Example 2. Given a plane $\pi$, point $P$ is on $\pi$, and point $Q$ is outside $\pi$. Find point $R$ in the plane $\pi$ such that the ratio $\frac{Q P+R P}{Q R}$ is maximized.
Let $R$ be a point on $\pi$, and the angles of $\triangle P G R$ be $\alpha, \beta, \gamma$. Then, by the Law of Sines, we have
$$
\frac{Q P+R P}{Q R}=\frac{\sin \gamma+\sin \beta}{\sin \alpha}=\frac{\cos \frac{\beta-\gamma}{2}}{\sin \frac{\alpha}{2}} .
$$

The numerator is maximized when $\beta=\gamma$, and the denominator is minimized when $\alpha$ is the smallest angle $\delta$ between $Q P$ and $\pi$. Therefore, the maximum value of $\frac{Q P+P R}{Q P}$ is $\frac{1}{\sin \frac{\delta}{2}}$, which is achieved when $R$ lies on the projection of $P Q$ onto $\pi$, and $P R = P Q$.

In particular, when the projection of $P Q$ onto $\pi$ is $P$, $R$ can be any point on the circumference of the circle centered at $P$ with radius $P Q$.