Inscribed Square Problem

Importance: Medium ✭✭

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Keywords:	simple closed curve
	square

Recomm. for undergrads: no

Posted	by:	dlh12
	on:	April 10th, 2008

Conjecture Does every Jordan curve have 4 points on it which form the vertices of a square?

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 simple) and $f(0)=f(1)$ (i.e., $f$ is closed).

Bibliography

[M] Meyerson, M.D., Equilateral triangles and continuous curves, Fund. Math. 110, (1980), 1--9.

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Already solved

On February 25th, 2021 Anonymous says:

This was proven in https://arxiv.org/pdf/2005.09193.pdf

inscription of squares in simple closed curves

On March 3rd, 2010 Anonymous says:

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.

in all simple closed curves there are 4n points

On November 8th, 2010 Anonymous says:

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

On November 25th, 2009 Anonymous says:

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

Yes

On June 7th, 2010 Anonymous says:

Yes. Walter Stromquist, Inscribed squares and square-like quadrilaterals in closed curves, Mathematika 36: 187-197 (1989).

no.

On June 2nd, 2010 Anonymous says:

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

On June 7th, 2010 Anonymous says:

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

On April 28th, 2008 Anonymous says:

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

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