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

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{x}& =& \cos \alpha \cos (\alpha -\theta )\\ \multicolumn{1}{c}{y}& =& \sin \alpha \cos (\alpha -\theta ).\end{array}}$

We see that

${\displaystyle \frac{y}{x}=\frac{\sin \alpha \cos (\alpha -\theta )}{\cos \alpha \cos (\alpha -\theta )}=\tan \alpha .}$

and so $P$ lies on the straight line $y=x\tan \alpha$.

The position $(x,y)$ depends only on $\cos (\alpha -\theta )$, $\alpha$ being a constant, and $\theta$ a variable. The distance of the point $P$ from $O$ is given by

${\displaystyle {\mathrm{OP}}^2=x^2+y^2={\cos }^2(\alpha -\theta )({\cos }^2\alpha +{\sin }^2\alpha )={\cos }^2(\alpha -\theta ).}$

Hence $\mathrm{OP}=\cos (\alpha -\theta )$ which is a maximum when $\cos (\alpha -\theta )=1$, that is when $\alpha =\theta$. This occurs when $\mathrm{OQPR}$ is a rectangle as shown in the diagram. "


We get an even simpler method of solution by using the fact that the angles $\mathrm{QOR}$ and $\mathrm{QPR}$ are both 90 degrees so that $\mathrm{OQPR}$ is a cyclic quadrilateral with $\mathrm{PR}$ as a chord. We have $\angle \mathrm{POR}=\angle \mathrm{PQR}=\alpha$ because these two angles are subtended by the same chord of the circle. This shows that $\angle \mathrm{POR}$ is constant and hence that the locus of $P$ is the straight line $y=x\tan \alpha .$

What can you say about the locus of $P$ if the triangle $\mathrm{PQR}$ is not a right angled triangle?