Problem 969

Question 94, Solution: As shown in the figure, establish a plane with 0 as the origin. Let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{C}}$. According to the conditions, $A$ and $C$ are both on the unit circle centered at 0, $|O B|=2$, and $\angle B O A=60^{\circ}$.
Extend $A O$ to intersect $\odot 0$ at $D$, and take the midpoint of $O D$ as $E$, then $\left|c+\frac{\vec{a}}{2}\right|=|E C|$.
On $O B$, take point $F$ such that $|O F|=\frac{1}{2}$, then $\frac{O F \mid}{|O C|}=\frac{|O C|}{|O B|}=\frac{1}{2}$, hence $\triangle O C F \sim \triangle O B C$, so $\frac{1}{2}|\vec{c}-\vec{b}|=\frac{|B C|}{2}=$ $|\mathrm{FC}|$. Therefore, we know:
$$\left|\vec{c}+\frac{\vec{a}}{2}\right|+\frac{1}{2}|\vec{c}-\vec{b}|=|\mathrm{EC}|+|\mathrm{FC}|$$
(1) First, find the minimum value. In fact, according to Ptolemy's theorem:
$$|\mathrm{EC}| \cdot|\mathrm{OF}|+|\mathrm{FC}| \cdot|\mathrm{OE}| \geq|\mathrm{EF}| \cdot|\mathrm{OC}| \Rightarrow|\mathrm{EC}|+|\mathrm{FC}| \geq \sqrt{3}$$

Equality holds if and only if $\angle C O D=60^{\circ}$.
(2) Next, find the maximum value. In fact,
$$\begin{array}{l} (|\mathrm{EC}|+|\mathrm{FC}|)^{2} \leq 2\left(|\mathrm{EC}|^{2}+|\mathrm{FC}|^{2}\right) \\ =2\left[\left(\mathrm{OC}{ }^{2}+\mathrm{OE}^{2}-2 \mathrm{OC} \cdot \mathrm{OE} \cdot \cos \angle \mathrm{COE}\right)+\left(\mathrm{OC}^{2}+\mathrm{OF}^{2}-2 \mathrm{OC} \cdot \mathrm{OF} \cdot \cos \angle \mathrm{COF}\right)\right] \\ =2\left(2+\frac{1}{2}-\cos \angle \mathrm{COE}-\cos \angle \mathrm{COF}\right) \\ \leq 2\left(2+\frac{1}{2}+2 \cos \frac{\angle \mathrm{BOD}}{2}\right)=7 \end{array}$$

Thus, $|\mathrm{EC}|+|\mathrm{FC}| \leq \sqrt{7}$, equality holds if and only if $\angle C O D=\angle C O B=120^{\circ}$.