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I was studying folding of circles out of boredom, when I noticed that folding a circle never concludes with a form where the circle intercepted itself. If you fold a rectengular piece of paper for example, for some particular folding, the paper will intercept with itself, but the cirlce does not do that. I was able to proof that this is always the case, but I would really like to show that the circle is the only figure with this property, or isnt, i dont know. Can someone help? I dont know if there is a field which studys such concepts, I am new to mathematics of this kind.

L0G1C-08
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    Intercept? As far as I can think, maybe you're trying to fold by the diagonals of the rectangle, and you're getting a part of the paper which isn't covered by the rectangle after folding it. Right? – Gwen May 12 '24 at 12:57
  • @ACB I'm surprised that no one used the area argument that I gave below in the question that you linked to. I think this argument works in that question, too? – Yi Jiang May 12 '24 at 13:21
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    @YiJiang The premises of the questions are different. Having a non-convex folding is not immediately equivalent to having a self-intersecting boundary fold. Indeed, I can picture shapes and folds where the boundaries intersect (without superimposing locally), but the folded shape is still convex. I don't think this really qualifies as a duplicate; it's a related question with different answers. – Theo Bendit May 12 '24 at 14:40
  • @ACB Yes thank you very much, and to all other users, thank you! – L0G1C-08 May 12 '24 at 16:26

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Hint: Consider a line that cuts the shape into two parts that has equal areas. If the reflection of the boundary of first part w.r.t this line doesn't intersect the boundary of the second part, then it either lies within the second part, or lies outside the second part, or overlaps exactly with the second part. Since the areas are the same, it must be the third.

So, any line that cuts the area evenly must be an axis of symmetry. Is a circle the only shape that has this property? Consider one such line, then another that's perpendicular to it. (Reflection in x)*(Reflection in y) becomes rotation in 2D, so it must have $180^\circ$ rotation symmetry w.r.t the intersection of these two lines. So, any line through this point cuts its area into two equal halves. Thus, it's symmetric w.r.t any line that passes through this point. So, if the distance from one point on the boundary to this point is $r$, then any other point must have the same distance. By definition it's a circle.

Yi Jiang
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