We consider (LRC) .

**Conjecture** Suppose runners having distinct constant integer speeds start at a common point (origin) and run laps on a circular track with circumference 1. Then, there is a time when no runner is closer than from the origin.

Case of runners is discussed here. Here we look at case .

Let denote the distance of x to the nearest integer.

**Case ** . At time all runners are at least from the origin. Done.

**Case ** . There is time such that runners are at least from the origin – see case . At time runners are at the same positions as runners at time . If at time , we are done. Otherwise, since and 5 are co-prime, . Done.

**Case ** .

*Definition* Sector .

There is a time such that runners are at least from the origin – see case . Since and 5 are co-prime those runners visit all sectors once at times , and runners 1 and 2 will be at the same position. There 5 such times, during 2 times runner 3 will be at sectors and during 2 times runner 4 will visit the same sectors. Therefore, there is . Done.

Two previous cases follow from this post.

Now difficult case.

**Case ** .

Rearrange speeds that . If there more than one way choose the one with maximum sum. We have 2 sub-cases: , .

. The runner 1 is faster than either runner 2 or 3. Therefore, they meet at distance greater than .Now runner 4.

- this runner is slower than either runner 2 or 3 (call this runner ) which is at the same position as runner 4. Therefore, it will meet runner 1 later than runner , which meet runner 1 at distance larger than . Done.
- . This runner should move fast to reach runner 1 at distance . So, it will cover at least of the lap. Then, looking at opposite direction in time in the same time it will cover the same distance, and will be at least from the origin when runner 1 reach distance If it is moving even faster to meet runner 1 at it will miss runner 1 from the opposite side. Namely, it has to cover at least the whole lap, and therefore meet runner 1 at the opposite side before runner 1 reaches . Done.

- This is the most interesting case.
At times runners 2 and 3 are the same distance from the origin on the opposite sides from the origin.

Runner 1 exhibit repeated motion at . It makes cycles, where stands for greatest common divisor. The number of different positions it attain evenly distributed along the track is . If half of the time runner 1 is away from the origin. With larger there is bigger fraction, with the limit . Let is the time when runner 1 is distant. At the same time runners 2 and 3 same distance on the opposite sides from the origin. Now at times runner 1 stays at the same position, runners 2 and 3 are on the opposite position relative to the origin, and runner 4 visiting all sectors once each time. Therefore, there are total 2 moments when runner 4 in , and there are total, possibly different 2 times runner 2 in , on the other hand when runner 2 in sector 0 runner 3 is in sector 4 and vice verse. Therefore, there is . Done.

PS. If you find error or better exposition leave comment. If you do not want to be in the list of those talking to crackpot trisector, leave a note in the comment, the comment will not appear public, but the content will be implemented in the text.

PPS. There is a much shorter prove for in Barajas and Serra(2008). Algebraically it is short, but I think it lucks an intuition. I have filling this is quite general approach, and cases and bring new ingredients (non prime , and competition between 2 pairs of paired runners ). May be we need to wait for the new ingredient until .