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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.

illustration2

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.

5by1_rectangle

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]
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  • 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. $\endgroup$ Commented Jan 7, 2022 at 22:12
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    $\begingroup$ @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
    Commented Jan 7, 2022 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
    Commented Jan 7, 2022 at 22:35
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    $\begingroup$ @Taozi: So you are not talking about circles but rather meaning discs ... $\endgroup$
    – Maksim
    Commented Jan 7, 2022 at 22:41
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    $\begingroup$ @Sergey Sadly that's not true. Think of a rectangle folded such that you match two corners when folding - you obtain a pentagon. $\endgroup$ Commented Jan 7, 2022 at 23:33

3 Answers 3

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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.

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8
  • $\begingroup$ (+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
    Commented Jan 8, 2022 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
    Commented Jan 8, 2022 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
    Commented Jan 8, 2022 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
    Commented Jan 8, 2022 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
    Commented Jan 8, 2022 at 16:43
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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.

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3
  • $\begingroup$ 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. $\endgroup$ Commented Jan 25, 2022 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
    Commented Jan 25, 2022 at 15:29
  • $\begingroup$ Ah, I had misread the original folding procedure. Thanks! $\endgroup$ Commented Jan 25, 2022 at 15:33
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I came here from this question, which I found a simple and concise proof for the following:

Consider any simply connected shape on a 2D Euclidean plane. If, for any line that intersects this shape, the reflection of the parts of the boundary that lie on one side of the line do not intersect the parts of the boundary that lie on the other side, then the shape must be a circle. Here, two curves intersect means that there is at least one point that lies on both curves, and at least one point that lies on one curve but not another. (So completely overlapping does not count as intersecting.)

(I don't understand why that question got closed. Last time I checked, the consensus seems to be that these are related but different questions. If anyone can re-open it, please do. But anyway, here I'm going to reduce this problem to that one.)

If it's true that the reflection of the half can't intersect with the other half here as well, then we're done. Is it possible that the reflection intersects with the other half while the shape remains convex? Unfortunately, it's true. A simple example is matching the vertex of a rectangle to the opposite one on the diagonal. The resulting shape is still convex, although the boundaries intersect.

But, notice that, if you move the line slightly above, it'll no longer be convex. So, the only remaining task is, to make this argument rigorous. (Which takes way more effort than proving the previous statement, though.)

So, let's suppose that: There is a line, $L$, that cuts the boundary into two parts, $B_u$ and $B_d$, the upper and lower parts. The reflection of $B_d$, $B'_d$, intersects with $B_u$, thus there must be a point $p_i$ which lies on both $B_u$ and $B'_d$, and another point $p_d$ which is on $B'_d$ but not on $B_u$.

Due to symmetry, w.l.o.g, suppose the line $L$ is horizontal, and suppose $p_d$ is to the left of $p_i$ and under $B_u$. enter image description here Here $B'_d$ and $B_u$ may intersect at a single point $p_i$, or they may overlap on a curve where $p_i$ lies on. It doesn't matter which case it is.

Now, we want to show that, if we move the line $L$ slightly above, then the reflection of $B_d$ will also move slightly above, which will cause the shape to be non-convex.


The following is just the thought process which is not rigorous, but it shows the direction.

To do that, consider the point where the two curves splits. If we zoom in enough, the boundaries should appear linear, like this:

enter image description here

Then, if we move the blue dotted line a little bit above, it will intersect the red solid line, which results in the shape being concave.

So, is that true that if we zoom in enough, the curve appears linear?


Now, we introduce the concept of "left tangent line".

First, we prove that, for any point $p$ on one of the curves, for any angle $\delta$, there exists a distance $d$ and an angle $\theta$ such that, for any point $p_1$ that's to the left of $p$ which is less than $d$ from $p$ horizontally, the line that connect $p$ and $p_1$ must make an angle within $[\theta,\theta+\delta]$ with the $-x$ axis.

Translation to plane words: if we zoom in enough, the curve should appear to be almost linear (If we connect any two points, the angle of the slope of the line should be within a small interval).

The proof is actually very simple. Suppose it's not true. There's an angle $\delta$ such that there's no distance that satisfies this condition. Then, we choose an arbitrary point, $p_1$, to the left of $p$. Suppose the angle of the line through $p_1$ and $p$ is $\theta$. Because the shape is convex, the angle between $-x$ axis and the line that connects $p$ and any point $q$ on the curve between $p_1$ and $p$ must be greater than $\theta$. Then, there must be a point $p_2$ where the line through $p_2$ and $p$ makes an angle that is at least $\theta+\delta$. Then, we must be able to find another point $p_3$, which makes an angle $\theta+2\delta$. We can keep doing that. Since $\delta$ is finite, if we do this enough times, the line will be rotated for more than $\pi$, so it's impossible that the point is to the left of $p$, contradiction.

So, we proved that the left slope, thus the left tangent line, must exist, which does not necessarily equal the right slope or right tangent line, which is obvious considering the vertex of a polygon.

Now, we are able to limit the tangent line to an arbitrarily small angle. How do we proceed?

enter image description here Basically, for a small region to the left of the point $p_s$ where the two curves split, we are able to limit the upper curve to be inside the solid triangle, and the lower curve inside the dashed triangle.

If the two triangles don't intersect, then we can move the lower triangle a little bit above, so that they intersect:

enter image description here

Suppose the line that connects $q'_d$ and $p'_s$ intersects with the line that connects $q_u$ and $p_s$ at $A$, and the line $p_u - p'_s$ and $q_u - p_s$ intersect at $B$.

Since $p_u$ is on the curve, every point below the line $p_u - p'_s$ (and above horizontal line $L$) must be inside the shape, because it must remain convex. But the upper boundary is inside the solid triangle $q_u p_u p_s$, and the reflection of the lower boundary is inside the dashed triangle $q'_d p'_d p'_s$. Thus, every point inside the triangle $ABp'_s$ is not inside the shape. Contradiction.

Now, we still have one assumption: We can find a small enough distance such that the two triangles don't intersect. Unfortunately, this is not always true. It's possible that the two curves have the same tangent line at $p_s$, i.e, they're tangent at $p_s$.

But notice that: we don't really need to use the point where the two curves split. We can move the reflected curve up and down. If that point doesn't work, we can move the curve and let them intersect at another point, where they are not tangent.

So, we want to show that, there exist an $x$ coordinate $x_0$, such that the corresponding points $A_u$ and $A_d$ with $x$-coordinate $x_0$ on the curves $B_u$ and $B'_d$ respectively, have different left slope.

Assume that this is not true: For any $x$ coordinate, the two points on the two curves with this $x$ coordinate has the same left slope. Then we can just start from the point where they intersect, and every pair of points to the left of it on the two curves must overlap. Thus the two curves to the left of the intersecting point must overlap, contradicting the fact that there exist a point that is to the left of the intersecting point, and lies on one curve but not the other.

(Update: Here it's missing a special case: the intersecting point $p_s$ lies on a vertical part of the curves. Example: We're folding a rectangle along one of its edges. So, to make the argument work, we must add another condition: If the two curves only intersect at a vertical part of them, then consider the line that's parallel to it which cuts the shape into two equal areas, as I did in the answer to the other question. Then, under this condition, if they still intersect, they must intersect at a different point that's not on a vertical part of the curves, because otherwise, if the two curves still only intersect on the vertical part, the boundary must overlap completely, or the shape is not convex. This implies that the shape must be symmetric w.r.t this line, which reduces to the result of other question.)

Now it becomes clear how we construct the proof.

There exist a pair of points $A_u$, $A_d$ with the same $x$ coordinate $x_0$ that lies on the two curves and have different left slopes. Suppose the difference in the angle of the tangent lines is $\theta$. Next, we want to make sure that the two triangles don't overlap, thus we choose an angle $\delta<\theta$, and there must exist a distance $d_u$ such that any point on $B_u$ that has $x$ coordinate within $[x_0-d_u, x_0)$ must satisfy that the line connecting it and $p_u$ makes an angle within $[\alpha,\alpha+\delta]$ with the $-x$ axis. There must exist a distance $d_d$ for the reflected lower curve $B'_d$ as well, and a corresponding $\beta=\alpha-\theta$. We choose the minimum of the two distances, $d$.

Then, we move the line $L$ vertically by half of the vertical distance of $A_u$ and $A_d$, so that they overlap afterwards. Let's call the point $p_s$.

Now the situation is just as before, but we have guaranteed that the two triangles don't overlap.

enter image description here

Suppose the vertical distance between $p_u$ and $q_d$ is $h$. Moving the folding line $L$ above by $\frac{h}{4}$ will result in the reflected lower curve rising by $\frac{h}{2}$, thus the two triangles must intersect as shown in the figure above, thus causing the shape to be concave.

Since the reflected lower boundary can't intersect with the upper boundary, it reduces to the question that I linked to in the beginning. Thus the boundary must be a circle (the shape is a disk). The proof is complete.

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