The other day I attempted to prove under rather lax conditions that

~~There exists a time t s.t. all the moving runners are exactly at a distance 1/2 from the still runner.~~

Turns out, I can't (and I think my claim is provably false). The trouble is with any runner who's speed is a factor of 2. But I think I can show that the strategy works absent such a runner. So let me update the claim.

If the runners' speeds are odd relative primes then there exists a time t s.t. all the moving runners are exactly at a distance 1/2 from the still runner.

In fact, let us describe an algorithm for calculating this time. As before, we consider time in discrete steps. Consider

*pmax*, the maximum runner speed in the set {

*p1*,

*p2*, ..,

*pn*}.

Scale the track to be of length 2

*pmax*. while keeping the runners' speeds constant. (You can think of this as our changing the units of time.) Consider the track to consist of 2

*pmax*bins of length 1.

As before, label the bins from 0 to 2

*pmax*-1. The position (bin)

*bi*of any runner with speed

*pi*at time

*t*is

bi=t pimod 2pmax

Observe at time

*t*= 1, and every odd time thereafter, the fastest runner (with speed

*pmax*) is directly across the starting point (at bin

*pmax*).

Similarly, for each runner

*i*, find the smallest positive solution to

pmax= ti pimod 2pmax

Because the

*pi*'s are odd coprimes there always exist solutions for

*ti*. And these

*ti 's*are always odd. Note as with the fastest runner, any odd multiple of

*ti*is also a solution.

The time

*t1/2*when all runners are directly across the starting point then is

t1/2= ∏ti;ti≠ti-1,ti≠ti-2, ..

where the product is taken over

*'s. And after renormalizing our track length to 1*

__distinct__tiT1/2= (1/ 2pmax) ∏ti

Examples:

1. {3, 5, 7, 11} gives

*T1/2*= 11/22 = 1/2

(Here, the distinct

*ti*'s are 1 and 11.)

2. {7, 11, 13, 15} gives

*T1/2*= (13 . 15) / (2 . 15) = 13/2

~

So the runner with speed 2 is special.

Now one thing to note is that if this argument is correct, then we can arrive at

*approximate*solutions for the case where there is an even speed runner, by scaling the speeds to a high odd factor and then tweaking these values to be odd coprimes.

#### Addenda:

__August 15, 2019__

The claim can be relaxed so that the only requirement is that the runners' speeds be odd. This is because adding a runner to the race whose speed divides the speed of any the original runners does not change

*T1/2*. So the updated claim is

If the runners' speeds are odd~~relative primes~~then there exists a time t s.t. all the moving runners are exactly at a distance 1/2 from the still runner.