Ok, here's my 2nd attempt, I quickly gave up on trying to do a fallacious 1=2 proof so I'm just going throw my creative idea out here and see what sticks...
Let t1 = the time required to run the first 500m at velocity v1.
Let t2 = the time required to run the second 500m.
Thus the total time to run the track is t1 + t2.
v1 * t1 = 500m
2v1 * (t1 + t2) = 1000m
Therefore, 2v1 * t1 = 1000m = 2v1 * (t1 + t2)
t1 = t1 + t2 --> t2 = 0
So it takes zero time to run the second 500m, which seems to imply one must run at infinite speed. After all, the runner needs to be at the 500m mark and the 1000m mark at the same time... wait a second!
Is there a place on Earth (the physical hint) where it is possible for the 500m mark and 1000m mark to be in the same spot (or apart by less than the width of a runner's feet)? Indeed there is. There are actually two different regions on Earth where this can happen: near the North Pole, and near the South Pole.
The runner and the track should both start at a point that is [500/(2*pi)] m away from either the North Pole or South Pole. Then runner and track must both go either straight due west or straight due east. The track itself is just a simple line or a designated thin strip and as long as the runner's foot touches the track it should be considered on-track.
Once the runner has run straight due west or due east for 500m, he/she will have also completed a full circle of radius [500/(2* pi)]m and arrived back at their starting point. This starting point also happens to be where the 1000m mark (and end of the track) is. Which means that once the runner has covered 500m, they have also reached the 1000m mark at that very same moment. That satisfies the t2 = 0 requirement.
Therefore, the velocity the runner needs to run at is simply v1 the whole time. So, v2 = v1, and the runner is going both straight due west (or due east) and running a circle near one of Earth's poles at the same time.
wow I'm impressed! here is a 100% upvote :)
anyway, I do not agree that running straight west/east is running "straight". By straight, I mean a straight line define on the 3D Cartesian space, but not a straight line defined on a sphere's surface ^^
I'm completely happy with things getting deeper :)
Nice try!
Thanks. Now I'm really curious to find out what your unconventional, "outside the box" solution is.
:) go find out!