text:
\section{Introduction}
Let $\Omega$ be a bounded subset of $\mathbb{R}^2$. The Cheeger constant of $\Omega$, introduced by Jeff Cheeger in \cite{cheger} in connection with the first eigenvalue of the Laplacian, is defined as
\begin{equation}
    \label{chee}
    h(\Omega):=\inf\left\{\frac{P(E)}{\abs{E}} \,  : \, E \, \text{measurable and} \, E\subseteq\Omega\right\},
\end{equation}
where $P(E)$ is the perimeter of $E$ in the sense of De Giorgi and $\abs{E}$ is the area of $E$.
The minimum in \eqref{chee} is achieved when $\Omega$ has Lipschitz boundary, see as a reference \cite{parini}, and the set $E$ that realizes this minimum is called a \emph{Cheeger set} of $\Omega$ and it is denoted by $C_\Omega$. For the properties of the Cheeger constant and for an introductory survey, see for example \cite{caselles,algo,parini}. In particular, in the case of planar convex sets, the authors in \cite{caselles} prove that the Cheeger set is unique, while in \cite{algo} the authors give a characterization for the Cheeger constant.

The problem of finding the Cheeger constant of a domain has been widely considered and has several applications (see \cite{parini} for a general overview). One of the possible interpretations of the Cheeger constant can be found for instance in the context of maximal flow and minimal cut problems: in particular, in \cite{strang}, the authors solve the problem of computing exact continuous optimal curves and surfaces for image segmentation in $2D$ and $3D$ reconstruction from a stereo image pair and this has applications in medical image process (see \cite{appleton}). The Cheeger problem appears also in the study of plate failure under stress (see \cite{keller}).
For this reason, it is useful to have estimates of the Cheeger constant in terms of geometric quantities that can be easily computed.

In the present paper, we are interested in describing all possible inequalities involving the Cheeger constant of a given open, bounded and convex set $\Omega\subset\mathbb{R}^2$ and two among the following geometrical quantities: the area $\abs{\Omega}$, the perimeter $P(\Omega)$, the inradius $r(\Omega)$, the circumradius $R(\Omega)$, the minimal width $\omega(\Omega)$ and the diameter $d(\Omega)$. So, we aim to study the associated Blaschke--Santal\'o diagrams of these triplets and collect them all in one single paper together with new conjectures.

A Blaschke--Santal\'o diagram is a tool that allows to visualize all the possible inequalities between three geometric quantities. More precisely, if we consider three shape functionals $(J_1,J_2,J_3)$, we want to find a system of inequalities describing the set
$$\{(J_1(\Omega), J_2(\Omega))|\  J_3(\Omega)=1, \, \Omega \in \mathcal{K}^2\},$$
where we denote by $\mathcal{K}^2$ the class of non-empty sets in $\mathbb{R}^2$ that are open, bounded and convex.
This kind of diagram was introduced by Blaschke in \cite{blaschke2}, in order to investigate all possible relations between the volume, the surface area and the integral mean curvature in the class of compact convex sets in $\mathbb{R}^3$.
Following the idea of Blaschke, Santal\'o in \cite{santalo} proposed the study of these diagrams for all the triplets of the following geometrical quantities: area, perimeter, inradius, circumradius, minimal width and diameter; these diagrams were studied under the constraint of convexity and six of them are still not completely solved. We refer to the introduction in \cite{delyon2} for the accurate state of art.
Moreover, for classical results about Blaschke--Santal\'o diagram, we refer for example to \cite{cifre3,cifre4,cifre2,cifre_salinas, cifre, cifre_gomis,santalo} and for more recent results we recall \cite{MR3653891,branden,delyon,delyon2,bs-numerics, FL21, LZ}.

In  \cite{ftJMAA} and \cite{ftouhi_cheeger}  the author studies the Blaschke--Santal\'o diagram involving the Cheeger constant.
More precisely, in \cite{ftJMAA}, it is studied the Blaschke--Santal\'o diagram between the Cheeger constant, the area and the inradius, and it is proved that, if $\Omega$ in $\mathcal{K}^2$, then
\begin{equation}\label{eq:hra}
    \frac{1}{r(\Omega)}+\frac{\pi r(\Omega)}{|\Omega|}\leq h(\Omega) \leq \frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}},
    \end{equation}
where the upper bound in \eqref{eq:hra} is achieved by (and only by) sets which are homothetical to their form body (see Definition \ref{formbody}), for instance circumscribed sets, meanwhile the lower one is achieved by (and only by) stadiums. Then, in \cite{ftouhi_cheeger}, it is studied the diagram between Cheeger constant, area and perimeter and it is proved that if $\Omega\in\mathcal{K}^2$, then
  \begin{equation}\label{eq:hpa}
    \frac{P(\Omega)+\sqrt{4\pi|\Omega|}}{2|\Omega|}\leq h(\Omega)\leq \frac{P(\Omega)}{|\Omega|},
    \end{equation}
where the upper bound is achieved by any set which is Cheeger of itself (in particular stadiums), meanwhile the lower one is achieved, for example, by circumscribed polygons. We also recall that in \cite{henrot_luc} the maximization problem of the Cheeger constant among sets of constant width is studied.

Now we state the main results of the paper.
In order to do that, we need to define the following classes of admissible sets (we refer to \cite[Table 2.1]{inequalities_convex} for the associated constraints):
\begin{enumerate}
    \item $\displaystyle{\mathcal{K}^2_{r,P}=\{\Omega \in \mathcal{K}^2: \, r(\Omega)=r, \, P(\Omega)=P\}}$, where $P\ge 2\pi r$;
    \vspace{2mm}
    \item $\mathcal{K}^2_{d,r}=\{\Omega \in \mathcal{K}^2: \, r(\Omega)=r, \,  d(\Omega)=d\}$, where $d\ge 2 r$;
    \vspace{2mm}
    \item $\mathcal{K}^2_{r,R}=\{\Omega \in \mathcal{K}^2: \, r(\Omega)=r, \, R(\Omega)=R\}$, where $R\ge r$;
    \vspace{2mm}
     \item $\mathcal{K}^2_{\omega,d}=\{\Omega \in \mathcal{K}^2: \, \omega(\Omega)=\omega, \, d(\Omega)=d\}$, where $\omega\le d$;
    \vspace{2mm}
       \item $\mathcal{K}^2_{R,\omega}=\{\Omega \in \mathcal{K}^2: \, R(\Omega)=R, \, \omega(\Omega)=\omega\}$, where $2R\ge \omega$;
    \vspace{2mm}

    \item $\mathcal{K}^2_{\omega,P}=\{\Omega \in \mathcal{K}^2: \, \omega(\Omega)=\omega, \, P(\Omega)=P\}$, where $P\ge \pi \omega$;
    \vspace{2mm}
    \item $\mathcal{K}^2_{A,\omega}=\{\Omega \in \mathcal{K}^2: \, \abs{\Omega}=A, \, \omega(\Omega)=\omega\}$, where $\sqrt{3}A\ge \omega^2$;
    \vspace{2mm}
    \item $\mathcal{K}^2_{d,R}=\{\Omega \in \mathcal{K}^2: \, d(\Omega)=d, \, R(\Omega)=R\}$, where $\sqrt{3}R\le d< 2R$;
    \vspace{2mm}
    \item $\mathcal{K}^2_{\omega,r}=\{\Omega \in \mathcal{K}^2: \, \omega(\Omega)=\omega, \, r(\Omega)=r\}$, where $2r< w\le 3r$;
    \vspace{2mm}
    \item $\mathcal{K}^2_{A,R}=\{\Omega \in \mathcal{K}^2: \, \abs{\Omega}=A, \, R(\Omega)=R\}$, where $A\le \pi R^2$.
    \vspace{2mm}
    \item $\mathcal{K}^2_{R,P}=\{\Omega \in \mathcal{K}^2: \, R(\Omega)=R, \, P(\Omega)=P\}$, where $4 R<P\le 2\pi R$;
    \vspace{2mm}
     \item $\mathcal{K}^2_{P, d}=\{\Omega \in \mathcal{K}^2: \,  P(\Omega)=P, \, d(\Omega)=d\}$, where $2 d<P\leq \pi d$.
  \vspace{2mm}
    \item $\mathcal{K}^2_{A,d}=\{\Omega \in \mathcal{K}^2: \, \abs{\Omega}=A, \, d(\Omega)=d\}$, where $\pi d^2\ge 4A$
\end{enumerate}
\newpage
Firstly, let us state the following existence result.
\begin{theorem}\label{th:existence}
Let $\Omega\in \mathcal{K}^2$, then the minimization and the maximization shape optimization problems of the Cheeger constant $h(\Omega)$ admit a solution in the classes of sets defined in $(1)-(13)$.
 \end{theorem}

In the following Theorem, we consider the classes of sets for which we are able to  explicitly state the solution of the maximum and the minimum problem, which is equivalent to completely solve the relative Blaschke--Santaló diagrams.  For the precise definitions of the below-mentioned extremal sets see Section \ref{ext} and for the explicit bounds see Propositions \ref{prop_hrP}, \ref{prop_hdr} and \ref{prop_hRr}.
\begin{theorem}\label{th2}
The following results hold
\begin{enumerate}[label=(\roman*)]
\item The maximum and the minimum in $\mathcal{K}^2_{r,P}$ are achieved respectively by circumscribed sets and stadiums.
 \vspace{2mm}
 \item The maximum in $\mathcal{K}^2_{d,r}$ is achieved by symmetrical two-cup bodies; moreover, there exists $D_0>0$ such that if $d\geq r D_0$ the minimum in $\mathcal{K}^2_{d,r}$ is achieved by symmetrical spherical slices, while, if $d<rD_0$, the minimum is achieved by regular smoothed nonagons.
    \vspace{2mm}
 \item The maximum and the minimum in $\mathcal{K}^2_{r,R}$ are achieved respectively by two-cup bodies and symmetrical spherical slices.
\end{enumerate}
\end{theorem}

In the following Theorem we analyse the classes of sets, for which we can partially solve the related Blaschke--Santaló diagrams, that are $\mathcal{K}^2_{\omega,d}$ $\mathcal{K}^2_{R,\omega}$, $\mathcal{K}^2_{\omega, P}$ and $\mathcal{K}^2_{A,\omega}$. See Propositions \ref{prop_hdw}, \ref{prop_hRw}, \ref{prop_hwP} and \ref{prop_hAw} for the explicit bounds.

\begin{theorem}\label{th3}
The following results hold
\begin{enumerate}[label=(\roman*)]
\item If $\omega\le \sqrt{3}/2 d$, the maximum in $\mathcal{K}^2_{\omega,d}$ is achieved by subequilateral triangles. Moreover, the minimum is achieved by symmetrical spherical slices

\item The minimum in
$\mathcal{K}^2_{R,\omega}$ is achieved by symmetrical spherical slices, while, if $\omega\le 3/2 R$, the maximum is achieved by subequilateral triangles.

\item The minimum in $\mathcal{K}^2_{\omega,P}$ is achieved by stadiums, while, if $\omega\le \frac{P}{2\sqrt{3}}$, the maximum is achieved by  subequilateral triangles.

\item The maximum in $\mathcal{K}^2_{A,\omega}$ is achieved by subequilateral triangles, while, if $\omega\le \frac{2\sqrt{A}}{\sqrt{\pi}}$, the maximum is achieved by stadiums.
    \end{enumerate}
\end{theorem}
In the class $\mathcal{K}^2_{d, R}$ we have only solved the maximization problem, while in the class $\mathcal{K}^2_{R,\omega}$ the minimization one. See Propositions \ref{prop_hRd} and \ref{prop_hRw} for the explicits bounds.
\begin{theorem}\label{th4}
The following results hold
\begin{enumerate}[label=(\roman*)]
\item The maximum in $\mathcal{K}^2_{d,R}$ is achieved by subequilateral triangles.
\item The minimum in $\mathcal{K}^2_{\omega, r}$ is achieved by subequilateral triangles.
\end{enumerate}
\end{theorem}

As far as the remaining classes of sets, that are $\mathcal{K}^2_{A,R}, \mathcal{K}^2_{R,P}, \mathcal{K}^2_{P,d}$ and $\mathcal{K}^2_{A,d}$, we are not able to identify the extremal sets. We refer to Propositions \ref{prop_hAR}, \ref{prop_hRP}, \ref{prop_hPd}, \ref{prop_hdA} for bounds that are not sharp in the Blaschke--Santaló sense, which means that they do not correspond to parts of the boundary of the diagram.

The paper is organized as follows. In Section \ref{sec2} we state the preliminary results, the definitions used throughout the paper and the known inequalities relating the Cheeger constant to one of the geometric quantities taken into consideration. Section \ref{secnum} is dedicated to the description of the numerical methods used to compute the functionals and to approach the Blaschke--Santal\'o diagrams. In Section \ref{sec3} we prove the main theorems and in Section \ref{secult} we state new conjectures. In the final Appendix, for the convenience of the reader, we have collected all the inequalities proved throughout the paper.

\section{Notations and Preliminaries}\label{sec2}
Throughout this article, $\norma{\cdot}$ will denote the Euclidean norm in $\mathbb{R}^2$,
 while $(\cdot)$ is the standard Euclidean scalar product for $n=2$. We denote by $P(\Omega)$ the perimeter of $\Omega$ and by $|\Omega|$ the volume of $\Omega$. Moreover, $B_r$ is the ball of radius $r>0$ centered at the origin, while $\mathbb{S}^1$ is the unit sphere in $\mathbb{R}^2$. In the following, we work with the class of sets $\mathcal{K}^2$ defined as
\begin{equation*}
    \mathcal{K}^2:=\{ \Omega\:|\: \Omega \;\, \text{is an open, bounded and convex set of  }\; \mathbb{R}^2\}\setminus \{\emptyset\}.
\end{equation*}
We also define the following function, known as cardinal sinus,
\begin{equation}\label{sinc}
    {\rm sinc}:x\in\mathbb{R}\longmapsto {\rm sinc}(x)=\frac{\sin(x)}{x},
\end{equation}
and we will denote its inverse, whenever it is defined, as ${\rm arcsinc}(x)$.
\subsection{Classical results ad preliminary lemmas }

We provide the classical definitions and results that we  need in the following.
\begin{definition}
    \label{minksum}
    Let $\Omega,K\subset \mathbb{R}^2$ two convex bounded sets with non-empty interior. We define the \emph{Minkowski sum} $(+)$ and \emph{difference} $(\sim)$ as
    \begin{equation*}
        \label{sum}
        \Omega+K:=\{x+y \, : \, x\in \Omega, \, y\in K\},
    \end{equation*}
    \begin{equation*}
        \label{diff}
        \Omega\sim K:=\{x\in \mathbb{R}^2 \, : \, x+K\subseteq \Omega\}.
    \end{equation*}
\end{definition}

We recall now the definition of Hausdorff distance.
\begin{definition}
Let $\Omega,K\subset\mathbb{R}^2$ two non-empty compact sets, we define the Hausdorff distance between $\Omega$ and $K$ as
\begin{equation*}
\label{disth}
 d_{\mathcal{H}}(\Omega,K)=\inf \left\{  \varepsilon>0  \; :\; \Omega\subset K+B_{\epsilon} \right\}.
 \end{equation*}
\end{definition}
 Note that, in the case that $\Omega$ and $K$ are convex sets, we have that $d_\mathcal{H}(\Omega,K)=d_\mathcal{H} (\partial \Omega, \partial K)$.

Let $\{\Omega_k\}_{k\in\mathbb N}$ be a sequence of non-empty, open, bounded convex subsets of $\mathbb{R}^2$, we say that $\Omega_k$ converges to $\Omega$ in the Hausdorff sense and we denote
\[
\Omega_k\stackrel{\mathcal H}{\longrightarrow} \Omega
\]
if and only if $d_{\mathcal H}(\Omega_k,\Omega)\to 0$ as $k\to \infty$. We recall that, by the Blaschke selection theorem (see as a reference \cite{schneider}, Theorem $1.8.7$), every bounded sequence of convex sets has a subsequence that converges in the Hausdorff sense to a convex set.

Let us now recall the following definitions:
\begin{definition}
  Let $\Omega\in\mathcal{K}^2$. The \emph{distance function from the boundary of} $\Omega$ is the function $ d(\cdot, \partial \Omega):\Omega \to [0,+\infty[$ defined as
 $$d(x,\partial\Omega)=\inf_{y\in\partial\Omega}\norma{x-y}.$$
  The \emph{inradius} $r(\Omega)$ of $\Omega$ is defined as	$$r(\Omega)=\sup_{x\in \Omega} d(x,\partial\Omega).$$
    Finally, the \emph{circumradius} $R(\Omega)$ is defined as
    $$R(\Omega)= \min_{x\in\Omega}\max_{y\in\partial\Omega} \norma{x-y}.$$
   \end{definition}
We need now to introduce the support function of a convex set.
\begin{definition}\label{support}
Let $\Omega\in \mathcal{K}^2$. The \emph{support} function of $\Omega$ is defined as
    \begin{equation*}
        p_\Omega(y)=\max_{x\in \Omega} (x\cdot y), \qquad y\in \mathbb{R}^2.
    \end{equation*}

\end{definition}

 In this paper, we will also consider the minimal width (or thickness) of a convex set, that is to say the minimal distance between two parallel supporting hyperplanes. More precisely, we have
\begin{definition}
    Let $\Omega\in\mathcal{K}^2$. The width of $\Omega$ in the direction $y \in \mathbb{R}$ is defined as
    \begin{equation*}
        \omega_{\Omega}(y)=p_\Omega(y)+p_\Omega( -y)
    \end{equation*}
    and the \emph{minimal width} of $\Omega$ as
\begin{equation*}
    \omega(\Omega)=\min\{  \omega_{\Omega}(y)\,|\; y\in\mathbb{S}^{1}\}.
\end{equation*}
\end{definition}

We introduce now the inner parallel set of a convex set $\Omega$.                                                  \begin{definition}
    \label{inner_parallel}
    Let $\Omega$ be an open, bounded and convex set, the \emph{inner parallel set} of $\Omega$ at distance $t\in [0, r(\Omega)]$ is
    \begin{equation*}
        \Omega_{-t}=\{x\in \Omega \, : \, d(x,\partial\Omega)\ge t\}.
    \end{equation*}
   \end{definition}

 \begin{remark}\label{rem:in}
 We remark that, by definition, we have that
$$\Omega_{-t}=\Omega \sim tB_1.$$
Moreover, we observe that for any $y\in \mathbb{S}^1$ and for every $\Omega,K\in \mathcal{K}^2$ , one has, see e.g.  \cite{schneider},
 \begin{equation*
     p_{\Omega\sim K}( y)\le p_\Omega(y)-p_K(y),
 \end{equation*}
 so, in the case $K=tB_1$, this reads
 \begin{equation}\label{diff_parall}
     p_{\Omega_{-t}}(y)\le p_\Omega(y)-t.
 \end{equation}
 Moreover, as it is observed in \cite[Proposition  3.2]{jahn},
 one has
\begin{equation}
    \label{circpar}
    R(\Omega+ K)\le R(\Omega)+R(K)
\end{equation}
and, if $K=tB_1$, equality holds in \eqref{circpar}.
\end{remark}

We are now in a position to prove the following Lemma.
\begin{lemma}\label{lem:diameter_inner_set}
Let $\Omega\in\mathcal{K}^2$. We have for every $ t\in [0,r(\Omega)]$:
\begin{align}
\label{inr}
     &r(\Omega_{-t})= r(\Omega)-t, \\
    \label{eq:diameter}
     &d(\Omega_{-t})\leq d(\Omega)-2t,\\
     \label{width}
     &\omega (\Omega_{-t})\leq \omega (\Omega)-2t,\\
     \label{circumradius}
     &R(\Omega_{-t})\leq R(\Omega)-t,\\
     \label{perimeter}
     &P(\Omega_{-t})\leq P(\Omega)-2\pi t.
\end{align}
\end{lemma}
\begin{proof}

      The proof of \eqref{inr} can be found in \cite[Lemma 1.4]{larson}.

     Let us now prove \eqref{eq:diameter}.
Let $x_t,y_t\in \Omega_{-t}$ be two diametrical points of $\Omega_{-t}$ (i.e., such that $\|x_t-y_t\|=d(\Omega_{-t})$). We denote by $x,y\in \Omega$ the points corresponding to the intersection of the line containing $x_t$ and $y_t$ with the boundary of $\Omega$. We have
$$d(\Omega)\ge \|x-y\|=\|x-x_t\|+\|x_t-y_t\|+\|y_t-y\|=\|x-x_t\|+d(\Omega_{-t})+\|y_t-y\|\ge d(\Omega_{-t})+2t,$$
where the last inequality is a consequence of the fact that $x_t,y_t\in \Omega_{-t}=\{x\in \Omega\ |\ d(x,\partial \Omega)\ge t\}$. \\
The proof of \eqref{width} follows directly from the definition of width and \eqref{diff_parall}.

 We prove now \eqref{circumradius}.
\begin{comment}
Let $\overline{x}$ be the center of the circumcircle of $\Omega$: {\color{red} it coincides with the center of the circumcirle of $\Omega_{-t}$. \ilias{I don't think so, you can take for example a triangle}} By definition, one has
$$R(\Omega)=\max_{y\in\partial\Omega}\norma{\overline{x}-y}$$
and
\begin{equation}
    \label{cc}
    R(\Omega_{-t})=\max_{y\in\partial\Omega_{-t}} \norma{\overline{x}-y}.
\end{equation}
Let $y_t\in \partial \Omega_t$ be the point that realizes the maximum in \eqref{cc} and $y\in\partial \Omega$ the point obtained by the intersection of the line containing $\overline{x}$ and $y_t$ with the boundary of $\Omega$.So,
$$R(\Omega)=\max_{y\in\partial\Omega}\norma{\overline{x}-y}\ge \norma{\overline{x}-y}=\norma{\overline{x}-y_t}+\norma{y_t-y}\ge R(\Omega_{-t})+t.$$
\end{comment}
 As observed in Remark \ref{rem:in}, and, in particular, by formula \eqref{circpar}, for every $\Omega\in \mathcal{K}^2$, we have that $R(\Omega+tB_1)=R(\Omega)+t$. Thus, we have $$R(\Omega_{-t})= R(\Omega_{-t}+tB_1)-t\leq R(\Omega)-t.$$
  The last inequality follows from the inclusion $\Omega_{-t}+t B_1 \subset \Omega$.
\item For \eqref{perimeter}, we use a similar method as above:
$$P(\Omega_{-t}) = P(\Omega_{-t}+t B_1) - t P(B_1)= P(\Omega_{-t}+t B_1) -2\pi t \leq P(\Omega)-2\pi t.$$
\end{proof}

Finally, the following Lemma will play a key role in the proof of the main theorem.
\begin{lemma}\label{lem:main}
Let $\Omega\in \mathcal{K}^2$. We assume that there exists a continuous function $g_\Omega:[0,r(\Omega)]\rightarrow \mathbb{R}$ such that
\begin{equation}\label{kawohl}
  \forall t\in [0,r(\Omega)],\ \ \ \ |\Omega_{-t}|\leq g^\Omega(t),\ \ \ \text{(resp. $|\Omega_{-t}|\ge g^\Omega(t)$)}.
\end{equation}
We then have the inequality
\begin{equation}\label{kawohl2}
    h(\Omega)\ge \frac{1}{t_{g^\Omega}}\ \ \ \ \ \text{(resp. $h(\Omega)\leq \frac{1}{t_{g^\Omega}}$)},
\end{equation}
where $t_{g^\Omega}$ is the smallest (resp. largest) solution to the equation $g^\Omega(t) = \pi t^2$ on $[0,r(\Omega)]$.
\end{lemma}
\begin{proof}
From Theorem 1 in \cite{algo}, we know that there exists a unique $t=t_\Omega>0$ such that $\abs{\Omega_{-t}}=\pi t^2$ and $\displaystyle{h(\Omega)=1/t_\Omega}$. So, it is clear that, if there exists a function $g(t)$ such that \eqref{kawohl} holds,  the smallest (resp. largest) solution  to $g^\Omega(t)=\pi t^2$, that we denote by $t_{g^\Omega}$, must satisfy
\eqref{kawohl2}.
\end{proof}

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            \draw (7,0) node[right] {$t$};
            \draw [-latex] (0,-.5) -- (0,5);

            \draw [domain=0:7, red ,line width=0.4mm] plot(\x,{.09*(\x)^2});

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            \draw [domain=0:6.1] plot(\x,{1.2*(sqrt(2.7^2-\x)-.03*\x^2)+.75});

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        \path[name intersections={of = line1 and line3, by = P2}];
        \filldraw[black] (P2) circle(1.5pt);

            \path[name path = line4] (P1) --++ (-90:3);

            \path[name path = line5] (0,0) --++ (0:11);

            \path[name intersections={of = line4 and line5, by = P3}];
          \draw[black, dashed] (P1) -- (P3);

          \draw (P1 |- 0,0) --++ (-90:0.1);
          \draw[black] (P1 |- 0,0) node[below] {$\frac{1}{h(\Omega)}$};

            \path[name path = line6] (P2) --++ (-90:5);

            \path[name intersections={of = line5 and line6, by = P4}];
          \draw[black, dashed] (P2) -- (P4);

          \draw[black] (P4 |- 0,0) --++ (-90:0.1);
          \draw[black] (P4 |- 0,0) node[below] {$t_{g^\Omega}$};

          \path[name path = line7] (0,0) --++ (90:7);

          \path[name intersections={of = line2 and line7, by = P5}];
          \draw[black] (P5) --++ (0:0.1);
          \draw[black] (P5) --++ (180:0.1);
          \draw (P5) node[left] {$|\Omega|$};

          \path[name intersections={of = line3 and line7, by = P6}];
          \draw[black] (P6) --++ (0:0.1);
          \draw[black] (P6) --++ (180:0.1);
          \draw (P6) node[left] {$g^{\Omega}(0)$};

          \path[name intersections={of = line2 and line5, by = P7}];
          \draw[black] (P7) --++ (90:0.1);
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          \draw (P7) node[below] {$r$};

            \draw[red] (8.5,3.5)--  (9,3.5);
            \draw(9,3.5) node[right] {Graph of the function $t\longmapsto \pi t^2$.};

            \draw (8.5,3) --  (9,3);
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            \draw [green] (8.5,2.5) --  (9,2.5);
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            \draw (8.2,3.8) --  (15,3.8);
        \end{tikzpicture}

\subsection{Inequalities relating the Cheeger constant to one geometric quantity}
In the following paragraph we recall some known inequalities relating the Cheeger constant to one of the geometric quantity taken into account, obtained just  combining classical  results.
\begin{proposition}\label{two}
Let  $\Omega\in\mathcal{K}^2$. We have:
\begin{enumerate}
    \item $h(\Omega)\ge 2\sqrt{\frac{\pi}{|\Omega|}}$;
    \item $h(\Omega)\ge\frac{4\pi}{P(\Omega)}$;
    \item $h(\Omega)\geq \frac{4}{d(\Omega)} $ ;
    \item $\frac{1}{r(\Omega)}\leq h(\Omega)\leq \frac{2}{r(\Omega)}$;
    \item $h(\Omega)\geq \frac{2}{R(\Omega)} $.
\end{enumerate}
In $(1)-(2)-(3)-(5)$ the equality is achieved by balls, while the upper bound in $(4)$ is achieved by balls and the lower bound by thinning vanishing stadiums.
\end{proposition}
\begin{proof}
\begin{enumerate}
    \item We have, using the classical isoperimetric inequality $P(E)\ge 2\sqrt{\pi|E|}$,
    \begin{equation}\label{h1}
        h(\Omega)= \inf_{\underset{E\subset \Omega,\  |E|>0}{E\text{ is measurable }}}\frac{P(E)}{|E|}\ge \inf_{\underset{E\subset \Omega,\  |E|>0}{E\text{ is measurable }}} \frac{2\sqrt{\pi}\sqrt{|E|}}{|E|}= \inf_{\underset{E\subset \Omega,\  |E|>0}{E\text{ is measurable }}} \frac{2\sqrt{\pi}}{\sqrt{|E|}}= \frac{2\sqrt{\pi}}{\sqrt{|\Omega|}}.
    \end{equation}

    \item Using \eqref{h1} and the classical isoperimetric inequality, we have
    $$h(\Omega)\ge \frac{2\sqrt{\pi}}{\sqrt{|\Omega|}} \ge \frac{4\pi}{P(\Omega)}. $$

    \item Using \eqref{h1} and the isodiametric  inequality $|\Omega|\leq \frac{\pi}{4}d(\Omega)^2$, we have
    $$h(\Omega)\ge \frac{2\sqrt{\pi}}{\sqrt{|\Omega|}} \ge \frac{2\sqrt{\pi} }{\sqrt{\frac{\pi}{4}d(\Omega)^2}}=\frac{4}{d(\Omega)}.$$

    \item For the upper bound, we have
    $$h(\Omega)= \inf_{\underset{E\subset \Omega,\  |E|>0}{E\text{ is measurable }}}\frac{P(E)}{|E|}\leq \frac{P(D_{r(\Omega)})}{|D_{r(\Omega)}|}= \frac{2}{r(\Omega)},$$
    where $D_{r(\Omega)}$ is a ball of radius $r(\Omega)$ inscribed in $\Omega$.
    \\
     For the lower bound, we have
    $$h(\Omega)= \inf_{\underset{E\subset \Omega,\  |E|>0}{E\text{ is measurable }}}\frac{P(E)}{|E|} \ge \inf_{\underset{E\subset \Omega,\  |E|>0}{E\text{ is measurable }}}\frac{1}{r(E)}\ge \frac{1}{r(\Omega)},$$
    where we use the inequalities $|E|<r(E)P(E)$ (see \cite{inequalities_convex}) and $r(E)\leq r(\Omega)$.
    \item Using \eqref{h1} and the inequality $|\Omega|\leq \pi R(\Omega)^2$ (see \cite{inequalities_convex}), we have

    $$h(\Omega)\ge \frac{2\sqrt{\pi}}{\sqrt{|\Omega|}} \ge \frac{2}{R(\Omega)}.$$
\end{enumerate}
\end{proof}

\subsection{Extremal sets and their properties} \label{ext
In this Section, we describe special planar  shapes that appear in the statement of the main results.

Firstly, we recall the definition of the form body of convex set $\Omega$, following \cite{schneider}. A point $x\in\partial\Omega$ is called \emph{regular} if the supporting hyperplane at $x$ is uniquely defined. The set of all regular points of $\partial\Omega$ is denoted by ${\rm reg}(\Omega)$.
We also let $U(\Omega)$ denote the set of all outward pointing unit normals to $\partial\Omega$ at points of ${\rm reg}(\Omega)$.

\begin{definition}\label{formbody}
The form body $\Omega^\star$ of a set $\Omega\in\mathcal{K}^2$ is
defined as
$$\Omega^\star =\bigcap_{u\in U(\Omega)} \{x\in \mathbb{R}^2: \, (x,u)\le 1\}.$$
\end{definition}

\begin{definition}
\label{circumscribed}
A circumscribed set is a set which is homothetic to its form body.
\end{definition}

We observe that a  polygon whose incircle touches all its sides is a particular circumscribed set.
\begin{definition}\label{stadium}
A \emph{stadium} $\mathcal{R}$ is defined as the convex hull of the union of two balls in $\mathbb{R}^2$ with the same radius (see Figure \ref{fig:stad}).
\end{definition}

\begin{figure}[ht]
\centering
    \begin{tikzpicture}
\draw [line width=0.4pt,dashed] (-2,0) circle (1cm);
\draw [line width=0.4pt, dashed] (2,0) circle (1cm);
\draw [line width=2pt] (-2,1)-- (2,1);
\draw [line width=2pt] (-2,-1)-- (2,-1);
\draw [shift={(-2,0)},line width=2pt]  plot[domain=1.5707963267948966:4.71238898038469,variable=\t]({1*1*cos(\t r)+0*1*sin(\t r)},{0*1*cos(\t r)+1*1*sin(\t r)});
\draw [shift={(2,0)},line width=2pt]  plot[domain=-1.5707963267948966:1.5707963267948966,variable=\t]({1*1*cos(\t r)+0*1*sin(\t r)},{0*1*cos(\t r)+1*1*sin(\t r)});
\end{tikzpicture}
\hspace{-2em}\caption{Stadium} \label{fig:stad}
\end{figure}

\begin{definition}\label{def:slice}
The \emph{symmetrical spherical slice} $\mathcal{S}$ of diameter $d$ and width $\omega$ is the convex set obtained by the intersection of a ball of radius $d/2$ and a strip of width $\omega<d$ centered at the center of the ball (see Figure \ref{fig:slice}).
\end{definition}

\begin{figure}[ht]\label{fig:slice}
\centering
\begin{tikzpicture}[scale=0.7]
\draw [line width=0.4pt,dashed] (0,0) circle (3cm);
\draw [shift={(0,0)},line width=2pt]  plot[domain=2.8017557441356713:3.481429563043915,variable=\t]({1*3*cos(\t r)+0*3*sin(\t r)},{0*3*cos(\t r)+1*3*sin(\t r)});
\draw [shift={(0,0)},line width=2pt]  plot[domain=-0.3398369094541218:0.339836909454122,variable=\t]({1*3*cos(\t r)+0*3*sin(\t r)},{0*3*cos(\t r)+1*3*sin(\t r)});
\draw [line width=2pt] (-2.8284271247461903,1)-- (2.82842712474619,1);
\draw [line width=2pt] (-2.8284271247461903,-1)-- (2.82842712474619,-1);
\end{tikzpicture}
\caption{Symmetrical spherical slice.}
\end{figure}

\begin{definition}
A \emph{two-cup body} $\mathcal{C}$ is the convex hull of a ball in $\mathbb{R}^2$ with two points that are symmetric with respect to the center of the ball (see Figure \ref{fig:cup}).
\end{definition}

\begin{figure}[ht]
\centering
    \begin{tikzpicture}[scale=0.7]
\draw [line width=0.4pt,dashed] (0,0) circle (2cm);
\draw [line width=2pt] (-4,0)-- (-1,1.7320508075688774);
\draw [line width=2pt] (1,1.7320508075688772)-- (4,0);
\draw [line width=2pt] (4,0)-- (1,-1.7320508075688774);
\draw [line width=2pt] (-1,-1.7320508075688772)-- (-4,0);
\draw [shift={(0,0)},line width=2pt]  plot[domain=1.0471975511965976:2.0943951023931957,variable=\t]({1*2*cos(\t r)+0*2*sin(\t r)},{0*2*cos(\t r)+1*2*sin(\t r)});
\draw [shift={(0,0)},line width=2pt]  plot[domain=4.1887902047863905:5.235987755982988,variable=\t]({1*2*cos(\t r)+0*2*sin(\t r)},{0*2*cos(\t r)+1*2*sin(\t r)});
\end{tikzpicture}
\caption{Two-cup body.} \label{fig:cup}
\end{figure}

\begin{definition}
A \emph{subequilateral triangle} $T_I$ is an isosceles triangle with two equal angles greater than $\pi/3$.
\end{definition}

The following class of sets (introduced in \cite{yamanouti}, see also \cite{santalo}) represents a way to pass in a continuous manner from the equilateral triangle to the Reuleaux triangle.

\begin{definition}\label{yamanouti}
A \emph{Yamanouti set} $Y$ is a set obtained by an equilateral triangle by constructing on any side an arc of circle centered in the opposite vertex and with radius less or equal than the side itself. The Yamanouti set is the convex hull of the set obtained in this way (see Figure \ref{yama}).
\end{definition}

\begin{figure}[ht]
	\centering
	\begin{tikzpicture}[scale=1.5]

	\draw [line width=0.4pt,dashed] (-2,0)-- (0,3.4641016151377553);
	\draw [line width=0.4pt,dashed] (2,0)-- (0,3.4641016151377553);
	\draw [line width=0.4pt,dashed] (2,0)-- (-2,0);
	\draw [line width=2pt] (0,3.4641016151377553)-- (-0.8325821135072361,2.533076897813977);
	\draw [line width=2pt] (0,3.4641016151377553)-- (0.8325821135072389,2.5330768978139733);
	\draw [line width=2pt] (2,0)-- (1.61,1.186549619695696);
	\draw [line width=2pt] (2,0)-- (0.7774178864927629,-0.2555249023719164);
	\draw [line width=2pt] (-2,0)-- (-0.77741788649276,-0.2555249023719166);
	\draw [line width=2pt] (-2,0)-- (-1.61,1.1865496196956953);
	\draw [shift={(0,3.4641016151377553)},line width=2pt]  plot[domain=4.506350634077912:4.918427326691467,variable=\t]({1*3.8*cos(\t r)+0*3.8*sin(\t r)},{0*3.8*cos(\t r)+1*3.8*sin(\t r)});
	\draw [shift={(2,0)},line width=2pt]  plot[domain=2.4119555316847165:2.824032224298272,variable=\t]({1*3.8*cos(\t r)+0*3.8*sin(\t r)},{0*3.8*cos(\t r)+1*3.8*sin(\t r)});
	\draw [shift={(-2,0)},line width=2pt]  plot[domain=0.3175604292915215:0.7296371219050757,variable=\t]({1*3.8*cos(\t r)+0*3.8*sin(\t r)},{0*3.8*cos(\t r)+1*3.8*sin(\t r)});
	\end{tikzpicture}
	\caption{Yamanouti set}
	\label{yama}
\end{figure}

\begin{definition}
A \emph{symmetrical lens} $\mathcal{L}$ is defined as the intersection of two  balls of the same radius in $\mathbb{R}^2$ (see Figure $\ref{Symmetrical lens}$).
\end{definition}

\begin{figure}[h]
    \centering
    \begin{tikzpicture}
    \draw [shift={(0,0)},line width=2pt]  plot[domain=0.5235987755982988:2.6179938779914944,variable=\t]({1*3*cos(\t r)+0*3*sin(\t r)},{0*3*cos(\t r)+1*3*sin(\t r)});
    \draw [shift={(0,3)},line width=2pt]  plot[domain=3.665191429188092:5.759586531581287,variable=\t]({1*3*cos(\t r)+0*3*sin(\t r)},{0*3*cos(\t r)+1*3*sin(\t r)});
    \end{tikzpicture}
    \caption{Symmetrical lens.}
    \label{Symmetrical lens}
\end{figure}

In \cite{delyon2} the authors define the smoothed regular nonagon as follows.
\begin{definition}\label{def:nonagon}
Let $r>0$ and $2 r<d<2\sqrt{3} r$. The \emph{smoothed regular nonagon} of inradius $r$ and diameter $d$, that we denote by $\mathcal{N}$, is the convex set enclosed in an equilateral triangle $T_E$ of inradius $r$, following the construction below.
Let $\eta_i$ the normal angles to the sides of $T_E$ and let
$$\tau:= (3+\sqrt{d^2-3 r^2})/2, \,\, \, \, \text{and}\, \, \, \,  h:=\sqrt{d^2-\tau^2}.$$
We define now the points $A_i, B_i, M_i$, for $i=1,2,3$:
$$A_i:=\begin{pmatrix}
\cos{\eta_i}+h \sin{\eta_i} \\
\sin{\eta_i}-h\cos{\eta_i}
\end{pmatrix}, \, \, B_i:=\begin{pmatrix}
\cos{\eta_i}-h \sin{\eta_i} \\
\sin{\eta_i}+h\cos{\eta_i}
\end{pmatrix}, \, \, \, M_i:=(1-\tau)\times\begin{pmatrix}
\cos{\eta_i} \\
\sin{\eta_i}
\end{pmatrix}.$$
We obtain $K_E(d)$ as follows (see Figure \ref{fig:non}):
\begin{itemize}
    \item the points $A_i, B_i$ and $M_i$, for $i=1,2,3$, belong to $\partial \mathcal{N}$;
    \item $\arc{B_1M_3}$ and $\arc{M_1A_3}$ are diametrically opposed arcs of the same circle of diameter $d$, the same for the pairs $\arc{B_2M_1}$ and $\arc{M_2A_1}$, $\arc{M_2 B_3}$ and $\arc{M_3A_2}$;
    \item the boundary contains the segments $\overline{A_iB_i}$, for $i=1,2,3$, and the contact point $I_i$ with the incircle is the middle of the corresponding segment.
\end{itemize}
\end{definition}

\definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666}

\begin{figure}[h]
\centering
\begin{tikzpicture}[scale=2]
\draw [line width=0.5pt,dashed] (-1.7320508075688776,-1)-- (0,2);
\draw [line width=0.5pt,dashed] (0,2)-- (1.7320508075688772,-1);
\draw [line width=0.5pt,dashed] (1.7320508075688772,-1)-- (-1.7320508075688776,-1);
\draw [shift={(0.26031766672006906,0.1502944749556473)},line width=2pt]  plot[domain=1.7941968844018048:2.412567118938454,variable=\t]({1*1.175*cos(\t r)+0*1.175*sin(\t r)},{0*1.175*cos(\t r)+1*1.175*sin(\t r)});
\draw [shift={(0.26031766672006906,0.1502944749556473)},line width=2pt]  plot[domain=4.917815739437731:5.536185973974379,variable=\t]({1*1.175*cos(\t r)+0*1.175*sin(\t r)},{0*1.175*cos(\t r)+1*1.175*sin(\t r)});
\draw [shift={(0,-0.30058894991129426)},line width=2pt]  plot[domain=-0.3001982179913911:0.31817201654525773,variable=\t]({1*1.175*cos(\t r)+0*1.175*sin(\t r)},{0*1.175*cos(\t r)+1*1.175*sin(\t r)});
\draw [shift={(0,-0.30058894991129426)},line width=2pt]  plot[domain=2.8234206370445354:3.441790871581184,variable=\t]({1*1.175*cos(\t r)+0*1.175*sin(\t r)},{0*1.175*cos(\t r)+1*1.175*sin(\t r)});
\draw [shift={(-0.2603176667200691,0.15029447495564732)},line width=2pt]  plot[domain=3.888591986795:4.506962221331649,variable=\t]({1*1.175*cos(\t r)+0*1.175*sin(\t r)},{0*1.175*cos(\t r)+1*1.175*sin(\t r)});
\draw [shift={(-0.2603176667200691,0.15029447495564732)},line width=2pt]  plot[domain=0.7290255346513395:1.3473957691879885,variable=\t]({1*1.175*cos(\t r)+0*1.175*sin(\t r)},{0*1.175*cos(\t r)+1*1.175*sin(\t r)});
\draw [line width=2pt] (-1.1160254037844388,0.06698729810778098)-- (-0.616025403784439,0.9330127018922192);
\draw [line width=2pt] (0.616025403784439,0.9330127018922192)-- (1.1160254037844388,0.06698729810778058);
\draw [line width=2pt] (-0.5,-1)-- (0.5,-1);
\begin{scriptsize}
\draw [fill=black] (0,-1) circle (1.5pt);
\draw[color=black] (0.06912757788806292,-1.1312016285860344) node {$I_1$};
\draw [fill=uuuuuu] (-0.8660254037844388,0.5) circle (1.5pt);
\draw[color=uuuuuu] (-0.8959352773734842,0.6739581877547015) node {$I_3$};
\draw [fill=uuuuuu] (-0.616025403784439,0.9330127018922192) circle (1.5pt);
\draw[color=uuuuuu] (-0.6409693831911335,1.1110425777815864) node {$A_3$};
\draw [fill=uuuuuu] (0.8660254037844389,0.5) circle (1.5pt);
\draw[color=uuuuuu] (0.9797200571107441,0.6739581877547015) node {$I_2$};
\draw [fill=uuuuuu] (-1.1160254037844388,0.06698729810778098) circle (1.5pt);
\draw[color=uuuuuu] (-1.241795246763608,0.20597972252004358) node {$B_3$};
\draw [fill=uuuuuu] (0.616025403784439,0.9330127018922192) circle (1.5pt);
\draw[color=uuuuuu] (0.688330463759486,1.1110425777815864) node {$B_2$};
\draw [fill=uuuuuu] (1.1160254037844388,0.06698729810778058) circle (1.5pt);
\draw[color=uuuuuu] (1.1891563273319607,0.24597972252004358) node {$A_2$};
\draw [fill=uuuuuu] (0.5,-1) circle (1.5pt);
\draw[color=uuuuuu] (0.5699534414605375,-1.1394134781704879) node {$B_1$};
\draw [fill=uuuuuu] (-0.5,-1) circle (1.5pt);
\draw[color=uuuuuu] (-0.4316982856844117,-1.1194134781704879) node {$A_1$};
\draw [fill=uuuuuu] (0,1.296095379298727) circle (1.5pt);
\draw[color=uuuuuu] (0.06912757788806292,1.4752795694706569) node {$M_1$};
\draw [fill=uuuuuu] (1.1224515242003252,-0.6480476896493634) circle (1.5pt);
\draw[color=uuuuuu] (1.2982622521241877,-0.57338833606587083) node {$M_3$};
\draw [fill=uuuuuu] (-1.1224515242003252,-0.6480476896493633) circle (1.5pt);
\draw[color=uuuuuu] (-1.2909011715558347,-0.49338833606587083) node {$M_2$};
\end{scriptsize}
\end{tikzpicture}
\caption{Smoothed regular nonagon}\label{fig:non}
\end{figure}

We recall now some of the Blaschke--Santaló
sharp inequalities that we will need in the sequel between three of the following geometric quantities: perimeter, area, inradius, circumradius, diameter and width. On the other hand, in the rest of the paper, when needed, we will directly refer to the results contained in the survey paper \cite{inequalities_convex}.

Firstly, let us consider the diagram $(A,d,r)$. We have the following two theorems.

\begin{theorem}[\cite{cifre_salinas} \& \cite{delyon2} Theorem 1]
Let $\Omega\in \mathcal{K}^2$. Then, it holds
\begin{equation}
    \label{del_2cap}
    \abs{\Omega}\ge r(\Omega)\sqrt{d^2(\Omega)-4r^2(\Omega)}+r^2(\Omega)\left(\pi- 2\arccos{\left(\frac{2r(\Omega)}{d(\Omega)}\right)}\right),
\end{equation}
where equality holds if and only if $\Omega$ is a two-cup body.
\end{theorem}

\begin{theorem}[\cite{delyon2}, Theorem 2] \label{delthm}
Let $\Omega\in \mathcal{K}^2$. Then, it holds
\begin{equation}\label{del:non}
     \abs{\Omega}\le \psi\left(d\left(\Omega\right),r(\Omega)\right),
\end{equation}
where
\begin{equation}\label{func_del}
   \psi(d,r):=\begin{cases}
    \displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{d^2-3r^2}-r)+\frac{3d^2}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{d}\right)}\right)}, & \text{if} \, \, \,  d\le r D^*\vspace{1mm} \\
    \displaystyle{r\sqrt{d^2-4r^2}+\frac{d^2}{2}\arcsin{\left(\frac{2r}{d}\right)}}, & \text{if} \, \, \, d\ge r D^*
\end{cases}
\end{equation}
and $D^*$ is the unique number in $[2,2\sqrt{3}]$ for which the two expression of the function $\psi(d,r)$ are equal.

Moreover, if $d(\Omega)\le r(\Omega)D^*$, we have equality in \eqref{del:non} if and only if $\Omega$ is a regular smoothed nonagon,  while, if $d(\Omega)>r(\Omega)D^*$, we have equality if and only if $\Omega$ is a symmetrical spherical slice
\end{theorem}

As far as the diagram $(A, \omega, R)$, we recall the following.
\begin{theorem}[\cite{cifre_salinas}, Theorem 3] \label{cifrsalth}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}
\label{henk}
\abs{\Omega}\leq \chi(\omega(\Omega), R(\Omega)),
\end{equation}
where
\begin{equation}\label{henkchi}
\chi(\omega(\Omega), R(\Omega)):=\frac{\omega(\Omega)}{2} \sqrt{4 R(\Omega)^2-\omega(\Omega)^2} + 2R(\Omega)^2 \arcsin{\frac{\omega(\Omega)}{2R(\Omega)}},
\end{equation}
and equality in \eqref{henk} holds if and only if $\Omega$ is a symmetrical spherical slice.
\end{theorem}

\begin{theorem}[\cite{cifre_salinas}, Theorem 6]
Let $\Omega\in\mathcal{K}^2$. Then, if $\omega(\Omega)\leq \frac{3}{2}R(\Omega)$, it holds
\begin{equation}\label{ARw_low}
   16\abs{\Omega}^6\geq R^2(\Omega) \omega^2(\Omega)\left(16 \abs{\Omega}^4-R^2(\Omega)\omega^6(\Omega)\right)
\end{equation}
and equality holds if and only if $\Omega$ is a subequilateral triangle
\end{theorem}

We recall the following inequality from the diagram $(R,r,\omega)$.

\begin{theorem}[\cite{cifre_gomis}, Theorem 2]\label{cifre_gomis_1_thm}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{cifre_gomis_1_eq}
    \left(4 r(\Omega)-\omega(\Omega)   \right)\left( \omega(\Omega)-2 r(\Omega)\right)\leq \frac{2 r^3(\Omega)}{R(\Omega)}
\end{equation}
and equality holds if and only if $\Omega$ is an isosceles triangle.
\end{theorem}

\begin{theorem}[\cite{santalo}, Section 10]\label{santalo_thm_1}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{santalo_eq_1}
    \omega(\Omega)\leq R(\Omega)+r(\Omega),
\end{equation}
where equality is achieved by any set obtained by an equilateral triangle of circumradius $R(\Omega)$ by replacing the edges by three equal circular arcs, in particular each arc of circle is centered on the height relative to the same side.
\end{theorem}

The following theorem deals with the $(A, r, R)$ diagram.

\begin{theorem}[\cite{cifre_salinas}, Theorems 1 and 2]\label{inr_circ_thm}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{secondhRr}
    \abs{\Omega}\ge 2r(\Omega)\left(\sqrt{R(\Omega)^2-r(\Omega)^2}+r\arcsin{\frac{r(\Omega)}{R(\Omega)}}\right),
\end{equation}
and equality in \eqref{secondhRr} holds if and only if $\Omega$ is a two-cup body.
Moreover, we have
\begin{equation}\label{cifreRr}
    \abs{\Omega}\le \varphi(R(\Omega), r(\Omega)),
\end{equation}
where
\begin{equation}
    \label{cifreArR}
    \varphi(R(\Omega), r(\Omega)):= 2\left(r\sqrt{R(\Omega)^2-r(\Omega)^2}+R^2(\Omega)\arcsin{\frac{r(\Omega)}{R(\Omega)}}\right),
\end{equation}
and equality in \eqref{cifreArR} holds if and only if $\Omega$ is a symmetrical spherical slice
\end{theorem}

Now we quote two inequalities concerning the $(A,w,r)$ and $(P,w,r)$ diagrams.

\begin{theorem}[\cite{cifre_salinas}, Theorem 5] \label{cifre_salinas_thm_rw}
Let $\Omega \in \mathcal{K}^2$. Then, it holds
\begin{equation} \label{cifre_salinas_rwA}
    (\omega(\Omega)-2r(\Omega))^2(4r(\Omega)-\omega(\Omega))\abs{\Omega}^2\le r^4(\Omega)\omega^3(\Omega),
\end{equation}
\begin{equation} \label{cifre_salinas_rwP}
    (\omega(\Omega)-2r(\Omega))^2(4r(\Omega)-\omega(\Omega))P^2(\Omega)\le 4 r(\Omega)^2\omega^3(\Omega)
\end{equation}
In both inequalities, equality holds if and only if $\Omega$ is a subequilateral triangle
\end{theorem}
We recall this result from the $(d,w,r)$ diagram.
\begin{theorem}[\cite{cifre4}, Theorem 1-2]\label{cifre_wrd_thm}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
  \begin{equation}\label{cifre_wrd_eq_1}
        d^2(\Omega)(\omega(\Omega)-2r(\Omega))^2(4r(\Omega)-\omega(\Omega))\le 4r^4(\Omega)\omega(\Omega),
    \end{equation}
    where equality holds if and only if $\Omega$ is a subequilateral triangle $T_I$, and
    \begin{equation}\label{cifre_wrd_eq_2}
        \omega(\Omega)-r(\Omega)\leq \dfrac{\sqrt{3}}{3} d(\Omega),
    \end{equation}
    and equality holds if  $\Omega$ is a Yamanouti set
\end{theorem}

Finally, as far as the diagram $(A,P,R)$, we have

\begin{theorem}[\cite{santalo}]
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{santalo_APR}
   |\Omega|<\frac{P(\Omega)\left(P(\Omega)-4R(\Omega)\cos{\arcsinc{\left(\frac{4R(\Omega)}{P(\Omega)}\right)}}\right)}{8\arcsinc{\left(\frac{4R(\Omega)}{P(\Omega)}\right)}},
\end{equation}
where the function ${\rm arcsinc}$ is the inverse of the function defined in \eqref{sinc} and the equality is attained by symmetrical lenses.
\end{theorem}

\section{Numerical results and Blaschke--Santal\'o diagrams}\label{secnum}
In this Section, we introduce numerical tools, that we use to obtain more information on the diagrams and state some conjectures.

\subsection{Generation of random convex polygons}
We want to provide a numerical approximation of the diagrams studied in Section \ref{sec3}. To do so, a natural idea is to generate a large number of convex sets (more precisely convex polygons) for each we compute the involved functionals. Nevertheless, the task of (properly) generating random convex polygons is quite challenging and interesting on its own. The main difficulty is that one wants to design an efficient and fast algorithm that allows obtaining a uniform distribution of the generated random convex polygons. For more clarification, let us discuss two different (naive) approaches:
\begin{itemize}
    \item one easy way to generate random convex polygons is by rejection sampling. We generate a random set of points in a square; if they form a convex polygon, we return it, otherwise we try again. Unfortunately, the probability of a set of $n$ points uniformly generated inside a given square to be in convex position is equal to $p_n = \left(\frac{\binom{2n-2}{n-1}}{n!}\right)^2$, see \cite{random_polygon}. Thus, the random variable $X_n$ corresponding to the expected number of iterations needed to obtain a convex distribution follows a geometric law of parameter $p_n$, which means that its expectation is given by $\mathbb{E}(X_n)=\frac{1}{p_n}=\left(\frac{n!}{\binom{2n-2}{n-1}}\right)^2 $. For example, if $N=20$, the expected number of iterations is approximately equal to $2.10^9$, and since one iteration is performed in an average of $0.7$ seconds, this means that the algorithm will need about $50$ years to provide one convex polygon with $20$ sides;
    \item another natural approach is to generate random points and take their convex hull. This method is quite fast, as one can compute the convex hull of $N$ points in a $\mathcal{O}(N\log(N))$ time (see \cite{MR475616} for example), but it is not quite relevant since it yields to a biased distribution.
\end{itemize}

In order to avoid the issues stated above, we use an algorithm presented in \cite{sander}, that is based on the work of P. Valtr \cite{random_polygon}, where the author computes
the probability of a set of $n$ points uniformly generated inside a given square to be in convex position. The author remarks (in Section 4) that the proof yields a fast and
non-biased method to generate random convex sets inside a given square. We also refer to \cite{sander} for a nice description of the steps of the method and a beautiful
animation where one can follow each step, one will also find an implementation of Valtr's algorithm in Java that we decided to translate in Matlab.
To obtain the different figures below, we generate $10^5$ random convex polygons of unit area and number of sides between $3$ and $30$, for which we compute the involved functionals. We then obtain
clouds of dots that provide approximations of the diagrams. This approach has been used in several works, we refer for example
to \cite{AH11}, \cite{ftJMAA} and \cite{FL21}.

\subsection{About the computation of the functionals}
Let us give few details on the numerical computation of the functionals involved in the paper:
\begin{itemize}
    \item The \textbf{Cheeger constant} is computed by using a Beniamin Bogosel's code \cite{zbMATH07173414} based on the characterization of the Cheeger sets of planar convex sets given in \cite{kawohl} and the toolbox Clipper, a very good implementation of polygon offset computation by Agnus Johnson.
    \item The \textbf{inradius} is also computed by using the tootbox Clipper and the fact that $r(\Omega)$ is the solution to the equation $|\Omega_{-t}|=0$.
    \item The \textbf{diameter} is computed via a fast method of computation, which consists of finding all antipodal pairs of points and looking for the diametrical one between them. This is classically known as Shamos algorithm \cite{MR805539}.
    \item The \textbf{area} is computed by using Matlab's function \texttt{"polyarea"}.
    \item The \textbf{minimum width} of a polygon $\Omega$ of vertices $\{A_1,\cdots,A_N\}$ is computed by using the following formula
    $$\omega(\Omega)=\min_{i\in \llbracket 1,N \rrbracket} \max_{j\in \llbracket 1,N \rrbracket} d(A_j,(A_iA_{i+1})),$$
    where $d(A_j,(A_iA_{i+1}))$ corresponds to the distance between the point $A_j$ and the line $(A_iA_{i+1})$ (with the convention $A_{N+1}:=A_1$).
    \item The \textbf{circumradius} of a convex set $\Omega$ can be written as follows
    $$R(\Omega) = \min_{c\in \Omega}\max_{x \in \Omega} \|c-x\|.$$
    It is computed by using Matlab's routine \texttt{"fminmax"}.
\end{itemize}

\section{Proof of the main Results} \label{sec3}
\subsection{Proof of Theorem \ref{th:existence} }
We start this Section by proving the existence results stated in Theorem \ref{th:existence}.

\begin{proof}[Proof of the existence]
Let us consider the minimization problem of the Cheeger constant in the classes of sets $(1)-(13)$; the maximization problem can be dealt in an analogous way.\\
For all of these classes of sets, in order to prove the existence of the solution, we consider a minimizing sequence $(\Omega_k)_{k\in\mathbb N}$ and we prove that it satisfies the hypothesis of the Blaschke Selection Theorem (see Theorem $1.8.7$ in \cite{schneider}), that is to say its boundedness up to translations. Concerning the class of sets involving a diameter or a circumradius constraint, it is clear that, up to a translation, the minimizing sequence is contained in a sufficiently big ball. So it remains to study the problem in $\mathcal{K}^2_{r,P}$, $\mathcal{K}^2_{A,\omega}$, $\mathcal{K}^2_{r,\omega}$ and $\mathcal{K}^2_{\omega,P}$. Concerning $\mathcal{K}^2_{r,P}$ and $\mathcal{K}^2_{\omega,P}$, we know from \cite{inequalities_convex} that $P=P(\Omega_k)>2 d(\Omega_k)$, for every $k$, so the sequence of the diameters $d(\Omega_k)$ is equibounded and, consequently, there exists a sufficiently big ball containing the sequence $(\Omega_k)_{k}$.
As far as $\mathcal{K}^2_{r,\omega}$, it is possible to prove the boundness of the minimizing sequence whenever $\omega(\Omega_k)>2r(\Omega_k)$, indeed it holds (see \cite{inequalities_convex})
$$d(\Omega_k)\le \frac{\omega^2(\Omega_k)}{2(\omega(\Omega_k)-2r(\Omega_k))}.$$
For the last class $\mathcal{K}^2_{A,\omega}$, from \cite{inequalities_convex}, we know that, if $2\omega(\Omega_k)\leq \sqrt{3} d(\Omega_k)$, then $$2\abs{\Omega_k}\ge \omega(\Omega_k)d(\Omega_k),$$and, also in this case, the boundedness follows.

So, for every class of sets considered, the Blaschke selection theorem ensures us that, up to a subsequence, $\Omega_k$ converges in the Hausdorff sense to a set $\Omega^*$; it remains only to prove that this set belongs to the relative class of admissible sets. We observe that all the considered constraints are stable for the Hausdorff convergence. In particular, in \cite{delyon2} it is proved the stability of the inradius, in \cite{pierre} the stability of the diameter and in \cite{schneider} the stability of area, perimeter and width. It only remains to show that the circumradius is continuous with respect to the Hausdorff distance in the class of admissible sets having a circumradius constraint. Since $R(\Omega_k)=R$, $\forall k\in\mathbb N$, we have $\Omega_k\subseteq B_R$. Using the stability of the Hausdorff convergence for the inclusion (see \cite{pierre}, Prop $2.2.17$), we have that $\Omega^*\subseteq B_R$, and consequently $R(\Omega^*)\le R$. By contradiction, let us suppose that $R(\Omega^*)<R$, so there exists $\overline{R}>0$ such that $R(\Omega^*)<\overline{R}<R$ and so $\Omega^*\subseteq B_{\overline{R}}$. Therefore, by the Hausdorff convergence, for sufficiently large $k$, $\Omega_k\subseteq B_{\overline{R}}$, but this would imply $R\le \overline{R}$, that is absurd.

In order to conclude, we observe that in all the above cases, the set $\Omega^*$ cannot degenerate to a segment. If we are working in a class of sets involving an inradius, area or width constraint, then, it is clear that, thanks to the continuity of the inradius, width and area under Hausdorff convergence, there exists, up to a translation, a sufficiently small ball contained in the minimizing sequence.
On the other hand,
if we consider the minimization problem in $\mathcal{K}^2_{d,R}$, $\mathcal{K}^2_{P,d}$ and $\mathcal{K}^2_{R,P}$, the inradius can be bounded from below by a positive quantity.
In \cite[Section 9]{santalo} it is proved that
\begin{equation*}
    r(\Omega_k)\geq \dfrac{d^2(\Omega_k)\sqrt{4 R^2(\Omega_k)-d^2(\Omega_k)}}{2R(\Omega_k)\left(2 R(\Omega_k)+\sqrt{4 R^2(\Omega_k)-d^2(\Omega_k)}\right)};
\end{equation*}
in \cite[Section 3]{henk} it is proved that
\begin{equation*}
    r(\Omega_k)\geq \frac{P(\Omega_k)}{4}-\frac{d(\Omega_k)}{2}
\end{equation*}
which also yields to
\begin{equation*}
    r(\Omega_k)\geq \dfrac{ P(\Omega_k)}{4}-R(\Omega_k),
\end{equation*}
because $d(\Omega_k)\leq 2R(\Omega_k)$.

Recalling now that the Cheeger constant is continuous with respect to the Hausdorff convergence when the sets do not degenerate to a segment (see \cite[Proposition 3.1]{reverse_cheeger}), the existence part of the theorem is proved.
\end{proof}

\subsection{Proof of Theorem \ref{th2}}
The following paragraphs of this Section are dedicated to the proof of the explicit bounds and their sharpness in the Blaschke--Santaló sense.

\subsubsection{The triplet $(h,r,P)$}
\begin{proposition}\label{prop_hrP}
Let $\Omega\in \mathcal{K}^2$. Then, it holds
\begin{equation}\label{eq:hrP}
\frac{1}{r(\Omega)} +\frac{\pi}{P(\Omega)-\pi r(\Omega)} \leq h(\Omega)\leq \frac{1}{r(\Omega)}+\sqrt{\frac{2\pi }{P(\Omega)r(\Omega)}},
\end{equation}
where equalities are achieved by circumscribed sets in the upper bound and by the stadiums in the lower bound.
\end{proposition}

\begin{proof}
We combine the classical convex geometric inequalities (see \cite{bonnesen2, bonnesen, inequalities_convex})
\begin{equation}\label{pra}
    \frac{P(\Omega)r(\Omega)}{2}\leq |\Omega|\leq r(\Omega)(P(\Omega)-\pi r(\Omega)),
\end{equation}
with the estimates \eqref{eq:hra} to obtain the optimal inequalities \eqref{eq:hrP}.

 The upper bound in \eqref{eq:hra} is an equality for circumscribed sets, since both the upper bound in \eqref{eq:hra} and the lower bound in \eqref{pra} are equalities for circumscribed sets.
 The lower bound is achieved by stadiums, since both the lower bound in \eqref{eq:hra} and the upper bound in \eqref{pra} are achieved with equality sign on this class of sets.
\end{proof}

\subsubsection{The triplet $(h,d,r)$}
In the following we will denote by $\mathcal{S}_{r,d}$ the symmetrical spherical slice of inradius $r$ and diameter $d$ and we will denote by $\mathcal{N}_{r,d}$ the regular smoothed nonagon of same inradius and diameter, see Definitions \ref{def:nonagon} and \ref{def:slice}
\begin{proposition}\label{prop_hdr} Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{up_hdr}
    h(\Omega)\leq \frac{1}{r(\Omega)}+ \sqrt{\frac{\pi}{r(\Omega)\sqrt{d^2(\Omega)-4r^2(\Omega)}+r^2(\Omega)\left(\pi-2\arccos{\left(\frac{2r(\Omega)}{d(\Omega)}\right)}\right)}},
\end{equation}
where equality is achieved if and only if $\Omega$ is a symmetric two-cup body. Moreover, we have
\begin{equation}
    \label{hdr_low}
    h(\Omega) \ge      \dfrac{1}{t_{g_1^\Omega}}
\end{equation}
where $t_{g_1^\Omega}$ is the smallest solution to the equation
$$g_1^\Omega(t):=\psi(d(\Omega)-2t,r(\Omega)-t)=\pi t^2$$ on the interval $[0, r(\Omega)]$ and the function $\psi$ is defined in \eqref{func_del}. Moreover, there exists $D_0$ such that the problem $$\min\{h(\Omega)\ |\ \Omega\in\mathcal{K}^2_{d,r}\}$$ is solved by $\mathcal{N}_{r,d}$ if $d<rD_0$ and by $\mathcal{S}_{r,d}$ if $d\ge r D_0$.
\end{proposition}

\begin{proof}
Let us start by proving \eqref{up_hdr}. We just need to combine the upper bound in \eqref{eq:hra}, which is sharp on circumscribed sets (see Definition \ref{circumscribed}), and \eqref{del_2cap}, which is an equality for and only for two-cup bodies, that are particular circumscribed sets.

Let us now prove \eqref{hdr_low}.
For any fixed parameter $r>0$, we consider the function

\begin{equation*}
   \Psi_r(x) =\begin{cases}
   f_r(x):= \displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{x^2-3r^2}-r)+\frac{3x^2}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{x}\right)}\right)}, & \text{if} \, \, \,  x\le r D^*\vspace{1mm} \\
    g_r(x):=\displaystyle{r\sqrt{x^2-4r^2}+\frac{x^2}{2}\arcsin{\left(\frac{2r}{x}\right)}}, & \text{if} \, \, \, x\ge r D^*
\end{cases}
\end{equation*}
defined for $x\ge 2r$. Let us prove that the function $\Psi_r$ is strictly increasing.
We have for $x<r D^*$
$$f_r'(x) = \frac{3 \sqrt{3} r x}{2 \sqrt{x^2-3 r^2}} + 3 x \left(\frac{\pi }{3}-\arccos\left(\frac{\sqrt{3} r}{x}\right)\right)-\frac{3 \sqrt{3} r}{2 \sqrt{1-\frac{3 r^2}{x^2}}}= 3 x \left(\frac{\pi }{3}-\arccos\left(\frac{\sqrt{3} r}{x}\right)\right),$$
and, for $x>r D^* $,
$$g_r'(x) = -\frac{r}{\sqrt{1-\frac{4 r^2}{x^2}}}+\frac{r x}{\sqrt{x^2-4 r^2}}+x \arcsin\left(\frac{2 r}{x}\right)=x \arcsin\left(\frac{2 r}{x}\right)>0. $$
Thus, the function $f_r$ is increasing on $[2r,2\sqrt{3}r]$ and is decreasing on $[2\sqrt{3}r,+\infty)$ and the function $g_r$ is increasing on $[2r,+\infty)$. Moreover, we have by Theorem \ref{delthm} that $D^*\leq 2\sqrt{3}$. So, the function $f_r$ is increasing on the sub-interval $[2r,r D^*]$ and, since $f_r(r D^*)=g_r(r D^*)=\Psi_r(D^*)$, the function $\Psi_r$  is increasing on $[2r,+\infty)$.

Let $t\in [0,r(\Omega)]$, we apply the result of Theorem \ref{delthm} on the convex set $\Omega_{-t}$. Thus, we have
$$|\Omega_{-t}|\leq \Psi_{r(\Omega_{-t})}(d(\Omega_{-t})) = \Psi_{r(\Omega)-t}(d(\Omega_{-t}))\leq \Psi_{r(\Omega)-t}(d(\Omega)-2t)=:g_1^\Omega(t),$$
where we use the monotonicity of the function $\Psi_{r(\Omega)-t}$ and \eqref{inr} and \eqref{eq:diameter}.

Now, using Lemma \ref{lem:main}, we have the following bound for the Cheeger constant
$$h(\Omega)\ge \frac{1}{t_{g_1^\Omega}},$$
where $t_{g_1^\Omega}$ is the smallest solution to the equation $g_1^\Omega(t)=\pi t^2$ on the interval $[0,r(\Omega)]$.

Let $r>0$ and $d\ge 2r$.
We consider two cases:
\begin{itemize}
    \item If $d>r D^*$, we have for every $t\in[0,r)$
    \begin{equation}\label{eq:proof_hdr}
    |(\S_{r,d})_{-t}|=|\S_{r-t,d-2t}|=\Psi_{r-t}(d-2t),
    \end{equation}
    where the first equality is a consequence of the equality $(\S_{r,d})_{-t}=\S_{r-t,d-2t}$ and the second one is a consequence of \cite[Theorem 2]{delyon2} and the following estimate
    $$d((\S_{r,d})_{-t})=d-2t>rD^*-2t>rD^*-tD^*=(r-t)D^*=r((\S_{r,d})_{-t})D^*,$$
    where we used $D^*\approx 2,3888>2$ (see \cite[Theorem 2]{delyon2}). Thus, we have by \eqref{eq:proof_hdr}
    $$ h(\S_{r,d})=\frac{1}{t_{g_1^\Omega}},$$
    with
    $$r(\S_{r,d})= r,\ \ d(\S_{r,d})= d.$$
    \item If $d<r D^*$, we take $t^*:=\frac{D^*r-d}{2+D^*}$ as the value for which the graphs of the functions $t\longmapsto |(\mathcal{N}_{r,d})_{-t}|$ and $t\longmapsto |(\S_{r,d})_{-t}|$ intersect each other, see Figure \ref{fig:discuss_1}. We note that $(\mathcal{N}_{r,d})_{-t}=\mathcal{N}_{r-t,d-2t}$ for every $t\in [0,t^*]$.

    As shown in the figures below, we have the following  cases:
    \begin{itemize}
        \item If $t^*>\frac{1}{h(\mathcal{N}_{r,d})}$, we have $h(\mathcal{N}_{r,d})=\frac{1}{t_{g_1^\Omega}}$.

        \item If $t^*<\frac{1}{h(\mathcal{N}_{r,d})}$, we have $h(\S_{r,d})=\frac{1}{t_{g_1^\Omega}}$.

        \item If $t^*=\frac{1}{h(\mathcal{N}_{r,d})}$, we have $h(\mathcal{N}_{r,d})=h(\S_{r,d})=\frac{1}{t_{g_1^\Omega}}$.
    \end{itemize}
\end{itemize}

\begin{center}
\begin{tikzpicture}
\draw[->] (-.5,0) -- (7,0);
\draw (7,0) node[right] {$t$};
\draw [->] (0,-.5) -- (0,5);

\draw [domain=0:7, red,line width=0.4mm] plot(\x,{.09*(\x)^2});

\draw [domain=0:6.1, green] plot(\x,{sqrt(2.7^2-\x)-.03*\x^2});

\draw [domain=1.3:6.1, dashed, blue, line width=0.8mm] plot(\x,{sqrt(2.7^2-\x)-.03*\x^2});

\draw (4.2,0) node[below] {\tiny  $ \frac{1}{h(\mathcal{S}_{r,d})}$};
\draw (3.9,-0.1) -- (3.9,0.1);

\draw (3.1,0) node[below] {\tiny $ \frac{1}{h(\mathcal{N}_{r,d})}$};

\draw (3.25,-0.1) -- (3.25,0.1);
\draw [dotted] (3.25,0) -- (3.25,1);

\draw [dotted] (3.9,0) -- (3.9,1.3);

\draw (1.3,0) node[below] {$t^*$};
\draw (1.3,-0.1) -- (1.3,0.1);

\draw (0,4) node[left] {$|\mathcal{N}_{r,d}|$};
\draw (-0.1,4) -- (0.1,4);

\draw (0,2.7) node[left] {$|\mathcal{S}_{r,d}|$};
\draw (-0.1,2.7) -- (0.1,2.7);

\draw (6.1,0) node[below] {$r$};
\draw (6.1,-0.1) -- (6.1,0.1);

\draw [dotted] (1.3,0) -- (1.3,2.4);

\draw(0,4) .. controls (2,1.3) and (2.5,1)  .. (6.1,0);

\draw[ dashed, blue, line width=0.8mm](0,4) .. controls (1,2.7)  .. (1.3,2.4);

\draw (1.3,2.4) node {\Large $\bullet$};

\draw[ dashed, blue, line width=0.8mm](8.5,4) --  (9,4);
\draw (9,4) node[right] {Graph of the function $t\longmapsto \Psi_{r-t}(d-2t)$.};

\draw[line width=0.4mm, red](8.5,3.5) --  (9,3.5);
\draw(9,3.5) node[right] {Graph of the function $t\longmapsto \pi t^2$.};

\draw (8.5,3) --  (9,3);
\draw(9,3) node[right] {Graph of the function $t\longmapsto |(\mathcal{N}_{r,d})_{-t}|=|\mathcal{N}_{r-t,d-2t}|$.};

\draw [green] (8.5,2.5) --  (9,2.5);
\draw(9,2.5)  node[right] {Graph of the function $t\longmapsto |(\mathcal{S}_{r,d})_{-t}|=|\mathcal{S}_{r-t,d-2t}|$.};

\draw (8.2,2.2) --  (8.2,4.3);
\draw (8.2,2.2) --  (17.5,2.2);
\draw (17.5,2.2) --  (17.5,4.3);
\draw (8.2,4.3) --  (17.5,4.3);
\end{tikzpicture}

\vspace{1cm}
\begin{tikzpicture}
\draw[->] (-.5,0) -- (7,0);
\draw (7,0) node[right] {$t$};
\draw [->] (0,-.5) -- (0,5);

\draw [domain=0:7, red,line width=0.4mm] plot(\x,{.09*(\x)^2});

\draw [domain=0:6.1, green] plot(\x,{sqrt(2.7^2-\x)-.03*\x^2});

\draw [domain=4.8:6.1, dashed, blue, line width=0.8mm] plot(\x,{sqrt(2.7^2-\x)-.03*\x^2});

\draw (3.5,0) node[below] {\tiny  $ \frac{1}{h(\mathcal{S}_{r,d})}$};
\draw (3.9,-0.1) -- (3.9,0.1);

\draw (4.25,-.3) node[below] {\tiny $ \frac{1}{h(\mathcal{N}_{r,d})}$};

\draw (4.15,-0.1) -- (4.15,0.1);
\draw [dotted] (4.15,0) -- (4.15,1.5);

\draw [dotted] (3.9,0) -- (3.9,1.3);

\draw (4.8,0) node[below] {$t^*$};
\draw (4.8,-0.1) -- (4.8,0.1);

\draw (0,4) node[left] {$|\mathcal{N}_{r,d}|$};
\draw (-0.1,4) -- (0.1,4);

\draw (0,2.7) node[left] {$|\mathcal{S}_{r,d}|$};
\draw (-0.1,2.7) -- (0.1,2.7);

\draw (6.1,0) node[below] {$r$};
\draw (6.1,-0.1) -- (6.1,0.1);

\draw [dotted] (4.8,0) -- (4.8,1);

\draw(0,4) .. controls (3,3.8) and (5,0)  .. (6.1,0);

\draw[ dashed, blue, line width=0.8mm](0,4) .. controls (1.6,3.84) and (3,2.84) .. (4.8,.9);

\draw (4.8,.9) node {\Large $\bullet$};
\end{tikzpicture}
\label{fig:discuss_1}
\end{center}

This ends the proof of the existence of optimal sets.
\end{proof}

\begin{remark}
We note that the symmetrical slices and the smoothed nonagons are not the only sets solving the shape optimization problem $\min\{h(\Omega)\ |\ \Omega\in \mathcal{K}^2_{d,r}\}$. Indeed, if for example we consider  a spherical slice $\mathcal{S}$ and denote by $C_\S$ its Cheeger set, we have $h(\S)=h(C_\S)$ and by the explicit characterization of the Cheeger sets given in \cite[Theorem 1]{kawohl}, we have
$$d(C_\S)=d(\S)$$
and
$$r(C_\S)=r\left(\S_{-\frac{1}{h(\S)}}+\frac{1}{h(\S)}B_1\right)=r\left(\S_{-\frac{1}{h(\S)}}\right)+\frac{1}{h(\S)}= r(\S)-\frac{1}{h(\S)}+\frac{1}{h(\S)}=r(S),$$
meanwhile $\S\ne C_\S$, which proves the non-uniqueness.
\end{remark}

\begin{remark}
We give the following explicit lower bound.
In \cite{inequalities_convex}, it is proved that
\begin{equation*}\label{scotty}
   |\Omega|< 2 d(\Omega)r(\Omega).
\end{equation*}
By applying the strategy of Lemma \ref{lem:main}, we obtain that
\begin{equation*}
    h(\Omega) \geq \dfrac{4-\pi}{d(\Omega)+2 r(\Omega)-\sqrt{(d(\Omega)+2r(\Omega))^2-2(4-\pi)d(\Omega)r(\Omega)}}.
\end{equation*}
\end{remark}

\subsubsection{The triplet $(h,r,R)$ }
\begin{proposition}\label{prop_hRr}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{hrRlow}
    h(\Omega) \ge \frac{1}{t_{g_2^\Omega}},
\end{equation}
where $t_{g_2^\Omega}$ is the smallest solution of the equation
 $${g_2^\Omega}(t):=2\left(\left(r-t\right)\sqrt{(R(\Omega)-t)^2-(r(\Omega)-t)^2}+(R(\Omega)-t)^2\arcsin{\left(\frac{r(\Omega)-t}{R(\Omega)-t}\right)}\right)=\pi t^2.$$
 The equality in \eqref{hrRlow} is achieved if and only if $\Omega$ is a symmetrical spherical slice. Moreover, it holds
\begin{equation}
    \label{hrRup}
    h(\Omega)\le \frac{1}{r(\Omega)}+ \sqrt{\frac{\pi }{2r(\Omega)\left(\sqrt{R(\Omega)^2-r(\Omega)^2}+r(\Omega)\arcsin\left(\frac{r(\Omega)}{R(\Omega)}\right)\right)}},
\end{equation}
where the equality in \eqref{hrRup} is achieved by two-cup bodies.
\end{proposition}

\begin{proof}
In order to prove \eqref{hrRlow}, we apply the result of Lemma \ref{lem:main}. Let us introduce the function $$\varphi:(R,r)\longmapsto 2\left(r\sqrt{R^2-r^2}+R^2\arcsin{\frac{r}{R}}\right),$$ which is increasing with respect to the first variable, indeed
$$\frac{\partial\varphi}{\partial R}(R,r)=2R \arcsin\left(\frac{r}{R}\right)>0$$
By applying  \eqref{cifreRr} (where the equality holds only for symmetrical spherical slices), we have for every $t\in [0,r(\Omega)]$
$$|\Omega_{-t}|\leq \varphi(R(\Omega_{-t}),r(\Omega_{-t}))=\varphi(R(\Omega_{-t}),r(\Omega)-t)\leq \varphi(R(\Omega)-t,r(\Omega)-t)=:{g_2^\Omega}(t),$$
where the last inequality is a consequence of the monotonicity of the function $R\longmapsto \varphi(R,r)$ and the fact that  $R(\Omega_{-t})\leq R(\Omega)-t$ (see as a reference \ref{lem:diameter_inner_set}). Finally, we conclude by applying the result of Lemma \ref{lem:main}.
  In order to prove \eqref{hrRup}, we combine the upper bound in \eqref{eq:hra}
and the inequality \eqref{secondhRr}.

As far as the sharpness of \eqref{hrRup} is concerned, we observe that \eqref{eq:hra} is sharp on circumscribed sets (see Definition \ref{circumscribed}) and \eqref{secondhRr} is attained by symmetric two-cup bodies, that are also circumscribed sets.
\end{proof}

\begin{remark}
We can give the following explicit lower bound
\begin{equation*}
h(\Omega)\ge \frac{4-\pi}{2(R(\Omega)+r(\Omega))-\sqrt{4(R(\Omega)+r(\Omega))^2-4(4-\pi)R(\Omega)r(\Omega)}},
\end{equation*}
which can be obtained starting from
$$\abs{\Omega}\le 4R(\Omega)r(\Omega),$$
see \cite{henk}, and the strategy from Lemma \ref{lem:main}.
\end{remark}

\subsection{Proof of Theorem \ref{th3}}
\subsubsection{The triplet $(h,\omega,d)$}
\begin{proposition}\label{prop_hdw}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{hdwlow}
h(\Omega)\ge \frac{1}{t_{g_3^\Omega}}
\end{equation}
where $t_{g_3^\Omega}$ is the smallest solution to
$${g_3^\Omega}(t):=\frac{\omega(\Omega)-2t}{2}\sqrt{(d(\Omega)-2t)^2-(\omega(\Omega)-2t)^2}+ \frac{(d(\Omega)-2t)^2}{2}\arcsin\left(\frac{\omega(\Omega)-2t}{d(\Omega)-2t}\right)=\pi t^2.$$
The equality in \eqref{hdwlow} is achieved by symmetrical spherical slices.
Moreover,
\begin{itemize}
    \item if $\omega(\Omega)\le \frac{\sqrt{3}}{2}d(\Omega)$, it holds
\begin{equation}\label{hdwup}
   h(\Omega)\le  h(T_I),
\end{equation}
where $T_I$ is the subequilateral triangle such that $\omega(T_I)=\omega(\Omega)$ and $d(T_I)=d(\Omega)$. The equality in \eqref{hdwup} is achieved by the isosceles triangle $T_I$,
\item and if $\frac{\sqrt{3}}{2}d(\Omega)\leq\omega(\Omega) \leq d(\Omega)$, we have
\begin{equation}\label{hdwup_1}
h(\Omega)\le \frac{\sqrt{3}}{\sqrt{3}\omega(\Omega)-d(\Omega)}+\sqrt{\frac{2\pi}{\pi \omega(\Omega)^2-\sqrt{3}d(\Omega)^2+6\omega(\Omega)(\tan\left(\arccos(\frac{\omega(\Omega)}{d(\Omega)})\right)-\arccos(\frac{\omega(\Omega)}{d(\Omega)}))}}.
\end{equation}
The equality in \eqref{hdwup_1} is achieved by equilateral triangles.
\end{itemize}

\end{proposition}

\begin{proof}
Let us start by proving the lower bound \eqref{hdwlow}, by using the strategy given in Lemma \ref{lem:main}.
For every $\Omega\in\mathcal{K}^2$, it holds (see \cite{kubota} and also \cite{inequalities_convex}) \begin{equation*}
    \label{Adw_up}
    \abs{\Omega}\le \frac{\omega(\Omega)}{2}\sqrt{d^2(\Omega)-\omega^2(\Omega)}+ \frac{d^2(\Omega)}{2}\arcsin\left(\frac{\omega(\Omega)}{d(\Omega)}\right)
\end{equation*}
with equality if and only if $\Omega$ is a symmetrical spherical slice. If we denote by
$$f(d,w)=\frac{w}{2}\sqrt{d^2-w^2}+\frac{d^2}{2}\arcsin\left(\frac{w}{d}\right),$$
we have
$$\frac{\partial f}{\partial d}(d,w)= d\arcsin\left(\frac{w}{d}\right)>0,$$
$$\frac{\partial f}{\partial w}(d,w)=\sqrt{d^2-w^2}>0.$$
Thus, using Lemma \ref{lem:main}, we have
$$\abs{\Omega_{-t}}\le f(d(\Omega_{-t}), \omega(\Omega_{-t}))\le f(d(\Omega)-2t, \omega(\Omega)-2t)$$
and
$$h(\Omega)\ge \frac{1}{t_{g_3^\Omega}},$$
where $t_{g_3^\Omega}$ is the smallest solution to
$${g_3^\Omega}(t):=f(d(\Omega)-2t, \omega(\Omega)-2t)=\pi t^2.$$
In order to prove the upper bound \eqref{hdwup}, we consider the following minimization problem for the area in the class of convex sets with given diameter and width, studied in \cite{sholander} and \cite{inequalities_convex}:
\begin{enumerate}[label=(\roman*)]
\item if $2\omega(\Omega)\le \sqrt{3}d(\Omega)$, then
\begin{equation}\label{min1}
   2\abs{\Omega}\ge \omega(\Omega)d(\Omega),
\end{equation}
where equality is achieved by triangles;
\item if $\sqrt{3}d(\Omega)\le 2\omega(\Omega)\le 2 d(\Omega)$, then
\begin{equation}\label{min2}
    \displaystyle{2\abs{\Omega}\ge\pi \omega^2(\Omega)-\sqrt{3}d^2(\Omega)+6\omega^2(\Omega)\left(\tan\left(\arccos\left(\frac{\omega(\Omega)}{d(\Omega)}\right)\right)-\arccos\left(\frac{\omega(\Omega)}{d(\Omega)}\right)\right)}=|T_Y|,
\end{equation}
  where the equality is achieved by the Yamanouti set $T_Y$ such that $\omega(T_Y)=\omega(\Omega)$ and $d(T_Y)=d(\Omega)$.
\end{enumerate}

 Moreover, if we consider the minimization problem of the inradius in the class of convex set with given diameter and width, we have from Theorem \ref{cifre_wrd_thm} (see \eqref{cifre_wrd_eq_1} and \eqref{cifre_wrd_eq_2}):
 \begin{equation}\label{min3}
   r(\Omega)\geq  \begin{cases}
 r(T_I), & \text{if}\ \, 2\omega(\Omega)\le \sqrt{3}d(\Omega)\\
 r(T_Y) & \text{if}\ \, \sqrt{3}d(\Omega)\le 2\omega(\Omega)\le 2 d(\Omega).
 \end{cases}
  \end{equation}

So combining \eqref{eq:hra} with the estimates \eqref{min1}, \eqref{min2} and \eqref{min3}, we obtain \eqref{hdwlow}

 \begin{equation*}
   h(\Omega)\leq  \begin{cases}
\frac{1}{r(T_I)}+\sqrt{\frac{\pi}{|T_I|}}= h(T_I), & \text{if}\ \, 2\omega(\Omega)\le \sqrt{3}d(\Omega)\\
 \frac{1}{r(T_Y)}+\sqrt{\frac{\pi}{|T_Y|}} & \text{if}\ \, \sqrt{3}d(\Omega)\le 2\omega(\Omega)\le 2 d(\Omega).
 \end{cases}
  \end{equation*}
The explicit formula given in the inequality \eqref{hdwup_1} is obtained by using \eqref{min2} and $r(T_Y)=\omega(T_Y)-\frac{d(T_Y)}{\sqrt{3}}$, see \cite[Theorem 2]{cifre4}.
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[scale=.4]{fig_hwd.eps}
    \includegraphics[scale=.5]{fig_hwd_zoom.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,\omega,d)$.}
    \label{fig:hwd}
\end{figure}
\newpage
We are also able to give an explicit, but not sharp, lower bound of the Cheeger constant in terms of the width and the diameter.
\begin{remark}
Let $\Omega\in \mathcal{K}^2$. Then, it holds
\begin{equation}\label{lr}
    h(\Omega)>\frac{1}{\omega(\Omega)}+\frac{1}{d(\Omega)}+\sqrt{\left(\frac{1}{\omega(\Omega)}+\frac{1}{d(\Omega)}\right)^2-\frac{4-\pi}{\omega(\Omega)d(\Omega)}},
\end{equation}
where equality is asymptotically achieved by a sequence of thin collapsing rectangles.
In order to prove \eqref{lr}, it is enough to consider the inequality
$$\abs{\Omega}\le \omega(\Omega) d(\Omega)$$
and to use the strategy of Lemma \ref{lem:main}.
\end{remark}

\subsubsection{ The triplet $(h, R,\omega)$}
\begin{proposition}\label{prop_hRw}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{low_hRw}
  h(\Omega)\ge \frac{1}{t_{g_4^\Omega}},
\end{equation}
where $t_{g_4^\Omega}$ is the smallest solution of the equation
$$g_4^\Omega(t):=  \frac{(\omega(\Omega)-2t)}{2} \sqrt{4 \left(R(\Omega)-t\right)^2-\left(\omega(\Omega)-2t\right)^2} + 2(R(\Omega)-t)^2 \arcsin{\left(\frac{\omega(\Omega)-2t}{2(R(\Omega)-t)}\right)}= \pi t^2$$
on $[0,r(\Omega)]$.

The equality in \eqref{low_hRw} is achieved by symmetrical spherical slices.
Moreover, it holds
\begin{equation}\label{upper_hRw}
h(\Omega)\leq h(T_I), \qquad \text{ if  } \, \omega(\Omega)\le \frac{3}{2}R(\Omega),
\end{equation}
where $T_I$ is the subequilateral triangle such that $R(T_I)=R(\Omega)$ and $\omega(T_I)=\omega(\Omega)$. The equality in \eqref{upper_hRw} is realized by the subequilateral triangle $T_I$.
\end{proposition}

\begin{proof}
Let us start by proving the lower bound \eqref{low_hRw}, by using the strategy given in Lemma \ref{lem:main}.
Let us recall the function defined in \eqref{henkchi}: $$\chi:(\omega, R)\longmapsto \frac{\omega}{2} \sqrt{4 R^2-\omega^2} + 2R^2 \arcsin{\frac{\omega}{2R}}.$$
We have, for every $R,\omega>0$,
$$\frac{\partial \chi}{\partial R}(\omega, R) = 4R \arcsin{\frac{\omega}{2 R}}\ge 0 \ \ \ \text{and}\ \ \ \ \frac{\partial \chi}{\partial \omega}(\omega, R) = \sqrt{4R^2-\omega^2}\ge 0.$$
Thus, using Theorem \ref{cifrsalth}, we have, for every $t\in [0,r(\Omega))$,
$$|\Omega_{-t}|\leq \chi(\omega(\Omega_{-t}), R(\Omega_{-t}))\leq \chi(\omega(\Omega)-2t ,R(\Omega)-t):=g_4^\Omega(t),$$
where we use \eqref{width} and \eqref{circumradius}.
By Lemma \ref{lem:main}, we have
$$h(\Omega)\ge \frac{1}{t_{g_4^\Omega}},$$
where $t_{g_4^\Omega}$ is the smallest solution to the equation $g_4^\Omega(t)=\pi t^2$ on the interval $[0,r(\Omega)]$.

Let us now prove the upper bound \eqref{upper_hRw}.
We recall the inequality  \eqref{eq:hra}:
\begin{equation}
 \label{kf}
    h(\Omega)\le \frac{1}{r(\Omega)}+ \sqrt{\frac{\pi}{\abs{\Omega}}},
\end{equation}
where equality is achieved for instance by circumscribed sets,  in particular, by triangles.
In order to prove \eqref{upper_hRw}, we consider the following facts:
\begin{enumerate}[label=(\roman*)]
    \item In \eqref{ARw_low}, it is proved that, if $\Omega\in\mathcal{K}^2$,
    \begin{equation}\label{firsty}
        \abs{\Omega}\ge \abs{T_I} \quad \text{if} \, \, \omega(\Omega)\le \frac{3}{2}R(\Omega),
    \end{equation}
  where $T_I$ is the subequilateral triangle with given width and circumradius and
  nothing is known if $\omega(\Omega)\ge 3/2R(\Omega)$;
    \item In Theorem \ref{cifre_gomis_1_thm}, see \eqref{cifre_gomis_1_eq}, and Theorem \ref{santalo_thm_1}, see \eqref{santalo_eq_1}, it is proved that if
    \begin{equation}\label{lasty}
       r(\Omega) \geq \begin{cases}
             r(T_I) & \text{if} \, \, {\omega(\Omega)\le \frac{3}{2}R(\Omega)}\\
             \omega(\Omega)-R(\Omega) & \text{if} \, \,  {\omega(\Omega)\ge \frac{3}{2}R(\Omega)},
        \end{cases}
    \end{equation}
    where, if $\omega(\Omega)\ge \frac{3}{2}R(\Omega)$, equality is achieved by any set obtained by an equilateral triangle of circumradius $R(\Omega)$ by replacing the edges by three equal circular arcs, in particular each arc of circle is centered on the height relative to the same side.
\end{enumerate}

So, if $\omega(\Omega)\le \frac{3}{2}R(\Omega)$, combining \eqref{kf} with \eqref{firsty} and \eqref{lasty}, we get the thesis.
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[scale=.4]{fig_hw_circumradius.eps}
    \includegraphics[scale=.3]{fig_hw_circumradius_zoom.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,\omega,R)$.}
    \label{fig:hwR}
\end{figure}

In the following Remark, we give some not-sharp bounds, that are explicit.
\begin{remark}

We can prove that
\begin{equation}\label{ns}
     h(\Omega)\le \frac{3}{\omega(\Omega)}+\sqrt{\frac{\pi}{\sqrt{3}R(\Omega)\omega(\Omega)}}.
\end{equation}
We recall the following inequalities, proved in \cite{inequalities_convex}:
\begin{equation}
    \label{Arw}
    \abs{\Omega}\ge \sqrt{3}R(\Omega)\omega(\Omega)\ \ \ \text{and}\ \ \
    \omega(\Omega)\le 3 r(\Omega).
\end{equation}
By combining these estimates and the upper bound in \eqref{eq:hra}, we have
$$h(\Omega)\leq \frac{3}{\omega(\Omega)}+\sqrt{\frac{\pi}{\sqrt{3}R(\Omega)\omega(\Omega)}}.$$
Since the equality in \eqref{eq:hra} is achieved by circumscribed sets, while the  equalities in \eqref{Arw} are achieved by equilateral triangles, that are particular circumscribed sets, we have the equality in \eqref{ns} for equilateral triangles.

Moreover, another not-sharp upper bound can be obtained by using the strategy from Lemma \ref{lem:main}, starting from
$$\abs{\Omega}< 2 R(\Omega)\omega(\Omega),$$
which is asymptotically achieved by a sequence of rectangles with circumradius that goes to infinity (see \cite{henk}). We get
$\abs{\Omega_{-t}}\le 2 R(\Omega_{-t})\omega(\Omega_{-t})\le 2(R(\Omega)-t)(\omega(\Omega)-2t)$
and, consequently,
$$ h(\Omega)\ge\frac{4-\pi}{(2R(\Omega)+\omega(\Omega))-\sqrt{(2R(\Omega)+\omega(\Omega))^2-(4-\pi)(2R(\Omega)\omega(\Omega))}}.$$
\end{remark}

\subsubsection{The triplet $(h, \omega, P)$}
\begin{proposition}\label{prop_hwP}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}
    \label{hwp_low}
    h(\Omega)\ge \frac{2}{\omega(\Omega)}+\frac{2\pi}{2P(\Omega)-\pi \omega(\Omega)},
\end{equation}
where the equality is achieved by stadiums.
Moreover, if $P(\Omega)\ge 2\sqrt{3}\omega(\Omega)$, it holds
\begin{equation}
    \label{hwp_up}
    h(\Omega)\le h(T_I),
\end{equation}
where $T_I$ is the subequilateral triangle such that $P(T_I)=P(\Omega)$ and $\omega(T_I)=\omega(\Omega)$. The equality in \eqref{hwp_up} is achieved by the subequilateral triangle $T_I$.
\end{proposition}

\begin{proof}
 The inequality \eqref{hwp_low} is a consequence of \eqref{hAw_low} and the inequality
$$|\Omega|\leq \frac{\omega(\Omega)}{2}\left(P(\Omega)-\frac{\pi\omega(\Omega)}{2}\right),$$
which is an equality for stadiums (see for example \cite{inequalities_convex} and the references therein).

Let us now assume that $P(\Omega)\ge 2\sqrt{3}\omega(\Omega)$. In order to prove \eqref{hwp_up}, we recall the inequality \eqref{cifre_salinas_rwP}

$$    P(\Omega)\le \sqrt{\frac{4 r^2(\Omega)\omega^3(\Omega)}{(\omega(\Omega)-2r(\Omega))^2(4r(\Omega)-\omega(\Omega))}}=:f_{\omega(\Omega)}(r(\Omega))$$
By direct computations, we prove that the continuous function
$$f_{\omega(\Omega)}: r\longmapsto \sqrt{\frac{4 r^2 \omega(\Omega)^3}{(\omega(\Omega)-2r)^2(4r-\omega(\Omega))}}$$
is strictly increasing on $\left[\frac{\omega(\Omega)}{3}, \frac{\omega(\Omega)}{2}\right)$. Let us denote by $g_{\omega(\Omega)}$ the inverse function of $f_{\omega(\Omega)}$, which is also continuous and strictly increasing. We have
$$r(\Omega)\ge g_{\omega(\Omega)}(P(\Omega))=r(T_I),$$
where $T_I$ is any subequilateral triangle such that $\omega(T_I) = \omega(\Omega)$ and $P(T_I)=P(\Omega)$. Moreover, since $P(\Omega)\ge 2\sqrt{3}\omega(\Omega)$, we have by the results of \cite{yamanouti}
$$|\Omega|\ge |T_I|,$$
see also \cite{inequalities_convex}.
Finally, we have $$h(\Omega)\leq \frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}}\leq \frac{1}{r(T_I)}+\sqrt{\frac{\pi}{|T_I|}}=h(T_I).$$
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[scale=.4]{fig_hpw.eps}
    \includegraphics[scale=.5]{fig_hpw_zoom.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,\omega,P)$.}
    \label{fig:hpw}
\end{figure}

\subsubsection{The triplet $(h, |\cdot| , \omega)$}
\begin{proposition}\label{prop_hAw}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{hAw_low}
    h(\Omega)\geq \frac{2}{\omega(\Omega)}+\frac{\pi \omega(\Omega)}{2 \abs{\Omega} },
\end{equation}
where equality is achieved by stadiums. Moreover, it holds
\begin{equation}\label{hAw_up}
    h(\Omega)\leq h(T_I),
\end{equation}
where $T_I$ is a subequilateral triangle such that $|\Omega|=|T_I|$ and $\omega(\Omega)=\omega(T_I)$. The equality in \eqref{hAw_up} is achieved by the subequilateral triangle $T_I$.
\end{proposition}

\begin{proof}
Let us prove the inequality \eqref{hAw_low}. We recall the lower bound in \eqref{eq:hra}
$$h(\Omega)\ge \frac{1}{r(\Omega)}+\frac{\pi r(\Omega)}{2|\Omega|},$$
which is an equality if and only if $\Omega$ is a stadium. Since the function $r\longmapsto  \frac{1}{r}+\frac{\pi r}{2|\Omega|}$ is strictly decreasing and $r(\Omega)\leq \frac{\omega(\Omega)}{2}$ (where equality holds for stadiums), we obtain \eqref{hAw_low}.

Let us now prove inequality \eqref{hAw_up}. We recall the inequality \eqref{cifre_salinas_rwA}
$$    |\Omega|\le \sqrt{\frac{ r^4(\Omega)\omega^3(\Omega)}{(\omega(\Omega)-2r(\Omega))^2(4r(\Omega)-\omega(\Omega))}}=:f_{\omega(\Omega)}(r(\Omega))$$
By direct computations, we prove that the continuous function
$$f_{\omega(\Omega)}: r\longmapsto \sqrt{\frac{r^4 \omega(\Omega)^3}{(\omega(\Omega)-2r)^2(4r-\omega(\Omega))}}$$
is strictly increasing on $\left[\frac{\omega(\Omega)}{3}, \frac{\omega(\Omega)}{2}\right)$. Let us denote by $g_{\omega(\Omega)}$ the inverse function of $f_{\omega(\Omega)}$, which is also continuous and strictly increasing. We have
$$r(\Omega)\ge g_{\omega(\Omega)}(|\Omega|)=r(T_I),$$
where $T_I$ is any subequilateral triangle such that $\omega(T_I) = \omega(\Omega)$ and $|T_I|=|\Omega|$. Thus, we have
$$h(\Omega)\leq \frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}}\leq\frac{1}{r(T_I)}+\sqrt{\frac{\pi}{|T_I|}}=h(T_I),$$with equality if and only if $\Omega=T_I$.
\end{proof}

\begin{figure}[h]
    \centering
   \includegraphics[scale=.5]{fig_hwa.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,\omega,|\cdot|)$.}
    \label{fig:hwa}
\end{figure}

\begin{remark}\label{rk:hAw}
One may use classical convex geometry inequalities to obtain simpler bounds than the implicit one given in \eqref{hAw_up}. Indeed, if we combine the inequalities in \eqref{eq:hra} and the following classical ones
\begin{equation*}\label{eq:rw}
    \frac{2}{\omega(\Omega)}\leq \frac{1}{r(\Omega)}\leq \frac{2}{\omega(\Omega)}+ \frac{\omega(\Omega)}{\sqrt{3}|\Omega|}
\end{equation*}
where the lower bound is realized in particular by stadiums, and the upper bound by equilateral triangles (see for example \cite{inequalities_convex} and the references therein), we obtain the following inequalities
\begin{equation}\label{hAw_up1}
h(\Omega)\leq \frac{2}{\omega(\Omega)}+ \frac{\omega(\Omega)}{\sqrt{3}|\Omega|}+\sqrt{\frac{\pi}{|\Omega|}}
\end{equation}
and
\begin{equation}\label{hAw_up2}
    h(\Omega)\leq \frac{2}{\omega(\Omega)-\frac{\omega(\Omega)^3}{4|\Omega|}}+\sqrt{\frac{\pi}{|\Omega|}}.
\end{equation}
The bound \eqref{hAw_up1} is achieved by equilateral triangles and \eqref{hAw_up2} is asymptotically achieved by a sequence of thin subequilateral triangles.
\end{remark}

\subsection{Proof of Theorem \ref{th4}}
\subsubsection{The triplet $(h,  d,R)$}
\begin{proposition}\label{prop_hRd}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{hRdup}
    h(\Omega)\le \frac{2R(\Omega) \left(2R(\Omega)+ \sqrt{4R(\Omega)^2-d(\Omega)^2}\right)}{d(\Omega)^2\sqrt{4R(\Omega)^2-d(\Omega)^2}}+ \sqrt{\frac{4\pi R(\Omega)^2}{d(\Omega)^3 \sqrt{4R(\Omega)^2-d(\Omega)^2}}},
\end{equation}
where the equality is achieved by subequilateral triangles.
\end{proposition}
\begin{proof}
The inequality \eqref{hRdup} is obtained by combining the upper bound in \eqref{eq:hra},
which is an equality for circumscribed sets (in particular subequilateral triangles), and
$$
    r(\Omega)\ge \frac{d(\Omega)^2\sqrt{4R(\Omega)^2-d(\Omega)^2}}{2R(\Omega)\left(2R(\Omega)+\sqrt{4R(\Omega)^2-d(\Omega)^2}\right)}\ \ \ \text{and}\ \ \     |\Omega|\ge \frac{d(\Omega)^3\sqrt{4R(\Omega)^2-d(\Omega)^2}}{4R(\Omega)^2},
$$
respectively proved in \cite{santalo} and \cite{cifre2}, where the equalities hold only for subequilateral triangles.
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[scale=.44]{fig_hdR.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,R,d)$.}
    \label{fig:hdR}
\end{figure}
\newpage
\subsubsection{The triplet $(h, \omega, r)$}
\begin{proposition}\label{prop_hwr}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{ineq:hwr_lower}
h(\Omega)\ge \frac{1}{r(\Omega)}+\frac{1}{r(\Omega)}\sqrt{\pi\left(1-\frac{2r(\Omega)}{\omega(\Omega)}\right)\sqrt{\frac{4r(\Omega)}{\omega(\Omega)}-1}},    \end{equation}
where the equality is achieved by subequilateral triangles.
\end{proposition}

\begin{proof}
The proof of \eqref{ineq:hwr_lower} is inspired by the proof of \cite[Theorem 5]{cifre_salinas}. It is known that the incircle of a set $\Omega$ meets the boundary of $\Omega$ either in two diametrically opposite points, or in three points that form the vertices of a triangle, see \cite{bonnesen}. In the first case, we have $\omega(\Omega)=2r(\Omega)$, thus the inequality \eqref{ineq:hwr_lower} is equivalent to $h(\Omega)\ge \frac{1}{r(\Omega)}$, which is proved in Proposition \ref{two}. In the second case, let us denote by $T$ a triangle formed by three lines of support common to $\Omega$ and the incircle.
We have $r(\Omega)=r(T)$ and, by $\Omega \subset T$ and the monotonicity with respect to the inclusion, we get
\begin{equation}\label{h}
   h(\Omega)\ge h(T)
\end{equation}
and
\begin{equation}\label{omega}
\omega(\Omega)\leq \omega(T).
\end{equation}
So, inequality \eqref{ineq:hwr_lower} is equivalent to the following one
$$
\frac{1}{r(\Omega)h(\Omega)}f\left(\frac{r(\Omega)}{\omega(\Omega)}\right)\leq 1,
$$
where $f:x\in\left[\frac{1}{3},\frac{1}{2}\right]\longmapsto 1+\sqrt{\pi\left(1-2x\right)\sqrt{4x-1}}$. We observe that the function $g:x\in\left[\frac{1}{3},\frac{1}{2}\right]\longmapsto \left(1-2x\right)\sqrt{4x-1}$ is decreasing. Indeed,
$$g'(x) = \frac{4(1-3x)}{\sqrt{4x-1}}\leq 0.$$
Thus, $f$ is also decreasing on $\left[\frac{1}{3},\frac{1}{2}\right]$. Then, since $\frac{r(\Omega)}{\omega(\Omega)}\ge \frac{r(T)}{\omega(T)}$ by \eqref{omega}, we have
$$f\left(\frac{r(\Omega)}{\omega(\Omega)}\right) \leq  f\left(\frac{r(T)}{\omega(T)}\right).$$
Moreover, we have, by \eqref{h}, $$\frac{1}{r(\Omega)h(\Omega)}\leq \frac{1}{r(T)h(T)}.$$
Thus, we have
$$\frac{1}{r(\Omega)h(\Omega)}f\left(\frac{r(\Omega)}{\omega(\Omega)}\right) \leq \frac{1}{r(T)h(T)}f\left(\frac{r(T)}{\omega(T)}\right)=\frac{1}{1+r(T)\sqrt{\frac{\pi}{|T|}}}f\left(\frac{r(T)}{\omega(T)}\right),$$
where we used the equality $h(T) = \frac{1}{r(T)}+\sqrt{\frac{\pi}{|T|}}$, which holds because $T$ is a triangle, see \cite{ftJMAA}.
Now, we use the inequality $$|T|\leq \frac{r(T)^2}{\left(1-\frac{2r(T)}{\omega(T)}\right)\sqrt{\frac{4r(T)}{\omega(T)}-1}},$$
which is an equality if and only if $T$ is a subequilateral triangle, see \cite[Theorem 5]{cifre_salinas}. We then have
$$\frac{1}{r(\Omega)h(\Omega)}f\left(\frac{r(\Omega)}{\omega(\Omega)}\right) \leq\frac{1}{1+r(T)\sqrt{\frac{\pi}{|T|}}}f\left(\frac{r(T)}{\omega(T)}\right)\leq 1,$$
which ends the proof.
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[scale=.40]{fig_hw_inradius.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,\omega,r)$.}
    \label{fig:hwr}
\end{figure}

\begin{remark}It is possible to give a not sharp upper bound
\begin{equation*}\label{ineq:hwr_upper}
h(\Omega) \leq \frac{1}{r(\Omega)}+ \frac{\sqrt{\pi\sqrt{3}}}{\omega(\Omega)}
\end{equation*}
 In order to prove it, we combine the upper bound in \eqref{eq:hra} and $\omega(\Omega)^2\leq \sqrt{3}|\Omega|$, see \cite{inequalities_convex}.
\end{remark}

\subsection{Some not-sharp bounds}
The following four diagrams remain unsolved. Nevertheless, we provide some partial results.

\subsubsection{The triplet $(h,R,|\cdot|)$}
\begin{proposition}\label{prop_hAR}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{RA_low}
    h(\Omega)\ge \frac{1}{2R(\Omega)}+\frac{\pi R(\Omega)}{2|\Omega|}+\sqrt{\frac{\pi}{|\Omega|}},
\end{equation}
where equality is achieved by balls. Moreover, it holds
\begin{equation}
    \label{RA_up}
    h(\Omega)<\frac{1}{R(\Omega)}+\frac{4R(\Omega)}{|\Omega|},
\end{equation}
where equality is achieved by a sequence of thinning stadiums.
\end{proposition}
\begin{proof}
Let us start by proving \eqref{RA_low}.
Combining the lower bound in \eqref{eq:hpa} and the inequality (see \cite{inequalities_convex})$$|\Omega|\leq R(\Omega)(P(\Omega)-\pi R(\Omega)),$$ we obtain the claimed result.

As far as \eqref{RA_up}, we have
$$h(\Omega) \leq \frac{P(\Omega)}{|\Omega|}<\frac{\frac{|\Omega|}{R(\Omega)}+4R(\Omega)}{|\Omega|}=\frac{1}{R(\Omega)}+\frac{4R(\Omega)}{|\Omega|},$$
where we use the inequality (see as a reference \cite{inequalities_convex})
\begin{equation}\label{ARP_scott}
    |\Omega|>R(\Omega)(P(\Omega)-4R(\Omega)).
\end{equation}
The equality in \eqref{ARP_scott}, and so the equality in \eqref{RA_up}, are asymptotically  achieved by thinning stadiums.
\end{proof}

\begin{figure}[h]
    \centering
    \includegraphics[scale=.42]{fig_hRA.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,R,|\cdot|)$.}
    \label{fig:hRA}
\end{figure}

\subsubsection{The triplet $(h, P,R)$}
\begin{proposition}\label{prop_hRP}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{PR_up}
h(\Omega)<\frac{P(\Omega)}{R(\Omega)(P(\Omega)-4R(\Omega))},
\end{equation}
where equality is achieved by thinning stadiums. Moreover, it holds
\begin{equation}
    \label{PR_low}
    h(\Omega) \ge \frac{4\arcsinc{\left(\frac{4R(\Omega)}{P(\Omega)}\right)}}{P(\Omega)-4R(\Omega)\cos{\left(\arcsinc{\left(\frac{4R(\Omega)}{P(\Omega)}\right)}\right)}} + \sqrt{\frac{8\pi \arcsinc{\left(\frac{4R(\Omega)}{P(\Omega)}\right)}}{P(\Omega)\left(P(\Omega)-4R(\Omega)\cos{\left(\arcsinc{\left(\frac{4R(\Omega)}{P(\Omega)}\right)}\right)}\right)}},
\end{equation}
where equality is achieved by balls.
\end{proposition}

\begin{proof}
The inequality \eqref{PR_low} is a direct consequence of the inequalities $h(\Omega)\leq \frac{P(\Omega)}{|\Omega|}$ and $|\Omega|>R(\Omega)(P(\Omega)-4R(\Omega))$.\\
As far as \eqref{PR_up}, we combine the lower bound in \eqref{eq:hpa} and inequality \eqref{santalo_APR}.
The equality in \eqref{ARP_scott}, and so the equality in \eqref{PR_up}, is  asymptotically achieved by  thinning stadiums, while the equality in the upper bound in \eqref{eq:hpa} is achieved by circumscribed set, the equality in \eqref{santalo_APR} is achieved by symmetrical lenses, so, the equality in \eqref{PR_up} is achieved by balls.
\end{proof}

\begin{remark}
We note that one can use the inequality (see \cite{inequalities_convex})
$$|\Omega|\leq 2R(\Omega)(P(\Omega)-2R(\Omega))$$
and the Lemma \ref{lem:main} to obtain the following explicit bound:
$$h(\Omega)\ge \frac{P(\Omega)+2(\pi-2)R(\Omega)+\sqrt{P(\Omega)^2-2\pi P(\Omega)R(\Omega)+4\pi(\pi-1)R(\Omega)^2}}{2R(\Omega)(P(\Omega)-2R(\Omega))}.$$
\end{remark}
\begin{figure}[h]
    \centering
    \includegraphics[scale=.38]{fig_hP_circumradius.eps}
   \includegraphics[scale=.5]{fig_hP_circumradius_zoom.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,P,R)$.}
    \label{fig:hp_circumradius}
\end{figure}

\pagebreak
\subsubsection{The triplet $(h, P, d)$}
\begin{proposition}\label{prop_hPd}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
\begin{equation}\label{ineq:hpd_lower}
    h(\Omega) \ge \frac{4 \arcsinc{\left(\frac{2d(\Omega)}{P(\Omega)}\right)}}{P(\Omega)-2d(\Omega)\cos{\left(\arcsinc{\left(\frac{2d(\Omega)}{P(\Omega)}\right)}\right)}} + \sqrt{\frac{8\pi\arcsinc{\left(\frac{2d(\Omega)}{P(\Omega)}\right)}}{P(\Omega)(P(\Omega)-2d(\Omega)\cos{\left(\arcsinc{\left(\frac{2d(\Omega)}{P(\Omega)}\right)}\right)}}},
\end{equation}
where equality is achieved for balls. Moreover, if $P(\Omega) \in (2d(\Omega),3d(\Omega)]$, it holds
\begin{equation}\label{ineq:hpd_up1}
    h(\Omega) < \frac{4}{P(\Omega)-2d(\Omega)} + \sqrt{\frac{4\pi}{(P(\Omega)-2d(\Omega))\sqrt{P(\Omega)(4d(\Omega)-P(\Omega))}}},
\end{equation}
and, if $P(\Omega) \in [3d(\Omega),\pi d(\Omega)]$,
\begin{equation}\label{ineq:hpd_up2}
    h(\Omega) < \frac{4}{P(\Omega)-2d(\Omega)} + \sqrt{\frac{4\pi}{\sqrt{3} d(\Omega)(P(\Omega)-2d(\Omega))}}.
\end{equation}
\end{proposition}
\begin{proof}
The inequality \eqref{ineq:hpd_lower} is obtained by combining the lower bound in \eqref{eq:hpa}
and the following inequality (see \cite{inequalities_convex})
$$|\Omega|<\frac{P(\Omega)\left(P(\Omega)-2d(\Omega)\cos{\arcsinc{\left(\frac{2d(\Omega)}{P(\Omega)}\right)}}\right)}{8\arcsinc{\left(\frac{2d(\Omega)}{P(\Omega)}\right)}}.$$
The upper bounds \eqref{ineq:hpd_up1} and \eqref{ineq:hpd_up2} are obtained by combining the upper bound in \eqref{eq:hra} and the inequalities (see \cite{inequalities_convex})
$$ P(\Omega)<2d(\Omega)+4r(\Omega)$$
and
 \begin{equation*}\label{min4}
   4|\Omega|\geq  \begin{cases}
 (P(\Omega)-2d(\Omega))\sqrt{4P(\Omega)(d(\Omega)-P(\Omega))}, & \text{if}\ \, P(\Omega) \in (2d(\Omega),3d(\Omega)]\\
 \sqrt{3} d(\Omega)(P(\Omega)-2d(\Omega)) & \text{if}\ \, P(\Omega) \in (3d(\Omega),\pi d(\Omega)].
 \end{cases}
  \end{equation*}
\end{proof}

\begin{figure}[h!]
    \centering
    \includegraphics[scale=.31]{fig_hPd.eps}
        \includegraphics[scale=.35]{fih_hPd_zoom1.eps}
    \includegraphics[scale=.35]{fig_hPd_zoom2.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,P,d)$.}
    \label{fig:hpd}
\end{figure}

\newpage

\subsubsection{The triplet $(h,  |\cdot|, d)$}
\begin{proposition}\label{prop_hdA}
Let $\Omega\in\mathcal{K}^2$. Then, it holds
$$ h(\Omega)> \frac{d(\Omega)}{|\Omega|}+\sqrt{\frac{\pi}{|\Omega|}},$$
where the equality is asymptotically achieved by thin vanishing symmetrical two-cup bodies. Moreover, it holds
$$h(\Omega) \leq  \frac{4}{d(\Omega)}+\frac{2d(\Omega)}{|\Omega|},$$
and
$$h(\Omega) < \frac{2 d(\Omega)}{|\Omega|}+\sqrt{\frac{\pi}{|\Omega|}},$$
where, in both cases, the equality is asymptotically achieved by  thin vanishing rectangles.
\end{proposition}
\begin{proof}
We have $$h(\Omega) \ge \frac{P(\Omega)}{2|\Omega|}+\sqrt{\frac{\pi}{|\Omega|}} > \frac{2 d(\Omega)}{2|\Omega|}+ \sqrt{\frac{\pi}{|\Omega|}}, $$
where we use the lower bound in \eqref{eq:hpa} and $P(\Omega)>2 d(\Omega)$.

Moreover, we have
$$h(\Omega) \leq \frac{P(\Omega)}{|\Omega|} \leq  \frac{4}{d(\Omega)}+\frac{2d(\Omega)}{|\Omega|},$$
where we use the inequality $P(\Omega)\leq \frac{4|\Omega|}{d(\Omega)}+ 2d(\Omega)$ (see \cite{inequalities_convex}).
On the other hand, we have
$$h(\Omega) \leq  \frac{1}{r(\Omega)}+\sqrt{\frac{\pi}{|\Omega|}}<\frac{2 d(\Omega)}{|\Omega|}+ \sqrt{\frac{\pi}{|\Omega|}},$$
where we have use the upper bound in \eqref{eq:hra} and the inequality $|\Omega|<2d(\Omega) r(\Omega)$.
\end{proof}

\begin{figure}[h]
    \centering
  \includegraphics[scale=.44]{fig_hdA.eps}
   \includegraphics[scale=.44]{fig_hdA_zoom.eps}
    \caption{Blaschke--Santal\'o diagram of the triplet $(h,d,|\cdot|)$.}
    \label{fig:hdA}
\end{figure}
\newpage
\section{Conclusions and conjectures}\label{secult}
In this last Section we collect all the conjectures that we can deduce from the numerical approximation of the Blaschke--Santaló diagrams that we have plotted.
\begin{conjecture}
As far as the diagram $(h, \omega, d)$ is concerned, we conjecture that, if $\sqrt{3}/2 d(\Omega)\le\omega(\Omega)\le d(\Omega)$, then for all $\Omega\in \mathcal{K}^2$
$$h(\Omega)\le h(Y)$$
where $Y$ is the Yamanouti set with $\omega(Y)=\omega(\Omega) $ and $d(Y)=d(\Omega)$ (see Figure \ref{fig:hwd}).
\end{conjecture}

\begin{conjecture} As far as the diagram $(h,  R, \omega)$ is concerned, we conjecture that, if $\omega(\Omega)\ge \frac{3}{2}R(\Omega)$, then, for all $\Omega\in \mathcal{K}^2$,
\begin{equation*}
    h(\Omega)\leq h(\Bar{T}),
\end{equation*}
where $\Bar{T}$ is a set obtained by an equilateral triangle of circumradius $R(\Omega)$ by replacing the edges by three equal circular arcs, as described in Theorem \ref{santalo_thm_1} (see Figure \eqref{fig:hwR}).
\end{conjecture}

\begin{conjecture} As far as the diagram $(h, \omega, P)$ is concerned, we conjecture that, if $\pi\omega(\Omega)\le P(\Omega)\le 2\sqrt{3}\omega(\Omega)$, then, for all $\Omega\in \mathcal{K}^2$,
\begin{equation*}
    h(\Omega)\leq h(Y),
\end{equation*}
where $Y$ is the Yamanouti set with $P(Y)=P(\Omega)$ and $\omega(Y)=\omega(\Omega)$. (see Figure \eqref{fig:hpw}).
\end{conjecture}

\begin{conjecture} As far as the diagram $(h, d, R)$ is concerned, we conjecture that
\begin{equation*}
    h(\Omega)\geq h(N),
\end{equation*}
where $N$ is the set described as follows (see \cite{cifre2}).
Let $\Gamma$ and $\gamma$ be the circumcircle and the incircle of a constant width set: they are concentric and $d=\omega= R + r$. The extremal set can be constructed in the following way: an equilateral triangle $PQR$ is inscribed in the circle $\Gamma$, and now we take the circular arcs of radius $R+r$ drawn about the three vertex points. These arcs touch $\gamma$ at the opposite points $\Bar{P}, \Bar{Q}, \Bar{R}$ of $P, Q, R$, respectively. Furthermore, we construct three circles of radius $(R + r)/2$ that have the sides of the triangle as chords and whose centers lie inside the triangle. The required constant width set has 3-fold symmetry and it is formed by nine arcs
of the six constructed circles (see Figure \ref{fig:hdR}).
\end{conjecture}

\begin{conjecture}
As far as the diagram $(h,|\cdot|, R)$ is concerned, we conjecture that
$$h(\Omega)\le h(\mathcal{S})$$
and
$$h(\Omega)\ge h(\mathcal{C}),$$
where $\mathcal{S}$ is the symmetrical spherical slice with $\abs{\mathcal{S}}=\abs{\Omega}$ and $R(\mathcal{S})=R(\Omega)$ and $\mathcal{C}$ is the two-cup body with $\abs{\mathcal{C}}=\abs{\Omega}$ and $R(\mathcal{C})=R(\Omega)$ (see Figure \ref{fig:hRA}).
\end{conjecture}

\begin{conjecture}
As far as the diagram $(h,  R, P)$ is concerned, we conjecture that
$$h(\Omega)\le h(\mathcal{S})$$
and
$$h(\Omega)\ge h(\mathcal{Q}),$$
where here $\mathcal{S}$ is the symmetrical spherical slice with $R(\mathcal{S})=R(\Omega)$ and $P(\mathcal{S})=P(\Omega)$ and $\mathcal{Q}$ the quasi-lens with $R(\mathcal{Q})=R(\Omega)$ and $P(\mathcal{Q})=P(\Omega)$ , which is defined as the convex hull of the Cheeger set of a symmetrical lens and its two diametrical points. (see Figure \ref{fig:hp_circumradius}).
\end{conjecture}

\begin{conjecture}
As far as the diagram $(h, P, d)$ is concerned, we conjecture that
$$h(\Omega)\leq h(T_I) \quad \text{if }\ \  P(\Omega)\le 3d(\Omega)$$
and
$$h(\Omega)\ge h(\mathcal{Q}),$$
where $T_I$ is the subequilateral triangle with $R(T_I)=R(\Omega)$ and $P(T_I)=P(\Omega)$ and $Q$ is the quasi-lens set with $R(Q)=R(\Omega)$ and $P(Q)=P(\Omega)$ (see Figure \ref{fig:hpd}).
\end{conjecture}

\begin{conjecture}
As far as the diagram $(h, |\cdot|,d )$ is concerned, we conjecture that

$$h(\Omega)\ge h(\mathcal{C}),$$
and
$$h(\Omega)\le \begin{cases}
h(\mathcal{S}), \,  \text{if } d>\frac{2}{\sqrt[4]{3}}\sqrt{\abs{\Omega}}\\
h(\mathcal{N}), \, \text{if } d(\Omega) < \frac{2}{\sqrt[4]{3}}\sqrt{\abs{\Omega}},
\end{cases}$$
 where $\mathcal{C}$ is the two-cup body with $|\mathcal{C}|=|\Omega|$ and $d(\mathcal{C})=d(\Omega)$ ,  $\mathcal{S}$ the spherical slice with $|\mathcal{S}|=|\Omega|$ and $d(\mathcal{S})=d(\Omega)$ and $\mathcal{N}$ the smoothed nonagon with $|\mathcal{N}|=|\Omega|$ and $d(\mathcal{N})=d(\Omega)$ (see Figure \ref{fig:hdA}).
\end{conjecture}
\newpage

\section{Appendix: Summary tables with the results}
In this first table we summarize the results relative to the diagrams that are completely solved.
\begin{figure}[!h] \label{table1}
    \centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Param. & Condition & Inequality  & Extremal sets & Ref. \tabularnewline
\hline
$h,P,A$ &  & $h\leq \frac{P}{A}$ & Cheeger to itself & \cite{ftouhi_cheeger}  \tabularnewline
 &  & $h\ge \frac{P+\sqrt{4\pi A}}{2A}$ &  circumscribed sets &  \tabularnewline \hline
$h,r,A$ &  & $h\leq \frac{1}{r}+\sqrt{\frac{\pi}{A}}$ &  circumscribed sets & \cite{ftJMAA}  \tabularnewline
&  & $h\ge \frac{1}{r}+\frac{\pi r}{A}$ &  stadiums &  \tabularnewline\hline
$h,r, P$ &  & ${h\leq \frac{1}{r}+\sqrt{\frac{2\pi }{P r}}}$  & circumscribed sets & Prop. \ref{prop_hrP}  \tabularnewline
 &  & ${h\ge \frac{1}{r}+\frac{\pi }{P -\pi r}}$ &  stadiums &  \tabularnewline

\hline
$h,d,r$  &  & ${h\leq \frac{1}{r}+ \sqrt{\frac{\pi}{r\sqrt{d^2-4r^2}+r^2(\pi-2\arccos{\left(\frac{2r}{d}\right)})}}}$ & two-cup bodies & Prop. \ref{prop_hdr} \tabularnewline
 &  & ${h\ge \frac{1}{t_{g_1^\Omega}} }$ \textcolor{blue}{$(i)$} & spherical slices/smoothed nonagons &
 \tabularnewline
\hline
$h, r, R$ &  & $h\le \frac{1}{r}+ \sqrt{\frac{\pi }{2r\left(\sqrt{R^2-r^2}+r\arcsin\left(\frac{r}{R}\right)\right)}}$ &  two-cup bodies & Prop. \ref{prop_hRr}  \tabularnewline
 &  & $h\ge\frac{1}{t_{g_2^\Omega}}$ \textcolor{blue}{$(ii)$} &  spherical slices &  \tabularnewline
\hline
\end{tabular}

    \label{fig:polygons}
\end{figure}
\begin{footnotesize}
\newline
\textcolor{blue}{$(i)$} $t_{g_1^\Omega}$ is the smallest solution on $[0, r(\Omega)]$ to $$g_1^\Omega(t):=\psi(d(\Omega)-2t,r(\Omega)-t)=\pi t^2, $$ where
\begin{equation*}
   \psi(d,r):=\begin{cases}
    \displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{d^2-3r^2}-r)+\frac{3d^2}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{d}\right)}\right)}, & \text{if} \, \, \,  d\le r D^*\vspace{1mm} \\
    \displaystyle{r\sqrt{d^2-4r^2}+\frac{d^2}{2}\arcsin{\left(\frac{2r}{d}\right)}}, & \text{if} \, \, \, d\ge r D^*.
\end{cases}
\end{equation*}
and $D^*$ is the unique number in $[2,2\sqrt{3}]$ for which the two espression of the function $\psi(d,r)$ are equal.
\newline
 \textcolor{blue}{$(ii)$}  $t_{g_2^\Omega}$ is the smallest solution on $[0, r(\Omega)]$ to  $${g_2^\Omega}(t):=2\left((r-t)\sqrt{(R(\Omega)-t)^2-(r(\Omega)-t)^2}+(R(\Omega)-t)^2\arcsin{\left(\frac{r(\Omega)-t}{R(\Omega)-t}\right)}\right)=\pi t^2.$$
\end{footnotesize}

\newpage

In this second table we summarize the results of the partially solved Blaschke--Santal\'o diagrams.

\begin{figure}[!h] \label{table3}
    \centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Param. & Condition & Inequality  & Extremal sets & Ref. \tabularnewline
\hline
$h, \omega, d$ & $\omega\le \sqrt{3}/2d$ & $h\le \frac{1}{r(\omega, d)}+\sqrt{\frac{2\pi}{\omega d}} $ \textcolor{blue}{$(iii)$} &  subequilateral triangles & Prop. \ref{prop_hdw}     \tabularnewline
& $\sqrt{3}/2d\le \omega\le d$ & $  h\le \frac{\sqrt{3}}{\sqrt{3}\omega-d}+\sqrt{\frac{2\pi}{\pi \omega^2-\sqrt{3}d^2+6\omega(\tan\left(\arccos(\frac{\omega}{d})\right)-\arccos(\frac{\omega}{d}))}}$  & equilateral triangles  &  \tabularnewline
 &  & $h(\Omega)\ge \frac{1}{t_{g_3^\Omega}}$ \textcolor{blue}{$(iv)$}& spherical slices & \tabularnewline
\hline
$h,  R, \omega$ & $\omega\le 3/2 R$ & $h\leq \frac{1}{r(\omega, R)}+\sqrt{\frac{\pi}{A(\omega, R)}}$ \textcolor{blue}{$(v)$} & subequilateral triangles & Prop. \ref{prop_hRw}  \tabularnewline
 &  & $h\ge\frac{1}{t_{g_4^\Omega}}$ \textcolor{blue}{$(vi)$} & spherical slices &  \tabularnewline
\hline
$h, \omega, P$ &  $P\geq 2\sqrt{3}\omega$ & $h\le \frac{1}{r(\omega, P)}+\sqrt{\frac{\pi}{A(\omega, P)}} $ \textcolor{blue}{$(vii)$} &  subequilateral triangles & Prop. \ref{prop_hwP}  \tabularnewline
 &  & $  h\ge \frac{2}{\omega}+\frac{2\pi}{2P-\pi \omega}$ &  stadiums &  \tabularnewline
\hline
$h, A, \omega$ &  & $ h\leq \frac{1}{r(\omega, A)}+ \sqrt{\frac{\pi}{A}}$ \textcolor{blue}{$(viii)$} &  subequilateral triangles & Prop. \ref{prop_hAw}  \tabularnewline
 &  & $ h\geq \frac{2}{\omega}+\frac{\pi \omega}{2 A }$ &  stadiums &  \tabularnewline
\hline
$h,  d, R$ &  & $h\leq \frac{2R(2R+\sqrt{4R^2-d^2})}{d^2\sqrt{4R^2-d^2}}+\sqrt{\frac{4\pi R^2}{d^3\sqrt{4R^2-d^2}}}$ &  subequilateral triangles & Prop. \ref{prop_hRd}  \tabularnewline
\hline
$h,\omega,r$ &  & $h\geq \frac{1}{r}+\frac{1}{r}\sqrt{\pi\left(1-\frac{2r}{\omega}\right)\sqrt{\frac{4r}{\omega}-1}}$ &  subequilateral triangles & Prop. \ref{prop_hwr}
\tabularnewline
\hline

\end{tabular}

    \label{fig:polygons1}
\end{figure}
\begin{footnotesize}
 \textcolor{blue}{$(iii)$} $r(\omega,d)$ is given by
   \begin{equation*}
        d^2\left(\omega-2r(\omega,d)\right)^2(4r(\omega,d)-\omega)= 4r^4(\omega,d)\;\omega.
    \end{equation*}
 \newline
\textcolor{blue}{$(iv)$} $t_{g_3^\Omega}$ is the smallest solution to
$${g_3^\Omega}(t):=f(d(\Omega)-2t, \omega(\Omega)-2t)=\pi t^2, $$
where
$$f(d,w)=\frac{w}{2}\sqrt{d^2-w^2}+\frac{d^2}{2}\arcsin\left(\frac{w}{d}\right)$$
 \newline
\textcolor{blue}{$(v)$} $r(\omega,R)$ is  given by
\begin{equation*}
    \left(4 r(\omega,R)-\omega   \right)\left( \omega-2 r(\omega,R)\right)= \frac{2 r^3(\omega,R)}{R}
\end{equation*}
and $A(\omega,R)$ is given by
\begin{equation*}
   16A(\omega,R)^6= R^2 \omega^2\left(16 A(\omega,R)^4-R^2\omega^6\right).
\end{equation*}
\newline
\textcolor{blue}{$(vi)$} $t_{g_4^\Omega}$ is the smallest solution on $[0, r(\Omega)]$ to
$$g_4^\Omega(t):= \chi(\omega(\Omega)-2t ,R(\Omega)-t)=\pi t^2,$$
where $$\chi(\omega, R):=\frac{\omega}{2} \sqrt{4 R^2-\omega^2} + 2R^2 \arcsin{\frac{\omega}{2R}}.$$
\newline
\textcolor{blue}{$(vii)$} $r(\omega, P)$ is given by
\begin{equation*}
    (\omega-2r(\omega, P))^2(4r(\omega, P)-\omega)P^2= 4 r(\omega, P)^2\omega^3
\end{equation*}
and $A(\omega, P)$ is the middle root of the equation
 $$128 P A(\omega, P)^3 - 16\omega(5P^2+\omega^2)A(\omega, P)^2 + 16\omega^2 P^3 A(\omega, P)- \omega^3P^4=0$$
\newline
\textcolor{blue}{$(viii)$} $r(\omega, A)$ is given by
\begin{equation*}
    (\omega-2r(\omega, A))^2(4r(\omega, A)-\omega)A^2= r^4(\omega,A) \;\omega^3.
\end{equation*}
\end{footnotesize}

\begin{figure}[!h] \label{table5}
    \centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Param. & Condition & Inequality  & Extremal sets & Ref. \tabularnewline
\hline
$h, A, R$ &  & $   h<\frac{1}{R}+\frac{4R}{A}$ &  thinning stadiums & Prop. \ref{prop_hAR}     \tabularnewline
 &  &   $ h\ge \frac{1}{2R}+\frac{\pi R}{2A}+\sqrt{\frac{\pi}{A}}$ & balls & \tabularnewline
\hline
$h, R, P$ && $ h<\frac{P}{R(P-4R)}$ & thinning stadiums  & Prop. \ref{prop_hRP}      \tabularnewline
 & &  $  h \ge \frac{4\arcsinc{\left(\frac{4R}{P}\right)}}{P-4R\cos{\left(\arcsinc{\left(\frac{4R}{P}\right)}\right)}} + \sqrt{\frac{8\pi \arcsinc{\left(\frac{4R}{P}\right)}}{P\left(P-4R\cos{\left(\arcsinc{\left(\frac{4R}{P}\right)}\right)}\right)}}$& balls&  \tabularnewline
\hline

$h, P, d$ & $2d<P<3d$ & $  h < \frac{4}{P-2d} + \sqrt{\frac{4\pi}{(P-2d)\sqrt{P(4d-P)}}}$ &   & Prop. \ref{prop_hPd}     \tabularnewline
& $3 d\leq P\leq \pi d$ & $   h < \frac{4}{P-2d} + \sqrt{\frac{4\pi}{\sqrt{3} d(P-2d)}}$  &   &  \tabularnewline
 &  &  $ h \ge \frac{4 \arcsinc{\left(\frac{2d}{P}\right)}}{P-2d\cos{\left(\arcsinc{\left(\frac{2d}{P}\right)}\right)}} + \sqrt{\frac{8\pi\arcsinc{\left(\frac{2d}{P}\right)}}{(P-2d\cos{\left(\arcsinc{\left(\frac{2d}{P}\right)}\right)}}}$& balls & \tabularnewline
\hline
$h, A, d$ && $ h\leq  \frac{4}{d}+\frac{2d}{A}$ &   & Prop. \ref{prop_hdA}     \tabularnewline
&  & $  h< \frac{2 d}{A}+\sqrt{\frac{\pi}{A}}$  &   &  \tabularnewline
 &  &  $h> \frac{d}{A}+\sqrt{\frac{\pi}{A}}$& thinning two-cup & \tabularnewline
\hline

\end{tabular}

\end{figure}
Finally in this  last table we have  summarized the not Blaschke--Santaló sharp inequalities that we have found.

\newpage
\begin{comment}
\section{Appendix: Table with the results}

\begin{figure}[!h] \label{table2}
    \centering
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Param. & Condition & Inequality  & Extremal sets & Ref. \tabularnewline
\hline
$h,r, P$ &  & ${h\leq \frac{1}{r}+\sqrt{\frac{2\pi }{P r}}}$  & circumscribed sets & Prop. \ref{prop_hrP}  \tabularnewline
 &  & ${h\ge \frac{1}{r}+\frac{\pi }{P -\pi r}}$ &  stadiums &  \tabularnewline

\hline
$h,d,r$  &  & ${h\leq \frac{1}{r}+ \sqrt{\frac{\pi}{r\sqrt{d^2-4r^2}+r^2(\pi-2\arccos{\left(\frac{2r}{d}\right)})}}}$ & two-cup bodies & Prop. \ref{prop_hdr} & \tabularnewline
 &  & ${h\ge \frac{1}{t_{g_1^\Omega}} }$ \textcolor{blue}{$(*)$} & spherical slices/smoothed nonagons &
 \tabularnewline
\hline
$h, \omega, R$ & $\omega\le 3/2 R$ & $h\leq h(T_I)$ & subequilateral triangles & Prop. \ref{prop_hRw} & \tabularnewline
 &  & $h\ge\frac{1}{t_{g_2^\Omega}}$ \textcolor{blue}{$(**)$} & spherical slices &  &\tabularnewline
\hline
$h, R, r$ &  & $h\le \frac{1}{r}+ \sqrt{\frac{\pi }{2r\left(\sqrt{R^2-r^2}+r\arcsin\left(\frac{r}{R}\right)\right)}}$ &  two-cup bodies & Prop. \ref{prop_hRr} & \tabularnewline
 &  & $h\ge\frac{1}{t_{g_3^\Omega}}$ \textcolor{blue}{$(***)$} &  spherical slices &  &\tabularnewline
\hline

$h, A, \omega$ &  & $ h\leq h(T_I)$ &  subequilateral triangles & Prop. \ref{prop_hAw} & \tabularnewline
 &  & $ h\geq \frac{2}{\omega}+\frac{\pi \omega}{2 A }$ &  stadiums &  &\tabularnewline
\hline
$h, \omega, P$ &  $P\geq 2\sqrt{3}\omega$ & $h\le h(T_I) $ &  subequilateral triangles & Prop. \ref{prop_hwP} & \tabularnewline
 &  & $  h\ge \frac{2}{\omega}+\frac{2\pi}{2P-\pi \omega}$ &  stadiums &  &\tabularnewline
\hline

$h, d, \omega$ & $\omega\le \sqrt{3}/2d$ & $h\le h(T_I) $ &  subequilateral triangles & Prop. \ref{prop_hdw} &     \tabularnewline
& $\sqrt{3}/2d\le \omega\le d$ & $  h\le \frac{\sqrt{3}}{\sqrt{3}\omega-d}+\sqrt{\frac{2\pi}{\pi \omega^2-\sqrt{3}d^2+6\omega(\tan(\delta-\delta)}}$ \textcolor{red}{$(\star)$} &  Yamanouti sets &  &  \tabularnewline
 &  & $h(\Omega)\ge \frac{1}{t_{g_4^\Omega}}$ \textcolor{blue}{$(****)$}& spherical slices & &\tabularnewline
\hline

$h,P,A$ &  & $h\leq \frac{P}{A}$ & Cheeger to itself & \cite{ftouhi_cheeger} &  \tabularnewline
 &  & $h\ge \frac{P+\sqrt{4\pi A}}{2A}$ &  circumscribed sets & & \tabularnewline
\hline
$h,r,A$ &  & $h\leq \frac{1}{r}+\sqrt{\frac{\pi}{A}}$ &  circumscribed sets & \cite{ftJMAA} & \tabularnewline

 &  & $h\ge \frac{1}{r}+\frac{\pi r}{A}$ &  stadiums &  &\tabularnewline
\hline
$h,\omega,r$ &  & $h\geq \frac{1}{r}+\frac{1}{r}\sqrt{\pi\left(1-\frac{2r}{\omega}\right)\sqrt{\frac{4r}{\omega}-1}}$ &  subequilateral triangles & Prop. \ref{prop_hwr} & \tabularnewline
\hline

$h,R,d$ &  & $h\leq \frac{2R(2R+\sqrt{4R^2-d^2})}{d^2\sqrt{4R^2-d^2}}+\sqrt{\frac{4\pi R^2}{d^3\sqrt{4R^2-d^2}}}$ &  subequilateral triangles & Prop. \ref{prop_hRd} & \tabularnewline
\hline

\end{tabular}
    \caption{Table of the best known inequalities.}
    \label{fig:polygons}
\end{figure}

\textbf{Comments:}
\newline
\textcolor{red}{$(\star)$} We set $\delta=\arccos(\frac{\omega}{d})$
\newline
\textcolor{blue}{$(*)$} $t_{g_1^\Omega}$ is the smallest solution on $[0, r(\Omega)]$ to $$g_1^\Omega(t):=\psi(d(\Omega)-2t,r(\Omega)-t)=\pi t^2, $$ where
\begin{equation}\label{func_del}
   \psi(d,r):=\begin{cases}
    \displaystyle{\frac{3\sqrt{3}r}{2}(\sqrt{d^2-3r^2}-r)+\frac{3d^2}{2}\left(\frac{\pi}{3}-\arccos{\left(\frac{\sqrt{3}r}{d}\right)}\right)}, & \text{if} \, \, \,  d\le r D^*\vspace{1mm} \\
    \displaystyle{r\sqrt{d^2-4r^2}+\frac{d^2}{2}\arcsin{\left(\frac{2r}{d}\right)}}, & \text{if} \, \, \, d\ge r D^*.
\end{cases}
\end{equation}
and $D^*$ is the unique number in $[2,2\sqrt{3}]$ for which the two espression of the function $\psi(d,r)$ are equal.
\newline
\textcolor{blue}{$(**)$} $t_{g_2^\Omega}$ is the smallest solution on $[0, r(\Omega)]$ to
$$g_2^\Omega(t):= \chi(\omega(\Omega)-2t ,R(\Omega)-t=\pi t^2,$$
where $$\chi(\omega, R):=\omega \sqrt{4 R^2-\omega^2} + 4R^2 \arcsin{\frac{\omega}{2R}}.$$
\newline
 \textcolor{blue}{$(***)$}  $t_{g_3^\Omega}$ is the smallest solution on $[0, r(\Omega)]$ to  $${g_3^\Omega}(t):=2\left(r\sqrt{(R(\Omega)-t)^2-(r(\Omega)-t)^2}+(R(\Omega)-t)^2\arcsin{\left(\frac{r(\Omega)-t}{R(\Omega)-t}\right)}\right)=\pi t^2.$$
\newline
\textcolor{blue}{$(****)$} $t_{g_4^\Omega}$ is the smallest solution to
$${g_4^\Omega}(t):=f(d(\Omega)-2t, \omega(\Omega)-2t)=\pi t^2, $$
where
$$f(d,w)=\frac{w}{2}\sqrt{d^2-w^2}+d^2\arcsin\left(\frac{w}{d}\right)$$

\end{comment}

\medskip\noindent

{\bf Acknowledgements}:
The authors would like to thank Jimmy Lamboley for useful discussions.

The first author is supported by the Alexander von Humboldt-Professorship program and by the project ANR-18-CE40- 0013 SHAPO  financed by the French Agence Nationale de la Recherche (ANR).

The third author is supported by an Alexander von Humboldt  Research Fellowship.

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redpajama_set_name: RedPajamaArXiv