I seek a two-dimensional shapes $S$, bounded by a Jordan curve,
that optimally balances its isoperimetric ratio $r(S)$
against what I call its invisibility index $iv(S)$.
Define the *isoperimetric ratio* $r(S)$ of $S$ to be
$4 \pi A / L^2$, where $A$ is the area of $S$ and $L$ its
perimeter. This ratio is in $(0,1]$
and achieves $1$ for $S$ a disk.
See, e.g., the
Wikipedia article on the isoperimetric inequality.
Define *invisibility index* $iv(S)$
to be the probability that that a pair $(x,y)$ of random points
in $S$
(chosen uniformly and independently)
are invisible to one another in the sense that
the segment $xy$ includes a point strictly exterior to $S$.
($iv(S)$ is $1$ minus the
Beer convexity of $S$.)

Q. What shape (or shapes) $S$ maximize the product $P(S) = r(S) \cdot iv(S)$?

If $S$ is a disk, $r(S)=1$ and $iv(S)=0$ so $P(S)=0$. If $S$ is a thin spiral, then $r(S)$ approaches $0$ and $iv(S)$ approaches $1$ so $P(S)$ approaches $0$. In between, $P(S) > 0$.

I've computed $P(S)$ for the very narrow class of
symmetric `L`

s, unit squares with a square removed
from one corner, as illustrated below.

^{ Two symmetric Ls with different parameters $a$. Origin at lowerleft corner. }

These shapes are determined by one parameter $a$ as illustrated. Among this class of shapes, it appears that the maximum product $P(S)$ is achieved when $a \approx \frac{1}{4}$, the left shape above. Plots $r(\,)$, $iv(\,)$, and $P(\,)$ are shown below. The isoperimetric ratio for a square is $r(1) = \pi/4 \approx 0.79$.

^{ Red: $r(a)$. Blue: $iv(a)$. Green: Product $P(a)$. }

**Update**. Seems like Gerhard Paseman's figure-8
, with $r=\frac{1}{2}$, $iv=\frac{1}{2}$, $P=\frac{1}{4}$,
is the extreme shape. (In comments I mistakenly said $iv=\frac{1}{4}$.)

the most nonconvex disk:P $\endgroup$3more comments