The exact estimate is $\frac{\pi}{4}$.

Indeed, but this estimate allows for different solutions then the one you are thinking of Mikhail!

I don't understand the aim of this remark. Anyway, the initial problem is quite easy. Construct two strips of width 1, which contain the polygon and form angle $\pi/3$ (strips $AB$-$CD$ and $AD$-$BC$). Their intersection is a rombus $ABCD$. We assume that $AC$ is a vertical line. Consider a horizontal strip of width 1, which contain the polygon. Guess: for what position of that strip the area of the thick hexagon is maximal

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An estimate $\pi/4$ is very well known. I guess, a lot of participants were familiar with it.

I realized that I don't know any elementary proof for this fact, though I saw the estimate for several times, but everytime without a solution. Anyway, my proof is the following. Consider a line with polar angle $\varphi$ and let $r(\varphi)$ be a length of projection of the figure to this line. Let's calculate $I=\int_0^\pi r(\varphi)d\varphi$. Since $r(\varphi)\leq 1$ we get $I\leq \pi$. On the other hand $I=P$, where $P$ is the perimetr of the figure . It means that $P\leq \pi$. And we know that the maximal area for all figures of perimetr at most $\pi$ has a circle of radius $\frac{1}{2}$. I.e. I reduced this problem to a really well known not-easy one (isoperimetrical problem). That's all.

Fedor Petrov wrote: An estimate $\pi/4$ is very well known. I guess, a lot of participants were familiar with it. I think you are giving the Romanian team too much credit (and you did give the Russian team last year too little ). It seems that only one student knew this estimate, and only a couple more solved the problem.

Valentin Vornicu wrote: Fedor Petrov wrote: An estimate $\pi/4$ is very well known. I guess, a lot of participants were familiar with it. I think you are giving the Romanian team too much credit (and you did give the Russian team last year too little ). It seems that only one student knew this estimate, and only a couple more solved the problem. I guess it is a very funny joke. Is my solution correct? If so, then I don't understand...

What don't you understand Mikhail?

I don't understand low results on this problem.

Myth wrote: I don't understand low results on this problem. It's simple: the Romanian team is not that good as you think it is

I guess that Mikhail does not understand, why such easy problem was so poorely solved by Romanian students. I do not understand this, too, but, to be true, I often do not understand such things. By the way, an estimate $\pi/4$ may be proved without reduction to isoperimeter problem: let our (convex) set lie in an upper half-plane and contain the origin. Its boundary has equation $r=r(\phi)$ in polar coordinates, where $\phi$ moves from 0 to $\pi$. An area equals $\frac{1}{2}\int_0^{\pi}r^2(\phi)d\phi$. But we have $r^2(\phi)+r^2(\phi+\pi/2)\le 1 \, (0\le \phi\le \pi/2)$ by Pythagor theorem. Now just integrate the last inequality on $[0,\pi/2]$.

Oh, my God! I have just remembered! I even saw a picture for this solution, but I completely forgot it

I guess most of them can solve it if the problem states the exact estimate $\pi/4$, the bound $\sqrt{3}/2$ is too weak that most people don't know how to arrive Anyway, this problem can be easily done by isoperimetric theorem.

is there an elementary solution to this one?

Which one do you mean? Isoperimetric problem has a simple solution if you know that such figure exists.

i meant to the initial problem. is there any solution without integrals? i'm also interested in a simple solution of the isoperimetric problem

I gave an elementary solution for the initial problem in the very beginning.

hey, sorry 4 mi ignorance, but wat is the isoperimetric problem?

From all figures of given perimetr find the the figure having maximal area.

i didn't understand the elementary solution, which seems to be only for a hexagon

No, it works for every polygon. We construct a hexagon of area $\sqrt{3}/2$, which contains our polygon inside.

Myth wrote: From all figures of given perimetr find the the figure having maximal area. Which is circle and i have just find a nice prove for it ha-ha .see if it is right? Lemma one: If a shape has mor than two lines that the shape is symmetric due to these lines then, these lines are concurrent. Lets name the figure $\mathcal L$.Consider two points like $A$ and $B$ on its premeter such that these points devide the premeter of the figure into two equal parts. Lets name these parts of premeter $\mathcal E$ and $\mathcal L$. So the lengh$ \mathcal E$=lengh$ \mathcal Q $=$\frac12 $ premeter of $\mathcal G$. Consider these areas: 1.Area between the line $AB$ and $\mathcal E$=$S[\mathcal E AB]$ 2.Area between the line $AB$ and $\mathcal Q$=$S[\mathcal QAB]$ Now If $S[\mathcal EAB] \ge S[\mathcal QAB]$ then delete the $\mathcal Q$ from the figure and replace it by the symmetric of $\mathcal E$ WRT $AB$ ,And we will reach a new figure with same premeter of our first shap but with a greater area : Equal premeters:lenght$ \mathcal E$+lenght$\mathcal Q=2$lengh$\mathcal Q$ Bigger Area: $2S[\mathcal E AB] \ge S[\mathcal E AB]+S[\mathcal QAB]$. by these argumets ,in the shape that satisfies the condition of the isopremeter problem , for every two points like $(A,B)$ that besicts the premeter,$AB$ bisects the Area and $AB$ is a symmetric line for the shape . By using lemma one we can see that all of this lines are congurrent . So we have changed the problem to this : Find all of the shapes that are symmetric wrt $\infty$ congurrent lines.

lomos_lupin, I don't want to disappoint you-but your proof to isoperimetric theorem is completely wrong. Nevermind, I had also the wrong guess. See http://www.mathlinks.ro/Forum/viewtopic.php?t=29793 for more details.

Next post please,this is edited.

Rest of post 23 Let the symmetric lines be $l_1,...,l_n$ and they be concurrent at $O$ ,its clear that every point of shape has a same distance from the $O$ so the shape is a circle with center of $O$. Sweet home

Could you elaborate your idea and give a formal proof?

Ok look at post 23 ,I have edited it.

That area is less than $\pi /2-\sqrt3/2 < \pi /4$.The proof is not very complicate,if you want it just let me know.

I don't understand. If the polygon has all sides and diagonals length <1, isn't it contained in a circle of diameter 1, which has area pi/4? What am I assuming that is false?

Everything. Is an equilateral triangle of side-length $1$ contained in a circle of diameter $1$? The best result is Jung's theorem, stating that a plane figure where the distances between any two of its points are at most $1$ is contained in a circle of radius $\dfrac {1} {\sqrt{3}}$ (which in fact is best precisely for the triangle). But that has too large an area for the problem; $\dfrac{\pi}{3} > \dfrac {\sqrt{3}}{2}$, unfortunately

Of course, I was not thinking properly. Now my next question is: can we improve the bound in this problem by using more stripths of worth 1, as in myth's solution?

What do you mean by "improve the bound"? the best bound was shown to be $\pi/4$, right?

I mean the bound of sqrt(3)/2.

Doesn't exist an elementary solution?