Ratio of diameter to area of a set

Question

It may be quite a basic and common thing but I haven't found much after a while of searching and I failed to figure that myself...

Let's have a (connected) set $M$ and let $\text{diam}(M)$ be its diameter. How big can his area be?

Or in other words - evaluate this expression: $$\sup_{M}\left\{\frac{\text{area}(M)}{\text{diam}^{2}(M)}\right\}$$

For example let $M$ be a square with side $a$. Then:

$$\frac{\text{area}(\text{square})}{\text{diam}^{2}(\text{square})} = \frac{a^{2}}{\left( a\sqrt{2} \right)^{2}} = \frac{1}{2}$$

I'm really looking forward to see a (sketch of a) proof of such thing because no matter how simple it looks I just don't even know where to start...

Check the book Geometric Inequalities by Burago and Zalgaller, I remember seeing it there. I think the square is the answer and the proof uses Besikovitch inequality. — Moishe Kohan, Jan 12 '14 at 20:43
Well, you can make a circumscribed circle around the square for example. The area will increase while the diameter will remain the same... Anyway, that book you mentioned looks really good, I'll take a look at it. — Jeyekomon, Jan 12 '14 at 21:25
Yes, of course. I should have thought a bit more before typing. — Moishe Kohan, Jan 12 '14 at 21:37
If anyone is interested, I found a generalization of this question. Look at the Isoperimetric inequalities. — Jeyekomon, Sep 29 '20 at 12:14

score 3 · Accepted Answer · answered Jan 13 '14 at 18:44

I found a great one-page proof at http://www.math.ntnu.no/~hanche/blog/isodiametric.pdf.

They assume that one point on the boundary is at the origin and that the shape lies in the upper half plane. They then use polar coordinates for the area:

$\int_0^{\pi} 1/2 (f(\theta))^2 d\theta$ and rewrite it as

$1/2\int_0^{\pi/2} (f(\theta))^2 +(f(\theta+\pi/2))^2 d\theta$

The integrand is the squared length between the points at $\theta$ and $\theta+\pi/2$ on the boundary, which is at most $d^2$. So the greatest the area can be is $(\pi d^2)/4$.

The link has a great picture.

It's interesting for me that the inequality has its own name... And the source you provided is great! Thank you. — Jeyekomon, Jan 14 '14 at 15:24

Ratio of diameter to area of a set

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