Here "fold" means "fold a piece of paper (along a straight crease)". The sketch below shows that one can always find a fold by which an ellipse or rectangle loses convexity. But it seems a circle remains convex no matter how the crease is chosen?

I am not sure how to deal with a problem like this, where the shape is generic and a convenient coordinate system cannot be defined.

Update: to be precise, the circle means "disk", which includes both the border and the interior. Same for all other shapes.


Update Jan. 13 I am thrilled by the number of upvotes. Here is an interesting experiment. Starting simple, let's use an upright rectangle centered at the origin, and reflect it w.r.t. all creases passing the origin (the experiment can only use a finite subset of such creases). The area union of all reflected shapes appears to be a disk.


This observation seems to allow the following general statement: a 2d shape, convex or not, transforms into a new shape by folding. The union of all new shapes is a disk.

To be honest, this statement itself sounds like a question in need of a proof. But I feel it is highly related with the original question. So I update it here along with the Mathematica code for plotting the figure (). Enjoy!

foldAlongMiddleAxis[mya_, mytheta_] := 
 Block[{a = mya, cs = Cos[mytheta], 
   ss = Sin[mytheta]},
  polPrime1 = 
   Polygon[{{cs, ss}, {cs, ss} + a {-ss, cs}, {-cs, -ss} + 
      a {-ss, cs}, {-cs, -ss}}];
  polPrime2 = Polygon[{{1, 0}, {1, a}, {-1, a}, {-1, 0}}]; 
  Graphics[{{Opacity[0], EdgeForm[Gray], polPrime2}, {Opacity[0], 
     EdgeForm[Gray], polPrime1}}, AspectRatio -> Automatic]]

plotAllFoldsAndTrajO[a_, n_] := Show[{
   Graphics[{LightPink, Opacity[0.9], EdgeForm[Gray], 
     Polygon[{{1, -a}, {1, a}, {-1, a}, {-1, -a}}]}],
   Table[foldAlongMiddleAxis[a, mytheta], {mytheta, 
     Range[0, 2 \[Pi], \[Pi]/n]}],
   ParametricPlot[{{Cos[\[Theta]], Sin[\[Theta]]} + 
      a {-Sin[\[Theta]], Cos[\[Theta]]}}, {\[Theta], 0, 2 \[Pi]}, 
    PlotStyle -> {Black, Thick}]

plotAllFoldsAndTrajO[5, 50]
  • 2
    $\begingroup$ Seems to be true. You're essentially asking which shapes, when folded along any line, don't have a crease. This seems to be the set of shapes that are symmetrical w.r.t. that line, and that, for any line. The circle is the only one AFAIK. $\endgroup$ Jan 7 at 22:12
  • 2
    $\begingroup$ @StefanLafon I am not sure your last sentence is correct. For example, a rectangle can remain convex for many folds (e.g., fold a corner a little bit inside). Those folds do not require any symmetrical properties. $\endgroup$
    – Taozi
    Jan 7 at 22:31
  • 1
    $\begingroup$ @Maksim You can't remove the folded part and the part it overlaps, they are still part of the original paper. $\endgroup$
    – Taozi
    Jan 7 at 22:35
  • 1
    $\begingroup$ @Taozi: So you are not talking about circles but rather meaning discs ... $\endgroup$
    – Maksim
    Jan 7 at 22:41
  • 2
    $\begingroup$ @Sergey Sadly that's not true. Think of a rectangle folded such that you match two corners when folding - you obtain a pentagon. $\endgroup$ Jan 7 at 23:33

2 Answers 2


This isn't a complete answer, more like a broad sketch of a proof, but it's too long for a comment.

First, we can see that the region must be convex, since if it is not convex, a very tiny fold (think of the fold as being a line segment joining two points on the boundary of the region, so if a fold is "tiny" the length of that segment is very small) will not appreciably alter the shape, thus it will remain non-convex.

Second, it is intuitive (but no proof) that a convex polygon cannot work, because in such a shape, there must be an interior angle where two adjacent sides meet and has measure less than $\pi$. Then since a polygon must have at least 3 sides, there is at least one other side not adjacent to this angle, which must form a triangle with the other two sides (the angle cannot point "the wrong way" because the region is convex). Therefore, there is a line of reflection between this side and the vertex of the subtending angle that reflects the vertex across to the other side, and this results in a non-convex shape.

Then we can generalize further to exclude any convex shapes that have a polygonal vertex, since a similar argument to the above shows that there must be a line that will reflect that vertex in a way that will make it "poke out" of any smooth curve.

So we know the region must be smooth. Then I believe the remaining argument is to show that the boundary must have constant curvature. This strikes me as a previously solved problem but I don't have a literature reference.

  • $\begingroup$ (+1) Another possible idea for the next step after smoothness is established... Start with two parallel tangents alongside an arbitrary direction, so that the shape is between them. Consider reflections about a third parallel drawn at equal distance between the two. The points of contact must be symmetric about the axis for the fold to be convex, then - maybe - it follows that the entire figure must be symmetric about the axis. In that case the problem would reduce to How many non-infinite plane curves with infinite reflectional symmetry?. $\endgroup$
    – dxiv
    Jan 8 at 3:14
  • $\begingroup$ This may work: with a smooth shape, pick a fold direction along which it is not symmetrical (i.e. no axis in this direction makes a symmetry). Mostly likely this means there are two points on the border, P and fold(P), with a non-symmetrical tangent. Then place the folding axis in this direction so that it sends P to fold(P). It creates locally a non-convex shape (like the self-intersection point in each figure in the question). $\endgroup$
    – Armin Rigo
    Jan 8 at 9:19
  • $\begingroup$ Your paragraph claiming "we can generalize further to exclude any convex shapes that have a polygonal vertex" is wrong, because there may be more than one non-smooth point so it is not at all easy to prove that you can make that corner poke out when folded. $\endgroup$
    – user21820
    Jan 8 at 16:09
  • $\begingroup$ @user21820HATESSMOKING-HATS Such a boundary that does not contain any isolated non-smooth points would be pathological. You are welcome to construct such a counterexample and demonstrate it meets the conditions of the problem, as at no point did I claim that my reasoning constitutes a formal proof, merely a possible approach to one. $\endgroup$
    – heropup
    Jan 8 at 16:42
  • $\begingroup$ No I'm not saying there is a counter-example; all I said was "it is not at all easy to prove". And yes, maybe a proof can be obtained along the lines you gave, but you need to fix your paragraph because it wrongly says "any smooth curve" as if assuming that there is only one non-smooth point that you are using to poke out the other side. $\endgroup$
    – user21820
    Jan 8 at 16:43

Sketch Proof

The boundary of the original convex shape is differentiable almost everywhere. Let $A,B,C$ be three points on the boundary where the boundary is differentiable and let $t_A,t_B,t_C$ be the respective tangents.

Folding along the perpendicular bisector of $AB$ shows that $t_A$ and $t_B$ must be equally but oppositely inclined to $AB$. Similarly for $t_B$ and $t_C$ and for $t_C$ and $t_A$. Then $t_A,t_B,t_C$ are tangents to the circle through $A,B,C$.

This circle is therefore determined by $A,B$ alone and so contains every point on the boundary where the boundary is differentiable. It therefore contains every point on the boundary.

  • $\begingroup$ What does it mean for $t_A$ and $t_B$ to be oppositely inclined to $AB$? As stated I don't understand how this follows from the starting assumption or why it implies that they are tangents to the circle. $\endgroup$ Jan 25 at 15:15
  • $\begingroup$ The reflection in the perpendicular bisector of $AB$, must map $t_A$ onto $t_B$ . Otherwise, the convex hull of points arbitrarily close to the image of $t_A$ and arbitrarily close to $t_B$ would contain a disc about $B$ entirely contained in the new convex region. This new region would therefore contain points on $AB$ extended which is impossible. $\endgroup$
    – user502266
    Jan 25 at 15:29
  • $\begingroup$ Ah, I had misread the original folding procedure. Thanks! $\endgroup$ Jan 25 at 15:33

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