This is a puzzle from P. Winkler: "Show that, given any closed curve in the plane, there is a square containing the curve, all four sides of which touch the curve."

I was NOT able to solve it quickly (i.e. in less than 30 minutes :D), and then I was impatient and looked at the solution presented on the linked page (I regret it, because it ended the fun of searching for a solution). I would like to see different solutions, so guys please try to solve it without looking at the solution there. :D

Can you prove it? Please show your proof!

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    $\begingroup$ I think it should work this way (sketch): Pick an orientation for the lake. Easy to circumscribe a rectangle. Now rotate the rectangle you made adjusting the sides, adjusting the sides as you go to preserve tangency. Eventually you get back to the original with width and length exchanged...so there had to be a square along the way. $\endgroup$ – lulu Jul 19 '15 at 15:02
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    $\begingroup$ Oh, sorry. Didn't look at the linked solution. Obviously equivalent (more or less identical) to my comment. Nothing new in my solution, but at least it appears to work. $\endgroup$ – lulu Jul 19 '15 at 15:14

Define the function $f:S^1\to \mathbb{R}$, where $S^1$ is the unit circle (in other words all possible directions on the plane). For a fixed $x\in S^1$ define the rectangle $R(x)$ as follows: there are to lines $l_1$ and $l_2$ that have the same direction as $x$ and are touching our curve, also there are two lines $l_3$ and $l_4$ that have perpendicular direction with respect to $x$ and are touching our curve. The rectangle generated by these lines let's call $R(x)$ and define $$f(x) = \frac{\text{length of the side of $R(x)$ which has $x$ direction}}{\text{length of the side of $R(x)$ which perpendicular to $x$ direction}}.$$

Note, that if for some $x_1\in S^1$ we have $f(x_1)<1$ ($f(x_1)>1$), then for perpendicular direction $x_2$ we would have $f(x_2)>1$ ($f(x_2)>1$). Since the curve itself is continuous map, we have that the function $f$ is a continous function, therefore for some direction $x_0$ we have $f(x_0)=1$, which means that $R(x_0)$ is a square touching all sides with the curve.

  • $\begingroup$ I am struggling a bit with the argument that $f$ is continuous. What if the curve is so irregular (and concave) that it is not possible to find a "center" about which the curve could be expressed as a function in polar coordinates? Do we then look at the convex hull rather than the curve itself? $\endgroup$ – Marconius Jul 23 '15 at 0:25

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