A kind of elementary problems still with several open questions comes from the following problem:

Let $P$ be a simple polygon and let $\gamma : [0,1] \to \mathbb{R}^2$ be a simple closed curve. Does $\gamma$ contain an inscribed polygon (that is a polygon whose vertices belong to $\gamma([0,1])$) similar to $P$?

A nice introduction to such problems can be found here. In this note, we present some theorems related to this subject.

Theorem 1: Let $\gamma : [0,1] \to \mathbb{R}^2$ be a simple closed curve and $T$ be any triangle. Then $\gamma$ has an inscribed triangle similar to $T$.

Proof. The idea of the proof is in fact really simple, but many notations are needed. The main idea is summarized here and the different steps of the construction are drawn in the following picture:

Let $C_{int}$ be the bounded connected component of $\mathbb{R}^2 \backslash \Gamma$, where $\Gamma= \gamma([0,1])$, and $C_s=C+ (0,s)$ for all $s \geq 0$.

Because $\Gamma$ is compact, $r= \min \{ s >0 \mid C_s \cap \Gamma \neq \emptyset\}$ is well-defined. Let $PQR$ be a triangle similar to $T$ whose circumscribed circle is $C_r$ such that $P \in \Gamma$ and $QR$ is the longest side of $PQR$.

For all $s>0$, let $h_s$ be the homothety of ratio $s$ and centered at $P$. Because $\Gamma$ is compact, $t= \min \{ s \geq 0 \mid h_s(Q) \in \Gamma \ \text{or} \ h_s(R) \in \Gamma\}$ is well-defined. To fix the notations, say $h_t(Q) \in \Gamma$. For convenience, let $P=P_0$, $Q_0=h_t(Q)$ and $R_0=h_t(R)$.

Because $\overline{C_{int}}$ is compact, there exist $A,B \in \partial C_{int}=\Gamma$ such that $d(A,B)= \max\limits_{x,y \in \overline{C_{int}}} d(x,y)$. Let $p_0,q_0,a_0,b_0 \in [0,1]$ such that $\gamma(p_0)=P_0$, $\gamma(q_0)=Q_0$, $\gamma(a_0)=A$ and $\gamma(b_0)=B$.

Now, let $P(t)= \gamma((1-t)p_0+ta_0)$ and $Q(t)= \gamma((1-t)q_0+tb_0)$.

For each $t \in [0,1]$, we may chose a point $R(t)$ so that $P(t)Q(t)R(t)$ be similar to $P_0Q_0R_0$ and $t \mapsto R(t)$ be continuous. (There are only two possibles points to make our triangle similar to $P_0Q_0R_0$; by orienting the segment $P(t)Q(t)$, it is possible to fix our choice (taking the right point or the left point) to make the function $t \mapsto R(t)$ continuous.)

Now, notice that $d(R(1),Q(1)) \geq d(P(1),Q(1)) = d(A,B)$ and $Q(1) \in \Gamma$, so $R(1) \in \mathbb{R}^2 \backslash C_{int}$ by definition of $A$ and $B$. Therefore, there exists $t \in [0,1]$ such that $R(t) \in \partial C_{int}= \Gamma$.

Finally, $P(t)Q(t)R(t)$ is an inscribed triangle similar to $T$. $\square$

Theorem 2: Every simple closed curve in $\mathbb{R}^2$ has an inscribed rectangle.

Lemma 2.1: A homeomorphism $f : \partial \mathbb{D} \to \partial \mathbb{D}$ can be extended into a homeomorphism $g : \mathbb{D} \to \mathbb{D}$.

Proof. Let $g : x \mapsto \left\{ \begin{array}{cl} \displaystyle \|x\| \cdot f \left( \frac{x}{\|x\|} \right) & \text{if} \ x \neq 0 \\ 0 & \text{otherwise} \end{array} \right.$. Because $\|g(x)\|=\|x\|$ for all $x \in \mathbb{D}$, $g$ is continuous; moreover, $g$ is clearly an injective extension of $f$. Then, for all $x \in \mathbb{D}\backslash \{0\}$,

$\displaystyle g \left( \|x\| f^{-1} \left( \frac{x}{\|x\|} \right) \right)= \|x \| \cdot \left\| f^{-1} \left( \frac{x}{\|x\|} \right) \right\| \cdot f\left( f^{-1} \left( \frac{x}{\|x\|} \right) \right) =x$,

so $g$ is surjective. Finally, because $\mathbb{D}$ is compact, we deduce that $g$ is a homeomorphism. $\square$

Lemma 2.2: There is no continuous injection $P^2 \mathbb{R} \hookrightarrow \mathbb{R}^3$.

It is not a trivial result; we refer to a course on algebraic topology or on topology of surfaces for a rigorous proof.

Proof of theorem 2. Let $\gamma : [0,1] \to \mathbb{R}^2$ be a simple closed curve and let

$\Gamma = \gamma([0,1])$ and $X= \{ \{P,Q \} \mid P,Q \in \Gamma, \ P \neq Q \}$.

Our first goal is to define a natural topology on $X$.

Let $\sim$ be the equivalent relation on $\Gamma \times \Gamma \backslash \Delta$ (where $\Delta$ is the diagonal, ie. $\Delta= \{ (x,x) \mid x \in \Gamma\}$) defined by: $(P_1,Q_1) \sim (P_2,Q_2) \Leftrightarrow \{P_1,Q_1\} = \{P_2,Q_2\}$. Then, we use the following bijection to endow $X$ with a topology:

$\left\{ \begin{array}{ccc} (\Gamma \times \Gamma \backslash \Delta) / \sim & \to & X \\ (P,Q) & \mapsto & \{P,Q\} \end{array} \right.$.

Moreover, if $h : \Gamma \to \mathbb{S}^1$ is a homeomorphism, then the homeomorphism $h \times h : \Gamma \times \Gamma \backslash \Delta \to \mathbb{T} \backslash \Delta$ induces a homeomorphism from $X$ onto the quotient of $\mathbb{T} \backslash \Delta$ by the reflection with respect to $\Delta$, that is

Thus, $X$ is homeomorphic to $\mathbb{M} \backslash \partial \mathbb{M}$ where $\mathbb{M}$ is the Möbius strip.

Now let $f : \left\{ \begin{array}{ccc} X & \to & \mathbb{R}^3 \\ \{P,Q\} & \mapsto & \left( \frac{P+Q}{2}, d(P,Q) \right) \end{array} \right.$.

Because $f$ is just the quotient map associated to the restriction of $\varphi : \left\{ \begin{array}{ccc} \mathbb{R}^2 & \to & \mathbb{R}^3 \\ (P,Q) & \mapsto & \left( \frac{P+Q}{2}, d(P,Q) \right) \end{array} \right.$ to $\Gamma \times \Gamma \backslash \Delta$, we deduce that $f : X \to \mathbb{R}^3$ is continuous.

Moreover, if $\{P_n,Q_n \} \to \{ P,P \} \in \Delta \subset X$ then

$f(\{P_n,Q_n \})= \left( \frac{P_n+Q_n}{2},d(P_n,Q_n) \right) \to (P,0) \in \Gamma$,

so $f$ can be extended to a continuous function $f: \mathbb{M} \to \mathbb{R}^3$ such that $f(\partial \mathbb{M})= \Gamma$.

Let $C_{int} \subset \mathbb{R}^2$ be the bounded connected component of $\mathbb{R}^2 \backslash \Gamma$ and let $Y$ be the quotient of $M \coprod C_{int}$ by the identification $f_{|\partial \mathbb{M}} : \partial \mathbb{M} \to \Gamma$. If $g : C_{int} \subset Y \to C_{int} \subset \mathbb{R}^2$ is a homeomorphism extending $f_{|\partial \mathbb{M}}$, then $f \coprod g$ induces a continuous map $\phi : Y \to \mathbb{R}^3$. (Such a homeomorphism $g$ exists according to lemma 2.1.)

Notice that if $f$ is injective then $\phi$ so is and that $Y$ is homeomorphic to the projective plane $P^2 \mathbb{R}$:

However, there does not exist any continuous injection $P^1 \mathbb{R} \hookrightarrow \mathbb{R}^3$ (lemma 2.2), so there exist $\{P_1,Q_1\} \neq \{ P_2,Q_2\}$ such that $f(\{P_1,Q_1\})= f(\{P_2,Q_2\})$.

In particular, the segments $[P_1,Q_1]$ and $[P_2,Q_2]$ meet at their midpoints and have the same length. Therefore, the points $P_1,P_2,Q_1,Q_2$ define an inscribed rectangle. $\square$

Theorem 3: Every simple closed curve in $\mathbb{R}^2$ that is symmetric to the origin has an inscribed square.

Proof. Let $\gamma : [0,1] \to \mathbb{R}^2$ be a simple closed curve, $\Gamma= \gamma([0,1])$ and $R$ be the rotation of angle $\pi/2$ centered at the origin. We suppose $\Gamma$ symmetric to the origin, that is $P \in \Gamma$ implies $-P \in \Gamma$.

It is sufficient to prove that $\Gamma \cap R(\Gamma) \neq \emptyset$. Indeed, if $P \in \Gamma \cap R(\Gamma)$ then the points $P$, $-P$, $R(P)$ and $-R(P)$ defines an inscribed square.

Because $\Gamma$ is compact, $d( \cdot ,0)$ attains a maximum (resp. a minimum) on $\Gamma$ at some point $P_{\infty}$ (resp. $P_0$). If $P_0$ or $P_{\infty}$ belong to $\Gamma \cap R(\Gamma)$ we are done, so suppose that $P_0,P_{\infty} \notin \Gamma \cap R(\Gamma)$.

Let $C_{int}$ (resp. $C_{\infty}$) denote the bounded (resp. undbounded) connected component of $\mathbb{R}^2 \backslash \Gamma$. Because $R$ is an isometry, $R(P_0)$ (resp. $R(P_{\infty})$) is as closed (resp. as far) to $0$ than any point of $\Gamma$. Therefore, $P_0 \in C_{int}$ and $P_{\infty} \in C_{\infty}$. Because $R \circ \gamma$ is a continuous path connecting $R(P_0)$ and $R(P_{\infty})$ we deduce that $R(\Gamma)$ meets $\partial C_{int}= \Gamma$. $\square$