# A conjecture of parallelogram inside convex and central symmetric curve

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 $r(a)r(b)\sin|a-b|\leq C$, and we want to bound is $S=\int_{\theta}{r(\theta)}^2$. Second, Without lose of generality, We assume $r(0)=r(90)=1,C=1$, and assume $Q$ is in the boundary of $Z=\{(x,y)|-1\leq x,y\leq 1\}$.
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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What are some of the "many reasons" ? –  a little don Nov 4 '10 at 16:55
Also, if you could edit the question with a brief sketch of proof of the 4/(4+pi) bound, it might help others modify it to give better bounds. –  Aryabhata Nov 4 '10 at 17:03