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Assume Q is a convex central symmetric curve, whose area is $\displaystyle S$. The area of the maximum parallelogram inside Q is $\displaystyle S'$.

How to prove the conjecture that $\displaystyle \frac{S'}{S} \ge \frac{2}{\pi}=0.6366\dots$?

For example, If Q is an ellipse, $\displaystyle S'=2ab$, $\displaystyle S=\pi ab$. If Q is a regular hexagon, $\displaystyle \frac{S'}{S}= \frac{2}{3}$.

It's trivial that $\displaystyle \frac{S'}{S} \ge \frac{1}{2}$, and I know how to prove $\displaystyle \frac{S'}{S} \ge \frac{4}{4+\pi}=0.56\dots$

From many reason, I believe this conjecture is true. Denote MAP="Maximum Area Parallelogram": For any Q and any direction $\theta$, let $P(Q,\theta)$ be the area of MAP which have a corner in this direction. $S'=max\{P(Q,\theta)\}$. In order to make $\frac{S'}{S}$ smallest, We need keep the largest one of ${P(Q,\theta)}$ small while S is a constant. Ellipse just keeps everyone in ${P(Q,\theta)}$ average. This is very special, I don't think there will be other curve having this property. On the other hand, distribute equally always lead to the min-max in our knowledge.

About the $\frac{4}{4+\pi}$ lowerbound, the idea is as follows: First, use polar function $r(\theta)$ to describe the curve. The condition is that $r(a)*r(b)*sin|a-b|<=C$, and we want to bound is $S=\Integral_{\theta}{r(\theta)}^2$. Second, Without lose of generality, We assume $r(0)=r(90)=1,C=1$, and assume $Q$ is in the boundary of $Z=\{ (x,y)|-1\le x,y\le 1\}$. Third, let $a=r(\theta)$ and $b=r(\theta+90)$ and find a bound for $(a^2+b^2)$ by Cauchy-Inequality. and it will give a bound for the area $S$.

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have you try symmetrization with respect to line? for analogous problem with maximal triangle it works –  Fedor Petrov Nov 4 '10 at 22:53
    
I' think I've know a way to prove it. But it's not pretty at all. And also need to fix several flaws... Still expecting for a genius proof~ –  galois Nov 7 '10 at 11:51
    
By the way, it was just a modification of the proof proposed by Agol here. Thank you very much! –  galois Nov 7 '10 at 11:55

1 Answer 1

This is an old result in convex geometry. See

E. Sas, ¨Uber ein Extremumeigenschaft der Ellipsen, Compositio Math. 6 (1939) 468– 470.

A. M. Macbeath, An extremal property of the hypersphere, Proc. Cambridge Philos. Soc. 47 (1951) 245–247.

Fedor Petrov's remark was right on target. Steiner symmetrization gives an easy proof (see Macbeath).

By the way, the equality case holds only for ellipses, which explains the titles of the papers above.

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