For any curve of unit length there exists a closed rectangle with area at most $A$ that covers the curve.

Find the minimum possible value of $A$ such that for any curve of unit length there exists a closed rectangle with area at most $A$ that covers the curve.

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The case of closed curves was solved by Emo Welzl in Worm in a box. – user53153 Jan 2 '13 at 6:17

I'm sure tis problem has been dealt with before. Nevertheless, there goes:

Theorem. Any plane curve $\gamma$ of length $L(\gamma)\leq 1$ can be covered by a rectangle $R$ of area $A(R)\leq{1\over4}$. The constant ${1\over4}$ is optimal.

Sketch of proof. There are two points $P$, $Q\in\gamma$ with $$|PQ|=\max_{{\bf z}_1,\ {\bf z}_2\in\gamma}|{\bf z}_1-{\bf z}_2|=:2d\ .$$ Assume $P=(-d,0)$, $Q=(d,0)$. Then for all points ${\bf z}=(x,y)\in\gamma$ one has $|x|\leq d$.

Assume there are points $(x,y)\in\gamma$ with $y>0$ and put $h:=\max_{{\bf z}\in\gamma} y\$; so there is a point $H=(x',h)\in\gamma$. If $$(*) \qquad y\geq0 \quad \forall\ {\bf z}\in\gamma$$ then $\gamma$ is contained in the rectangle $R:=[-d,d]\times[0,h]$ of area $A(R)=2d h$. Since $|PH|+|HQ|\leq 1$ we conclude that the maximal possible value of $h$ is when $x'=0$ (consider ellipses with foci $P$ and $Q$). It follows that $h\leq\sqrt{{1\over4}-d^2}$, so that $$A(R)\leq 2d\ \sqrt{{1\over4}-d^2}\leq {1\over4}\ ,$$ where the $\max$ is attained when $d={1\over4}\sqrt{2}$. The extremal configuration is an L-shaped curve $\gamma$ with two legs of length ${1\over2}$.

(Edit: The following is somewhat fishy.) If condition $(*)$ is not fulfilled then we have to consider two points $H'=(x',h')$ and $H''=(x'',-h'')\in\gamma$ of maximal $y>0$ and minimal $y<0$. The curve $\gamma$ is then contained in the rectangle $R=[-h'',h']\times[-d,d]$. Since now the distance $|H'H''|\geq h'+h''$ weighs in also it should not be too difficult to show that under these circumstances $A(R)\leq{1\over4}$.

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