Answer: $\sqrt{3}$
Extend $O A$ to $D$ and $O C^{\prime}$ to $E$ such that $A D=O A^{\prime}$ and $C^{\prime} E=O C$. Since $O D=O E=2$ and $\angle D O E=60^{\circ}$, we have $O D E$ is an equilateral triangle. Let $F$ be the point on $D E$ such that $D F=O B$ and $E F=O B^{\prime}$. Clearly we have $\triangle D F A \cong \triangle O B A^{\prime}$ and $\triangle E F C^{\prime} \cong O B^{\prime} C$. Thus the sum of the areas of triangles $A O C^{\prime}, B O A^{\prime}$, and $C O B^{\prime}$ is the same as the sum of the areas of triangle $D F A, F E C^{\prime}$, and $O A C^{\prime}$, which is at most the area of triangle $O D E$. Since $O D E$ is an equilateral triangle with side length 2 , its area is $\sqrt{3}$. Equality is achieved when $O C=O A^{\prime}=0$.
9. [6] Segments $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$, each of length 2, all intersect at a point $O$. If $\angle A O C^{\prime}=\angle B O A^{\prime}=$ $\angle C O B^{\prime}=60^{\circ}$, find the maximum possible value of the sum of the areas of triangles $A O C^{\prime}, B O A^{\prime}$, and $C O B^{\prime}$.