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Topologically each point on the edge of a square can be mapped uniquelly (including the corners?) to a point on the edge of a circle.

It seems it might be possible to deform in 3D space the square and the disc so that they can be joined at the edge without any gaps. A colleague managed to do that in the real world with paper objects. Obviously this is not a precise result, so it may be wrong.

As a commenter said, the deformation should be isometric.

Can you make Mathematica solve this problem and generate a 3D surface showing a isometrically deformed square and disc joined at the edge?

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Pictures taken from http://www.unitaryflow.com/2015/03/round-squares-exist.html

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    $\begingroup$ You need more constraints on your problem. I can just generate you e.g. a cube with a face cut off, and assert that "this is a deformed circle." $\endgroup$ – djp Mar 14 '15 at 22:12
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    $\begingroup$ @djp: Given the example of paper cutouts, I assume that the deformation is meant to be isometric. See also: dForm $\endgroup$ – Rahul Mar 15 '15 at 4:53
  • $\begingroup$ @Rahul interesting. I wish the OP had a little more of "this is what I've tried" etc. $\endgroup$ – djp Mar 15 '15 at 10:47
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circ := {Cos[#], Sin[#]} &

quad = Interpolation[
  Table[{f, {Cos[f], Sin[f]}}, {f, 0, 2 π, π/2}], 
  InterpolationOrder -> 1]

ParametricPlot3D[{quad[t] z + circ[t] (1 - z), z} // Flatten, 
  {t, 0, 2 π}, {z, 0, 1}, 
  Mesh -> None, PlotStyle -> Opacity[0.8], ColorFunction -> "Rainbow"]

enter image description here

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  • $\begingroup$ Your solution has the original square, disc, and adds an interpolating wall between them. This is not what was requested. The solution must not add a extra surface (the wall) $\endgroup$ – Meh Mar 14 '15 at 21:41

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