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Inscribed square problem wants that we know "Does every Jordan curve admit an inscribed square?"

From my amateur viewpoint it seems that the concept of Jordan curve can be straightforwardly generalized to $\mathbb R^n$ so as to define those curves so that they are images of continuous functions $f$ from $[0,1]$ to $\mathbb R^n$ in such a way that those functions are injective on $[0,1)$ and $f(0)=f(1)$

On Wikipedia, it is written:

" It is known that for any triangle T and Jordan curve C, there is a triangle similar to T and inscribed in C.[9][10] Moreover, the set of the vertices of such triangles is dense in C.[11] In particular, there is always an inscribed equilateral triangle"

But, I do not have an access to articles [9], [10], [11] so do not know do they only discuss Jordan curves in the plane.

I would just like to know is it known:

Do all Jordan curves in all spaces $\mathbb R^n$ admit an inscribed equilateral triangle?

Of course, avoid the non-interesting case $n=1$

If i had the access to [9] then I could know are there discussed equilateral triangles only for Jordan curves in $\mathbb R^2$ or in all $\mathbb R^n$ spaces.

What is known about this question of mine?

The "Inscribed square problem" page is linked because the quote from Wikipedia is on that page and also the links to those articles are on that page.

Now I am thinking that I am asking something very trivial here and am in a state of not knowing should I even post this question, but, I will post the question and face the consequences.

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    $\begingroup$ I think curves in $\mathbb R^n$ with $n > 2$ are not called "Jordan curves". $\endgroup$
    – Gerald Edgar
    Commented Jan 6, 2018 at 12:23
  • $\begingroup$ I imagine the answer is yes and it's perhaps not too difficult to prove. An embedded circle in $\mathbb R^n$ without any inscribed equilateral triangles you can think of as having a relatively "thick" tubular neighbourhood relative to its diameter. That's where I'd start. $\endgroup$
    – Ryan Budney
    Commented Feb 5, 2018 at 6:30
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    $\begingroup$ Here it is shown that if there is at least one point at which the curve is differentiable, such a triangle exists. $\endgroup$
    – Wojowu
    Commented Feb 5, 2018 at 13:04

2 Answers 2

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Theorem 5 in [9] proves that a simple closed curve with arbitrary metric contains vertices of an equilateral triangle. In particular, this applies to any simple closed curve in $\mathbb{R}^n$.

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  • $\begingroup$ Usually the word "metrizable" applies only to topological spaces. (So either all spaces homeomorphic to some space X are metrizable, or they all are not.) Perhaps what is meant here is a "rectifiable" simple closed curve? $\endgroup$
    – Daniel Asimov
    Commented Dec 11, 2023 at 20:19
  • $\begingroup$ Right; the author uses the notion "metrized". I changed the formulation. $\endgroup$
    – Jan Kyncl
    Commented Dec 12, 2023 at 3:23
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The papers only treat the planar case, and the methods are very specifically planar, so it looks like your question is open.

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