Is circle the only Jordan curve with this property?

Question

When I was thinking about one problem that has to do with Jordan curves the problem which I am going to describe now, arose in my mind.

And here it goes. It is known that for every $n\geq3$ the circle contains regular convex polygon with $n$ sides, or to try to be more precise and to describe my thoughts more clearly, for every $n\geq3$ the circle contains vertices of at least one regular convex polygon with $n$ sides.

The problem which arose in my mind is does the converse of this also holds, to state it directly : If $C$ is a Jordan curve with property that for every $n\geq3$ there is at least one regular convex polygon with $n$ sides which has vertices on $C$, is then $C$ a circle?

I believe that this is solved somewhere and would like that someone either gives me a link to the paper which contains the proof of this statement or that he/she proves it here in the answer, with arguments that are as elementary as possible.

Answer 1

No, this is not true. Without getting into great details of construction, here is the idea:

We will have a chain of circles, of ever decreasing sizes, such that the circles "converge" to the black point. We will cut small arcs in these circles, and we want them to be small enough so that n-th circle has vertices of n+2-gon. Now we use these cuts to connect the circles.

It's quite straightforward to see that this is a Jordan curve (e.g. we can map $(1/4,3/4)$ to the large circle, $(1/8,1/4)$ and $(3/4,7/8)$ to upper and lower parts of second circle etc. and just take care of connections between circles).

Answer 2

The answer is no.

For $r, \epsilon_0, \epsilon_1>0$, let $C_{r, \epsilon_0, \epsilon_1}$ be a circle of radius $r$ with two disjoint arcs of lengths $\epsilon_0$ and $\epsilon_1$ deleted. Note that for any $n$, there is a $q_n>0$ such that for all $r, \epsilon_0, \epsilon_1>0$, if ${\epsilon_0\over r}, {\epsilon_1\over r}< q_n$, then $C_{r, \epsilon_0, \epsilon_1}$ contains a regular $n$-gon.

Now let $C_i=C_{2^{-i}, {2^{-i^2}}, 2^{-(i+1)^2}}$; we can glue the $C_i$s by identifying deleted arcs of the same length. This produces a Jordan curve (well, once we add a point at the "end" to close it off, and fill in the unclosed arc at the "beginning") which contains a regular $n$-gon for each $n$.

EDIT: Wojowu's answer is essentially the same as mine, and has a nice picture of what this looks like.

---

By contrast, if we ask for the $n$-gons to have the same radius $r$ (as opposed to radii going to $0$), then the answer is yes: suppose $J$ is a Jordan curve with this propery. Let $C_i$ be a circle of radius $r$ such that $C_i\cap J$ contains a regular $i$-gon, and let $c_i$ be the center of $C_i$. Then since $J$ is compact, there is an infinite sequence $\{a_0<a_1<. . .\}$ such that $c_{a_i}$ converges - say, to $c$. Now since $J$ is closed, $J$ must contain the circle of radius $r$, centered at $c$; but since $J$ is a Jordan curve this means $J$ is precisely that circle.

Note that this also works if we just demand that the radii of the $n$-gons are bounded below, above $0$.

Also, as an immediate corollary, we get that you can't avoid something like my/Wojowu's construction:

The only smooth Jordan curve which contains a regular $n$-gon for every $n$ is the circle.