# 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 $r(a)*r(b)*sin|a-b|<=C$, 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