## The runner

Puzzlers all over the World United!

### The runner

One person starts to run from point A to point B. He runs at a different speed throughout the route and arrives at point B in the late evening. On the following day, the runner again starts running, this time from point B to point A. He, again, runs at different speeds, but arrives at point A after running for the exactly same amount of time as he did in the previous day. Can you prove that there was a moment during the day when the runner was at the same place during both days?
DArk0n3

Posts: 76
Joined: Sat Oct 06, 2012 4:20 pm

### Re: The runner

Excellent puzzle, it’s known as “Tourist Puzzle” and it have good application on many known theorems. Last time I met this was on the Nash Theorem, stating that every game with finite strategies, has Nash Equilibrium point.
Imagine that the first day the tourist movement is described by a function F(x), and the second day function is G(x). The X itself represents the distance from the starting point of the first day and the F(x) and G(x) equal the exact time of this moment.

So F(X) will start in point (0,p) : because the distance from the starting point is 0 and let’s say it’s exactly p o’clock. Since the starting point it can moves whatever it likes, but Y axis should be increasing, because time is like that, always increasing – no coming back. The F(X) final point will be (S,q) , where S Is the distance between the starting point and the final point, and q is the final time.

Don’t be afraid of the variables, they only help us to explain this in a short manner. Now what happen the other day? Well, it’s obvious that G(X) will start in point (S,p), because the distance from the starting point of the first day is still S and the starting time is p (because he start to travel back at the same time he did in the first day). Then again G(X) can moves in a manner whatever it likes, but the final point should be (0,q) – because he is going back to the starting position – 0, and the time of finishing his walking is exactly q o’clock.

Now if you draw this function on the ordinary Euclidian space, you are going to see that there is no way this functions to not cross. And they will cross on only one point – this point will have the same coordinate (m,n), where m represents the distance from the starting point and n is the exact time they will cross. Therefore – not only such point exist (the tourist to be on the same point in the same time – two different day), but it appears it is only one. It’s not possible two or more of this points to exist.

Now the more interesting problem is coming from the Banach Space:

Imagine you have a map – describing only and perfectly the territory of a country – for example Bulgaria. Imagine you are visiting Bulgaria and suddenly the map fall from your pocket and without folded fragments. Can we say that there is a point from the actual territory of Bulgaria, and a point from the map representing that point of the actual territory, are one above other?

showmyiq

Posts: 390
Joined: Sat Sep 15, 2012 9:45 pm

### Re: The runner

Very good explanation. I will think on your development of the initial problem and share any conclusions I come across
DArk0n3

Posts: 76
Joined: Sat Oct 06, 2012 4:20 pm

### Re: The runner

Yes, it is very interesting puzzle!
Anyway, the tourist puzzle will be published in the tomorrow puzzle and you will take the credits of proposing it! Thanks for the share!

showmyiq