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


          TwoLs
          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 $r(1) = \pi/4 \approx 0.79$.
          IsoInvisibGraph
          Red: $r(a)$. Blue: $iv(a)$. Green: Product $P(a)$.


Update. Seems like Gerhard Paseman's figure-8 Fig8, with $r=\frac{1}{2}$, $iv=\frac{1}{2}$, $P=\frac{1}{4}$, is the extreme shape. (In comments I mistakenly said $iv=\frac{1}{4}$.)

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


          TwoLs
          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 $r(1) = \pi/4 \approx 0.79$.
          IsoInvisibGraph
          Red: $r(a)$. Blue: $iv(a)$. Green: Product $P(a)$.


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


          TwoLs
          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 $r(1) = \pi/4 \approx 0.79$.
          IsoInvisibGraph
          Red: $r(a)$. Blue: $iv(a)$. Green: Product $P(a)$.


Update. Seems like Gerhard Paseman's figure-8 Fig8, with $r=\frac{1}{2}$, $iv=\frac{1}{2}$, $P=\frac{1}{4}$, is the extreme shape. (In comments I mistakenly said $iv=\frac{1}{4}$.)

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Shape that balances isoperimetric ratio with invisibility

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


          TwoLs
          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 $r(1) = \pi/4 \approx 0.79$.
          IsoInvisibGraph
          Red: $r(a)$. Blue: $iv(a)$. Green: Product $P(a)$.