11.1. Two runners, starting simultaneously at constant speeds, run on a circular track in opposite directions. One of them runs the loop in 5 minutes, while the other takes 8 minutes. Find the number of different meeting points of the runners on the track, if they ran for at least an hour.

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# Solution.

Let the length of the track be $\mathrm{S}$ meters. Then the speeds of the runners are $\mathrm{S} / 5$ and $\mathrm{S} / 8$ meters per minute, respectively.

Assume the runners start from the same point. They will meet again after $\mathrm{S} /(\mathrm{S} / 5+\mathrm{S} / 8)=40 / 13$ minutes.

Now, let's determine when they will meet again at the starting point. If the first runner completes the circular track $\mathrm{n}$ times and the second runner completes it $\mathrm{m}$ times, then by comparing the time, we get $5 \mathrm{n}=8 \mathrm{~m}, \mathrm{n}=8 \mathrm{a}, \mathrm{m}=5 \mathrm{~b}$, where $\mathrm{a}, \mathrm{b}$ are natural numbers.

The smallest possible values of $\mathrm{a}$ and $\mathrm{b}$ are 8 and 5, respectively. Therefore, they will both return to the starting point for the first time after $5 \cdot 8=40$ minutes. They meet every $40/13$ minutes on the track. In total, there will be $40 : (40 / 13)=13$ different meeting points.

If they start from different points, they will meet no later than $40/13$ minutes. Subsequently, events will unfold similarly to case 1 with a period of 40 minutes. Since $40 + 40/13$ minutes is less than an hour, there will be exactly 13 meeting points in an hour.

Answer: 13 points.

## Grading Criteria:

7 points - Both cases of the runners' movement are considered and justified.

5 points - Only one case of the runners' movement is considered and justified.