Area and perimeter foliations on spaces of polygons

Aziz El Kacimi Alaoui; Abdellatif Zeggar

arXiv:1902.04455·math.DG·February 13, 2019

Area and perimeter foliations on spaces of polygons

Aziz El Kacimi Alaoui, Abdellatif Zeggar

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TL;DR

This paper characterizes families of star-shaped polygons with fixed area and perimeter as leaves of a foliation in the space of polygons, and explores geometric properties of convex polygons related to these measures.

Contribution

It provides a complete description of polygon families with given area and perimeter, linking geometric properties to these measures.

Findings

01

Families of star-shaped polygons form a foliation with prescribed area and perimeter.

02

Convex polygons' inscriptibility and regularity relate to their perimeter and area.

03

The geometric properties of polygons are analyzed in the context of the foliation structure.

Abstract

We describe all families of star-shaped n-polygons in the Euclidean plane with prescribed perimeter and area ; they are leaves of a foliation F on the space of star-shaped n-polygons. By the way, we study some geometric properties of convex polygons, for instance their inscriptibility in a circle and their regularity in relation with the perimeter and the area.

Figures6

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Equations160

M_{k}^{'} = M_{k} and ∣∣ M_{k}^{'} M_{ℓ}^{'} ∣∣ = ∣∣ M_{k} M_{ℓ} ∣∣ for (k, ℓ) \in {1, \dots, n}^{2} .

M_{k}^{'} = M_{k} and ∣∣ M_{k}^{'} M_{ℓ}^{'} ∣∣ = ∣∣ M_{k} M_{ℓ} ∣∣ for (k, ℓ) \in {1, \dots, n}^{2} .

\hbox{Isom}({\hbox{\mathb E}})\times\widetilde{\cal{P}}_{n}\longrightarrow\widetilde{\cal{P}}_{n},\;\big{(}f,(M_{1},\cdots,M_{n})\big{)}\longmapsto\big{(}f(M_{1}),\cdots,f(M_{n})\big{)}

\hbox{Isom}({\hbox{\mathb E}})\times\widetilde{\cal{P}}_{n}\longrightarrow\widetilde{\cal{P}}_{n},\;\big{(}f,(M_{1},\cdots,M_{n})\big{)}\longmapsto\big{(}f(M_{1}),\cdots,f(M_{n})\big{)}

P_{n} = P_{n} /Isom (\mathb E) .

P_{n} = P_{n} /Isom (\mathb E) .

\cases < M_{1}, \dots, M_{n} > is the equivalence class of (M_{1}, \dots, M_{n}) in P_{n}, \cr \cr x_{k} = M_{k} M_{k + 1} = ∣∣ M_{k} M_{k + 1} ∣∣ for k \in {1, \dots, n - 2}, \cr \cr t_{k} = M_{n} M_{k} = ∣∣ M_{n} M_{k} ∣∣ for k \in {1, \dots, n - 1} .

\cases < M_{1}, \dots, M_{n} > is the equivalence class of (M_{1}, \dots, M_{n}) in P_{n}, \cr \cr x_{k} = M_{k} M_{k + 1} = ∣∣ M_{k} M_{k + 1} ∣∣ for k \in {1, \dots, n - 2}, \cr \cr t_{k} = M_{n} M_{k} = ∣∣ M_{n} M_{k} ∣∣ for k \in {1, \dots, n - 1} .

\cases 0 < t_{k} < x_{k} + t_{k + 1} \cr 0 < x_{k} < t_{k} + t_{k + 1} \cr 0 < t_{k + 1} < x_{k} + t_{k}

\cases 0 < t_{k} < x_{k} + t_{k + 1} \cr 0 < x_{k} < t_{k} + t_{k + 1} \cr 0 < t_{k + 1} < x_{k} + t_{k}

\cases 0 < x < y + z \cr 0 < y < x + z \cr 0 < z < x + y .

\cases 0 < x < y + z \cr 0 < y < x + z \cr 0 < z < x + y .

\cases M_{n} = M_{0} = O the origin of \mathb E, \cr \cr M_{1} has coordinates (t_{1}, 0) with t_{1} > 0, \cr \cr for any k \in {1, \dots, n - 2}, the measure of the angle M_{k} O M_{k + 1} = (O M_{k}, O M_{k + 1}) is in] 0, π [.

\cases M_{n} = M_{0} = O the origin of \mathb E, \cr \cr M_{1} has coordinates (t_{1}, 0) with t_{1} > 0, \cr \cr for any k \in {1, \dots, n - 2}, the measure of the angle M_{k} O M_{k + 1} = (O M_{k}, O M_{k + 1}) is in] 0, π [.

p = t + x + s, a = \frac{1}{4} (t + x + s) (t - x + s) (t + x - s) (- t + x + s) (H \overset{e}{ˊ} ron’s formula)

p = t + x + s, a = \frac{1}{4} (t + x + s) (t - x + s) (t + x - s) (- t + x + s) (H \overset{e}{ˊ} ron’s formula)

r = \frac{a}{p}, sin (α) = \frac{x}{2 r} = \frac{p x}{2 a} = \frac{2 x ( t + x + s )}{( t + x + s ) ( t - x + s ) ( t + x - s ) ( - t + x + s )} .

r = \frac{a}{p}, sin (α) = \frac{x}{2 r} = \frac{p x}{2 a} = \frac{2 x ( t + x + s )}{( t + x + s ) ( t - x + s ) ( t + x - s ) ( - t + x + s )} .

\alpha(t,x,s)=\cases{\alpha^{\prime}(d(t,x,s))&\hbox{ if $x^{2}<t^{2}+s^{2}$}\cr{}\cr{\pi\over 2}&\hbox{ if $x^{2}=t^{2}+s^{2}$}\cr{}\cr\alpha^{\prime\prime}(d(t,x,s))&\hbox{ if $x^{2}>t^{2}+s^{2}$.}}

\alpha(t,x,s)=\cases{\alpha^{\prime}(d(t,x,s))&\hbox{ if $x^{2}<t^{2}+s^{2}$}\cr{}\cr{\pi\over 2}&\hbox{ if $x^{2}=t^{2}+s^{2}$}\cr{}\cr\alpha^{\prime\prime}(d(t,x,s))&\hbox{ if $x^{2}>t^{2}+s^{2}$.}}

Ω_{n} = {(t_{1}, x_{1}, \dots, t_{n - 2}, x_{n - 2}, t_{n - 1}) : k = 1, \dots, n - 2, (t_{k}, x_{k}, t_{k + 1}) \in V, k = 1 \sum n - 2 α (t_{k}, x_{k}, t_{k + 1}) < 2 π} .

Ω_{n} = {(t_{1}, x_{1}, \dots, t_{n - 2}, x_{n - 2}, t_{n - 1}) : k = 1, \dots, n - 2, (t_{k}, x_{k}, t_{k + 1}) \in V, k = 1 \sum n - 2 α (t_{k}, x_{k}, t_{k + 1}) < 2 π} .

a d (b^{2} + c^{2} - e^{2}) + b c (a^{2} + d^{2} - e^{2}) = 0.

a d (b^{2} + c^{2} - e^{2}) + b c (a^{2} + d^{2} - e^{2}) = 0.

0 < α < π, 0 < β < π and a^{2} + d^{2} - 2 a d cos α = e^{2} = b^{2} + c^{2} - 2 b c cos β .

0 < α < π, 0 < β < π and a^{2} + d^{2} - 2 a d cos α = e^{2} = b^{2} + c^{2} - 2 b c cos β .

\cases cos α = \frac{a ^{2} + d ^{2} - e ^{2}}{2 a d} \cr \cr cos β = \frac{b ^{2} + c ^{2} - e ^{2}}{2 b c} .

\cases cos α = \frac{a ^{2} + d ^{2} - e ^{2}}{2 a d} \cr \cr cos β = \frac{b ^{2} + c ^{2} - e ^{2}}{2 b c} .

(A, B, C, D) inscribable ⟺ \frac{b ^{2} + c ^{2} - e ^{2}}{2 b c} = - \frac{a ^{2} + d ^{2} - e ^{2}}{2 a d}

(A, B, C, D) inscribable ⟺ \frac{b ^{2} + c ^{2} - e ^{2}}{2 b c} = - \frac{a ^{2} + d ^{2} - e ^{2}}{2 a d}

Θ (u) = u_{1} u_{2} (u_{4}^{2} + u_{5}^{2} - u_{3}^{2}) + u_{4} u_{5} (u_{1}^{2} + u_{2}^{2} - u_{3}^{2}) for u = (u_{1}, u_{2}, u_{3}, u_{4}, u_{5}) \in \mathb R^{5},

Θ (u) = u_{1} u_{2} (u_{4}^{2} + u_{5}^{2} - u_{3}^{2}) + u_{4} u_{5} (u_{1}^{2} + u_{2}^{2} - u_{3}^{2}) for u = (u_{1}, u_{2}, u_{3}, u_{4}, u_{5}) \in \mathb R^{5},

γ_{k} (ω) = Θ (t_{k}, x_{k}, t_{k + 1}, x_{k + 1}, t_{k + 2}) for k \in {1, \dots, n - 3}

γ_{k} (ω) = Θ (t_{k}, x_{k}, t_{k + 1}, x_{k + 1}, t_{k + 2}) for k \in {1, \dots, n - 3}

γ (ω) = (γ_{1} (ω), \dots, γ_{n - 3} (ω)) where ω = (t_{1}, x_{1}, t_{2}, \dots, x_{n - 2}, t_{n - 1}) \in Ω_{n} .

γ (ω) = (γ_{1} (ω), \dots, γ_{n - 3} (ω)) where ω = (t_{1}, x_{1}, t_{2}, \dots, x_{n - 2}, t_{n - 1}) \in Ω_{n} .

γ : Ω_{4} ⟶ \mathb R, ω = (t_{1}, x_{1}, t_{2}, x_{2}, t_{3}) ⟼ γ (ω) = t_{1} x_{1} (t_{3}^{2} + x_{2}^{2} - t_{2}^{2}) + t_{3} x_{2} (t_{1}^{2} + x_{1}^{2} - t_{2}^{2})

γ : Ω_{4} ⟶ \mathb R, ω = (t_{1}, x_{1}, t_{2}, x_{2}, t_{3}) ⟼ γ (ω) = t_{1} x_{1} (t_{3}^{2} + x_{2}^{2} - t_{2}^{2}) + t_{3} x_{2} (t_{1}^{2} + x_{1}^{2} - t_{2}^{2})

\cases ω^{'} = (t_{1}, x_{1}, \dots, t_{n - 2}, x_{n - 2}, t_{n - 1}) \in Ω_{n} \cr \cr ω^{''} = (t_{n - 2}, x_{n - 2}, t_{n - 1}, x_{n - 1}, t_{n}) \in Ω_{4}

\cases ω^{'} = (t_{1}, x_{1}, \dots, t_{n - 2}, x_{n - 2}, t_{n - 1}) \in Ω_{n} \cr \cr ω^{''} = (t_{n - 2}, x_{n - 2}, t_{n - 1}, x_{n - 1}, t_{n}) \in Ω_{4}

\pmatrix{A(\omega^{\prime})&0\cr B(\omega^{\prime\prime})&C(\omega^{\prime\prime})}

\pmatrix{A(\omega^{\prime})&0\cr B(\omega^{\prime\prime})&C(\omega^{\prime\prime})}

A(\omega^{\prime})=\pmatrix{{{\partial\gamma_{1}}\over{\partial t_{1}}}(\omega^{\prime})&{{\partial\gamma_{1}}\over{\partial x_{1}}}(\omega^{\prime})&\cdots&{{\partial\gamma_{1}}\over{\partial t_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{1}}\over{\partial x_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{1}}\over{\partial t_{n-1}}}(\omega^{\prime})\cr{}\cr{{\partial\gamma_{2}}\over{\partial t_{1}}}(\omega^{\prime})&{{\partial\gamma_{2}}\over{\partial x_{1}}}(\omega^{\prime})&\cdots&{{\partial\gamma_{2}}\over{\partial t_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{2}}\over{\partial x_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{2}}\over{\partial t_{n-1}}}(\omega^{\prime})\cr{}\cr\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\cr{}\cr{{\partial\gamma_{n-3}}\over{\partial t_{1}}}(\omega^{\prime})&{{\partial\gamma_{n-3}}\over{\partial x_{1}}}(\omega^{\prime})&\cdots&{{\partial\gamma_{n-3}}\over{\partial t_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{n-3}}\over{\partial x_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{n-3}}\over{\partial t_{n-1}}}(\omega^{\prime})}

A(\omega^{\prime})=\pmatrix{{{\partial\gamma_{1}}\over{\partial t_{1}}}(\omega^{\prime})&{{\partial\gamma_{1}}\over{\partial x_{1}}}(\omega^{\prime})&\cdots&{{\partial\gamma_{1}}\over{\partial t_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{1}}\over{\partial x_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{1}}\over{\partial t_{n-1}}}(\omega^{\prime})\cr{}\cr{{\partial\gamma_{2}}\over{\partial t_{1}}}(\omega^{\prime})&{{\partial\gamma_{2}}\over{\partial x_{1}}}(\omega^{\prime})&\cdots&{{\partial\gamma_{2}}\over{\partial t_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{2}}\over{\partial x_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{2}}\over{\partial t_{n-1}}}(\omega^{\prime})\cr{}\cr\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\cr{}\cr{{\partial\gamma_{n-3}}\over{\partial t_{1}}}(\omega^{\prime})&{{\partial\gamma_{n-3}}\over{\partial x_{1}}}(\omega^{\prime})&\cdots&{{\partial\gamma_{n-3}}\over{\partial t_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{n-3}}\over{\partial x_{n-2}}}(\omega^{\prime})&{{\partial\gamma_{n-3}}\over{\partial t_{n-1}}}(\omega^{\prime})}

B (ω^{''}) = (0, \dots, 0, \frac{\partial Θ}{\partial u _{1}} (ω^{''}), \frac{\partial Θ}{\partial u _{2}} (ω^{''}), \frac{\partial Θ}{\partial u _{3}} (ω^{''})) and C (ω^{''}) = (\frac{\partial Θ}{\partial u _{4}} (ω^{''}), \frac{\partial Θ}{\partial u _{5}} (ω^{''})) .

B (ω^{''}) = (0, \dots, 0, \frac{\partial Θ}{\partial u _{1}} (ω^{''}), \frac{\partial Θ}{\partial u _{2}} (ω^{''}), \frac{\partial Θ}{\partial u _{3}} (ω^{''})) and C (ω^{''}) = (\frac{\partial Θ}{\partial u _{4}} (ω^{''}), \frac{\partial Θ}{\partial u _{5}} (ω^{''})) .

\cases \frac{\partial Θ}{\partial u _{4}} (ω^{''}) = 2 x_{n - 1} t_{n - 2} x_{n - 2} + t_{n} (x_{n - 2}^{2} + t_{n - 2}^{2} - t_{n - 1}^{2}) \cr \cr \frac{\partial Θ}{\partial u _{5}} (ω^{''}) = 2 t_{n} t_{n - 2} x_{n - 2} + x_{n - 1} (x_{n - 2}^{2} + t_{n - 2}^{2} - t_{n - 1}^{2})

\cases \frac{\partial Θ}{\partial u _{4}} (ω^{''}) = 2 x_{n - 1} t_{n - 2} x_{n - 2} + t_{n} (x_{n - 2}^{2} + t_{n - 2}^{2} - t_{n - 1}^{2}) \cr \cr \frac{\partial Θ}{\partial u _{5}} (ω^{''}) = 2 t_{n} t_{n - 2} x_{n - 2} + x_{n - 1} (x_{n - 2}^{2} + t_{n - 2}^{2} - t_{n - 1}^{2})

0 = x_{n - 1} \frac{\partial Θ}{\partial u _{4}} (ω^{''}) - t_{n} \frac{\partial Θ}{\partial u _{5}} (ω^{''}) = 2 x_{n - 1} t_{n - 2} (x_{n - 1}^{2} - t_{n}^{2})

0 = x_{n - 1} \frac{\partial Θ}{\partial u _{4}} (ω^{''}) - t_{n} \frac{\partial Θ}{\partial u _{5}} (ω^{''}) = 2 x_{n - 1} t_{n - 2} (x_{n - 1}^{2} - t_{n}^{2})

0 = t_{n} [(x_{n - 2} + t_{n - 2})^{2} - t_{n - 1}^{2}]

0 = t_{n} [(x_{n - 2} + t_{n - 2})^{2} - t_{n - 1}^{2}]

\cases p (ω) = perimeter of ω \cr \cr A (ω) = area of ω \cr \cr Ψ (ω) = (p (ω), A (ω)) .

\cases p (ω) = perimeter of ω \cr \cr A (ω) = area of ω \cr \cr Ψ (ω) = (p (ω), A (ω)) .

\cases p (ω) = t_{1} + x_{1} + x_{2} + \dots + x_{n - 2} + t_{n - 1} \cr \cr A (ω) = \frac{1}{4} f (ω_{1}) + \dots + \frac{1}{4} f (ω_{n - 2})

\cases p (ω) = t_{1} + x_{1} + x_{2} + \dots + x_{n - 2} + t_{n - 1} \cr \cr A (ω) = \frac{1}{4} f (ω_{1}) + \dots + \frac{1}{4} f (ω_{n - 2})

\cases ω_{k} = (t_{k}, x_{k}, t_{k + 1}) \in V for k \in {1, ..., n - 2} \cr \cr area of ω_{k} = \frac{1}{4} f (ω_{k}) for k \in {1, \dots, n - 2} (H \overset{e}{ˊ} ron’s formula) \cr \cr f (x, y, z) = (x + y + z) (- x + y + z) (x - y + z) (x + y - z) for (x, y, z) \in V .

\cases ω_{k} = (t_{k}, x_{k}, t_{k + 1}) \in V for k \in {1, ..., n - 2} \cr \cr area of ω_{k} = \frac{1}{4} f (ω_{k}) for k \in {1, \dots, n - 2} (H \overset{e}{ˊ} ron’s formula) \cr \cr f (x, y, z) = (x + y + z) (- x + y + z) (x - y + z) (x + y - z) for (x, y, z) \in V .

f (v) = s (v) (s (v) - 2 x) (s (v) - 2 y) (s (v) - 2 z) .

f (v) = s (v) (s (v) - 2 x) (s (v) - 2 y) (s (v) - 2 z) .

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Full text

Area and perimeter foliations on spaces of polygons

by

Aziz EL KACIMI ALAOUI & Abdellatif ZEGGAR

(February 2019)

We describe all families of star-shaped $n$ -polygons in the Euclidean plane with prescribed perimeter and area; they are leaves of a foliation ${\cal F}^{\;(\ast)}$ on the space ${\cal P}_{n}^{\ast}$ of star-shaped polygons. By the way, we study some geometric properties of convex polygons, for instance their inscriptibility in a circle and their regularity in relation with the perimeter and the area.

We denote by 𝔼 an Euclidean vector plane equipped with its canonical affine structure and an orientation given by an orthonormal basis. The word ‘isometry’ means a transformation of 𝔼 which preserves the distance; necessarily it is an affine transformation of 𝔼. The origin of 𝔼 is denoted $O$ . For $A,B\in{\hbox{\mathb E}}$ we denote $\overrightarrow{AB}$ the vector $B-A$ . Let $n\geq 3$ be a natural integer.

0. Introduction

A figure of 𝔼 is just a part of it. But usually this name is given only to a one having a certain peculiarity: we see it all (it is bounded) or at least we understand how it is made to guess its behavior when it escapes our view, like a straight line… and its outline has a little regularity. A polygon is an example of such figure: it is bounded and bordered by a finite number of segments called sides or edges. To each polygon, one can associate invariants, among which are two real numbers that play an important role: the perimeter and the area.

The polygons of the plane are numerous and their shapes and sizes are varied. So the question of their equivalence therefore arises naturally. But in which sense?

From the set point of view, two polygons $\wp$ and $\wp^{\prime}$ are always equivalent: there is a bijection of 𝔼 which sends one on the other. But as we are in a Euclidean plane, we would like this bijection to preserve properties related to the affine Euclidean structure. There are several notions of equivalence; here are some of them (those that will interest us directly). We will say that $\wp$ and $\wp^{\prime}$ are:

isometric if there exists an isometry $f:{\hbox{\mathb E}}\longrightarrow{\hbox{\mathb E}}$ such that $f(\wp)=\wp^{\prime}$ . We can superimpose them; and we can still do that without going out of the plan if $\wp$ and $\wp^{\prime}$ are directly isometric, that is $f$ preserves the orientation; 2. 2.

similar if there is a similarity $f:{\hbox{\mathb E}}\longrightarrow{\hbox{\mathb E}}$ such that $f(\wp)=\wp^{\prime}$ . In a way, one of them is an enlargement of the other (as for the photos); 3. 3.

equivalent (shortly) if they have the same area. In this case, we can always go from one to another by cutting and geometric gluing; 4. 4.

isoperimetric if they have the same perimeter.

The isometric equivalence is the strongest and implies all the others. So it is too rigid to be ‘useful’: two isometric polygons differ only in the positions they occupy in the plane. It is rather the equivalences 3 and 4 that will occupy us here.

This paper takes its origin from the following question ( $\ast$ ): Are there two non-isometric triangles with same perimeter and same area? It led us first to the construction of a foliation ${\cal F}^{\;(\ast)}$ on the space of triangles: each leaf consists of triangles having same perimeter and same area. Then we made a more general study for the space of star-shaped polygons; this space contains convex polygons for which some properties related to area and perimeter have also been studied. Some of them are certainly known, but little absent in the ‘geometric literature’, which motivated us to insert them in this text.

( $\ast$ ) It was asked by Geoffrey Letellier to his teacher Valerio Vassallo who put it in turn to the first author. The latter has built for this purpose the foliation ${\cal F}$ on the space of the triangles. But just to the question asked, Geoffrey himself responded by constructing a one parameter family of isosceles triangles having the same perimeter and the same area (his example is given in subsection 4.6).

The integer $n$ we consider in all this paper will be greater than or equal to $3$ .

1. Preliminaries

1.1. Definition. A non degenerate polygon of 𝔼 is an element $\wp=(M_{1},\cdots,M_{n})$ of ${\hbox{\mathb E}}^{n}$ such that:

(i) for $i\neq j$ the point $M_{i}$ is distinct from $M_{j}$ ;

(ii) for any $k\in\{1,\cdots,n\}$ , the oriented angle $\widehat{M_{k}}=(\overrightarrow{M_{k}M_{k+1}},\overrightarrow{M_{k}M_{k-1}})$ has its measure in $]0,2\pi[\setminus\{\pi\}$ .

By convention, we set $M_{0}=M_{n}$ and $M_{n+1}=M_{1}$ . The orientation of the angle $\widehat{M_{k}}$ is the same as the orientation of the trigonometric circle centered at the point $M_{k}$ .

The points $M_{k}$ and the segments $[M_{k}M_{k+1}]$ are respectively the vertices and the sides of the polygon $\wp$ . If $M_{i}$ and $M_{j}$ are two non successive vertices, that is $|i-j|>1$ , we say that the segment $M_{i}M_{j}$ is a diagonal of the polygon.

Recall that a polygon is said to be:

equilateral if all its sides have the same length;
inscribable if all its vertices are on a same circle;
regular if it is both inscribable and equilateral.

The set of all $n$ -polygons of the plane 𝔼 will be denoted $\widetilde{\cal{P}}_{n}$ . Let $f$ be an isometry of the Euclidean plane 𝔼. The image $\wp^{\prime}=(M^{\prime}_{1},\cdots,M^{\prime}_{n})=\left(f(M_{1}),\cdots,f(M_{n})\right)$ of a polygon $\wp=(M_{1},\cdots,M_{n})$ is a polygon of 𝔼 such that:

[TABLE]

So we have a natural action:

[TABLE]

where $\hbox{Isom}({\hbox{\mathb E}})$ is the group of the affine isometries of 𝔼. The quotient space of this action will be denoted:

[TABLE]

The elements of ${\cal P}_{n}$ are called geometric polygons of 𝔼.

A geometric polygon of 𝔼 is said to be equilateral (resp. inscribable) if it admits a representative which is equilateral (resp. inscribable).

For an element $\wp=(M_{1},\cdots,M_{n})$ of $\widetilde{\cal P}_{n}$ , we will use the following notations:

[TABLE]

The positive numbers $t_{1},x_{1},\cdots,x_{n-2},t_{n-1}$ are the lengths of the sides and $t_{2},t_{3},\cdots,t_{n-2}$ are the lengths of the diagonals from the vertex $M_{n}$ (see the picture bellow for the case of the hexagon). We have $(n-2)$ lengths of type $x_{k}$ and $(n-1)$ lengths of type $t_{k}$ .

Since an isometry preserves the distance in the Euclidean plane 𝔼, $(t_{1},x_{1},t_{2},x_{2},\cdots,t_{n-2},x_{n-2},t_{n-1})$ does not depend on the choice of the representative $(M_{1},\cdots,M_{n})$ of the geometric polygon $<M_{1},\cdots,M_{n}>$ .

[TABLE]

For any triangle $(M_{k},M_{n},M_{k+1})$ with $1\leq k\leq n-2$ , the lengths $t_{k},x_{k},t_{k+1}$ satisfy the inequalities:

[TABLE]

that is, $(t_{k},x_{k},t_{k+1})$ is an element of the open set ${\cal V}$ of ${\hbox{\mathb R}}^{3}$ consisting of the triplets $(x,y,z)$ satisfying:

[TABLE]

Moreover, one can verify by induction that, for the lengths of the sides of a polygon, each length is strictly smaller than the sum of the others. For instance, we have the following nice:

1.2. Theorem [Pen]. For any natural integer $n\geq 3$ and any $n$ -tuple $u=(u_{1},\cdots,u_{n})$ of positive real numbers, there exists a unique inscribable geometric polygon $<M_{1},\cdots,M_{n}>$ such that $M_{k}M_{k+1}=u_{k}$ for $k\in\{1,\cdots,n\}$ if and only if, for any $j\in\{1,\cdots,n\}$ , we have the inequality: $u_{j}<\displaystyle\sum_{k\neq j}u_{k}.$

1.3. Definition. A $n$ -polygon $\wp=(M_{1},\cdots,M_{n})$ of 𝔼 is convex if, for any $k\in\{1,\cdots,n\}$ , the oriented angle $\widehat{M_{k}}=(\overrightarrow{M_{k}M_{k+1}},\overrightarrow{M_{k}M_{k-1}})$ has its measure in $]0,\pi[$ . It is called star-shaped polygon with respect to the vertex $M$ if, for any vertex $N\in\{M_{1},\cdots,M_{n}\}$ the open segment $]MN[$ is contained in the interior of $\wp$ . Of course, any convex polygon is a star-shaped polygon with respect to any of its vertices.

Star-shaped polygons $(M_{1},\cdots,M_{n})$ with respect to the vertex $M_{n}$ form a subspace $\widetilde{\cal P}_{n}^{\ast}$ of $\widetilde{\cal P}_{n}$ , invariant under the action of $\hbox{Isom}({\hbox{\mathb E}})$ (on $\widetilde{\cal P}_{n}$ ).

One can easily see that any element of $\widetilde{\cal P}_{n}^{\ast}$ is isometric to a unique star-shaped polygon $\wp=(M_{1},\cdots,M_{n})$ with respect to $M_{n}$ such that:

[TABLE]

${\cal P}_{n}^{\ast}$ will denote the space of the star-shaped polygons satisfying these conditions.

[TABLE]

In the following subsection we will recall some metric properties in a triangle which will be very useful for determining the space of polygons on which we will define area and perimeter foliations in section 3.

1.4. Let $OMN$ be a non degenerate triangle ; we set $OM=t$ , $ON=s$ , $MN=x$ and $\alpha=\widehat{MON}=(\overrightarrow{OM},\overrightarrow{ON})$ . All the real numbers $t$ , $s$ , $x$ are positive and $\alpha\in]0,\pi[$ . For the triangle $OMN$ we denote $r>0$ the radius of its inscribed circle, $p$ its perimeter and $a$ its area. We have the following well known formulaes:

[TABLE]

and:

[TABLE]

For a given $d\in]0,1[$ there are two numbers $\alpha^{\prime}(d)\in]0,{\pi\over 2}[$ and $\alpha^{\prime\prime}(d)\in]{\pi\over 2},\pi[$ such that $\sin(\alpha^{\prime}(d))=\sin(\alpha^{\prime\prime}(d))=d$ . We take $d={{px}\over{2a}}$ (which depends on $(t,x,s)$ ) and we define the function $\alpha:(]0,+\infty[)^{3}\longrightarrow]0,\pi[$ by:

[TABLE]

It is well defined and continuous.

Now consider the open set $\Omega_{n}$ of $({\hbox{\mathb R}}_{+}^{\ast})^{2n-3}$ given by:

[TABLE]

Let ${\cal L}_{n}:{\cal P}_{n}^{\ast}\longrightarrow\Omega_{n}$ be the map ${\cal L}_{n}(\wp)={\cal L}_{n}(<M_{1},\cdots,M_{n}>)=(t_{1},x_{1},t_{2},x_{2},\cdots,t_{n-2},x_{n-2},t_{n-1})$ where the real numbers $t_{1},x_{1},t_{2},x_{2},\cdots,t_{n-2},x_{n-2},t_{n-1}$ are given by the formulaes (1.4).

From now on we will identify the geometric star-shaped polygons to the points of the open set $\Omega_{n}$ by the map ${\cal L}_{n}$ . This identification ${\cal P}_{n}^{\ast}\simeq\Omega_{n}$ will enables one to study easily some properties of the space of geometric star-shaped polygons.

Let $\varphi:X\rightarrow Y$ be any map. A nonempty subset of $X$ of the form $\varphi^{-1}\left(\{y\}\right)$ will be called the level set (level line, level surface, level manifold…) of $\varphi$ at level $y\in Y$ .

2. The geometric inscribable polygons for $n\geq 4$

First note that, following our definition 1.3, an inscribable polygon $(M_{1},\cdots,M_{n})$ is always convex. It is therefore a star-shaped polygon with respect to any of its vertices. We can therefore represent it by an element of ${\cal P}_{n}^{\ast}$ . In this way, the set $\Gamma_{n}$ of inscribable polygons of 𝔼 can be viewed as a subset of the open set $\Omega_{n}$ . (Recall that any regular geometric polygon is inscribable.)

2.1. Remark. The fact that a polygon $(M_{1},\cdots,M_{n})$ is inscribable is equivalent to the fact that each one of the $n-3$ quadrilaterals: $Q_{1}=(M_{n},M_{1},M_{2},M_{3}),\;\cdots,Q_{n-3}=(M_{n},M_{n-3},M_{n-2},M_{n-1})$ is inscribable.

2.2. Lemma. A convex quadrilateral $(A,B,C,D)$ is inscribable if and only if the distances $a=AB,\;b=BC,\;c=CD,\;d=DA\;$ and $\;e=BD\;$ satisfy the relation:

[TABLE]

Proof. Let $\alpha$ and $\beta$ denote the measures respectively of the angles $\widehat{A}$ and $\widehat{C}$ . Then (by the cosine’s law of Al-Kashi):

[TABLE]

We deduce:

[TABLE]

On the other hand, the quadrilateral $(A,B,C,D)$ is inscribable if and only if $\alpha+\beta=\pi$ or, equivalently, if $\cos\beta=-\cos\alpha$ . Hence:

[TABLE]

which is also equivalent to $ad(b^{2}+c^{2}-e^{2})+bc(a^{2}+d^{2}-e^{2})=0.$

Remark 2.1 and Lemma 2.2 make possible to realize $\Gamma_{n}$ as a level set of a differentiable map. More precisely, consider the maps $\Theta:{\hbox{\mathb R}}^{5}\longrightarrow{\hbox{\mathb R}}$ , $\gamma_{k}:\Omega_{n}\longrightarrow{\hbox{\mathb R}}$ and $\gamma:\Omega_{n}\longrightarrow{\hbox{\mathb R}}^{n-3}$ defined by:

[TABLE]

and:

[TABLE]

2.3. Proposition. The set $\Gamma_{n}$ of geometric inscribable polygons is given by $\Gamma_{n}=\gamma^{-1}\left(\{0\}\right)$ . Moreover, this set is a differentiable submanifold of dimension $n$ of the Euclidean space ${\hbox{\mathb R}}^{2n-3}$ .

Proof. Let $\omega=(t_{1},x_{1},....,t_{n-2},x_{n-2},t_{n-1})$ be an element of $\Omega_{n}$ represented by a polygon $(M_{1},\cdots,M_{n})$ (in ${\cal P}_{n}^{\ast}$ ). Then we have the following equivalences:

$\omega\in\Gamma_{n}\Longleftrightarrow\hbox{$ (M_{n},M_{k},M_{k+1},M_{k+2}) $}\hbox{ is inscribable for$ 1\leq k\leq n-3 $}$

$\Longleftrightarrow\Theta(t_{k},x_{k},t_{k+1},x_{k+1},t_{k+2})=0\;\;\hbox{for$ 1\leq k\leq n-3 $}$

$\Longleftrightarrow\gamma_{k}(\omega)=0\;\;\hbox{for$ 1\leq k\leq n-3 $}$

$\Longleftrightarrow\gamma(\omega)=0$

$\Longleftrightarrow\omega\in\gamma^{-1}\left(\{0\}\right).$

Now we will prove, by induction on $n\geq 4$ , that the map $\gamma$ has maximal rank at each point $\omega$ of $\Omega_{n}$ .

$\bullet\;$ ** The case $n=4$ .** We have $\gamma=\gamma_{1}$ and the map:

[TABLE]

has maximal rank at each point $\omega\in\Omega_{4}$ . Indeed, $\displaystyle{{\partial\gamma}\over{\partial t_{2}}}(\omega)=-2t_{2}(t_{1}x_{1}+t_{3}x_{2})\neq 0$ . So $\Gamma_{4}=\gamma^{-1}\left(\{0\}\right)$ is a codimension $1$ submanifold of the open set $\Omega_{4}$ and then a submanifold of dimension $4$ of ${\hbox{\mathb R}}^{5}$ .

$\bullet$ Heredity. Suppose that, for a fixed integer $n\geq 4$ , the map $\gamma$ has maximal rank at each point of $\Omega_{n}$ . If to each element $\omega=(t_{1},x_{1},\cdots,t_{n-2},x_{n-2},t_{n-1},x_{n-1},t_{n})\in\Omega_{n+1}$ we associate:

[TABLE]

then we can write $\omega=(\omega^{\prime},x_{n-1},t_{n})$ and $\gamma(\omega)=\left(\gamma(\omega^{\prime}),\gamma(\omega^{\prime\prime})\right).$ Note that here we are using the same notation $\gamma$ for three different maps. The Jacobian matrix of the map $\gamma:\Omega_{n+1}\subset{\hbox{\mathb R}}^{2n-1}\rightarrow{\hbox{\mathb R}}^{n-2}$ at each point $\omega\in\Omega_{n+1}$ is given by:

[TABLE]

where :

[TABLE]

which is the Jacobian matrix of the map $\gamma:\Omega_{n}\subset{\hbox{\mathb R}}^{2n-3}\rightarrow{\hbox{\mathb R}}^{n-3}$ at the point $\omega^{\prime}$ ,

[TABLE]

By the induction hypothesis, the matrix $A(\omega^{\prime})$ has rank $n-3$ . On the other hand, the two partial derivatives:

[TABLE]

do not vanish simultaneously since otherwise we will have:

[TABLE]

which implies $x_{n-1}=t_{n}$ . Taking into account this equality in the relation ${{\partial\Theta}\over{\partial u_{4}}}(\omega^{\prime\prime})=0$ we obtain :

[TABLE]

and then $x_{n-2}+t_{n-2}=t_{n-1}$ . This contradicts the inequality $t_{n-1}<x_{n-2}+t_{n-2}$ since these are the lengths of the sides of the non degenerate triangle $(M_{n+1},M_{n-2},M_{n-1})$ .

The Jacobian matrix of the map $\gamma:\Omega_{n+1}\subset{\hbox{\mathb R}}^{2n-1}\rightarrow{\hbox{\mathb R}}^{n-2}$ is then of rank $n-2$ . The proof by induction is then over.

We deduce that, for any $n\geq 4$ , the map $\gamma:\Omega_{n}\subset{\hbox{\mathb R}}^{2n-3}\rightarrow{\hbox{\mathb R}}^{n-3}$ has maximal rank and then the nonempty set $\Gamma_{n}=\gamma^{-1}\left(\{0\}\right)$ is a codimension $(n-3)$ submanifold of the open set $\Omega_{n}$ of ${\hbox{\mathb R}}^{2n-3}$ . This implies that $\Gamma_{n}$ is a submanifold of dimension $n$ of ${\hbox{\mathb R}}^{2n-3}$ .

3. The area and perimeter foliations

In all this section the space ${\cal P}_{n}^{\ast}$ of star-shaped polygons will be identified (as we have proceed until now) to the open set $\Omega_{n}$ .

3.1. We define the maps $p:\Omega_{n}\longrightarrow{\hbox{\mathb R}}$ , ${\cal A}:\Omega_{n}\longrightarrow{\hbox{\mathb R}}$ and $\Psi:\Omega_{n}\longrightarrow{\hbox{\mathb R}}^{2}$ by:

[TABLE]

For any $\omega=(t_{1},x_{1},t_{2},\cdots,t_{n-2},x_{n-2},t_{n-1})\in\Omega_{n}$ , we have:

[TABLE]

with:

[TABLE]

Setting $s(x,y,z)=x+y+z$ , we obtain for any $v=(x,y,z)\in{\cal V}$ :

[TABLE]

On the other hand, the maps $p,\;{\cal A}$ and $\Psi$ are clearly differentiable with gradient vectors $\nabla p(\omega)$ and $\nabla\cal{A}(\omega)$ given by:

[TABLE]

where the logarithmic derivative ${{\nabla f}\over f}$ is given at each point $v=(x,y,z)\in{\cal V}$ by:

[TABLE]

We then deduce:

[TABLE]

This gives the partial derivatives of ${\cal A}$ :

[TABLE]

and for $k\in\{2,\cdots,n-2\}$ :

[TABLE]

3.2. Theorem. We have the following assertions.

(1) The perimeter function $p$ and the area function ${\cal A}$ are submersions on $\Omega_{n}$ . Then the level sets of $p$ (resp. of ${\cal A}$ ) are leaves of a codimension $1$ foliation ${\cal F}_{p}$ (resp. ${\cal F}_{a}$ ) on $\Omega_{n}$ .

(2) For $\omega\in\Omega_{n}$ , the differential $d\Psi(\omega)$ is of rank $2$ if $\omega$ is not a regular polygon and of rank $1$ if $\omega$ is a regular polygon. Then the map $\Psi$ defines a codimension $2$ foliation ${\cal F}$ on the open set $\Omega_{n}$ of ${\hbox{\mathb R}}^{2n-3}$ which consists of non regular polygons $\omega$ of $\Omega_{n}$ .

Proof. Let $\omega=(t_{1},x_{1},\cdots,t_{n-2},x_{n-2},t_{n-1})$ be an element of $\Omega_{n}$ and $(M_{1},\cdots,M_{n})$ one of its representatives as a satr-shaped polygon.

Point (1)

$(\star)$ For any $\omega\in\Omega_{n}$ , $dp(\omega)\neq 0$ since ${{\partial p}\over{\partial t_{1}}}(\omega)=1\neq 0$ . Then $p$ is a submersion on $\Omega_{n}$ .

$(\star)$ For any $\omega\in\Omega_{n}$ , ${{\partial{\cal A}}\over{\partial t_{1}}}(\omega)\neq 0$ or ${{\partial{\cal A}}\over{\partial x_{1}}}(\omega)\neq 0$ . Indeed we have the implications:

[TABLE]

But the equality $t_{1}+x_{1}=0$ can not be satisfied. Then $d{\cal A}(\omega)\neq 0$ . This proves that ${\cal A}$ is a submersion on $\Omega_{n}$ .

Point (2)

$(\star)$ If the sides of the polygon are not all equal, there exist two successive sides having $M$ as common point and different lengths. One can suppose that $M=M_{1}$ . In these conditions, the lengths $t_{1}=M_{n}M_{1}$ and $x_{1}=M_{1}M_{2}$ are different. This implies $\displaystyle{{\partial{\cal A}}\over{\partial t_{1}}}(\omega)\neq{{\partial{\cal A}}\over{\partial x_{1}}}(\omega)$ since we have the implication ${{\partial{\cal A}}\over{\partial t_{1}}}(\omega)={{\partial{\cal A}}\over{\partial x_{1}}}(\omega)\Longrightarrow t_{1}=x_{1}.$ The Jacobian matrix of the map $\Psi$ at the point $\omega$ is:

[TABLE]

and then it is of rank $2$ since the $2\times 2$ -matrix of ${\cal J}(\Psi,\omega)$ consisting of the two first columns is invertible.

$(\star)$ Suppose the polygon is equilateral. We consider two cases:

$\bullet$ The condition $(C)$ below is satisfied.

[TABLE]

The matrix ${\cal J}(\Psi,\omega)$ has rank $2$ since it admits a matrix of order $2$ which is reversible.

$\bullet$ The condition $(C)$ is not satisfied. In this case the following condition non(C) is satisfied:

[TABLE]

and we have:

[TABLE]

This implies that ${\cal J}(\Psi,\omega)$ is of rank $1$ . We shall prove that, in this case, $\omega$ is regular polygon.

For $k\in\{2,\cdots,n-2\}$ , we have $\displaystyle{{\partial{\cal A}}\over{\partial x_{k-1}}}(\omega)={{\partial{\cal A}}\over{\partial x_{k}}}(\omega)$ which implies:

[TABLE]

Thus, taking into account the fact that the sides $x_{k}$ are all equal in this case, we obtain by factorization :

[TABLE]

where:

[TABLE]

Thus $\displaystyle(t_{k-1}-t_{k+1})(t_{k-1}t_{k+1}-x_{k}^{2}+t_{k}^{2})(t_{k-1}t_{k+1}+x_{k}^{2}-t_{k}^{2})=0.$

$(\star)$ If $t_{k-1}-t_{k+1}=0$ , taking into account the relation ${{\partial{\cal A}}\over{\partial t_{k}}}(\omega)=0$ , we obtain:

[TABLE]

This implies that the two triangles $(M_{n},M_{k-1},M_{k})$ and $(M_{n},M_{k+1},M_{k})$ are rectangular respectively at $M_{k-1}$ and $M_{k+1}$ . Consequently the quadrilateral $(M_{n},M_{k-1},M_{k},M_{k+1})$ is inscribable ( $\widehat{M}_{k-1}={\pi\over 2}=\widehat{M}_{k+1}$ ).

$(\star)$ If $t_{k-1}t_{k+1}-x_{k}^{2}+t_{k}^{2}=0$ , the convex quadrilateral $(M_{n},M_{k-1},M_{k},M_{k+1})$ has the following properties:

[TABLE]

The triangle $(M_{k-1},M_{n},M_{k+1})$ is then isosceles at the $M_{n}$ and $t_{k-1}=t_{k+1}$ . This implies, like in the preceding case, that the quadrilateral $(M_{n},M_{k-1},M_{k},M_{k+1})$ is inscribable.

$(\star)$ If $\displaystyle t_{k-1}t_{k+1}+x_{k}^{2}-t_{k}^{2}=0$ , then the convex quadrilateral $(M_{n},M_{k-1},M_{k},M_{k+1})$ is still inscribable since it satisfies the relation $\cos\widehat{M_{k-1}}=-\cos\widehat{M_{k+1}}$ . Indeed, we have:

[TABLE]

We have proved that, in all cases, the quadrilateral $(M_{n},M_{k-1},M_{k},M_{k+1})$ is inscribable for any $k\in\{2,\cdots,n-2\}$ . Then the polygon $(M_{1},\cdots,M_{n})$ is inscribable. Since the latter is equilateral, it is necessarily regular.

Finally the singular points of the map $\Psi$ are the regular polygons and this map induces a submersion on $\Omega_{n}^{\ast}$ whose level sets are leaves of a foliation ${\cal F}$ .

4. The example of triangles

It is the situation where we see things more concretely and where the drawings are more visible. The way to treat the topic in this section will be slightly different from the others.

4.1. The space of non degenerate triangles

To give oneself a non degenerate triangle (in any Euclidean finite dimensional space) is to give oneself three real numbers $x>0$ , $y>0$ and $z>0$ such that:

[TABLE]

which represent the lengths of the sides. Exceptionally in this section, we shall denote a triangle by $\langle xyz\rangle$ instead of $(X,Y,Z)$ where the points $X$ , $Y$ and $Z$ are the vertices. Indeed it is well known that $\langle xyz\rangle$ is isometric to $\langle x^{\prime}y^{\prime}z^{\prime}\rangle$ if $x=x^{\prime}$ , $y=y^{\prime}$ and $z=z^{\prime}$ . (For the moment we will make the difference between a triangle and another obtained by permutation of the three numbers representing it even if, geometrically, they are the same!) From now on, $\lambda$ will be the half perimeter $\lambda={{x+y+z}\over 2}$ .

The set of non degenerate triangles is thus the open set $\Omega_{3}\subset({\hbox{\mathb R}}_{+}^{\ast})^{3}$ given by (1.7). We will describe it explicitly. To inequalities (4.1) are associated three equations defining respectively three planes:

[TABLE]

In the slice $\{x+y+z=2\lambda\}$ of ${\hbox{\mathb R}}_{+}^{3}$ , $\Sigma_{1}$ , $\Sigma_{2}$ and $\Sigma_{3}$ are the sides of an equilateral triangle in ${\hbox{\mathb R}}_{+}^{3}$ whose vertices are $X_{\lambda}=(0,\lambda,\lambda)$ , $Y_{\lambda}=(\lambda,0,\lambda)$ and $Z_{\lambda}=(\lambda,\lambda,0)$ (see the picture bellow); the interior $P_{\lambda}$ of the convex hull of these three points represents the space of triangles $\langle xyz\rangle$ whose perimeter is $2\lambda$ .

When $\lambda$ varies to $\lambda^{\prime}$ , we obtain another $P_{\lambda^{\prime}}$ , image of $P_{\lambda}$ by the homothety centered at the origin and with ratio $k={{\lambda^{\prime}}\over\lambda}$ . Thus, the space $\Omega_{3}$ is foliated by these $P_{\lambda}$ ; $\Omega_{3}$ is in fact the open cone with vertex the origin and basis anyone of these plaques $P_{\lambda}$ , for instance $P_{1}$ :

[TABLE]

For a particular situation which will appear thereafter, we recall the following result which we have already established in the general case of polygons.

For a given family of triangles with prescribed perimeter, the maximum of the area is realized by the equilateral triangle.

4.2. The perimeter foliation ${\cal F}_{p}$

Each $P_{\lambda}$ (where $\lambda\in{\hbox{\mathb R}}_{+}^{\ast}$ ) is the level set $p(x,y,z)=2\lambda$ where $p$ is the perimeter function $p(x,y,z)=x+y+z$ . We have also seen that the level surface $P_{\lambda}$ is the interior of the convex hull of the triangle $X_{\lambda}Y_{\lambda}Z_{\lambda}$ .

Thus we have a foliation ${\cal F}_{p}$ on $\Omega_{3}$ whose leaves are the surfaces $P_{\lambda}$ ( $\lambda>0$ ). Of course, ${\cal F}_{p}$ is trivial since isomorphic to the product $P_{1}\times{\hbox{\mathb R}}_{+}^{\ast}$ .

4.3. The area foliation ${\cal F}_{a}$

The function ${\cal A}:({\hbox{\mathb R}}_{+}^{\ast})^{3}\longrightarrow{\hbox{\mathb R}}_{+}^{\ast}$ which associates to a triangle $\langle xyz\rangle$ its area is given by Héron formula:

[TABLE]

The foliation ${\cal F}_{a}$ by which we will be interested is the foliation whose leaves are the level surfaces of this function.

$\bullet$ The surface at level $s$ of the function ${\cal A}$ on the open set $\Omega_{3}$ is exactly the surface at level $s^{2}$ of the function $\Phi={\cal A}^{2}$ . The benefit of working with $\Phi$ instead of ${\cal A}$ is that there is no more square root, which simplifies the calculations, among others that of the differential which plays a fundamental role. We consider then the function:

[TABLE]

$\bullet$ The differential of $\Phi$ has the form:

[TABLE]

where the functions $A$ , $B$ and $C$ are given as follows:

[TABLE]

An easy but long computation shows that these three functions $A$ , $B$ et $C$ are zero simultaneously only if $x=y=z=0$ , which can not happen since $(0,0,0)$ is not in $\Omega_{3}$ .

$\bullet$ If we fix the perimeter $2\lambda$ , the area function $a$ is maximal, and so is the function $\Phi$ , when $x=y=z={2\over 3}\lambda$ ; at this point $\Phi$ is equal to ${{\lambda^{4}}\over{27}}$ . These are the values taken by the function $\Phi$ on the open half line $\Delta$ whose equation is $x=y=z$ .

Now let $\Omega_{3}^{\ast}$ be the open set $\Omega_{3}\setminus\Delta$ . At $u=(x,y,z)\in\Omega_{3}^{\ast}$ , the differential $d_{u}\Phi$ has rank $1$ ; then the set level of $\Phi$ passing through this point is a regular surface $A$ , in fact an algebraic surface of degree $4$ . Its equation is:

[TABLE]

Let $G$ be the subgroup of Isom $({\hbox{\mathb R}}^{3})$ (the full group of isometries of the Euclidean space ${\hbox{\mathb R}}^{3}$ ) generated by the rotation whose axis is $\Delta$ and angle ${{2\pi}\over 3}$ and the reflection $\sigma$ with respect to the plane of equation $x=y$ . (The restrictions of these elements to the plane of equation $x+y+z=2\lambda$ is the group of isometries of the equilateral triangle $X_{\lambda}Y_{\lambda}Z_{\lambda}$ .) It leaves the space $\Omega_{3}$ invariant and also its boundary $\partial\Omega_{3}$ , the half line $\Delta$ and the open sets $\Omega_{3}$ and $\Omega_{3}^{\ast}$ . Then it acts on $\Omega_{3}$ and fixes each leaf of ${\cal F}_{a}$ ; the same applies to the foliation ${\cal F}_{p}$ .

4.4. Let $\Psi:\Omega_{3}^{\ast}\longmapsto\left({\hbox{\mathb R}}_{+}^{\ast}\right)^{2}$ be the function :

[TABLE]

Up to a multiplicative factor, the matrix of its differential at $u=(x,y,z)$ is :

[TABLE]

where $A$ , $B$ and $C$ are the functions given by (5.6). It can be shown that these functions are equal only if $x=y=z$ ; then, for $u\in\Omega_{3}^{\ast}$ , $d_{u}\Psi$ has rank $2$ . Thus, the level sets of $\Psi$ are regular curves, leaves of a foliation ${\cal F}_{\ast}$ on $\Omega_{3}^{\ast}$ .

4.5. On $\Omega_{3}$ we have a singular foliation ${\cal F}={\cal F}_{p}\cap{\cal F}_{a}$ . Its leaves of dimension [math] are the points of the open half line $\{\left({2\over 3}\lambda,{2\over 3}\lambda,{2\over 3}\lambda\right):\lambda\in{\hbox{\mathb R}}_{+}\}$ . The other leaves are of dimension $1$ ; each one has equation $\Psi(u)=\hbox{constant}$ in the open set $\Omega_{3}^{\ast}$ . These curves define (by restriction) a foliation on each plaque $P_{\lambda}$ (leaf of ${\cal F}_{p}$ ). To see what it is, this plaque is projected orthogonally on the plane $z=0$ ; we obtain the foliation drawn on the picture below. We will explain what all this means.

[TABLE]

The interior of the triangle $XYZ$ is the projection (which we denote by $\Theta_{\lambda}$ ) on the plane $z=0$ of the set $P_{\lambda}$ of triangles $\langle x_{\lambda}y_{\lambda}z_{\lambda}\rangle$ with perimeter $2\lambda$ . Note that the boundary of $P_{\lambda}$ is an equilateral triangle while $XYZ$ is an isosceles and rectangle triangle. The foliation ${\cal F}$ on $P_{\lambda}$ is isomorphic to the foliation on the picture via the diffeomorphism $f:P_{\lambda}\longrightarrow\Theta_{\lambda}$ defined by $f(x,y,z)=(x,y,0)$ with inverse $f^{-1}(x,y,0)=(x,y,2\lambda-x-y)$ .

$\bullet$ The point $\omega$ with coordinates $\left({2\over 3}\lambda,{2\over 3}\lambda\right)$ corresponds to the equilateral triangle $\langle xxx\rangle$ with maximal area. As we easily imagine, an equilateral triangle may never be deformed to an other one having the same perimeter and the same area.

$\bullet$ The curves at the interior of $\Theta_{\lambda}$ are leaves of a foliation of $\Theta_{\lambda}\setminus\{\omega\}$ , each leaf corresponds to the set of triangles having the same area. It has $\lambda(2\lambda-x)(2\lambda-y)(x+y)=8c$ as equation where $c$ is a constant varying in the interval $\big{]}0,{{8\lambda^{4}}\over{27}}\big{[}$ .

$\bullet$ The piece $UZ$ of diagonal corresponds to isosceles triangles (for which $x=y$ ). In each leaf, there is exactly the projections of two isosceles triangles $\langle xxz\rangle$ and $\langle x^{\prime}x^{\prime}z^{\prime}\rangle$ .

4.6. Geoffrey Letellier constructed two lines of isosceles triangles : $\langle x_{\lambda}y_{\lambda}z_{\lambda}\rangle$ and $\langle x_{\lambda}^{\prime}y_{\lambda}^{\prime}z_{\lambda}^{\prime}\rangle$ where $\lambda\in{\hbox{\mathb R}}_{+}^{\ast}$ with $x_{\lambda}=y_{\lambda}={11\over 14}\lambda$ , $z_{\lambda}={3\over 7}\lambda$ and $x_{\lambda}^{\prime}=y_{\lambda}^{\prime}={4\over 7}\lambda$ , $z_{\lambda}^{\prime}={6\over 7}\lambda$ . They are such that, for any $\lambda\in{\hbox{\mathb R}}_{+}^{\ast}$ :

[TABLE]

For instance, the two isosceles triangles $x=y=11$ , $z=6$ and $x^{\prime}=y^{\prime}=8$ , $z^{\prime}=12$ have the same perimeter equal to $28$ and the same area equal to $12\sqrt{7}$ .

$\bullet$ Finally one can see on the picture that all the situation is invariant by the reflection $\sigma$ (symmetry with respect to the diagonal $x=y$ ) while that on the triangle $P_{\lambda}$ is invariant by the full group $G$ . $\diamondsuit$

5. Some results related to the perimeter and the area

The following well known classical results are among the most beautiful theorems that we can cite in Euclidean elementary geometry of the plane.

5.1. Theorem (Isoperimetric inequality). Among all the convex polygons with prescribed perimeter, the regular polygon is the one whose area is maximum.

For a sketch of proof, see for instance [Han]. In the same order of ideas, we also have the following theorem. Its proof is not difficult but it is a bit long and not immediate. (And the reader can even attempt to reproduce it himself!)

5.2. Theorem. Among all convex polygons whose sides have given lengths, the inscribable polygon is the one whose area is maximum.

Using the analytic expression of the function “area" ${\cal A}:\omega\in\Omega_{n}\longrightarrow\hbox{area}(\omega)\in{\hbox{\mathb R}}$ , we prove the following result related to the two theorems above. (It was also partially established, by a different method, in [Khi].)

5.3. Theorem. We have the following results.

(1) For any real number $L>0$ , the differentiable manifold $p^{-1}\left(\{L\}\right)$ consisting of all polygons with perimeter $L$ , is diffeomorphic to a convex open set $\Omega_{n,L}$ of ${\hbox{\mathb R}}^{2n-4}$ and the restriction ${\cal A}_{L}:\Omega_{n,L}\rightarrow{\hbox{\mathb R}}$ of ${\cal A}$ to $\Omega_{n,L}$ admits a critical point at the unique regular polygon $\omega_{L}$ of perimeter $L$ .

(2) The convex polygons whose sides have given lengths form a differentiable manifold diffeomorphic to an open convex set of ${\hbox{\mathb R}}^{n-3}$ and the restriction of the function ${\cal A}$ to this open set admits a critical point at its unique inscribable polygon.

Proof. Recall that, for any $\omega=(t_{1},x_{1},\cdots,t_{n-2},x_{n-2},t_{n-1})\in\Omega_{n}$ , we have:

[TABLE]

with:

[TABLE]

Setting $h(v)={1\over 4}\sqrt{f(v)}$ for $v\in{\cal V}$ , we obtain:

[TABLE]

for any $\omega=(t_{1},x_{1},\cdots,t_{n-2},x_{n-2},t_{n-1})\in\Omega_{n}$ .

Point (1)

Let $L\in]0,+\infty[$ . For any $\omega=(t_{1},x_{1},t_{2},...,t_{n-2},x_{n-2},t_{n-1})\in\Omega_{n}$ , we have:

[TABLE]

Then, by considering the affine map $T:{\hbox{\mathb R}}^{2n-4}\rightarrow{\hbox{\mathb R}}^{2n-3}={\hbox{\mathb R}}\times{\hbox{\mathb R}}^{2n-4}$ given by:

[TABLE]

where $t_{1}(u)=L-x_{1}-\cdots-x_{n-2}-t_{n-1}$ , we see that $p^{-1}\left(\{L\}\right)$ is naturally identified to the convex open set $\Omega_{n,L}=T^{-1}\left(\Omega_{n}\right)$ of ${\hbox{\mathb R}}^{2n-4}$ .

For any $u=(x_{1},t_{2},\cdots,t_{n-2},x_{n-2},t_{n-1})\in\Omega_{n,L}$ , we have:

[TABLE]

Set $\displaystyle t_{1}=t_{1}(u)$ and $\displaystyle\omega_{k}=(t_{k},x_{k},t_{k+1})$ for $k\in\{1,\cdots,n-2\}$ . The partial derivatives for any $(x,y,z)\in{\cal V}$ are:

[TABLE]

and for $k\in\{2,\cdots,n-2\}$ :

[TABLE]

$(\star)$ If the sides of the polygon are not all equal, there are two consecutive ones with a common vertex $M$ and different lengths.

By changing the numbering of the vertices, one can assume $M=M_{1}$ . In these conditions, the lengths $t_{1}=M_{n}M_{1}$ and $x_{1}=M_{1}M_{2}$ are different and this implies $\displaystyle{{\partial{\cal A}_{L}}\over{\partial x_{1}}}(u)\neq 0$ and that $u$ is not a critical point of ${\cal A}_{L}$ .

$(\star)$ If the lengths of all the sides are equal, then $t_{1}=x_{1}=x_{2}=...=x_{n-2}=t_{n-1}$ and a necessary condition for this polygon to be a critical point of ${\cal A}_{L}$ , is ${{\partial{\cal A}_{L}}\over{\partial t_{k}}}(u)=0$ for any $k\in\{2,\cdots,n-2\}$ , or:

[TABLE]

This implies:

[TABLE]

The development of this relationship leads to next equality:

[TABLE]

where:

[TABLE]

Thus $\displaystyle(t_{k+1}-t_{k-1})(t_{k+1}t_{k-1}-x_{k}^{2}+t_{k}^{2})(t_{k+1}t_{k-1}+x_{k}^{2}-t_{k}^{2})=0.$ The proof ends as that of the Theorem 3.2. We thus obtain the inscriptibility of all the quadrilaterals $(M_{n},M_{k-1},M_{k},M_{k+1})$ for $k\in\{2,\cdots,n-2\}$ and then the inscriptibility of the polygon $(M_{1},\cdots,M_{n})$ . But since the latter has all its sides of the same length, it is necessarily regular.

Finally the singular points of the map ${\cal A}_{L}$ are the regular polygons of $\Omega_{n,L}$ , that is, the unique regular polygon $\omega_{L}$ of perimeter $L$ .

Point (2)

Let $v=(\overline{t}_{1},\overline{x}_{1},\cdots,\overline{x}_{n-2},\overline{t}_{n-1})\in{\cal V}_{n}$ and let $F_{v}$ be the set of convex polygons whose sides are $\overline{t}_{1}$ , $\overline{x}_{1}$ , $\cdots$ , $\overline{x}_{n-2}$ , $\overline{t}_{n-1}$ . This is a convex open set is of ${\hbox{\mathb R}}^{n-3}$ .

On the other hand, the area function $F_{v}\buildrel{\cal A}\over{\rightarrow}{\hbox{\mathb R}}$ is given, for $t=(t_{2},t_{3},\cdots,t_{n-2})\in F_{v}$ , by:

[TABLE]

Setting:

[TABLE]

one can express the partial derivatives of ${\cal A}$ as follows:

[TABLE]

A critical point $t\in F_{v}$ of the area function must satisfy:

[TABLE]

Setting $\omega_{k-1}=(u,v,w)$ and $\omega_{k}=(w,s,r)$ , we obtain ${\alpha\over\beta}=0$ where:

[TABLE]

and $\beta=(r-s-w)(r+s-w)(r-s+w)(r+s+w)(-u-v+w)(u-v+w)(-u+v+w)(u+v+w)$ or:

[TABLE]

This implies:

[TABLE]

or:

[TABLE]

Then we deduce:

[TABLE]

which implies that the quadrilateral $(M_{n},M_{k-1},M_{k},M_{k+1})$ is inscribable for any index $k\in\{2,\cdots,n-2\}$ . Hence the polygon $t$ is inscribable.

Conversely, if $t\in F_{v}$ is inscribable then, one can prove easily that $t$ is a critical point of the area function.

References

[Han] Hansen, V.L. Shadows of the Circle. World Scientific (1998).

[Khi] Khimshiashvili, G. Cyclic polygons as critical points. Proc. of I. Vekua Institute of Applied Mathematics Vol. 58, (2008), 74-83.

[Pen] Penner, R.C. The Decorated Teichmüller Space of Punctured Surfaces. Commun. Math. Phys. 113, (1987), 299-339.

[Per] Perrin, D. Mathématiques d’école. Nombres, mesures et géométrie. Cassini, Paris, (2005).

[Ram] Ramon, P. Introduction à la géométrie différentielle discrète. Editions Ellipses, (2013).

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