Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: regular n-gon runners problem
Replies: 12   Last Post: Jul 25, 2013 4:26 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
quasi

Posts: 10,312
Registered: 7/15/05
Re: regular n-gon runners problem
Posted: Jul 23, 2013 7:25 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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 n-gon, 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*(s-r) = t*e = e*d/q so that d*e = k*q = k*(pi+5*e)
>whence (d-5*k)*e = k*pi so that pi/e = (d-5*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/(s-q) = 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



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.