The crumpled paper problem I saw a theorem recently (and have seen similarly phrased problems before), which I found quite surprising. Unfortunately, I'm not sure what its name is (or if it has one). I'll refer to it as the "crumpled paper problem". It goes something like this:

Suppose you were to take two pieces of paper, identical in size and shape, and place one on top of another such that all 4 corners and all 4 edges line up. Now, every point along the top sheet can be "mapped" to its corresponding point on the bottom sheet. If you were to then crumple up the top sheet of paper, and toss it on top of the bottom sheet, there necessarily exists a point on the top sheet that lies precisely above its "mapping point" on the bottom sheet, no matter how it was crumpled or where it was placed.

This result, to me, at least, is very surprising. Intuitively, it seems to me that it must be that there is some way to crumple the paper, or location to toss it, such that no points line up. I can sort've imagined progressively folding the paper (analogous to crumpling), and kind of see how this might be, however, it's still quite challenging.
My question is, how is this proven, does the problem have a name, and to what field of mathematics does this belong? I'm very curious how this can be shown.

I've added those tags which I thought may be relevant. Unfortunately, I have no clue to what precise area of mathematics this belongs to. Please do correct tags as necessary.
 A: It is all about continuity of the crumpling transform. The 1D case is easier to understand.
If you plot the abscissa of the points of the crumpled line vs. the abscissa of the corresponding point in the flat line, you get a continuous function in the range $[0,1]$. As the graph of the function splits the unit square horizontally and is continuous, it must meet the diagonal of the square.
(In other terms, the crumpled abscissa starts larger and ends smaller than the straight abscissa, so it must be equal somewhere.)

In the 2D case, you can plot both the crumpled abscissa and ordinate as a function of the flat abscissa and ordinate, which yield two surfaces in a unit cube. These surfaces meet the oblique planes $x=u$ and $y=v$ along two continuous curves, and these curves meet in at least one point.
This works even if the paper is elastic and can be stretched (i.e. topological matter).
A: The mathematical description is this: You have a rectangle $R = [a,b] \times [c,d]$. Crumpling a copy of $R$ and tossing it on the top of the bottom sheet $R$ means that you have an injection $\phi : R \to R \times [0,\infty) \subset \mathbb R^3$. Here each point $(u,v,w)$ in the crumpled image $C = \phi(R)$ is uniquely identified by its"coordinates" $(r,s) \in R$ which means that $(u,v,w) = \phi(r,s)$. Your question is

Does there exists a point $x \in R$ such $\pi(\phi(x)) = x$? Here $\pi : \mathbb R^3 \to \mathbb R^2, \phi(u,v,w) =(u,v)$ is the orthogonal projection onto the $(u,v)$-plane.
In other words, does the function
$$f : R \to R, f(x) = \pi(\phi(x))$$
have a fixed point?

Note that whether or not the crumpling function $\phi$ is continuous is hard to say. If crumpling causes ruptures, even if they are microscopically small, it will be not. But let us assume that $\phi$ is continuous. Then your question is answered in the affirmative by the Brouwer fixed point theorem in dimension $2$. I am not going to give a statement or a proof, search for "Brouwer" in this forum or look here.
A: Let's assume that when you crumple a sheet of paper, the fibers don't stretch, so any two points on the crumpled paper will be no further apart than they were on the uncrumpled sheet. We also assume that crumpling is a continuous function.
Now let $P_0$ be a point on the uncrumpled sheet that is as close as possible to the shadow $Q_0$ of its location on the crumpled version. (Such a point exists because crumpling is continuous.) Let $P_1$ be the midpoint of the line segment $P_0Q_0$ (on the uncrumpled sheet), and let $Q_1$ be the shadow of the location of $P_1$ on the crumpled sheet.
If $d$ is the distance from $P_0$ to $Q_0$, then $Q_1$ must be inside (or on) a circle of radius $d/2$ centered at $Q_0$ but outside (or on) a circle of radius $d$ centered at $P_1$. There is only one such point: $Q_1$ must be on the extension of the line segment $P_0Q_0$, with $Q_0$ as the midpoint of $P_1Q_1$.  But we can now replace $P_0$ and $Q_0$ with $P_1$ and $Q_1$ (because they are the same minimal distance apart). Iterating this procedure produces a sequence of points $P_0,P_1,P_2,\ldots$, all in a straight line at distance $d$ apart. If $d\gt0$, this is a contradiction.
