Example 9 Consider on the same plane I: two concentric circles with radii $R^{1} \cdot \mathrm{j} r$ $(R>r)$. Let $P$ be a fixed point on the circumference of the smaller circle, and $B$ be a moving point on the circumference of the larger circle. The line $B P$ intersects the smaller circle at another point $C$. The line $l$ through point $P$ perpendicular to $B P$ intersects the smaller circle at another point $A$ (if $l$ is tangent to the smaller circle at $P$, then $A=P$).
(1) Find the set of values taken by the expression $B C^{2}+C A^{2}+A B^{2}$;
(2) Find the locus of the midpoint of segment $A B$.
(29th IMO)
Solution: (1) As shown in the figure, construct the parallelogram $B C C^{\prime} B^{\prime}$, and let $B^{\prime} C^{\prime}$ pass through point $A$. Suppose the intersection point of $B^{\prime} C^{\prime}$ on the smaller side is $A^{\prime}$. Since the distance from $O$ to $BC$ is equal to the distance from $O$ to $AA^{\prime}$, we have $z_{B}+z_{C}+z_{A}+z_{A^{\prime}}=0$, which means $z_{A}+z_{B}+z_{C}=-z_{A^{\prime}}$. Therefore,
$$
\begin{array}{l}
B C^{2}+C A^{2}+A B^{2} \\
=\left(z_{B}-z_{B}\right)\left(\bar{z}_{C}-\bar{z}_{B}\right)+\left(z_{A}-z_{A}\right)\left(\bar{z}_{A}-\bar{z}_{B}\right) \\
+\left(z_{B}-z_{A}\right)\left(\bar{z}_{B}-\bar{z}_{A}\right) \\
= 4 R^{2}+2 r^{2}-\left(z_{B} \cdot \bar{z}_{A}+z_{A} \cdot \bar{z}_{B}+z_{\cdot} \cdot \bar{z}_{B}+z_{B} \cdot \bar{z}_{\cdot}+z_{A} \cdot \bar{z}_{C}+z_{C} \cdot \bar{z}_{A}\right) \\
= 6 R^{2}+3 r^{2}-\left(z_{A}+z_{B}+z_{B}\right)\left(\bar{z}_{A}+\bar{z}_{B}+\bar{z}_{1}\right) \\
= 6 R^{2}+2 r^{2} . \text { (final value) }
\end{array}
$$

When $A$ coincides with $P$, it is easy to verify that $B C^{2}+C A^{2}+A B^{2}$ also equals $6 R^{2}+2 r^{2}$. Therefore, the set of values for the expression is $\left\{6 R^{2}+2 r^{2}\right\}$.
(2) Since $P B B^{\prime} A$ is also a parallelogram, $AB$ intersects $Q$ which is the midpoint of $P B^{\prime}$, so $z_{L^{\prime}}+z_{T}=2 z_{Q}$, then $z_{k}=2 z_{Q}-z_{P}$.
Since $B^{\prime}$ is on the larger circle, we have $\left|z_{B}\right|=R$, thus
$\left|2 z_{Q}-z_{P}\right|=R$, which means $\left|z_{Q}-\frac{z_{P}}{2}\right|=\frac{R}{2}$.
This indicates that the trajectory is a circle with the midpoint of line segment $OP$ as the center and $\frac{R}{2}$ as the radius.