Consider an convex plane figure $F$. How to prove that there is an affine transformation $a$ such that $\sqrt{3}$ diameter$(a(F))^2\leq 4$ area$(a(F))$?

I found only one reference, to "Über einige Affininvarianten konvexer Bereiche", but unfortunately it is in German.

Added: formula (12) there looks like desirable. After I found a solution myself, I can understand German. The proof there in the pages 734 (corresponding to considering $D'$ below) and 735 (considering $D''$). The author estimated $f/d_u^2$, $f$ is an area (Flacheninhalt) and $d_u$ is a diameter(Durchmesser).

So, emergency over, thank you))

the proof is rewritten by me in http://arxiv.org/abs/1306.4688

  • $\begingroup$ Funny, I'm also looking at this reference and I can't even find the statement of this result. It is supposed to be in pages 745 and 746. $\endgroup$ – alvarezpaiva May 13 '13 at 19:24
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    $\begingroup$ Did you try to take the largest-area (Jones-Loewner) ellipsoid E in F and then apply affine transformation which sends E to the unit circle? $\endgroup$ – Misha May 13 '13 at 19:52
  • $\begingroup$ @Misha: this will give a similar bound, but I don't think it will be sharp. Behrend's bound is sharp with equality if and only if $F$ is a triangle. But of course one should check ... $\endgroup$ – alvarezpaiva May 13 '13 at 20:15
  • $\begingroup$ I guess $F$ must have positive area. This is trivialy false if $F$ is a line segment. $\endgroup$ – Cristos A. Ruiz May 13 '13 at 21:02
  • $\begingroup$ Yes, the Jones ellipse method gives $diam^2 \le \frac{8}{\pi} Area \approx 2.5 Area$ which is a bit worse than $diam^2 \le \frac{4}{\sqrt{3}} Area \approx 2.3 Area$. $\endgroup$ – Misha May 13 '13 at 22:54

The article of Scott and Awyong, "INEQUALITIES FOR CONVEX SETS" containes several inequalities relating the area, perimeter, width, diameter, inradius and circumradius of planar convex sets. I tried to obtain the desired inequality (all of them have precise references given). Denote by $A$ the area, and by $d$ the diameter. At least, in the case $p\ge 3d$ (and I hope this will help) we obtain $$ A \ge \frac{\sqrt{3}}{4}d(p-2d)\ge \frac{\sqrt{3}}{4} d^2. $$ I did not check if we can obtain the inequality in another way. Some references claim that the inequality has been proved by Behrend (the german article), others relate it to Kubota.


It seems that I found a proof. Consider a figure $A$. Consider a figure $F$ with minimal ratio $S/d^2$ among all affine transforms of $A$.

Lemma. There are two diameters of $F$ with angle at least $\pi/3$ between them. Note that Lemma implies the estimation because $\sin(\pi/3) =\sqrt 3/2$ and area of $F$ is at least $d^2\sin(\pi/3)/2$.


Consider a diameter $D$. Try to squeeze $F$ in the direction of $D$ and stretch out in the perpendicular direction. It is not possible, therefore there is an other diameter $D'$ with angle at least $\pi/4$ and less than $\pi/3$ with $D$. Well, among all pairs of diameters chose the pair $D,D'$ with the biggest shapr angle between them.

Now try to perform an affine shift parallel to $D$ in the direction decreasing $D'$.

It is not possible, therefore there is a diameter $D''$ which is a kind in a "symmetric" position with $D'$. So, now either the angle between $D$ and $D'$ is big enough, or the angle between $D'$ and $D''$.

  • $\begingroup$ Do you get the equality case this way? $\endgroup$ – alvarezpaiva May 15 '13 at 22:06
  • $\begingroup$ yes sure. Triangle with sides equal $d$. Its diameter is $d$, its area is $\sqrt{3}d^2/4$. But any figure spanned on two intervals of length $d$ and angle $\pi/3$ between them works as well. $\endgroup$ – Nikita Kalinin May 16 '13 at 18:16

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