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.

  • $\begingroup$ can you should an example? I'm not sure I understand the rules properly $\endgroup$
    – Marius
    Commented Aug 4, 2017 at 6:32
  • $\begingroup$ @Marius, take a square and fold it along diagonal. You will get a triangle with 45,45,90 degree angles. Consclusion: you can construct such a triangle this way. $\endgroup$
    – klm123
    Commented Aug 4, 2017 at 8:40
  • $\begingroup$ Ah. ok. I was overthinking it. $\endgroup$
    – Marius
    Commented Aug 4, 2017 at 8:59

2 Answers 2


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.

  • $\begingroup$ How can you "keep folding along any edge of the polygon" before the process is complete, and before there is a polygon to fold against? $\endgroup$
    – Rosie F
    Commented Aug 4, 2017 at 8:20
  • 1
    $\begingroup$ @RosieF: I probably interpreted the question incorrectly. I considered that the folds could be done "freehand" along any line needed. If that is not the case, then the questioner will need to be more specific as to which operations are allowed. As far as I can tell, it is still an open problem which polygons are possible using origami mathematics. $\endgroup$ Commented Aug 4, 2017 at 8:42
  • $\begingroup$ What is "the angle of the sticking out part of the paper"? Define it, please. P.S. and yes, you can fold along any line. $\endgroup$
    – klm123
    Commented Aug 4, 2017 at 8:42
  • $\begingroup$ @klm123: It is the difference between the angle that the paper shape makes at the corner you are working on, and the angle you want the polygon to have at that corner. If the goal polygon has a corner angle of $\alpha<90$, after the first edge is folded the paper has 180 degrees at that corner instead of $\alpha$, so it is $180-\alpha$ too much. After folding the second edge that excess becomes $180-2\alpha$. After folding that back along the first edge it is $180-3\alpha$. If that number is positive, then the paper is sticking out over the second edge by that much, so fold back again, etc. $\endgroup$ Commented Aug 4, 2017 at 8:58

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.

  • 1
    $\begingroup$ Origami is a superset of ruler and compass constructions; for example, you can trisect an angle. $\endgroup$
    – boboquack
    Commented Aug 4, 2017 at 8:30

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