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 $\alpha + \gamma = 30^\circ$.

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