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Runners on a Square Field (Posted on 2009-01-15) Difficulty: 3 of 5
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 The Solution Submitted by Praneeth    
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Getting on base Comment 4 of 4 |
For the runners to be the same distance apart, they need to be on opposite corners again. If we think of the slower runner having a twin that runs at the same speed as the slower runner, and starts where the faster runner does, we only need ask when the twin and faster runner will cross again (on a corner), because at that point, the slower runner will be opposite the twin, and thus on a corner opposite the faster runner.

From here, we can solve for when they are next on a corner. Consider the faster runner's speed (call it a) relative to the twin's speed (call it b). We can see the first time both runners will be on a corner again is 1/gcd(a,b). (clearly at time 1, they will both be on a corner since they have integer speeds).

We would like to know how far apart they are at this point (time=1/gcd(a,b). The faster runner is at a/gcd(a,b) mod 4 and the slower runner is at b/gcd(a,b) mod 4, so the difference of the distance travelled is (a-b)/gcd(a,b).

If this is 0 mod 4, they are both at the same corner. If it's 2 mod 4, the time to meet is twice 1/gcd(a,b), and if it's 1 or 3 mod 4, the time to meet is 4 times this. This is equivalent to 4 divided by its gcd with 4.

Thus if we let t be 1/gcd(a,b), then the time it takes for them to be the same distance apart again is t*4/gcd((a-b)*t, 4)

  Posted by Gamer on 2009-01-17 03:56:35
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