To 9. On a plane, there is a fixed point $P$, consider all possible equilateral triangles $ABC$, where $AP=3, BP=2$. What is the maximum length of $CP$? (1961 Autumn Competition)
Solve for Ling and Hui
$$
\begin{array}{l}
\angle A P B=\alpha . \\
\angle B A P=\beta,
\end{array}
$$

Given $A B^{2}=3^{2}+2^{2}$
$$
\begin{aligned}
- & 12 \cos \alpha, \\
\cos \beta & =\frac{3^{2}+A B^{2}-2^{2}}{6 A B}, \sin \beta=\frac{2 \sin \alpha}{A B} .
\end{aligned}
$$

From this, we get
$$
\begin{aligned}
\cos \left(\frac{\pi}{3}+\beta\right)= & \frac{1}{2} \cos \beta-\frac{\sqrt{3}}{2} \sin \beta \\
= & \frac{3^{2}+A B^{2}-2^{2}-2 \sqrt{3} \sin \alpha}{12 A B}=2 A B \\
= & \frac{3^{2}+3^{2}-12 \cos \alpha}{12 A B} . \\
& -\frac{2 \sqrt{3} \sin \alpha}{2 A B} .
\end{aligned}
$$
(Applying the Law of Cosines in $\triangle P A C$, we get
$$
\begin{aligned}
P C^{2} & =A C^{2}+3^{2}-6 A C \cos \left(\frac{\pi}{3}+?\right) \\
& =13-12 \cos \left(\frac{\pi}{3}+\alpha\right) .
\end{aligned}
$$

Clearly, when $a=\frac{2 \pi}{3}$, $P C$ reaches its maximum value of 5.