A thread of length $\ell$ is tied inside a rectangle between sides $M1$ and $M2$, parallel to the remaining sides (as in the figure). The thread is released from both sides, maintaining its length, and wraps arbitrarily inside the rectangle. Show that there is a point on the wound thread that maintains the same distance to the sides $M1$ and $M2$ as before.
I have already solved it, but I think it is interesting because it is a graphic way to see the fixed point theorem, and I'm looking for more problems like this. Do you know of some book where I can find similar problems?
Thanks a lot.
I can't find any similar problems, but can offer an attempt to prove the one you have....
With the obvious choice of $X, Y$ the initial position of the thread can be viewed as the line $[0, 1] \times \{1/2\}$ in the unit square $[0, 1] \times [0, 1]$ and can apply the Euclidean metric to this space..
The second position can be seen as a path joining its two endpoints, and therefore has a representation $p(t) = (x(t), y(t))$ a continuous function mapping the unit interval into the unit square.
In particular we can parameterise the path by arc length, where we know from the initial position that the total arc length $= 1$.
Then if we consider just the $x$ component, it is also a continuous function
$x: [0, 1] \to [0, 1]$ where the arc length $t$ in the $x(t)$ represents the distance from $M1$ in the initial position.
So, $x$ is a continuous function mapping the compact convex set $[0, 1]$ to (not necessarily onto) itself and so by the Brouwer Fixed Point Theorem
https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem#One-dimensional_case
it has a point with $x(t) = t$.
I.e. there is at least one point in the second position at the same distance from $M1$ as in in the initial position.