In [On the number of inscribed squares of a simple closed curve in the plane](https://arxiv.org/abs/0810.4806) 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.
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this is my question:



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> 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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Can one find a Jordan curve which has exactly one inscribed rectangle?
As Wojowu writes in the comments, Vaughan's argument (which is a paragraph long and can be read [here on page 71](http://topo.math.auburn.edu/tp/reprints/v06/tp06107.pdf)) 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](https://projecteuclid.org/download/pdf_1/euclid.ijm/1258138153); see the paper for the precise meaning of generic in this context.


