Mathematicians Cut Apart Shapes to Find Pieces of Equations(quantamagazine.org)
quantamagazine.org
Mathematicians Cut Apart Shapes to Find Pieces of Equations
https://www.quantamagazine.org/mathematicians-cut-apart-shapes-to-find-pieces-of-equations-20191031/
30 comments
It seems obvious that you cannot cut a circle into finitely many pieces and combine them into a square of the same area.
Hence, I suppose that in the starting statement of "If you have two flat paper shapes and a pair of scissors, can you cut up one shape and rearrange it as the other?", the concept "flat paper shapes" refers to polytopes [1]. That is, N-dimensional generalizations of polygons.
Otherwise the statement "For two-dimensional shapes, the answer is easy: Just determine their areas. If they’re the same, the shapes are scissors congruent." is wrong with the circle - square case as a counter example.
[1] https://en.wikipedia.org/wiki/Polytope
Hence, I suppose that in the starting statement of "If you have two flat paper shapes and a pair of scissors, can you cut up one shape and rearrange it as the other?", the concept "flat paper shapes" refers to polytopes [1]. That is, N-dimensional generalizations of polygons.
Otherwise the statement "For two-dimensional shapes, the answer is easy: Just determine their areas. If they’re the same, the shapes are scissors congruent." is wrong with the circle - square case as a counter example.
[1] https://en.wikipedia.org/wiki/Polytope
How is it obvious? I am a mathematician and do not find it obvious in the least...
As a non-mathematician, for a 2d circle on a flat plane:
- The edge of a circle is curved (convex)
- It does not matter how small a piece you cut of the edge, there will always remain a convex curve.
- The edge of a square has straight edges, so you cannot put this piece at the edge. This means the old outside edge will have to be on the inside.
- You cannot fit the convex edges to each other or to a straight edge.
- Cutting a concave edge from the inside of the circle to fit the convex edge from the outside to will not help as it will produce a new convex edge.
Ergo: there is no place to put the convex outside edge of the circle, so you cannot turn it into a square.
- The edge of a circle is curved (convex)
- It does not matter how small a piece you cut of the edge, there will always remain a convex curve.
- The edge of a square has straight edges, so you cannot put this piece at the edge. This means the old outside edge will have to be on the inside.
- You cannot fit the convex edges to each other or to a straight edge.
- Cutting a concave edge from the inside of the circle to fit the convex edge from the outside to will not help as it will produce a new convex edge.
Ergo: there is no place to put the convex outside edge of the circle, so you cannot turn it into a square.
So not so OBVIOUS, anyway. You need the notion of convexity (not so easy to give abstractly unless you already know the definition). Items 4 and 5 look clear but from there to obviousness...
Being visually intuitive is very different to being obvious.
Being visually intuitive is very different to being obvious.
This seems a bit pedantic. I'm certainly not a mathematician but it does seem obvious that you cannot cut a circle into any number of pieces and rearrange it to be a square, as a circle has curves which would prevent you from making the square a solid without gaps
No no; I’m trying to explain that “obviousness” is a bad idea to prove anything...
Like “an infinite te has an infinite branch”...
Like “an infinite te has an infinite branch”...
You could create a concave edge with a series of triangles (really, trapezoids).
Ps. I suppose this would really only work for an approximate circle, which is a polytope.
Ps. I suppose this would really only work for an approximate circle, which is a polytope.
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It’s not obvious because, as stated, it’s wrong: https://arxiv.org/abs/1612.05833 You can prove that the disk and the square are not scissors-congruent, based on a convexity invariant, but if you allow more exotic dissections you can construct an equidecomposition.
Lots of equations no squared circle. As a picture book kind of person I'm disappointed.
I cannot visualise what to do with the (convex) curvy pieces that would be cut from the circle's border.
There would be no concave pieces to slot them into - and I couldn't cut out new concavities for them without producing even more convex curves.
There would be no concave pieces to slot them into - and I couldn't cut out new concavities for them without producing even more convex curves.
Neither can I but... obviousness is very strong.
I also don't think it's obvious [1], but that it could seem obvious from the way some mathematics (ahem, calculus) is taught to students.
[1] My first question will be whether your "circle" is of cardinality of R or of N... Edit: Actually, my question would rather be whether you allow countable or arbitrary union.
[1] My first question will be whether your "circle" is of cardinality of R or of N... Edit: Actually, my question would rather be whether you allow countable or arbitrary union.
The statement included the word 'finitely many' thus excluding even countable union.
At which point, the cardinality of the circle stops mattering I think. And at any rate, just... come on, we are obviously talking about a disk of radius r > 0 in the plane of real numbers using the euclidean metric.
At which point, the cardinality of the circle stops mattering I think. And at any rate, just... come on, we are obviously talking about a disk of radius r > 0 in the plane of real numbers using the euclidean metric.
Conservation of parts that have a curve.
No matter how you cut it, you will round edges that don't fit together with any other edge.
Technically you could cut everything into infinitely many points and then reassemble them into another shape (especially given the volume is the same).
I bet there’s some other condition such as “small at infimum”, e.g. we can make as many cuts as we want we can’t really make infinitely many cuts.
I bet there’s some other condition such as “small at infimum”, e.g. we can make as many cuts as we want we can’t really make infinitely many cuts.
If you accept the axiom of choice, then you can cut up a sphere into finitely many pieces and reassemble them to a sphere with twice the volume. (The pieces are not measurable, naturally.)
As sets of real numbers, yes.
However, if you have an approapriate metric (in this case I think cuts are area preserving) then you can't. [1]
[1] https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox#.... Look at the comment about measures changing.
Edit: Oh, I see you did say this at the end.
However, if you have an approapriate metric (in this case I think cuts are area preserving) then you can't. [1]
[1] https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox#.... Look at the comment about measures changing.
Edit: Oh, I see you did say this at the end.
As they are not measurable, they do not possess a finite number of describing equations, so “cutting” is not a good analogy.
Nah, just enchant a broom, cut it in two, and let them take care of the rest.
(Alas, no, that will only give a countable set of cuts... but the image is pleasing unless you're a mouse)
(Alas, no, that will only give a countable set of cuts... but the image is pleasing unless you're a mouse)
From my post:
> "cannot cut a circle into finitely many pieces"
> "cannot cut a circle into finitely many pieces"
I think the line integral of curvature (changes sign based on concave / convexness) along perimeters remains the same under cutting / pasting; and it's different for squares and circles, so doesn't work out.
Seems like a nice approach, but is the line integral of curvature along the perimeter of a square defined? The corners are infinitely small line segments with infinite curvature.
Perhaps this works for squares with slightly rounded-off corners.
Perhaps this works for squares with slightly rounded-off corners.
The square is defined as a union of lines, and the line integral of a whole shape is the summation of that of the curves it is a union of with open intervals not including the "corners" (points of 'infinite curvature' as you say). I think this definition would suffice because we just need to show that the sum is closed over cutting, no?
Must the cuts be straight?
If we allow curved cuts, it becomes possible despite your counterexample, doesn't it?
You still have conservation of 'convex curvy bits'.
You can eliminate a convex curvy bit by mating it with a concave curvy bit, but cutting out a concave curvy bit will again produce a convex curvy bit.
Can't get rid of the curvy bits!
You can eliminate a convex curvy bit by mating it with a concave curvy bit, but cutting out a concave curvy bit will again produce a convex curvy bit.
Can't get rid of the curvy bits!
There's a fairly recent Numberphile video about this.
https://www.youtube.com/watch?v=eYfpSAxGakI
https://www.youtube.com/watch?v=eYfpSAxGakI
Didn't the topology folks figure some of this out a few years ago? Felt like scissors congruent was just a restatement of some of the basics in topological analysis, from what I (probably incorrectly) recall.
https://dmsm.github.io/scissors-congruence/