## inscribed triangle and square problem

I have found today this problem (or conjecture, if you wish):

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

There is a survey about partial results on the Quomodocumque blog, but I have no access to a library from where I am now, so I think trying to figure it out alone might be a nice idea.

Jordan’s theorem When I mention it I mean this: any Jordan curve $\Gamma$ has an interior and an exterior, which are open connected by arcs sets such that the one called interior is bounded and together they form the complement of $\Gamma$. In particular any path (i.e. continuous curve) starting inside $\Gamma$ and ending outside it must pass through $\Gamma$.

Easy part: the triangle case

I already previously had come in contact with the analogous problem where one looked for inscribed equilateral triangles instead of squares, and that is easier: the answer was yes, by some topological argument, as follows.

There is a ($2$-valued) function $c$ with values in $\mathbb R^2$ depending continuously on $a, b\in\Gamma$ such that $[a,b,c]$ is an equilateral triangle. Using this and Jordan’s theorem, (with some patience) one verifies that there are couples $(a,b), (a',b')$ such that $c, c'$ are respectvely in the interior  and in the exterior of $\Gamma$. Moving the couple $(a,b)$ in $(a',b')$, by Jordan’s theorem again one has a nice equilateral triangle $[a,b,c'']$ with vertices all on $\Gamma$.

The procedure which works for equilateral triangles also works for any kind of triangle (you just have to use the ordering on edge lenghts carefully in each step, and after some trial and error you manage the right combination).

The square case

This case is really square. The problem with the above continuity methods based on Jordan’s theorem seems to be that as 2 vertices wander on $\Gamma$ the other two may pass independently from one connected component of $\mathbb R^2\setminus \Gamma$ to the other.

1) If you suppose the curve is smooth (say $C^1$) then you might try to take an inscribed “half-square” triangle $[a,b,c]$ (i.e. isosceles with one $\pi/2$ angle in $b$) which exists by the above theorem, and try to perturb it still leaving it inscribed, seeing what happens to the $4$th vertex $d$ of the square. To help intuition, suppose that near $a$ and $b$ the graph of the curve is a piece of line (recall you assumed $\Gamma\in C^1$ so this is not far from true). Then perturbations of $a,b$ give a $2$-dimensional set of perturbations of $c$‘s if and only if the lines representing $\Gamma$ near $a,b$ do not form an angle of $3\pi/4$. If this happens, then by the inverse function theorem one can find for each $a$ one $b$ such that $c$ completing the half square is still on $\Gamma$.

Then you could try to reason like before, asking yourself if there are two squares $[abcd], [a',b',c',d']$ with $a,b,c, a',b',c'\in\Gamma$ and $d, d'$ in different connected components of $\mathbb R^2\setminus \Gamma$. You can construct them as follows:

– if $a,b\in \Gamma$ are very close then there are 2 vertices completing the square inside $\Gamma$ (this uses the regularity of $\Gamma$). Then you can move $b$ along $\Gamma$ until the first time when one of $c,d$ touch $\Gamma$.

– if $a',b'\in \Gamma$ realize the diameter of $\Gamma$ then you can move $b'$ towards $a'$ until the first hitting as above and you get the second square.

Now you move $a, a'$ closer together until they become the same point, and let $b,c,b',c'$ follow them staying on $\Gamma$ and forming half-squares as discussed earlier, and $d,d'$ follow them completing at each moment the respective squares. During all this procedure by Jordan’s theorem $d, d'$ have kept on staying on opposite components of the complement of $\Gamma$. However nothing tells that there is a general way of moving one onto the other while $3$ vertices of the cubes stay all the time on the curve.

2) Let’s use the $3\pi/4$ angle found above to try producing a counterexample. The first idea I get is starting from a regular octahedron and moving two adjacent sides parallel to themselves towards the center, leaving all the other sides supported on the same lines. Exercise: find a square inscribed in this figure!

(there is no square in the original octagon that “survives” on this new one. However:)

3) Lemma: in all convex curves there is an inscribed square: just take any small square completely inside and “blow it up”, immagining that the curve represents a “rigid hole” in outside which the square cannot pass (but it can slide if this helps continuing to “expand”). At a certain point you cannot blow anymore: the square has at least 3 vertices on the curve and has reached the maximal size. Now suppose 3 vertices $a,b,c$ of $[abcd]$ are on $\Gamma$ and $d$ isn’t. This implies that the tangent cone to the interior of $[abcd]$ in $a$ and $c$ do not contain the sides $[ad],[cd]$ else by convexity $d\in\Gamma$. Then it (is easy to see that it) is not true that you cannot blow anymore. $\square$

One can try to describe the problem in a phase space setting: squares are points in $\mathbb R^8$ belonging to a $4$-dimensional subspace $H^4$. To $\Gamma$ there corresponds a $4$-dimensional surface $\Gamma^4$, homeomorphic to the $4$-dimensional torus, which intersects for sure $H^4$ on the $2$-plane given by $P^2:=\{(a,a,a,a):a\in \mathbb R^2\}$. The question is if $\Gamma^4\cap H^4\subset P^2$ or if it must (for any $\Gamma$) contain also points outside this plane. I cannot immagine $8$-dimensional space well, so I cannot see if there is come topological condition which could help….