Vary $R$ on $AC$, since $P,Q$ move linearly, $(APQ)$ goes through a fixed point. Taking $R = A, R = C$, we see that this fixed point is $O$. Observe that $\angle PHQ = 180 - \angle PRQ = \angle ARP + \angle CRQ = \angle A + \angle C = 180 - \angle ABC$ so $(APQ)$ goes through $H$ too. Therefore, $\angle BOH = \angle BPH = \angle BPQ + 90 - \angle PQR = \angle B - \angle A + 90 = \angle (BO, AC)$, so $OH || BC$, as desired. $\blacksquare$

Nice problem, but not really original - USA TST 2012/1 can help for this because that configuration is really similar. Also there was a similar problem in ARO 2001 - https://artofproblemsolving.com/community/c6h162p567 Anyways, solution: We have that $H$ lies on $BPQ$, and we have that $OH || AC \iff$ $O$ lies on $(BPQ)$. I did it with a trig bash (as seen in above two problems, it is helpful). We just have to prove that triangles $OMP$ and $ONQ$ are similar, so we want $OM/MP=ON/NQ$, which, as I said, is easy to calculate with a bit of trig (I introduced $M$ and $N$ - midpoints $AB$ and $BC$).

I got a very long solution using Moving Points(again) and angle chasing---

Attachments:
Sharygin_Geometry_Olympiad P-8.pdf (104kb)

coordinate bashing works here as well

I like this problem, unfortunately it's only for 8-9 grades... Claim: $O,H\in \odot (BQP).$ Proof. Observe that $\measuredangle PHQ=\measuredangle QRP=\measuredangle QRC+\measuredangle ARP=\measuredangle RCQ+\measuredangle PAR=\measuredangle PBQ.$ Next, composition of rotation $C\stackrel{Q}{\mapsto} R\stackrel{P}{\mapsto} A\stackrel{O}{\mapsto} C$ is identical, so by three rotations theorem $\measuredangle POQ=\frac{1}{2} \measuredangle AOC=\measuredangle PBQ.$ Now we see $\angle (OH;QP)=\measuredangle HPQ+\measuredangle OBP\stackrel{mod \text{ } \pi}{\equiv} \measuredangle RQP+\measuredangle QCR=\angle (AC;QP),$ so done.

We'll denote $\angle BAC=\alpha$, $\angle ABC=\beta$ and $\angle BCA=\gamma$. We have that: $$\angle APR=180^{\circ}-\angle PAR-\angle PRA=180^{\circ}-2\angle PAR=180^{\circ}-2\alpha$$$$\angle CQR=180^{\circ}-\angle QCR-\angle QRC=180^{\circ}-2\angle QCR=180^{\circ}-2\gamma$$Let $k_{1}$ be the circle with center $P$ and radius $PR$, i.e. $k_{1}=(P,PR)$ and $k_{2}$ be the circle with center point $Q$ and radius $QR$, that is, $k_{2}=(Q,QR)$. Then let $k_{1}\cap k_{2}=(R,D)$. Notice that: $$\angle ADC=\angle ADR+\angle RDC=\frac{1}{2}\angle APR+\frac{1}{2}\angle RQC=$$$$=(90^{\circ}-\alpha)+(90^{\circ}-\gamma)=\beta = \angle ABC$$We showed that $\angle ABC=\angle ADC$ and since $B$ and $D$ are in the same half-plane with respect to the line $AC$ this means that $ADBC$ is cyclic. Now since $PR=PD$ (radiuses in $k_{1}$) and $QR=QD$ (radiuses in $k_{2}$) we have that $PQ$ is the perpendicular bisector of $DR$. Therefore $\angle PDQ=\angle PRQ$ and $DR\perp PQ\perp RH$ since $H$ is the orthocenter of $\triangle PQR\Longrightarrow RH\parallel RD\Longrightarrow H\in RD$. Note that: $$\angle PDQ=\angle PRQ=180^{\circ}-\angle PRA-\angle QRC=180^{\circ}-\alpha-\gamma=\beta=\angle PBQ$$Therefore $PDBQ$ is cyclic. Notice that $\angle PHQ=180^{\circ}-\angle PRQ=180^{\circ}-\angle PDQ\Longrightarrow H\in \odot (PDQ)=\odot(PDBQ)$ (we proved that $PDBQ$ is cyclic). Also, since we showed that $ADBC$ is cyclic, then $OD=OB$ we have that: $$\angle DOB=2\angle DAB=2\angle DAP=\angle DAP+\angle PDA=\angle DPB$$$$\Longrightarrow O\in \odot(PDQ)\Longrightarrow D,P,O,H,Q,B\text{ lie on a circle}$$Now let $BO\cap AC=A'$. We have that: $$\angle OA'C=\angle BA'C=180^{\circ}-\angle A'BC-\angle A'CB=180^{\circ}-(\angle ABC-\angle OBA)-\gamma=$$$$=180^{\circ}-\left(\beta-\frac{180^{\circ}-\angle BOA}{2}\right)-\gamma=180^{\circ}-\beta+90^{\circ}-\frac{2\gamma}{2}-\gamma=90^{\circ}+\alpha-\gamma$$$$180^{\circ}-\angle HOA'=\angle BOH=\angle BPH=180^{\circ}-\angle APR-\angle RPH=$$$$=180^{\circ}-(180^{\circ}-2\alpha)-(90^{\circ}-\angle PRQ)=2\alpha-90^{\circ}+\angle PDQ=$$$$=2\alpha-90^{\circ}+\beta=90^{\circ}-\gamma+\alpha=\angle OA'C$$$$\Longrightarrow \angle OA'C+\angle HOA'=180^{\circ}\Longrightarrow \boxed{OH\parallel AC}$$

MatBoy-123 wrote: I got a very long solution using Moving Points(again) and angle chasing--- Can you or anyone else please tell how to construct the diagram on Geogebra for thiis prob... Like I was facing a lot of problem for constructing $AP=PR$ and $CQ=QR$. Please

[asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair A = dir(120), B = dir(210), C = dir(330), R = (3B+C)/4, P = extension((B+R)/2,(B+R)/2+dir(90),A,B), Q = extension((C+R)/2,(C+R)/2+dir(90),A,C), H = orthocenter(P,Q,R), O = origin, R1 = 2*foot(R,P,Q)-R; draw(A--B--C--A, blue); draw(R--Q--P--R, blue); draw(circumcircle(A,P,Q)^^unitcircle, lightblue); draw(R--R1, lightblue+dotted); dot("$A$", A, dir(130)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$R$", R, dir(270)); dot("$P$", P, dir(180)); dot("$Q$", Q, dir(0)); dot("$R'$", R1, dir(90)); [/asy][/asy] Switch to $A$-labeling. Let $R'$ be the reflection of $R$ over $\overline{PQ}$. Since \[\angle PR'Q = \angle PRQ = 180^\circ - (\angle PRB + \angle QRC) = 180^\circ - (\angle B + \angle C) = \angle A\]and \[PR' = PR = PB, \quad QR' = QR = QC \implies \frac{R'P}{PB} = 1 = \frac{R'Q}{QC},\]we have $R' = (APQ) \cap (ABC) \setminus A$ by Miquel points. Thus $\triangle R'BC \sim \triangle R'PQ \sim \triangle RPQ$, so if $H'$ is the orthocenter of $\triangle R'BC$, we have \[d(O,\overline{BC}) = \frac 12 R'H' = \frac 12 RH \cdot \frac{d(R',\overline{BC})}{d(R,\overline{PQ})} = RH \cdot \frac{d(R',\overline{BC})}{RR'} = d(H,\overline{BC})\]as desired.

Nice problem. The main claim is $B,O,P,Q$ are cyclic

This problem is very similar to USA TST 2012 #1 and interestingly can be solved in the same way I solved that problem. Let $AC \cap \odot(PQR)=X$ and let $O$ be the miquel point be $\odot(ARP \cap BPQ \cap CRQ)$ Notice that $\triangle XQR ~ \triangle BAC$ and because $\angle POX = 180^{\circ} - \angle BAC = 180^{\circ}-\angle XOP$ and similar expressions gives us that $O$ must be the orthocenter of $\triangle PQX$. Now it is trivial to see that $OH \parallel AC$ by simply reflecting the orthocenters about $PQ$ and noticing and applying reims theorem $\blacksquare$

BVKRB- wrote: let $O$ be the miquel point be $\odot(ARP \cap BPQ \cap CRQ)$ I guess you ment let $O$ be the miquel point be $\odot(AXP \cap BPQ \cap CXQ)$

How cum has anyone not discovered that it's a right angled traingle