## On the lonely runner conjecture II

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.

We first show trivial proves for $n=2,3$ runners.

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

Let $\kappa_n = \inf \max \limits_{0 \leq t \leq 1} \min \limits_{ 1 \leq i \leq n} \| v_i t \|$. where the infimum is taken over all n-tuples $v_1, ... v_n$ of distinct positive integers.

In terms of $\kappa_n$ the LRC state that $\kappa_n= \frac{1}{n+1}$

Without loss of generality (wlog) we can assume $k < m \Leftrightarrow v_k < v_m$.

Case $n=2$. Wlog we can assume $v_1, v_2$ are relatively prime. At time $t_k= \frac{m}{v_1+v_2}, m=1..v_1+v_2-1$ two runners are at the same distance from the origin from different sides of the lap at distances proportional to $\frac{1}{v_1+v_2}$. Since they are relatively prime both runners visit all the points $\frac{m}{v_1+v_2}, m=1..v_1+v_2-1$. The largest distance is $\frac{ \left[ \frac{v_1+v_2}{2} \right] }{v_1+v_2}$, where $\left[ \circ \right]$ means integer part.

For example, $v_1=1, v_2= 2$, at times $\frac{1}{3}, \frac{2}{3}$ runners are at positions $(\frac{1}{3}, \frac{2}{3})$ and $(\frac{2}{3},\frac{1}{3})$ and the maximum distance from origin is $\frac{\left[ \frac{1+2}{2}\right]}{3}= \frac{\left[1.5 \right]}{3}= \frac{1}{3}$.

Another example, $v_1= 2, v_2= 3$, the maximum distance is $\frac{2}{5} > \frac{1}{3}$.

If $v_1, v_2 \mod 2 =1$ the maximum distance is $\frac{1}{2} > \frac{1}{3}$ at time $\frac{1}{2}$.

In general the maximum distance is $r_{1,2}= \frac{1}{2} \left(1- \frac{ (v_1+v_2) \mod 2 }{ v_1+v_2} \right)$ with the minimum for runners at speeds $v_1= 1, v_2= 2$. Therefore, $\kappa_2= \frac{1}{3}$.

Case $n=3$.

• First we assume simple case: $v_3 \mod ( v_1+v_2 )= 0$, e.g $v_3= q(v_1+v_2), q \in \mathbb{N}_+$.

At times $t= \frac{m}{v_1+v_2}$ runner 3 is at the origin. There is a time such that some time before it was the same distance with runner 1 and some time after it will be the same distance with runner 2. (The situation is reversed at times $1-t$). We are interested when that is happening for the time runners 1 and 2 are most distant from the origin.

For runners 1 and 3: They need to pass $r_{1,2}$ with speed $v_1+v_3$, so the time needed is $\frac{r_{1,2} }{v_1+v_3}$. Runner 3 passes $r_{1, 3}= \frac{v_3 r_{1,2} }{v_1+v_3}= r_{1,2} \frac{1}{1+v_1 / v_3 }$. Therefore, the closest point is reached when $\frac{v_1}{v_3}= \frac{v_1}{t (v_1+v_2) }$ is maximum.

$r_{1, 3}= \frac{q ( v_1+v_2 ) }{v_1+ q ( v_1+v_2 ) } \frac{1}{2} \left(1- \frac{ (v_1+v_2) \mod 2 }{ v_1+v_2} \right)$

If $(v_1+v_2) \mod 2 = 0$, $r_{2, 3} < r_{1, 3}$ around $r_{1,2}= \frac{1}{2}$

$r_{2, 3}= \frac{1}{2} \frac{q ( v_1+v_2 ) }{v_2+ q ( v_1+v_2 ) }$

Lemma 1. $\frac{q ( v_1+v_2 ) }{v_2+ q ( v_1+v_2 ) } > \frac{1}{2}$.
Proof. $\frac{q ( v_1+v_2 ) }{v_2+ q ( v_1+v_2 ) } - \frac{1}{2} =$ $\frac{1}{2} \frac{ 2 q ( v_1+v_2 ) - v_2 - q ( v_1+v_2 )}{v_2+ q ( v_1+v_2 ) } =$ $\frac{1}{2} \frac{ q v_1 + ( q - 1) v_2 }{v_2+ q ( v_1+v_2 ) } > 0$ $\blacksquare$

Now, consider the case when $(v_1+v_2) \mod 2 = 1$.

$r_{1, 3}= \frac{1}{2} \frac{q ( v_1+v_2 ) }{v_1+ q ( v_1+v_2 ) } \frac{ v_1+v_2 - 1 }{ v_1+v_2} = \frac{1}{2} \frac{q ( v_1+v_2 - 1 ) }{v_1+ q ( v_1+v_2 ) }$

Lemma 2. $\frac{q ( v_1+v_2 - 1 ) }{v_1+ q ( v_1+v_2 ) } \geq \frac{1}{2}$
Proof. $\frac{ ( v_1+v_2 - 1 ) }{v_1/q + v_1+v_2 } - \frac{1}{2} =$ $\frac{2 v_1+ 2 v_2 - 2 - v_1/q - v_1-v_2 }{2 \left[ v_1/ q+ ( v_1+v_2 ) \right] } =$ $\frac{ v_1 \left( 1- \frac{1}{q} \right)+ v_2 - 2 }{2 \left[ v_1/ q+ ( v_1+v_2 ) \right] }$. Since $v_2 \geq 2$ nominator is greater than 0, except when $v_2=2, v_1= 1, q= 1$ $\blacksquare$

Corollary 3. $r_{1, 3} \geq \frac{1}{4}$.
Corollary 4. $r_{1, 3} = \frac{1}{4}$ only for $v_1=1, v_2=2, v_3=3$.

• Case: $v_3 \mod ( v_1+v_2 ) \neq 0$.

Lemma 5. Fix time $t= \frac{m}{v_1+v_2}, m \in \{1, ..., v_1+v_2-1 \}$. Let $r_3$ be the distance from the origin of runner 3 at the moment $t$ and $r_{1,2} > r_3$ be the distance of the runners 1 and 2 from the origin at the same time. The maximal distance when runners 1 and 2 or 1 and 3 equidistant around time $t$ is greater than $\frac{r_3+r_{1, 2}}{2}$.
Proof. Either the runner 3 is running toward or away from the origin. e.g. the distance to the origin either decrease with time or increase. Let $v$ be the speed of the runner 1 or runner 2 moving in opposite direction, so that the difference in distances is decreasing moving either forward or backward in time. Since $v_3 > v$ runner 3 will cover greater distance away from the origin that other runner $\blacksquare$

Lemma 6. Fix the time $t= \frac{m}{v_1+v_2}$ when $r_{1,2}= \frac{1}{2} \left(1- \frac{ (v_1+v_2) \mod 2 }{ v_1+v_2} \right)$. Let $r'_3= \frac{m v_3 \mod (v_1+v_2) }{v_1+v_2}$, and $r_3= \min (r'_3, 1-r'_3)= \| r'_3\|$ – distance of runner 3 from the origin. $r_{1, 2} + r_3 > \frac{1}{2}$.
Proof. $r_3 \geq \frac{1}{v_1+v_2}$, $r_{1,2}+r_3 \geq \frac{1}{2} \left(1- \frac{ (v_1+v_2) \mod 2 }{ v_1+v_2} \right) + \frac{1}{v_1+v_2} =$ $\frac{1}{2} + \frac{2- (v_1+v_2) \mod 2 }{v_1+v_2} > \frac{1}{2}$ $\blacksquare$

Combining Lemma 5 and Lemma 6 shows $\max (r_{1,3}, r_{2,3}) > \frac{1}{4}$.

Overall, $\kappa_3= \frac{1}{4}$ only for runners $v_1=1, v_2=2, v_3=3$.