G43 (21-4, USA) Given a point $P$ on a plane $\pi$ and a point $Q$ outside $\pi$. Find a point $R$ on $\pi$ such that the ratio $\frac{Q P+P R}{Q R}$ is maximized.
Let the projection of $Q$ on $\pi$ be $T$, $TP=a$, $QT=h$, where $a$ and $h$ are known quantities. Let $PR=x$,
then
$$
\begin{array}{c}
QP=\sqrt{h^{2}+a^{2}}, \\
QR=\sqrt{QT^{2}+TR^{2}} \\
\geqslant \sqrt{h^{2}+(a-x)^{2}} .
\end{array}
$$

Thus,
$$
\frac{QP+PR}{QR} \leqslant \frac{\sqrt{h^{2}+a^{2}}+x}{\sqrt{h^{2}+(a-x)^{2}}} .
$$

Using local adjustment, when $PR=x$ is fixed, we prove that $R$ must lie on the ray $PT$ (with a separate discussion when $P$ and $T$ coincide).

As shown in Figure 4-42, extend $TP$ to $S$ such that $QP=PS$, thus $S$ is a fixed point. Let $\angle S=\frac{\theta}{2}$ be a fixed angle. By the Law of Sines,
$$
\frac{QP+PR}{QR}=\frac{RS}{RQ}=\frac{\sin \angle RQS}{\sin \frac{\theta}{2}} \leqslant \frac{1}{\sin \frac{\theta}{2}} .
$$

Equality holds if and only if $\angle RQS=90^{\circ}$.
Thus, $R$ is constructed as follows:
(1) When $P$ does not coincide with $T$, let $QP$ form an angle $\theta$ with $\pi$, and the projection of $\theta$ on $\pi$ is $T$. $R$ lies on the ray $PT$, satisfying $\angle QRP=90^{\circ}-\frac{\theta}{2}$. At this time, $\frac{QP+PR}{QR}$ reaches its maximum value $\frac{1}{\sin \frac{\theta}{2}}$.
(2) When $P$ coincides with $T$, $R$ is any point on the circle in the plane $\pi$ with $P$ as the center and $QP$ as the radius. In this case, the maximum value is undoubtedly $\frac{1}{\sin 45^{\circ}}=\sqrt{2}$.