# Prove that if 2000 m is run in 4:50, then one mile is run under 4:00

Prove that if 2000 m is run in 4:50, then one continuous mile is run under 4:00.

I was thinking -- yeah, this follows easily from the other question, where I asked that if 2 miles are run in 7:59, then one mile must be run under 4:00.

But then I read that paper that Arturo provided a link to, and as 2000 m is not an integer multiple of 1609.344 m (1 mile). Now I am not in fact sure whether this is indeed true. The author of the paper, R. P. Boas, uses the universal chord theorem for cases where the total distance is an integer multiple of the distance in question, but says that for non-integer cases. It is not necessarily the case that a distance less than the total is ever covered in (average speed)*(that distance) or better.

Can there be shed some light on this?

(This is a follow-up from my last question, but I believe it is somewhat more difficult.)

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Note that if you assume your maximum speed is bounded, you can say something along these lines. –  Alex Becker Feb 26 '12 at 5:55

It's not true. For a counterexample, run the first kilometer in 25 minutes, stand still and catch your breath for 4 hours, and then cover the second kilometer during the last 25 minutes.

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Are you sure that 4 minutes will be enough? :-) –  Brian M. Scott Feb 26 '12 at 5:33
@Brian: No, so I just changed the interpretation to a more realistic one. –  Henning Makholm Feb 26 '12 at 5:36
Aw, shucks! I kind of liked the speedy version. –  Brian M. Scott Feb 26 '12 at 5:38
Huh... okay thanks. I feel embarrassed now. It seems so obvious when you put it like that. –  Nick Feb 26 '12 at 6:12
@imintomath: Stopping is just the most extreme example. In general, run the first $400$ and last $400$ meters at the same constant speed $v$, and then use the remaining $290-800/v$ seconds to cover the middle 1200 meters in any way you like. Then, every continuous $1600$-meter interval of the run will have taken you $290-400/v$ seconds, which is more than $4$ minutes if only $v>8\;\rm m/s$. –  Henning Makholm Feb 26 '12 at 13:22