Mongolian Mathematical Olympiad

Denote $\angle BAP=\angle PAR=\angle RAC=\alpha$. Then $\angle BAC=3\alpha$. Denote $\angle BCQ=\angle QCR=\angle RCA=\gamma$. Then $\angle ACB=3\gamma$. Since $\angle ABC=90^{\circ }$, $\angle ACB+\angle BAC=180^{\circ }-90^{\circ }=90^{\circ }\Rightarrow 3\alpha +3\gamma =90^{\circ }$ and $\alpha +\gamma =30^{\circ }$.

$△ACR$ : $\angle ARC=180^{\circ }-(\alpha +\beta )=150^{\circ }$. Let $(CQ)\cap (AP)=S$. Since $AR$, $CR$ bisectors of the $△ASC$, $SR$ also bisector. $\angle ASR=\angle CSR=60^{\circ }\Rightarrow \angle ASQ=60^{\circ }$. It implies $△ASR=△ASQ$. Hence $SQ=SR$ and $△SQR$ is isosceles. Therefore $\angle SQR=\angle SRQ=30^{\circ }$. $△CRQ$ : $\gamma =180^{\circ }-(142^{\circ }+30^{\circ })=8^{\circ }\Rightarrow \alpha +\gamma =30^{\circ }$ and $\alpha =22^{\circ }$. Thus $\angle BAC=3\alpha =66^{\circ }$.