Problem 1571

1. Denote the angles of triangle $ABC$ as $\angle BAC = \alpha$, $\angle ABC = \beta$, and $\angle BCA = \gamma$.
2. Since $AP = PR$ and $CQ = QR$, we have:
$$\angle APR = 180^\circ - \angle PAR - \angle PRA = 180^\circ - 2\alpha$$
$$\angle CQR = 180^\circ - \angle QCR - \angle QRC = 180^\circ - 2\gamma$$
3. 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 $Q$ and radius $QR$, i.e., $k_2 = (Q, QR)$. Let $k_1 \cap k_2 = \{R, D\}$.
4. Notice that:
$$\angle ADC = \angle ADR + \angle RDC = \frac{1}{2} \angle APR + \frac{1}{2} \angle RQC = (90^\circ - \alpha) + (90^\circ - \gamma) = \beta = \angle ABC$$
5. Since $\angle ABC = \angle ADC$ and $B$ and $D$ are in the same half-plane with respect to the line $AC$, this means that $ADBC$ is cyclic.
6. Since $PR = PD$ (radii in $k_1$) and $QR = QD$ (radii 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 $\triangle PQR \Longrightarrow RH \parallel RD \Longrightarrow H \in RD$.
7. Note that:
$$\angle PDQ = \angle PRQ = 180^\circ - \angle PRA - \angle QRC = 180^\circ - \alpha - \gamma = \beta = \angle PBQ$$
Therefore, $PDBQ$ is cyclic.
8. 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).
9. Since we showed that $ADBC$ is cyclic, then $OD = OB$. We have that:
$$\angle DOB = 2\angle DAB = 2\angle DAP = \angle DAP + \angle PDA = \angle DPB$$
$$\Longrightarrow O \in \odot(PDQ) \Longrightarrow D, P, O, H, Q, B \text{ lie on a circle}$$
10. Now let $BO \cap AC = A'$. We have that:
$$\angle OA'C = \angle BA'C = 180^\circ - \angle A'BC - \angle A'CB = 180^\circ - (\angle ABC - \angle OBA) - \gamma$$
$$= 180^\circ - \beta + 90^\circ - \frac{2\gamma}{2} - \gamma = 90^\circ + \alpha - \gamma$$
11. We also have:
$$180^\circ - \angle HOA' = \angle BOH = \angle BPH = 180^\circ - \angle APR - \angle RPH$$
$$= 180^\circ - (180^\circ - 2\alpha) - (90^\circ - \angle PRQ) = 2\alpha - 90^\circ + \angle PDQ$$
$$= 2\alpha - 90^\circ + \beta = 90^\circ - \gamma + \alpha = \angle OA'C$$
12. Therefore:
$$\angle OA'C + \angle HOA' = 180^\circ \Longrightarrow \boxed{OH \parallel AC}$$