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

Let be a simple polygon and let be a simple closed curve. Does contain an inscribed polygon (that is a polygon whose vertices belong to ) similar to ?

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

**Theorem 1:** Let be a simple closed curve and be any triangle. Then has an inscribed triangle similar to .

**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 be the bounded connected component of , where , and for all .

Because is compact, is well-defined. Let be a triangle similar to whose circumscribed circle is such that and is the longest side of .

For all , let be the homothety of ratio and centered at . Because is compact, is well-defined. To fix the notations, say . For convenience, let , and .

Because is compact, there exist such that . Let such that , , and .

Now, let and .

For each , we may chose a point so that be similar to and be continuous. (There are only two possibles points to make our triangle similar to ; by orienting the segment , it is possible to fix our choice (taking the right point or the left point) to make the function continuous.)

Now, notice that and , so by definition of and . Therefore, there exists such that .

Finally, is an inscribed triangle similar to .

**Theorem 2:** Every simple closed curve in has an inscribed rectangle.

**Lemma 2.1:** A homeomorphism can be extended into a homeomorphism .

**Proof.** Let . Because for all , is continuous; moreover, is clearly an injective extension of . Then, for all ,

,

so is surjective. Finally, because is compact, we deduce that is a homeomorphism.

**Lemma 2.2:** There is no continuous injection .

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 be a simple closed curve and let

and .

Our first goal is to define a natural topology on .

Let be the equivalent relation on (where is the diagonal, ie. ) defined by: . Then, we use the following bijection to endow with a topology:

.

Moreover, if is a homeomorphism, then the homeomorphism induces a homeomorphism from onto the quotient of by the reflection with respect to , that is

Thus, is homeomorphic to where is the Möbius strip.

Now let .

Because is just the quotient map associated to the restriction of to , we deduce that is continuous.

Moreover, if then

,

so can be extended to a continuous function such that .

Let be the bounded connected component of and let be the quotient of by the identification . If is a homeomorphism extending , then induces a continuous map . (Such a homeomorphism exists according to lemma 2.1.)

Notice that if is injective then so is and that is homeomorphic to the projective plane :

However, there does not exist any continuous injection (lemma 2.2), so there exist such that .

In particular, the segments and meet at their midpoints and have the same length. Therefore, the points define an inscribed rectangle.

**Theorem 3:** Every simple closed curve in that is symmetric to the origin has an inscribed square.

**Proof.** Let be a simple closed curve, and be the rotation of angle centered at the origin. We suppose symmetric to the origin, that is implies .

It is sufficient to prove that . Indeed, if then the points , , and defines an inscribed square.

Because is compact, attains a maximum (resp. a minimum) on at some point (resp. ). If or belong to we are done, so suppose that .

Let (resp. ) denote the bounded (resp. undbounded) connected component of . Because is an isometry, (resp. ) is as closed (resp. as far) to than any point of . Therefore, and . Because is a continuous path connecting and we deduce that meets .