Here’s a conjecture that’s been open for more than a hundred years. Prove that every closed curve in the plane has an inscribed square. To unpack that for non-math readers: suppose I draw some loop on a piece of paper, which might be really complicated, cross over itself, zig-zag back and forth a lot, doesn’t matter — as long as it eventually returns to its starting point. Then the problem is to show that, no matter how crazy the loop, you can always find four points on the loop which form a perfect square.

It’s a fun exercise to try to convince yourself that this is even plausible! Also, the version of the problem with “square” replaced by “equilateral triangle” is much easier.

### Like this:

Like Loading...

*Related*

Well, it seems to me that the self-intersecting curves are not interesting. We can take any given non-self-intersecting portion and discard the rest (for example, take just the top loop of a figure 8) and inscribe the square just in the chosen portion.

The next move seems to be to treat the curve as a limiting case of some polygon, and then to show that a square can be inscribed in every polygon. It seems evident that this is true for convex polygons, so we need to figure out a suitable way to dispose of concave polygons.

Intuition takes me no further.

You’re right on so far. The case of arbitrary convex curves (not just polygons) is settled, I should say — the paper “A square inscribed in a convex figure,” by C.M.Christensen from 1950 apparently deals with this case (in Danish.) I might post on this once more — I have a few more possibly interesting comments to make about it.

I foresee a problem with carrying out your intuitive idea, John. Every closed curve is the limiting case of a sequence of polygons. (Meaning, the area between them approaches 0.)

But even if every single polygon in the sequence contains a perfect square (definitely not all polygons have a perfect square, but all convex polygons with 4+ sides have a inscribed quadrilateral that may share sides and all convex polygons with 8+ sides have an inscribed quadrilateral that does not necessarily share any side at all), there still is another condition to be met for there to definitively be a perfect square within the curve.

It is this: there must exist a sequence of squares inscribed inside the corresponding elements in the sequence of polygons of which there is a limiting perfect square that they “converge” to. If the squares inscribed inside the polygons are all over the place, then there may not be a perfect square in the limiting curve.

[...] inscribed squares 09Aug07 Another thought about the problem of showing that every curve has an inscribed square. My guess is that more is true: that in fact every closed curve in the plane has an odd number of [...]

A trivial comment on mindloop’s comment:

The only issue with the convergence that I see (assuming the approximates have squares) is when the squares in the terms have diameter tending to zero: if the squares in the approximates have diameter uniformly bounded below, you’ll get a limiting square by passing to a subsequence.

You might be able to get rid of this problem by rescaling, but then I’m not sure how to convert the limiting square into a square for the desired curve.

I seem to remember R. Schwartz saying that it suffices to work with polygons, but I don’t remember why. I should ask him.

Cheers.

I’m pretty sure it can’t be reduced to polygons, and my guess is that the reason is as Richard says, that you might indeed have a continuous family of squares as the polygons vary (though even this is not obvious to me!) but these squares might have a single point (square of side zero) as limit.