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In On the number of inscribed squares of a simple closed curve in the plane it is shown that

Theorem: For every positive integer $n$ there is a simple closed curve in the plane (which can be taken infinitely differentiable and convex) which has exactly $n$ inscribed squares.

this is my question:

Can one find a Jordan curve which has exactly one inscribed rectangle? More generally, what is the result if we use "rectangle" instead of "square" in the foregoing theorem?

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  • $\begingroup$ Can't you just apply an affine transformation to the curves constructed in that paper to turn the squares into rectangles of any aspect ratio you like? $\endgroup$
    – j.c.
    Commented Oct 28, 2017 at 10:40
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    $\begingroup$ @j.c. How do you guarantee there are no other rectangles? $\endgroup$
    – Wojowu
    Commented Oct 28, 2017 at 10:54
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    $\begingroup$ Following this argument should show that there are infinitely many rectangles. Any embedding of a Mobius strip with a flat boundary should have infinitely many self-crossing points. $\endgroup$
    – Wojowu
    Commented Oct 28, 2017 at 11:01
  • $\begingroup$ @Wojowu Thanks, I misunderstood the question. $\endgroup$
    – j.c.
    Commented Oct 28, 2017 at 17:14
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    $\begingroup$ thank you,someone had "objected" that out of all $n$ squares,only one is in general position. Each of the remaining $n-1$ many squares would either disappear,or duplicate, under a small perturbation. The following webpage by Mark J. Nielsen webpages.uidaho.edu/~markn/squares considers various cases, in particular THEOREM C: Every simple closed curve has at least one inscribed rectangle (it refers to a paper by Meyerson, which is also already referred to in the answer posted below). A survey about inscribed squares by Matschke, ams.org/notices/201404/rnoti-p346.pdf $\endgroup$
    – Mirko
    Commented Feb 22, 2018 at 0:36

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As Wojowu writes in the comments, Vaughan's argument (which is a paragraph long and can be read here on page 71) shows that there ought to be infinitely many rectangles inscribed in any Jordan curve. (The rectangles come from the double points of a real projective plane in 3-space, and I think — though can't remember the reference — that the preimage of the double set has a component which is nontrivial in first homology, which would imply it's infinite.)

You could fix a side length ratio in advance, and then ask whether there are Jordan curves with exactly one inscribed rectangle with that side length ratio. Probably no one has asked/answered this. Generically there are an even number of rectangles (with fixed side length ratio bigger than 1) inscribed in any Jordan curve, as opposed to an odd number of squares, so you'd have to make the pair of rectangles coincide.

EDIT: The preimage of the double set of a cone over the standard Möbius strip in $\mathbb{R}^3$ — take the standard strip near the unit circle in the $xy$-plane and the cone point at e.g. the origin — is a pair of open arcs whose closure is a generator of $H_1$. And this is roughly the picture corresponding to the projective plane you get from considering Vaughan's idea applied to an ellipse. So my recollection was wrong, but was nearly right for that example if you consider the closure of the preimage of the double set. But after being wrong once, I'm even less sure than I was before that I said something near the truth.

EDIT 2: It's true for the closure of the double set for generic maps. It's a special case of Theorem 2 in Double decker sets of generic surfaces in 3-space as homology classes; see the paper for the precise meaning of generic in this context.

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  • $\begingroup$ I'd be curious to see a proof of the fact that the preimage of the double points has nontrivial homology if you can dig up a reference or recall the idea. $\endgroup$
    – j.c.
    Commented Oct 28, 2017 at 20:38
  • $\begingroup$ @j.c. I wasn't quite right, see my edit. I'll try to look up the source of my (erroneous) idea but who knows what will turn up. $\endgroup$
    – Haggai Nuchi
    Commented Oct 29, 2017 at 3:16
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    $\begingroup$ @j.c. found it. see edit 2 above. $\endgroup$
    – Haggai Nuchi
    Commented Oct 29, 2017 at 4:10

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