# Inscribed Square Problem

\begin{conjecture} Does every \Def{Jordan curve} have 4 points on it which form the vertices of a square? \end{conjecture}

A Jordan curve is a continuous function $f$ from the closed interval $[0,1]$ to the plane $\mathbb{R}^{2}$ with the properties that $f$ is injective on the half-open interval $[0,1)$ (i.e., $f$ is \emph{simple}) and $f(0)=f(1)$ (i.e., $f$ is \emph{closed}).

## Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847}

[M] Meyerson, M.D., Equilateral triangles and continuous curves, Fund. Math. 110, (1980), 1--9. % % (Put an empty line between individual entries)

* indicates original appearance(s) of problem.

### in all simple closed curves there are 4n points

Would you clarify? An obtuse triangle has only one inscribed square, so this theorem is not true for n>=2. Do you have a reference to this theorem? Strashimir Popvassilev

### Is the conjecture known to

Is the conjecture known to be true for C^1-smooth curves?

### no.

If it were true for C^1 curves, then since a Jordan curve is compact, it may be weierstrass approximated by a series of C^1 curves (indeed by curves whose component functions are polynomials) such that the series converges uniformly to the given jordan curve. Then by assumption, each curve in the sequence contains 4 points forming a square, and the sequence of squares can be regarded as (eventually) a sequence in the (sequentially) compact space of the 4-fold product of any closed epsilon enlargement of the area bounded by the original jordan curve. It follows that the sequence of squares contains a convergent subsequence, which can be shown to be a square lying on the original jordan curve.

Thus, proving the C^1 case proves the general case.

### This is flawed

The approximation argument is flawed: the squares on approximating curves may have sides decreasing to 0, in which case the limiting "square" degenerates to a point. In fact, Stromquist's theorem covers a much wider class of curves than C^1, but not all continuous curves.

### Quantifier

Phrasing should be changed from "Does any..." to "Does every..."

## inscription of squares in simple closed curves

There is a theorem that says:

in all simple closed curves there are 4n points that are vertex of n squares (inf = > n > =1)

Jorge Pasin.