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Sue Liu of Madras
College, St Andrew's sent this solution to the problem. Without
loss of generality we can let the length of $QR$ be 1 unit, and
take a coordinate system with the origin at $O$ and axes along $OR$
and $OQ$.
If $\angle PQR = \alpha$, where $0 < \alpha < 90^\circ$,
then $PQ = \cos \alpha$ and $PR = \sin \alpha.$ Let $\angle QRO =
\theta $ where $0 \leq \theta \leq 90^\circ$. Then, from the right
angled triangles $PSQ$ and $PTR$, we have $\angle PRT = \angle QPS
= \alpha  \theta$, and hence we can write down the coordinates of
the point $P$.
