quasi
Posts:
10,903
Registered:
7/15/05


Re: regular ngon runners problem
Posted:
Jul 23, 2013 7:25 PM


James Waldby wrote: >quasi wrote: > >> Suppose n runners, n >= 3, start at the same time and place >> on a circular track, and proceed to run counterclockwise >> along the track (forever), each at a distinct positive >> constant speed. >> >> Conjecture: >> >> If there is an instant where the locations of the n runners >> are the vertices of a regular ngon, then the speeds of the >> runners, arranged in ascending order, form an arithmetic >> sequence. >> >> Remark: >> >> It's easy to see that the converse holds. > >If I haven't made a mistake, your easy proof of the converse >leads to an easy proof that pi/e is rational, which it's nice >to have an easy proof of. :)
Haha.
>Let q, r, s be the speeds of runners m, n, o, and let them be >equal to pi+5e, pi+6e, and pi+7e respectively. Let the track >be an equilateral triangle with unit sides.
Ok.
>Let d, g, f be the distances traveled at the first time t>0 when >the locations of the n runners are the triangle's vertices.
No, you fooled yourself by your choice of the shape of track.
Just because the track has the shape of an equilateral triangle, that doesn't mean that the only way for the locations of the runners be vertices of an equilateral triangle is for them to coincide with the vertices of the track.
Had you stayed with a circular track, you would probably not have made that error.
>So d, g, f are three different positive integers.
In fact, as the rest of your argument makes clear, d, g, f cannot all be rational.
>From d = t*q we have t = d/q. Let integer k = f  g > 0. >Now k = t*(sr) = t*e = e*d/q so that d*e = k*q = k*(pi+5*e) >whence (d5*k)*e = k*pi so that pi/e = (d5*k)/k, which is >rational.
In fact, using your notation, the first time t > 0 when the locations of the runners form an equilateral triangle is at time t = 2/(sq) = 1/e. At that time, the distances d,g,f are given by
d = q*t = (Pi + 5*e)*(1/e) = Pi/e + 5
g = r*t = (Pi + 6*e)*(1/e) = Pi/e + 6
h = s*t = (Pi + 7*e)*(1/e) = Pi/e + 7
which implies that each runner is exactly one unit of distance away (where distance is measured along the track) from each of the others. Thus, their locations form an equilateral triangle.
quasi

