On the lonely runner conjecture III


We consider (LRC) .

Conjecture Suppose n>1 runners having distinct constant integer speeds v_k 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 \frac{1}{n+1} from the origin.

Case of n=2,3 runners is discussed here. Here we look at case n=4.

Let \| x \| denote the distance of x to the nearest integer.

Case v_k \mod 5 \neq 0,\ \forall k. At time t=\frac{1}{5} all runners are at least \frac{1}{5} from the origin. Done.

Case v_k= 5 r_k, k=1..3, v_4 \mod 5 \neq 0. There is time t_1 such that runners r_k, k=1..3 are at least \frac{1}{4} from the origin – see case n=3. At time t_2= \frac{t_1}{5} runners v_1, v_2, v_3 are at the same positions as runners r_1, r_2, r_3 at time t_1. If at time t_2, \| v_4 t_2\| \geq \frac{1}{5} we are done. Otherwise, since v_4 and 5 are co-prime, \exists m : \| v_4 \frac{m}{5} \|= \frac{2}{5} \Rightarrow \| \frac{2}{5} - v_4 \left( t_2+\frac{m}{5} \right) \| \leq \frac{1}{5} . Done.

Case v_k= 5 s_k, k=1,2, v_k \mod 5 \neq 0, k=3,4.

Definition Sector S_k= \left( \frac{k}{5}, \frac{k+1}{5} \right), k=0..4 .
There is a time t_1 such that runners s_k, k=1,2 are at least \frac{1}{3} from the origin – see case n=2. Since v_k, k=3,4 and 5 are co-prime those runners visit all sectors once at times \frac{m}{5}, m=0..4, and runners 1 and 2 will be at the same position. There 5 such times, during 2 times runner 3 will be at sectors S_0, S_4 and during 2 times runner 4 will visit the same sectors. Therefore, there is m:\ \| v_k \left( \frac{t_1+m}{5} \right) \| > \frac{1}{5}. Done.

Two previous cases follow from this post.
Now difficult case.

Case v_1= 5 s_1,\ v_k \mod 5 \neq 0, k=2,3,4.

Rearrange speeds that v_2+v_3 = 5 s_2 . If there more than one way choose the one with maximum sum. We have 2 sub-cases: s_1 \geq s_2, s_1< s_2.

  1. s_1 \geq s_2
    \exists m:\ \| v_2 \frac{m}{5} \|=\| v_3 \frac{m}{5} \| = \frac{2}{5}. The runner 1 is faster than either runner 2 or 3. Therefore, they meet at distance greater than \frac{1}{5}.

    Now runner 4.

    1. \| v_4 \frac{m}{5} \| = \frac{2}{5} \Rightarrow this runner is slower than either runner 2 or 3 (call this runner r) which is at the same position as runner 4. Therefore, it will meet runner 1 later than runner r, which meet runner 1 at distance larger than \frac{1}{5}. Done.
    2. \| v_4 \frac{m}{5} \| = \frac{1}{5} \Rightarrow . This runner should move fast to reach runner 1 at distance \frac{1}{5}. So, it will cover at least \frac{3}{5} + l 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 \frac{2}{5} from the origin when runner 1 reach distance \frac{1}{5} If it is moving even faster to meet runner 1 at \frac{1}{5} 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 \frac{1}{5} . Done.
  2. s_1< s_2 This is the most interesting case.

    At times t=\frac{m}{5s_2}, m=1..5s_2-1 runners 2 and 3 are the same distance from the origin on the opposite sides from the origin.

    Runner 1 exhibit repeated motion at t. It makes 5 \gcd ( s_1,s_2) cycles, where \gcd stands for greatest common divisor. The number of different positions it attain evenly distributed along the track is n_1=\frac{s_2}{\gcd ( s_1,s_2)} \geq 2 . If n_1=2 half of the time runner 1 is away from the origin. With larger n_1 there is bigger fraction, with the limit \frac{3}{5}. Let t_1 \in t 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 t_1+\frac{m}{5}, m=0..4 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 S_0, S_4, and there are total, possibly different 2 times runner 2 in S_0, S_4, on the other hand when runner 2 in sector 0 runner 3 is in sector 4 and vice verse. Therefore, there is m:\ \| v_k \left( t_1 + \frac{m}{5} \right)\| \geq 1. Done.

\blacksquare

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 n=4 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 n=5 and n=6 bring new ingredients (non prime n+1, and competition between 2 pairs of paired runners ). May be we need to wait for the new ingredient until n=3 \times 5 -1 = 14 .

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One Response to On the lonely runner conjecture III

  1. Tim Roberts says:

    There is a prize of $1000 on offer for a valid solution to the Lonely Runner conjecture. For details see the Unsolved Problems web site at
    http://unsolvedproblems.org/

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