Pigeonhole Principle - round stadium

Question

10 runners are in a round stadium. All of them start running from the same point at the same time. Each one runs in a constant speed $\ r_i>0 $. Prove that for every $\epsilon>0$ there is time $\ t>1 $ such that the distance of all the runners from the starting point is at most $\epsilon$. It should be done by the pigeonhole principle. I marked the total distance around the stadium in $\ x$ . I said the possible times, $\ t_i$ ,that the distance of runner $\ i $ is at most $\epsilon$ are $\frac{x}{r_i}p_i<t_i\le\frac{\epsilon}{r_i}+\frac{x}{r_i}p_i$, when $\ p_i\in\mathbb N $ is the number of full rounds the runner has made. How can I show there are $\ p_i$'s such that all the intervals have a common intersection point? Thanks

Answer 1

Divide the track length $x$ into $\left\lceil\frac x\epsilon\right\rceil$ segments. Let $S$ denote the set of these segments, and note where the runners are in $S^{10}$ once per time unit. Since this set has a finite number of elements, the runners will at some point have to return to an element they'd visited before. Now shift the movement such that the first time they visited they were all at the starting point. Then, since each segment has length at most $\epsilon$, on their second visit they are all at a distance at most $\epsilon$ from the starting point.