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You have a squared piece of very thin paper. You can fold it and glue together to get a piece of paper with other shape. You can make it as many times as you want. Are there convex polygons, which you can't get this way? What are those?
Polygons size doesn't matter, only the shape (i.e. angles and relation between sides).
I don't know if this problem has a solution.
Note that I interpret the question such that folds could be done "freehand" along any line needed. In other words, it is as if you already have a goal polygon marked out on the paper, and can fold along its edges. The aim is to fold away all the extraneous paper onto the polygon.
If you also have to mark out the goal polygon from scratch using folding techniques, then you have a much more difficult problem. As far as I can tell, it is still an open problem which polygons are possible using origami mathematics.
I think it is fairly obvious that every convex polygon that has been marked out on the paper can be folded, but it is a little tricky to prove in a mathematically rigorous way.
The basic idea is:
Just keep folding along any edge of the polygon where there is paper sticking out. Every fold reduces the area of the paper that is sticking out.
The problem is that this is not a proof.
It does not exclude the possibility that an infinite number of folds are needed. The amount by which each fold reduces the area of extraneous paper might decrease with each fold such that the total amount remaining never reaches zero.
Here is a folding method that serves as a rigorous proof by construction:
After folding along any edge of the polygon, I can assume that the newly folded paper at most only sticks out at the two adjacent edges. If the part being folded is so large as to reach more distant edges, simply fold it up using parallel folds until it becomes a thin strip that is being folded over the polygon. Just make sure you make the strip thinner than the distance to the nearest corner.
Use the above trick on one edge of the polygon, and then on the next adjacent edge.
If the folded paper for the second edge sticks out over the first edge, you can fold that part back and forth at that corner. If the polygon angle at that corner is $\alpha$, then each fold across that corner reduces the angle that the outline of the paper makes there by $\alpha$, so after a finite number of folds it will fall fully inside the polygon.
Repeat the above procedure for each edge in turn, to cleanly construct the polygon one edge at a time. For the final edge you may have to do the corner folding action at both ends.
By classical means, you can construct two perpendicular lines, then mark out along each of those lines a series of equidistant points, and from those points construct lines parallel to your first two lines. So by classical means you can construct the squares that you give us in the first place. So if a polygon can be constructed by your method, it can be constructed by classical means. So if a polygon cannot be constructed by classical means, it can't be constructed by your means either. So for example, regular polygons of 7, 9, 11 etc. sides can't be constructed by your means.
Actually, I don't see any means to construct any angle whose tangent is irrational.
