HMMT February

Extend $OA$ to $D$ and $OC'$ to $E$ such that $AD=OA'$ and $C'E=OC$. Since $OD=OE=2$ and $\angle DOE=60^{\circ}$, we have $ODE$ is an equilateral triangle. Let $F$ be the point on $DE$ such that $DF=OB$ and $EF=OB'$. Clearly we have $\triangle DFA \cong \triangle OBA'$ and $\triangle EFC' \cong OB'C$. Thus the sum of the areas of triangles $AOC', BOA'$, and $COB'$ is the same as the sum of the areas of triangle $DFA, FEC'$, and $OAC'$, which is at most the area of triangle $ODE$. Since $ODE$ is an equilateral triangle with side length 2, its area is $\sqrt{3}$. Equality is achieved when $OC=OA'=0$.