62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Answer: yes.

Arrange the trains clockwise as following: $B_8,B_1,B_7,B_2,B_6,B_3,B_5$ and $B_4$. Now let's show how they should move to meet in time.

Let $B_8$ move through the station of $B_1$ (at this time $B_1$ stays at place and is waiting for $B_8$), and they arrive to $O$ in time $t_8=\frac{3}{8}$.

Let train $B_7$ move through the station of $B_2$ (at this time $B_2$ stays at place and is waiting for $B_7$), and they arrive to $O$ in time $t_7=\frac{3}{7}$.

Let train $B_6$ move through the station of $B_3$ (at this time $B_3$ stays at place and is waiting for $B_6$) and they arrive to $O$ in time $t_7=\frac{1}{2}$.

$B_5$ goes to $O$ right away in time $t_5=\frac{2}{5}$.

$B_4$ goes to $O$ right away in time $t_4=\frac{1}{2}$.

Now we see that two groups — $B_6+B_3$ and $B_4$ — arrive at the time $\frac{1}{2}$, but the groups $B_8+B_1$ and $B_7+B_2$ arrive faster than in $\frac{1}{2}$, and their speeds are larger than those of first groups. So, the group $B_8+B_1$ upon its arrival to $O$ goes to the group $B_6+B_3$, unites with it, and goes to the point $O$ with its speed. Similarly, group $B_7+B_2$ will get $B_4$ to $O$ faster.

It's possible to find the time of arrival of each group to $O$. For example, for the last group this time is $\frac{37}{77}<\frac{1}{2}$. Indeed, group $B_7+B_2$ arrives at $O$ at $\frac{3}{7}$, at this time $B_4$ has travelled $\frac{12}{7}$ in the direction of point $O$. The time before their meet is $\frac{2}{4+7}=\frac{2}{77}$, and it's equal to the time spent to arrive back to $O$.