{BQ=QC}{=}\measuredangle QCB+\measuredangle CBH=\measuredangle HCB+\measuredangle CBH=\measuredangle CHB=90^{\circ}\implies PX\perp AT$$ Thus $QY\perp CT\implies CT\perp QR$ and $PX\perp AT\implies RP\perp AT$. Now let $O$ be the center of $\odot(ABC)$. As $QB=QC$ and $OB=OC$, we get that $OQ$ is the perpendicualr bisector of $BC\implies OQ\perp BC$. Similarly $PB=PA$ and $OA=OB\implies OP\perp AB$. Thus we can say that the perpendicular from $\{C$ onto $RQ$,$B$ onto $QP$,$A$ onto $PR\}$ are all concurrent at $T$ which $\implies \{\triangle ABC,\triangle QRP\}$ are orthologic to each other. From this we can conclude too that the perpendicular from $\{P$ onto $AB$,$Q$ onto $BC$,$R$ onto $AC\}$ are all concurrent at the other orthology center of $\{\triangle ABC,\triangle QRP\}$. But we already know that $OQ\perp BC,OP\perp AB\implies O$ is our desired concurrency point. From this however we also get that $OR\perp AC \implies R$ lies on the perpendicular from $O$ onto $AC$, i.e., the perpendicular bisector of $AC$ from which we finally conclude $RA=RC$.

Ok, post became too long, so had to separately post $\textbf{part (b) SOLUTION}$, sorry. Now $\textbf{part (b) SOLUTION}$ using complex bash . $\textbf{LEMMA: }$If $AB$ is a chord of the unit circle, then a point $z$ lies on the perpendicular from $O$ onto $AB$ if and only if $$\overline{z}=\frac{z}{ab}.$$ $\textbf{PROOF: }$We have that $OZ \perp AB$. Using Perpendicularity Criterion, we have, \begin{align*} &\frac{0-z}{b-a} + \overline{\left(\frac{0-z}{b-a}\right)} = 0\\ \implies &\frac{-z}{b-a} + \left(\frac{-\overline{z}}{\overline{b}-\overline{a}}\right) = 0\\ \implies &\frac{-z}{b-a} + \left(\frac{-\overline{z}}{\frac{1}{b} - \frac{1}{a}}\right) = 0\\ \implies &\frac{-z}{b-a} + \left(\frac{-ab\overline{z}}{(a-b)}\right) = 0\\ \implies &\left(\frac{ab\overline{z}}{(b-a)}\right) = \frac{z}{b-a}\\ \implies &\frac{ab\overline{z}}{\cancel{b-a}} = \frac{z}{\cancel{b-a}}\\ \implies &\overline{z} = \frac{z}{ab}. \end{align*}. All the other lemmas are from EGMO complex bash chapter . So, as proved in the Part (a) Solution 1, $R \in \odot(POQ)$. Now we proceed with phantom points, let $S' = OB \cap AC$, and $S$ be the reflection of $S'$ over $OF$. $\textbf{CLAIM: }\overline{B - R - S}$ are collinear. First let $B'$ be the point such that $BCAB'$ is an isoceles trapezium. So, we get that $B'$ is the reflection of $B$ over $OF$. Reflecting over $OF$, we get the following maps, \begin{align*} B &\xleftrightarrow{} B'\\ R &\xleftrightarrow{} R\\ S &\xleftrightarrow{} S'.\\ \end{align*} So, proving $\overline{B - R -S}$ are collinear is the same as proving $\overline{B'-R-S'}$ are collinear. We move onto the complex plane. Assign $A,B,C$ with the complex numbers $a,b,c$ respectively. Firstly as $P$ lies on the perpendicular bisector of $AB$ respectively, using our Lemma we have, \begin{align} \overline{p} = \frac{p}{ab}. \end{align} Also, using Complex Chord Lemma, we have, \begin{align*} &p + ac\overline{p} = a+c\\ \implies &p + ac \cdot \left( \frac{p}{ab} \right) = a+c \tag*{[Using (1)]}\\ \implies &p\left( 1 + \frac{c}{b} \right) = a+c\\ \implies &\boxed{p = \frac{(a+c)b}{b+c}}. \end{align*} Similary, we also get $$\boxed{q = \frac{(a+c)b}{a+b}}.$$ Moreover, now we have two facts, $R \in \odot(POQ)$ and $OR \perp AC$. Now LaTeXing became too hard so the rest of the solution is handwritten lol .

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As $\overline{B-R-S}$ are collinear, we now have that if $S=BR\cap AC$, and $S'$ is the reflection of $S$ over $OF$, then $\overline{O-S'-B}$ are collinear. Now reflect over the line $OF$ to get $\overline{O-S-B'}$ are collinear. So, now as we have that $\overline{O-S-B'}$ are collinear, so proving $OS\perp MN$ is equivalent to proving $OB'\perp MN$. Let $X=OB'\cap MN$. So we have \begin{align*} \measuredangle SXN&=\measuredangle SCN+\measuredangle CNX+\measuredangle XSC\\ &= \measuredangle ACB+\measuredangle BNM+\measuredangle OSC\\ &= \measuredangle ACB+\measuredangle BQM+\measuredangle OSS'\\ &\stackrel{SS'\parallel BB'}{=}\measuredangle ACB+(\measuredangle BQP-\measuredangle MQP)+\measuredangle OB'B\\ &\stackrel{OB=OB'}{=}\measuredangle ACB + (\measuredangle BQC-\measuredangle MBP)+\measuredangle B'BO\\ &=\measuredangle ACB + \measuredangle BQC-\measuredangle ABP+(\measuredangle B'BA + \measuredangle ABO)\\ &\stackrel{PB=PA}{=}\measuredangle ACB + (\measuredangle BCQ+\measuredangle QBC)-\measuredangle PAB+(\measuredangle B'BA - \measuredangle OBA)\\ &\stackrel{QB=QC}{=}\measuredangle ACB + 2\measuredangle BCQ-\measuredangle PAB+(\measuredangle B'BA - \measuredangle OBA)\\ &\stackrel{BB'\parallel AC}{=}\measuredangle ACB + 2\measuredangle BCA-\measuredangle CAB+\measuredangle CAB - \measuredangle OBA\\ &=\measuredangle ACB - 2\measuredangle ACB- (90^{\circ} -\measuredangle ACB)\\ &=-\measuredangle ACB - 90^{\circ} +\measuredangle ACB\\ &=90^{\circ} \end{align*}So $\measuredangle SXN=90^{\circ}\implies\measuredangle OXN=90^{\circ}\implies OX\perp MN$ and we are done. OOF! That's all!

Part A Solution

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In b, there is a short solution by writing trig ceva in BQP with trig ratio lemma to find QS/PS and writing trig ratio in OQP to find QS’/PS’. They turn out to be equal.(of course you have to calculate the angles). Any synthetic solution?

Sketch of part b: you can define $A'$, the reflection of the orthocenter among $AC$ in $\triangle ABC$. Then $A',O,S$ turn out to be collinear by Tales and similarity. Let $L$ be the center of $(PQRO)$, then you can show that $LR \parallel A'O$ by the similarity of $PQR$, $CBA$ and angle chasing which finishes the problem.

Another (similar to the above one) approach for b) is to notice that by angle chasing the problem is equivalent to prove that if $H$ is the orthocenter of $\triangle ABC$ and $O'$ is the reflection of $O$ in $AC$, then $H, O', S$ are collinear (which follows by length bashing, i.e. calculating $\frac{SA} {SC}$ in two ways).

Sol:- $PO$ and $QO$ are exterior angle bisectors in $\Delta BPQ$ so $O$ is $B$ excenter in $BPQ$. Let $I$ be the incenter of $BPQ$ we have that $\angle IPQ=\angle A=\angle RQP$ and $\angle IQP=\angle C=\angle RPQ$ so $I,R$ are symmetric wrt perpendicular bisector of $BC$. Since $I$ is antipode of $O$ in $OPQ$ we get $OR \perp PQ$. Let $B'$ be on $(BPQ)$ such that $B'B \parallel AC$. Note that $R$ is incenter of $B'PQ$ and $M,N$ are the midpoint of arcs $B'Q,B'P$. So $B'R \perp MN$. $BI \cap AC=D$. Projecting $-1=(B,D;I,O)$ on $DS$ we get $RO$ bisects $DS \implies RDOS$ is a kite. So $BI$ and $OS$ are symmetric wrt $RO$ and we also have that $BI$ and $B'R$ are symmetric wrt perpendicular bisector of $BC$ and hence $B'R \parallel OS \implies OS \perp MN$.

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Using the diagram above, showing $\frac{AD}{DC}=\frac{CQ}{QA}$ is also very feasible with a sine bash.

Sol for (a): Through angle chasing we can see OR and OB are isogonal in OED. Since OB passes the circumcenter, OR must pass the orthocenter and hence OR and PQ are perpendicular

Very clean sol for (b): Let M' and N' be the reflections of M and N accross the perpendicular bisector of PQ. M' is the midpoint of arc BQ while N' is the midpoint of arc BP. Since O is the A-excenter of APQ, this indicates that OB and M'N' are perpendicular. Hence, it is enough to show that OR is the angle bisector of BOS, which is trivial.

The problem falls apart once you notice that triangles $RPQ$ and $BAC$ are orthologic.