Prove that if two miles are run in 7:59 then one mile MUST be run under 4:00.

I'm in an argument with someone who claims that a two mile in 7:59 does not imply that one mile (at some point within the two miles) was covered in under 4:00. This is obviously wrong, but I'm not sure how to create a proof showing otherwise.

-
There is a related problem. Suppose two miles were covered in $8$ minutes. Is there a one mile interval that took exactly $4$ minutes? –  André Nicolas Feb 25 '12 at 21:09
Yep, Daniel Komen. It's too bad his career was so short or I bet he would've gotten a few other records too. –  Nick Feb 25 '12 at 22:44
Here is an alternative view: the men's world records set in the Mexico 1968 Olympics were: 100m 9.95s; 200m 19.83s; 4x100m 38.24s. Neither the 200m nor the 4x100m were treated as having had faster 100m times because the fast 100m segments were not from a standing start reacting to a starter's signal. –  Henry Feb 25 '12 at 23:22
Related problem: Suppose you have fewer than two pennies. Then you don't have a separate penny in each hand. –  JDH Feb 26 '12 at 14:23
The interesting part is that if you replace 2 miles by $1.999...9$ miles, the problem is not true anymore ;) Actually it is "theoretically" possible to run 1.999..9 miles in 4:01, so that no mile is runed in under 4 minutes.... –  N. S. Jul 16 '12 at 16:29

If $x \geq 4$ and $y \geq 4$ then $x+y \geq 8.$

EDIT: on André's extra credit problem, use Beni's way of writing, time function $f,$ then define $g(m) = f(m+1) - f(m)$ with $0 \leq m \leq 1.$ We know $f(0) = 0, \; f(2) = 8.$ So, $g(0) + g(1) = 8.$ If both $g(0), g(1)$ are $4,$ we are done with André's problem. If one of the pair is above 4, the other is below 4. So, by the Intermediate Value Theorem, there is then some other $0 < m < 1$ such that $g(m) = 4.$

-
Keeping it simple. I guess that is just enough to prove it. –  Pedro Tamaroff Feb 25 '12 at 21:09
I like the simplicity, too. –  000 Feb 25 '12 at 21:13
Intermediate Value Theorem. That's the term I was looking for! –  Nick Feb 25 '12 at 22:45
Me? What did I do? –  Arturo Magidin Feb 25 '12 at 22:52
@Will: Now you can blame me... (-: –  Arturo Magidin Feb 26 '12 at 3:05

Consider when the runner passes the one-mile mark. If it is before 4:00 then he ran the first mile in less than 4 minutes. If it is after 3:59, then the second mile was covered in less than 4 minutes. But because 3:59 comes before 4:00 at least one of these cases (and possibly both) must be true.

-

The comment by André Nicolas is related to a very pretty theorem that deserves to be much better known. I first came across it in R.P. Boas's Traveler's Suprises, which appeared in The Two-Year College Mathematics Journal, 10 no. 2 (1979), pp. 82-88 (though I read it in the reprint that appeared in the highly recommended Lion hunting and other mathematical pursuits).

After giving the standard Mean Value Theorem problem ("if you travel in a differentiable way and your average speed is $50$ mph, is there some instant during which your instantaneous speed was exactly $50$ mph?") and a small variant ("Is there some small interval of time during which your average speed was also exactly 50 mph?"), he then gives two very similar but subtler questions:

1. If you travel for time $h$, more than one hour, and average 50 mph for the trip, is there necessarily some one continuous hour during which you covered exactly $50$ miles?

2. Suppose you run a considerable number $m$ of miles and average $8$ minutes per mile. Is there necessarily some one continuous mile (like a "measured mile" on a highway) that you covered in exactly $8$ minutes?

Boas provides the answers by using the Universal Chord Theorem. The theorem was first proven by Levy as a generalization of Rolle's Theorem in Sur une Généralisation du Théorème de Rolle, C. R. Acad. Sci., Paris, 198 (1934) 424-425.

If $f(x)$ is a continuous function, a horizontal chord of $f$ means a line segment with slope $0$ with both ends on the graph of $f$.

Here is Levy's version; you can easily see the similarity with Rolle's Theorem:

Theorem. The values $\alpha = 1, \frac{1}{2},\frac{1}{3},\ldots$ are precisely those for which the following statement is true:

If $f$ is a continuous, real valued function on the closed unit interval such that $f(0)=f(1)$, then the graph of $f$ has a horizontal chord of length $\alpha$.

Proof. Let $f$ be a continuous function defined on $[0,1]$, with $f(0)=f(1)$. Suppose $n$ is an integer greater than $1$, but that $f$ has no horizontal chord of length $\frac{1}{n}$. Let $g(x) = f(x+\frac{1}{n}) - f(x)$; then $g(x)$ is never zero on $[0,1-\frac{1}{n}]$, but is continuous, so it cannot change sign on that interval. If $g(x)\gt 0$, then \begin{align*} g\left(1 - \frac{1}{n}\right) &\gt 0\\ g\left( 1 - \frac{2}{n}\right) &\gt 0\\ &\vdots\\ g\left(1 - \frac{n}{n}\right) = g(0)&\gt 0 \end{align*} which is equivalent to \begin{align*} f(1) - f\left(1 - \frac{1}{n}\right)&\gt 0\\ f\left(1-\frac{1}{n}\right) - f\left(1-\frac{2}{n}\right) &\gt 0\\ &\vdots\\ f\left(\frac{1}{n}\right) - f(0) &\gt 0. \end{align*} Adding all of them up we conclude that $f(1)-f(0)\gt 0$, a contradiction. If $g(x)\lt 0$, a similar argument implies $f(1)-f(0)\lt 0$.

Therefore, $g(x)=0$ for some $x\in [0,1-\frac{1}{n}]$, but this implies that $f(x+\frac{1}{n}) =f(x)$; that is, the graph of $f$ has a chord of length $\frac{1}{n}$.

(Aside: You'll note this is exactly the argument Will Jaggy gives in his answer.)

Now assume that $\frac{1}{\alpha}$ is not an integer. Then Levy gives the following function: $$f(t) = t\sin^2(\alpha\pi) - \sin^2(\alpha\pi t).$$ Then $f(1)=f(0)$, but an easy calculation shows $f(t +\alpha) - f(t) = \alpha\sin^2(\alpha\pi)\neq 0$ for all $t$. $\Box$

If we assume that the travel in question 1 is continuous, and we assume that the travel in question 2 is such that time is an increasing continuous function of distance (no stopping, no backing up), then one can deduce the affirmative answer to both questions from the Universal Chord Theorem by using the same trick that one uses to deduce the Mean Value Theorem from Rolle's Theorem. I'll leave that derivation to the interested reader.

Question 2 also has an affirmative answer if we drop the assumption that the travel did not include stopping or backing up, but according to Boas the proof is much more difficult (it comes from a generalization of the Universal Chord Theorem proven by Hopf, that any continuous curve that has a chord of length $1$ has a parallel chord of length $\frac{1}{n}$ when $n$ is an integer.; he cites Hopf's Über die Sehnen ebener Kontinuen und die Schleifen geschlossener Wege. Comment. Math. Helv. 9 (1937), 303-319.

-
Very nice! Thanks for taking the time to explain that. –  Nick Feb 26 '12 at 3:19
Hi, Arturo, I see you corrected the other mention of your name and somehow put the funny mark over the e in Andre. There should be an umlaut on the U in Uber, I do not know how to put one. –  Will Jagy Feb 26 '12 at 3:26
@Will: &Uuml; should do it. For the e, I used &eacute; –  Arturo Magidin Feb 26 '12 at 3:48
If I could give it more than 1 point, I would do, because this seems to be the only valuable answer for this question (not the original, but Andre's extended version). –  Vadim Feb 19 '14 at 18:59

Proof by contradiction: If he/she had needed more than (or exactly) 4 minutes for the first and for the second mile, then he/she would have needed more than (or exactly) 8 minutes for both.

-
I think it is minutes and seconds, not hours and minutes. Or otherwise the runner is a tortoise... –  Henning Makholm Feb 25 '12 at 21:02
Hopefully, those are minutes –  mixedmath Feb 25 '12 at 21:03
Haha, you're right, thanks :) –  Michalis Feb 25 '12 at 21:03
@HenningMakholm If they are minutes, that runner better be in the Olympics though. –  Alex Becker Feb 25 '12 at 22:45
There is no first or second mile, as soon he starts not unless he is within the last mile distance there would be uncountably many 1 miles, how does this prove that all of any of those miles was not ran under 4 minutes? –  Arjang Feb 25 '12 at 23:32

You could model your case as follows: denote $f$ the time function, $f:[0,2] \to \Bbb{R}_+$, increasing, with $f(0)=0$, which shows the time at distance $x$. You know that $f(2)=7:59$.

If you suppose that $f(x+1)-f(x)\geq 4:00$ then $f(2)=f(2)-f(1)+f(1)-f(0) \geq 8:00$, and this is a contradiction.

-
Perfect! Thanks! –  Nick Feb 25 '12 at 21:07

The way you argue with these people is with pretty pictures.

-

I think the simplest way would be to prove it wrong by assuming that the person is right.

The person claims: 2 miles are done in 7:59, but each mile of those two miles is done in at least 4 minutes. If the person disagrees at this point, then you do not have to prove anymore. Otherwise you have two miles, each not less than 4 minutes. Therefore, in total it must be at least 8 minutes, which is against the assumption that two miles are done in 7:59.

-
I agree with the approach, but it needs a small correction: the negation of the claim is that each mile took at least four minutes, not more than four minutes. –  Brian M. Scott Feb 26 '12 at 5:06
Thanks, I agree. The corrections have been made. –  Michal B. Feb 26 '12 at 13:06
If the person disagrees at this point, then you do not have to prove anymore... This is taking the agreement of one individual (possibly threatened by a gun pointed at them) for a mathematical proof. –  Did Mar 23 '12 at 6:42

Just to throw a relevant keyword into the discussion, the proof follows directly from the generalisation of the pigeonhole lemma to continuous intervals (notice that the “generalisation” section on Wikipedia describes a different generalisation which is non-continuous):

Given an interval $a$ which we decompose into $n$ parts of sizes $x_1 … x_n$ with $\sum_{i=1}^n x_i=a$ then there must be at least one $x_i$ such that $x_i \geq \frac a n$.

(I believe the generalisation is due to Gene Myers but it’s pretty easy to prove anyway.)

Adapted to your example, then $a = 7{:}59\mathrm{\,min}$, $n = 2$ and one mile must be run in ${} \geq \frac a 2$. Conversely, the other mile must then be run in $\leq a - \frac a 2 = \frac a 2 < 4{:}00\mathrm{\,min}$. $\square$

-

Suppose to the contrary that both miles were run at a time greater than or equal to 4:00. Then put together, the two miles were covered in at least 8:00. We arrive at a contradiction. Thus our initial assumption was incorrect. One mile must have been run in less than 4:00.

-

You can prove a stronger result: Either the first or the second mile of the race was run in under four minutes.

Let $t_1, t_2$ be the times for the first and second half of the race. Proof by contradiction: Suppose $t_1 > 4$ and $t_2 > 4$. Then $t_1 + t_2 > 8$.

-

The argument can be explained geometrically.

Assume the time taken for each of the two miles is $x$ and $y$.

And the total time taken is: $x+y=7'59''00$ (1).

To fulfill (1) we have to meet the boundary shown in blue in Figure A.

This boundary is met if and only if $x \geq 3'59''30$ and $y \geq 3'59''30$.

Figure A: http://i.stack.imgur.com/BvIeT.png

-