The (recently solved) inscribed square problem or Toeplitz conjecture posits that every closed, plane continuous (Jordan) curve ${\it \Gamma}$ in $\mathbb{R}^2$ contains all vertices of some square. It appears this theorem was just proven!

Most examples resemble the one on the left, in which the natural continuous parameterization along ${\it \Gamma}$ intersects the square's points in sequence. Let's call such an inscription cyclic.

enter image description here

However, some Jordan curves contain the points of a square in non-sequential order, such as shown at the right. Let's call such an inscription acyclic.


  • Are there Jordan curves that admit only acyclic inscriptions (and not also cyclic inscriptions)?
  • Given the inscribed square problem was just answered in the affirmative, can one prove whether acyclic-only curves exist?
  • Alternatively, or additionally: can one provide an example of such an acyclic-only curve?

My conjecture is that there are no such acyclic-only Jordan curves. My first approach has been to assume that there is a given acyclic square inscription for a Jordan curve and then prove--invoking continuity assumptions and topological methods--that there must also be a cyclic inscription. Alas, such a proof for even this partial case has been elusive.

  • $\begingroup$ Each edit bumps this post to the top of the active list and another question off. Could we perhaps wrap up the editing at least for a while, so that people have a stable question to look at? $\endgroup$
    – Todd Trimble
    Jul 15 '17 at 3:57

The recent work by Jason Cantarella, Elizabeth Denne and John McCleary implies (but does not prove) that the answer to all questions is negative, as I conjectured. Specifically, under certain general conditions one can find a cyclic inscription, and thus precluding the existence of an acyclic-only Jordan curve.

  • 1
    $\begingroup$ I don't understand "implies (but does not prove) ...." If statement S has been proved, and it implies statement T, then statement T has been proved. Perhaps "suggests" would be better than "implies"? $\endgroup$ Oct 7 '20 at 22:18
  • $\begingroup$ That "recent" work was from three years ago and relied on some continuity assumptions that might not hold in every Jordan curve. The NEW (2020) result proves the case. $\endgroup$ Oct 7 '20 at 22:57
  • 1
    $\begingroup$ Whether it was three years ago or three minutes ago, I don't see how it could imply a result without proving that result. If I prove something that implies the Riemann Hypothesis, then, by golly, I have proved the Riemann Hypothesis! (needless to say, I've done no such thing) $\endgroup$ Oct 8 '20 at 6:13
  • $\begingroup$ There are innumerable cases in which a mathematical result "implies" another result but does not "prove" it. Andy Wile's first announced proof of Fermat's Last Theorem (correctly) broke through several conceptual blocks, and was hailed by elite mathematicians (and popular press) as a full solution. But it was not. That took more effort. But even during that final period in extending his proof most elite mathematicians expected the full proof would be available (on the correct parts), even if they admitted the proof hadn't yet been found. This is all that "implies" ("suggests") means. $\endgroup$ Oct 8 '20 at 16:25
  • 1
    $\begingroup$ I understood "implies" in the usual (for mathematical conversations) mathematical sense, and I understood "does not prove" in the usual mathematical sense, and I pointed out the contradiction there, and I suggested using "suggests" instead of "implies". "What could be clearer than that?" Well, you could have written "suggests" instead of "implies" – that would have been a lot clearer (to me, at any rate). Or you could have written "implies (in the informal sense, not the mathematical sense)" – that, too, would have been clearer. $\endgroup$ Oct 9 '20 at 22:15

I am not a member here and so could not provide a comment. The claim that it has recently been solved is inaccurate. Green and Lobb solve the $\textbf{smooth}$ version of the Rectangle Peg Problem. For the square case, that’s been known since about a 100 years now (I suppose Schnirelman was the first) or at least some decades now. Indeed, it is true for all continuously differentiable curves.

I suppose you can check a few results of Richard Schwarz, in the last couple of years, which may answer your questions regarding the (a)cyclicality.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.