3 added 197 characters in body

Let $\mathcal C_d$ denote the set of all $d-$dimensional convex compact subsets with barycenter at the origin of the $d$-dimensional Euclidean space $\mathbb E^d$. Given an element $C\in\mathcal C_d$ with boundary $\partial C$, define $\Sigma_0\subset \partial C$ by $\Sigma_0=\partial C\cap(-\partial C)$. The complement $\partial C\setminus \Sigma_0$ is open and can be partitioned into two open subsets $\Sigma_\pm$ defined by $$\Sigma_-=\lbrace P\in\partial C\vert -P\in\mathrm{Int}(C)\rbrace$$ and $$\Sigma_+=\lbrace P\in \partial C\vert -P\not\in C\rbrace\ .$$ Set $\rho(C)=\mathrm{Area}(\Sigma_+)/\mathrm{Area}(\Sigma_-)$ (using the convention $\rho(C)=1$ if $\Sigma_+=\Sigma_-=\emptyset$ which happens exactly if $C$ is centrally symmetric) where $\mathrm{Area}(\Sigma_\pm)$ is the $(d-1)$ dimensional area.

The function $\rho:\mathcal C_d\longrightarrow [1,\infty)$ is invariant under the action of the linear group on $\mathcal C_d$ and is bounded. What is its maximal value (and on which convex sets is it achieved)?

Added: My argument for the invariance under the linear group is flawed. I am also no longer sure that $\rho$ is bounded (my proof used invariance under the linear group).

(Added after a moments thought: One remedy is to replace $\rho(C)$ by $\mathrm{Vol}(C)/\mathrm{Vol}(C\cap(-C))$ which is obviously invariant under linear transformations and which is bounded.)

A probably naive guess for the maximum is the value of $\rho$ on a simplex. In dimension $2$, one gets $\rho(\Delta)=2$ if $\Delta\in\mathcal C_2$ is a triangle and in dimension $3$ one gets $\rho(\Delta)=3$ for $\Delta$ a simplex in $\mathcal C_3$. I ignore the value of $\rho$ on simplices of dimension $\geq 4$ (question http://mathoverflow.net/questions/60167/name-of-a-polytope is motivated by the computation of these values).

A similar invariant which has perhaps been studied is obtained by considering $\lambda_C=\max_L_+\mathrm{length}(L_+\cap C)/\mathrm{length}((-L_+)\cap C)$ where the maximum is over the set of all half-lines rooted at the origin. The invariant $\lambda(C)$ is also bounded and invariant under the action of the linear group on $\mathcal C_d$. (Here I am sure since this is obvious.)

2 added 213 characters in body

Let $\mathcal C_d$ denote the set of all $d-$dimensional convex compact subsets with barycenter at the origin of the $d$-dimensional Euclidean space $\mathbb E^d$. Given an element $C\in\mathcal C_d$ with boundary $\partial C$, define $\Sigma_0\subset \partial C$ by $\Sigma_0=\partial C\cap(-\partial C)$. The complement $\partial C\setminus \Sigma_0$ is open and can be partitioned into two open subsets $\Sigma_\pm$ defined by $$\Sigma_-=\lbrace P\in\partial C\vert -P\in\mathrm{Int}(C)\rbrace$$ and $$\Sigma_+=\lbrace P\in \partial C\vert -P\not\in C\rbrace\ .$$ Set $\rho(C)=\mathrm{Area}(\Sigma_+)/\mathrm{Area}(\Sigma_-)$ (using the convention $\rho(C)=1$ if $\Sigma_+=\Sigma_-=\emptyset$ which happens exactly if $C$ is centrally symmetric) where $\mathrm{Area}(\Sigma_\pm)$ is the $(d-1)$ dimensional area.

The function $\rho:\mathcal C_d\longrightarrow [1,\infty)$ is invariant under the action of the linear group on $\mathcal C_d$ and is bounded. What is its maximal value (and on which convex sets is it achieved)?

Added: My argument for the invariance under the linear group is flawed. I am also no longer sure that $\rho$ is bounded (my proof used invariance under the linear group).

A probably naive guess for the maximum is the value of $\rho$ on a simplex. In dimension $2$, one gets $\rho(\Delta)=2$ if $\Delta\in\mathcal C_2$ is a triangle and in dimension $3$ one gets $\rho(\Delta)=3$ for $\Delta$ a simplex in $\mathcal C_3$. I ignore the value of $\rho$ on simplices of dimension $\geq 4$ (question http://mathoverflow.net/questions/60167/name-of-a-polytope is motivated by the computation of these values).

(