Two runners start on opposite corners of a square field with side 1 km, and run around the edge with integral speeds in the clockwise direction. In terms of these speeds, when will they be the same distance apart as when they started?
See other commenters' later posts for correction to this.
When they are on opposite corners they are the farthest apart they can be, so this distance will be accomplished only when they are again on opposite corners.
Say the faster runner goes s2 km per unit time and the slower runner goes s1 km per unit time.
The ratio of their speeds is s2/s1, which may or may not be reduced to its simplest form. Let G = gcd(s1,s2), so the reduced fraction is (s2/G)/(s1/G).
It requires 1/s1 and 1/s2 units of time to go from one corner of the square to the next, so when the faster runner reaches the next corner, the slower runner has gone s1/s2 = (s1/G)/(s2/G) of the way. The next integral multiple of this that will also be an integer will be s2/G times the amount of time spent thus far, that is, (1/s2)*(s2/G) = 1/G. At time 1/G, the faster runner will have traversed s2/G sides of the square while the slower runner would have traversed s1/G sides. If (s2-s1)/G is odd then the runners are on adjacent corners of the square and need to spend an additional 1/G units of time running to get to opposite corners. If (s2-s1)/G is a multiple of 4 (such as when the faster runner is 5 times as fast as the slower) then 1/G is sufficient. If (s2-s1)/G is a multiple of 2 other than a multiple of 4, they would also have to go an additional 1/G time units, again making a total of 2/G, as the two runners would then be at the same corner.
To summarize: if (s2-s1)/gcd(s1,s2) is a multiple of 4 then the answer is 1/gcd(s1,s2), otherwise it's 2/gcd(s1,s2).
Edited on January 15, 2009, 10:41 pm
Posted by Charlie
on 2009-01-15 18:38:27