Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it Mongolia

If $\angle CBA = 90^\circ$ , $\angle BAP = \angle PAR = \angle RAC$ , $\angle BCQ = \angle QCR = \angle RCA$ , $\angle QRC = 142^\circ$ then find the angle $\angle BAC$ .

Solution

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
$\angle BAC = 3\alpha = 66^\circ$ .

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