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