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   Do squares always exist in curves?
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   Author  Topic: Do squares always exist in curves?  (Read 2220 times)
BMAD
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Do squares always exist in curves?  
« on: Jun 1st, 2014, 9:34pm »
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Draw a simple (no crossings) closed (not necessarily convex) curve on a sheet of paper.
Can you always find four points on the curve that form the four vertices of a square?
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towr
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Re: Do squares always exist in curves?  
« Reply #1 on: Jun 1st, 2014, 10:42pm »
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Can the curve be fractal?  
I'm thinking, start with any curve, and wherever there are points that form a square, replace a bit of a curve to swirl around those points.
 
I don't actually know if that would work.
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Grimbal
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Re: Do squares always exist in curves?  
« Reply #2 on: Jun 2nd, 2014, 1:22am »
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I think yes, there is.
 
The hard thing is to prove it.
 
Informally, the proof is to take a square and let 2 opposite corner A and C run around the loop.  Doing that, you rotate the square while trying to fit the 2 other corners B and D within the loop.  If you start with A and C on a horizontal line, and B and D can fit within the loop, then after 1/4 turn, A and C are on a vertical line, where BD originally was and now B and D are outside of the loop.
So, at some point (for some angle between AC and the horizontal), you go from "B and D both fit in the loop" to "B and D fit outside the loop".  At that point, it should be possible to fit both B and D on the loop.
 
There are plenty of loose ends, but it makes me feel such a square must exist.
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Re: Do squares always exist in curves?  
« Reply #3 on: Jun 2nd, 2014, 7:12am »
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on Jun 2nd, 2014, 1:22am, Grimbal wrote:
I think yes, there is.
 
The hard thing is to prove it.
 
Informally, the proof is to take a square and let 2 opposite corner A and C run around the loop.  Doing that, you rotate the square while trying to fit the 2 other corners B and D within the loop.  If you start with A and C on a horizontal line, and B and D can fit within the loop, then after 1/4 turn, A and C are on a vertical line, where BD originally was and now B and D are outside of the loop.
So, at some point (for some angle between AC and the horizontal), you go from "B and D both fit in the loop" to "B and D fit outside the loop".  At that point, it should be possible to fit both B and D on the loop.
 
There are plenty of loose ends, but it makes me feel such a square must exist.

It's probably better to define a rhombus for each direction:
 
1) pick a point A on the curve and project a line in the chosen direction.
 
2) For each point of intersection, C, find the midpoint of AC and project a line perpendicular to the chosen direction.
 
3) Each pair of intersections of the perpendicular with the curve, BD, defines a kite, ABCD.
 
4) I'm not 100% convinced, but I think you can then pick a continuous sequence of kites where A, C, B and D all move continuously around the curve (each point potentially reversing direction multiple times) so that the line AC sweeps across the entire width of the shape. At one side of the shape, AC will be left of the midpoint of BD; at the other, it'll be right of the midpoint. Since you have a continuous sequence, at some point AC must be the perpendicular bisector of BD and ABCD be a rhombus.
 
Once you have a set of rhombi for each direction, you can again find continuous sequences. If you can find one that returns to the original shape after a net 90-degree rotation, you're done.
 
 
 
So the holes are:
 
A) proving you can span the width of the shape with a continuous sequence of (not completely degenerate) kites
B) proving you can find a continuous sequence of rhombi that gives a 90 degree rotation
 
There's an additional assumption that the curve is smooth for the continuity arguments to work.
 
 
 
Of course, there's also the degenerate solution where you pick the same point 4 times and produce the size-0 square.
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rloginunix
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Re: Do squares always exist in curves?  
« Reply #4 on: Jun 2nd, 2014, 8:54am »
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I would suggest a Division (or Calculus) approach for a proof:
 
divide the whole into solvable parts, obtain the solution for each one, find a way to combine the solutions of the parts into the solution of the whole.
 
Whole = arbitrary smooth curve.
 
Part = inscribed-square-proved-to-exist curve (circles, for example).
 
Combine = approximate an arbitrary curve with a collection of inscribed-square-proved-to-exist curves.
 
Find a way = apply the idea of limit to the set of above (approximate) squares.
 
Haven't touched limits in years, sorry, but that's one reasonable way to attack this, I think.
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Re: Do squares always exist in curves?  
« Reply #5 on: Jun 2nd, 2014, 1:21pm »
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Wikipedia says it is an open problem.
 
http://en.wikipedia.org/wiki/Inscribed_square_problem
 
They even explain that the problem with a limit of successive approximations is that the solutions can converge to a simple point.
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rloginunix
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Re: Do squares always exist in curves?  
« Reply #6 on: Jun 2nd, 2014, 2:55pm »
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Bummer! (worked for compass-only regular pentagon construction ...)
 
Is this some sort of crowd sourcing style outstanding problems solution search?
 
I've seen something similar in this PBS Nova episode, skip forward to the 40-th minute. One gentlemen from the Carnegie Mellon University created a game called "fold it", posted it on the internet and in about 3 weeks the community of the online gamers solved the (highly visual, I take it) protein folding problem that the Ph.Ds and computers couldn't.
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Re: Do squares always exist in curves?  
« Reply #7 on: Jun 2nd, 2014, 4:48pm »
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I think the wikipedia link is enough to solve the puzzle affirmatively. After all, it's not like you can actually draw any of the curves that towr was talking about on paper.  Any curve you could draw on paper would be well-behaved enough to have an inscribed square.
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Re: Do squares always exist in curves?  
« Reply #8 on: Jun 3rd, 2014, 6:25am »
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The proof for a piecewise smooth curve resembles my outline above - it uses a different approach to prove there must be a rhombus, though it's plausible that the detailed proof that there must be an odd number of crossings also fills the hole in my argument.
 
The problem I have with the Wikipedia article's account of the proof is that it doesn't acknowledge the existence of my second hole (the requirement that the rhombi for different directions include a continuous rotating sequence - in particular, with a concave shape, with the short diagonal tangent to a concavity at both ends, with one concavity to the left and the other to the right, a small rotation one way will cause a discontinuity in the sequence of rhombi - in particular, the short diagonal could jump to being the long diagonal without an intermediate square)
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