# 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.

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 $iv=\frac{1}{4}$.)

• Does a (shape approximating a) figure 8 shape have measure (approaching) 1/4? Gerhard "Unsure Of The Numerical Calculation" Paseman, 2017.02.08. – Gerhard Paseman Feb 8 '17 at 15:42
• That might be the limiting figure, but the intermediate figures would be like barbells with small twisty handles. Except for necklace or beaded string shapes, I doubt I can improve much on that measure. Gerhard "Needs Coffee Before Needing Pearls" Paseman, 2017.02.08. – Gerhard Paseman Feb 8 '17 at 15:53
• If we consider $r(S)$ to be a measure of disk-ness of $S$, and $iv(S)$ to be a measure of nonconvexity, then you seem to be seeking the most nonconvex disk :P – Wojowu Feb 8 '17 at 16:16
• I'd go for a cardioid-like shape -maybe with a longer cusp – Pietro Majer Feb 8 '17 at 17:52
• For the L-shapes above with $a<1/2$, we have $A=2a(1-a)+a^2$, $L=4$, $$iv = \frac{2-4a+a^2-a^2\log(\frac{1}{a}-1)}{(2-a)^2}.$$ So their maximum is actually quite close to $a=2/7$, with $P=.113$. – Matt F. Feb 15 '17 at 2:38

A Pacman is a good candidate.

Take a unit circle and cut out a sector with angular width $\alpha$. Then:

\begin{align} A &= \pi - \frac{\alpha}{2}\\ L &= 2\pi - \alpha+2\\ iv &= \frac{(\pi-\alpha)^2}{(2\pi-\alpha)^2}\\ P &= \frac{2\pi(\pi-\alpha)^2}{(2\pi-\alpha+2)^2(2\pi-\alpha)} \end{align}

So $P$ approaches $\pi^2/(2\pi+2)^2$ or $0.144$ near $\alpha=0$. This is a Pacman whose mouth is nearly closed.

• Nice idea! Larger $P$ than the two-disk figure-8. – Joseph O'Rourke Feb 14 '17 at 23:33
• Really? I thought we had P for the figure 8 at 1/4=0.25. Gerhard "Can't Find Where Mistake Is" Paseman, 2017.02.14. – Gerhard Paseman Feb 15 '17 at 1:47
• @GerhardPaseman: I calculated $r=1/2$, $iv=1/4$, $P=1/8$. – Joseph O'Rourke Feb 15 '17 at 12:02
• I think @GerhardPaseman is right, his shape has iv=1/2 – Matt F. Feb 15 '17 at 15:42
• @MattF.: Oh, you are right. Apologies to @ GerhardPaseman. – Joseph O'Rourke Feb 19 '17 at 11:56