, 2011

a. Note that $R$ is the circumcentre of triangle $BHC$, hence $OR$ bisects $BC$. Also, $\angle BRC = 2\angle(180^\circ - \angle BHC) = 2\angle QRP = 2\angle A = \angle BOC$, so it follows that $R$ is the reflection of $O$ across the line $BC$.

Next, since $QR$ is perpendicular to $BH$, which is perpendicular to $AC$, we have that $AC \parallel QR$, and hence $\angle BQR = \angle A = \angle BOR$, which shows that $BQOR$ is a cyclic quadrilateral. Using the sine rule in triangle $BQR$ we see that the circumradius of triangle $BQR$ is equal to $\frac{BR}{2\sin BQR} = \frac{R}{2\sin A}$, where $R$ is the circumradius of triangle $ABC$. Similarly, $CPOR$ is a cyclic quadrilateral with circumradius $\frac{R}{2\sin A}$.

Let the line through $O$ perpendicular to $BH$ meet $BH$ in $K$ and $AB$ in $L$, let the foot of the altitude from $B$ be $E$ and let the midpoint of $BH$ be $M$. It is well known that $AH = OR$, and hence, since $AH \parallel OR$ and $QR \parallel OK \parallel AC$, it follows that $HE = MK \implies KE = MH = BM \implies AL = BQ$. Hence $L$ and $Q$ are symmetric about the midpoint of $AB$, and so $OQ = OL$. Hence, since $OL \parallel AC$, $\angle OQA = \angle OLQ = \angle A$.

Using the sine rule in triangle OAQ shows that it has circumradius
$\frac{AO}{2 \sin \angle AQO} = \frac{R}{2 \sin A}$. Similarly, $\angle OPA = \angle A$ and triangle OAP has circumradius $\frac{R}{2 \sin A}$.

These radii are equal to $R$ if and only if $\sin A = \frac{1}{2}$, i.e. if and only if $\angle A = 30^\circ$.

b. Since $QR$ is the perpendicular bisector of $BH$, we have that $\angle HQR = \angle BQR = \angle A$. Considering the opposite angles of these two angles and recalling that $\angle AQO = \angle A$, we have that $\angle HQO + 180^\circ = 3\angle A \Rightarrow \angle HQO = 3\angle A - 180^\circ$. Next, since $\angle OQA = \angle A = \angle OPQ$, it follows that $\angle POQ = 360^\circ - 3\angle A$. Hence $\angle POQ + HQO = 360^\circ - 3\angle A + 3\angle A - 180^\circ = 180^\circ$, which shows that $OP \parallel QH$. Hence, if $PQ$ passes through $O$, it also passes through $H$. In this case, $\angle HQO = 0^\circ = 3\angle A - 180^\circ$, implying that $\angle A = 60^\circ$.