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!
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.