Problem 4

If $P\ne X$, there is exactly one point $R$ on the ray $PX$ satisfying the right-angle condition. If $P=X$, write $h=PQ$; then the ratio is $\frac{h+PR}{\sqrt{h^2+PR^2}}$, whose maximum occurs exactly when $PR=h$. Thus every point of the circle in $\pi$ centered at $P$ with radius $PQ$ is maximizing.