Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
The case of closed curves was solved by Emo Welzl in Worm in a box. –  user53153 Jan 2 '13 at 6:17

1 Answer 1

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 $A(R)\leq{1\over4}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.