Circumscribing Squares Around Closed Curves I'm interested in solving the following problem (Problem 12-6 in "Introduction to Analysis" by Arthur Mattuck).

Let $C$ be a continuous closed curve, i.e., one without endpoints. Show convincingly that it is always possible to circumscribe a square around $C$, that is, find a square all four of whose sides touch $C$ (so that the square cannot be shrunk without rotation to a smaller rectangle enclosing $C$). (Try some sketches).

My attempt:
I can't see how to start here as I'm not really clear on the problem. The use of "touch" makes me think of tangents, but then the curve $C$ is only said to be continuous. I'm also not sure about the meaning of "shrunk". Is there a clear interpretation of the problem which I can't see? I would also appreciate a hint on how to get started on this problem.
Thanks.
 A: (I'm assuming the problem takes place in $\Bbb{R}^2$.)
The first thing I would do is reject the closed curve $C$, and replace it with a compact set $C$. Any (continuous) curve $r : [0, 1] \to \Bbb{R}^2$ has a compact image, as it is the continuous image of a compact set.
Next, consider the function
$$f : \Bbb{R}^2 \to \Bbb{R} : u \mapsto \sup_{c \in C} u \cdot c - \inf_{d \in C} u \cdot d = \sup_{c, d \in C} u \cdot (c - d).$$
Before I get into why I'm defining this function, I want to claim that it is continuous. Unfortunately, I don't have a particularly elementary proof in mind, but $f$ is a convex function as it is the pointwise supremum of a family of linear (and hence convex) functions $u \mapsto u \cdot (c - d)$, indexed by $(c, d) \in C^2$. This makes the function lower semicontinuous, and convex, and since we are in finite dimensions, this implies $f$ is continuous.
Now, I want to restrict $f$ to the unit circle $S_{\Bbb{R}^2}$, centred at $(0, 0)$. What does the function $f|_{S_{\Bbb{R}^2}}$ do? We start with a unit vector $u$, and fit two parallel lines, both perpendicular to $u$, to fit our set $C$ between as snugly as possible. The value of $f(u)$ is the distance between these two lines. Essentially, our problem reduces now to finding orthonormal vectors $e_1, e_2$ such that $f(e_1) = f(e_2)$, as the two lines perpendicular to $e_1$ and the two lines perpendicular to $e_2$ form a square which cannot be shrunk in the way the question describes.
Consider
$$g : \Bbb{R} \to \Bbb{R} : \theta \mapsto f(\cos(\theta), \sin(\theta)).$$
Clearly, $g(\theta + 2\pi) = g(\theta)$ for all $\theta$. Further, note that, since $f$ is an even function, i.e. $f(-u) = f(u)$, we can refine this:
$$g(\theta + \pi) = g(\theta)$$
for all $\theta \in \Bbb{R}$. Now, $g$ is a continuous, periodic function. We can therefore adapt an argument like this; we should be able to find some $\alpha$ such that $g(\alpha) = g(\alpha + \pi/2)$. This is what we want! We can then take $e_1 = (\cos(\alpha), \sin(\alpha))$ and $e_2 = (\cos(\alpha + \pi/2), \sin(\alpha + \pi/2))$ to be our orthonormal vectors parallel to the sides of our minimal square.
To establish the existence of $\alpha$, consider the function $h(\theta) = g(\theta) - g(\theta + \pi/2)$. If $h(0) = 0$, then we are done: take $\alpha = 0$. Otherwise $h(0) > 0$ or $h(0) < 0$. If the former, then $h(\pi/2) < 0$, so by the IVT, $h(\alpha) = 0$ for some $\alpha$ between $0$ and $\pi/2$. Similar logic works when $h(0) < 0$, and hence $h(\pi/2) > 0$. Either way, we are done.
Hopefully someone comes up with a more elementary proof!
A: This follows from the Intermediate Value Theorem. If $C$ is our closed curve in $\mathbb{R}^{2}$, let $l$ and $l'$ be two parallel lines so that $C$ is contained inside the open strip between them (for example, you may choose $l$ and $l'$ parallel to the $x$-axis). Now, translate them parallel to their original positions until they touch $C$.
Now, repeat the process with two lines perpendicular to $l$ to obtain a rectangle $ABCD$ circumscribed about $C$. We will show that for a proper choice of slope of $l$ the rectangle becomes a square.
Indeed, let $f_{1}(l)$ denote the length of the side of the rectangle, say WLOG $AD$, parallel to $l$, and $f_{2}(l)$ denote the length of the side $AB$ perpendicular to $l$. Clearly the rectangle would be a square if $f_{1}(l)-f_{2}(l) = 0$.
Let $l^{*}$ be a line perpendicular to $l$. The corresponding rectangle circumscribing $C$ with sides parallel and perpendicular to $l^{*}$ is just the rectangle $ABCD$, but now the side $AD$ is perpendicular to $l^{*}$ and the side $AB$ is parallel to $l^{*}$. That is, $f_{1}(l^{*}) = f_{2}(l)$ and $f_{2}(l^{*})=f_{1}(l)$. Therefore,
$$f_{1}(l^{*})-f_{2}(l^{*}) = f_{2}(l)-f_{1}(l) = -(f_{1}(l)-f_{2}(l)).$$
Now, as we rotate $l$ until it is parallel to $l^{*}$, the circumscribed rectangle changes continuously. That is, the difference $f_{1}(l)-f_{2}(l)$ depends continuously on $l$. But then, rotating from $l$ to $l^{*}$ makes this difference change sign. Therefore, the Intermediate Value Theorem tells us that at some point (i.e. for a suitable slope of $l$) this difference must be $0$, and so the rectangle becomes a square, as desired.
