smc

Let the speed of $A$ be $a$ and the speed of $B$ be $b$. The first time that $A$ and $B$ will be equidistant from $O$, $B$ will have not yet reached $O$. Thus, after two minutes, $B$'s distance from $O$ will be $500-2b$, and $A$'s distance from $O$ will be $2a$. Setting these expressions equal to each other and dividing by 2, we see that $a=250-b$. After another eight minutes (or after a total of ten minutes since $A$ was at $O$), $A$ and $B$ will again be equidistant from $O$, but this time $B$ will have passed $O$. The distance $A$ will be from $O$ is $10a$, and the distance $B$ will be from $O$ is $10b-500$. Setting these expressions equal to each other and dividing by 10, we see that $a=b-50$. Adding the two equations that we have obtained above, we see that $2a=250-b+b-50$, and so $a=100$. Substituting this value of $a$ into the second equation, we see that $100=b-50$, or $b=150$. Then, $\frac{a}{b}=\frac{100}{150}=\frac{2}{3}$, so the ratio of $A$'s speed to that of $B$ is $\boxed{2:3}$.