The South African Mathematical Olympiad Third Round

Let $AQ$ extended meet the circumcircle of triangle $ABC$ in $T$, and let the line through $Q$ and $R$ intersect $PC$ in $D$. See Figure 2: Figure 2 Put $\angle BAQ=\angle CAQ=\alpha$, so that also $\angle BCT=\alpha$ and $\angle TBC=\alpha$. Put $\angle BCA=\theta$, so that also $\angle BTA=\theta$. Put $\angle ACP=\varphi$, so that also $\angle ABP=\varphi$.

Then $\angle PBC=\angle PCB$ (since $BP=CP)=\theta +\varphi$. From $\angle PBT+\angle TCP=180^{\circ }$ (since $BPCT$ is a cyclic quadrilateral), we therefore have $2(\alpha +\theta +\varphi )=180^{\circ }$, giving $\alpha +\theta +\varphi =90^{\circ }$. Since triangle $QPD$ is isosceles ($QD\parallel BC$, so that $\angle PQD=\angle PBC=\angle PCB=\angle PDQ$), we have $PQ=PD$, forcing $BQ=CD$. Furthermore, chords $BT$ and $CT$ both subtend the same angle $\alpha$ at $A$, hence are equal in length. It follows that $△TBQ\equiv △TCD$, hence $\angle DTC=\angle QTB=\theta$. But since $\angle DRC=\angle BCR=\theta$ (recall that $QD\parallel BC$), we see that $RDCT$ is a cyclic quadrilateral. This implies that $\angle TRD=90^{\circ }$ (because $\angle TCD=\alpha +\theta +\varphi =90^{\circ }$). So $\angle TRQ=90^{\circ }$, and we also have that $RQBT$ is a cyclic quadrilateral. (Recall that $\angle QBT=90^{\circ }$.) It follows that $\angle QRB=\angle QTB=\theta$, and we thus have $\angle RBC=\theta$, since $QD\parallel BC$. Finally, $BR=RC$ follows from the fact that $\angle RBC=\angle RCB=\theta$.

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Alternative solution.

Let the line through $R$ and $Q$ intersect $AB$ and $PC$ in $S$ and $T$, respectively. See Figure 3: Figure 3 Put $\gamma =\angle PBC=\angle PCB$. Then also $\angle PAC=\gamma$. From the fact that $ST\parallel BC$, it follows that $\angle BQS=\angle QBC=\gamma$, so that $\angle PQR=\gamma$. Thus $PAQR$ is a cyclic quadrilateral. Hence $\angle RPQ=\angle RAQ=\alpha$, and since $\angle CPB=\angle CAP=2\alpha$, we have $\angle RPT=\alpha$. So $QR=TR$.

We see that $△BQR\equiv △CTR$ ($\angle BQR=\angle CTR=180^{\circ }-\gamma$, $QR=RT$ and $BQ=CT$ (from $PB=PC$ and $PQ=PT$)). Consequently, $BR=CR$.