I am strictly sure that we must prove there exists one point $O$ inside $P$ such that for each side $AB$ (corresponds an assigned area $S$), the area of OAB is not greater than a half of $S$. Whenever one can prove with only $N=3,4,5,6$ ($P$ is an $N$_polygon), one uses the Helly's lemma. That is clear? I will check and post if it is really true.

My proof goes by another way, the main idea is that you can "cancel" parallel sides. Pardon me that I am too tired to type it now. And thank Valentin for posting the problems

nmtruong1986 wrote: I am strictly sure that we must prove there exists one point $O$ inside $P$ such that for each side $AB$ (corresponds an assigned area $S$), the area of OAB is not greater than a half of $S$. Whenever one can prove with only $N=3,4,5,6$ ($P$ is an $N$_polygon), one uses the Helly's lemma. That is clear? I will check and post if it is really true. This is FALSE! You can find a counterexample for $n = 6$.

how did you do on this problem, Yufei?

Guys what do you think,is this a problem of geometry or combinatorics?I mean in the short list.

Tiks wrote: Guys what do you think,is this a problem of geometry or combinatorics?I mean in the short list. I think that is geometry because of one simple reason:On 99,9 percents of newer IMO (last 15 years) there were two problems from Geometry part of shortlist.

There are two solutions on the official IMO website. On this problem, only 10 IMO contestants scored $\geq 5$! They are CHN 4 CHN 6 DEU 6 FRA 5 KOR 6 (6 marks) LVA 3 (5 marks) MDA 1 POL 3 RUS 1 RUS 4

So this problem is probably more difficult than this one: http://www.mathlinks.ro/Forum/viewtopic.php?p=352683#p352683

Actually, there exists a fairly short solution. Draw the lines through all vertices of the polygon that divide the area in half. Look at the corresponding "butterflies". They cover the entire polygon (that is true for any set of lines). Now, for each butterfly ABOCD A __ B \ / \/ O /\ C/__\ D (Here AB is not necessarily parallel to CD, but I was too lazy to create a PNG picture), we clearly, have $|OA|\cdot |OB|=|OC|\cdot |OD|$ (here we use the fact that both AD and BC divide the area of the polygon into 2 equal parts). It follows from here that either $|OB|\ge |OC|$ or $|OA|\ge |OD|$. Thus either the area of BDC or that of ACD is at least twice that of OCD and similarly for the other side. Thus the sum of the areas of the triangles corresponding to $AB$ and $CD$ is at least twice the area of the butterfly. Now it is clear that it doesn't really matter whether to consider the whole sides of the original polygon or to subdivide each side into several smaller sides: the sum in question remains the same. The end.

fedja wrote: we clearly, have $|OA|\cdot |OB|=|OC|\cdot |OD|$ (here we use the fact that both AD and BC divide the area of the polygon into 2 equal parts). I fail to see why this is true.

The areas of the triangles $OAB$ and $OCD$ must be equal.

fedja wrote: The areas of the triangles $OAB$ and $OCD$ must be equal. why? the polygon isn't regular.

Because $AB$ and $CD$ divide the area in two equal parts. Now look at what is to the "right" of $AB$ and what is to the "right" of $CD$ on the picture. You have a big common part to the right of the broken line $BOD$ plus one triangle in one case and the other one in the other case. But the total area in both cases must be half of that of the polygon. So, disregarding the area of the common part, we see that the triangles must have the same area.

why there must be lines that divides the polygons area by 2. this isnt regular i think

I am quite sure that in some polygons you will not be able to find a line through 2 vertices that divides the area in half.

No, each line goes through one true vertex and some other point that may lie on some side ("fake vertex"). The point is that adding those fake vertices to the true vertices of the polygon doesn't change anything in the problem. .

One can always draw, from each existing vertex, a line that divides the area in half. If the other end is not a vertex of the polygon add it to the list of vertices. Adding these vertices does not affect anything in the problem (some angles will be 180, but that does not matter at all). EDIT: Well, I see that Fedja explained the same thing in the meantime.

fedja wrote: Look at the corresponding "butterflies". They cover the entire polygon (that is true for any set of lines). Why? The lines dividing the area are not necessary concurrent!

I don't think you need concurrency

A. S. K. wrote: fedja, have you been on IMO 2006? You should get a special prize for your solution... If I'm not mistaken, he's a professor.

If I remember correctly, the first official solution is very similar to fedja's solution. In fact, the subdivision of the sides at the end of the solution more or less generates the fake vertices you need for fedjas solution. The second solution mentioned here with minkowski sums is more or less the second official solution - in fact, the "associate" as they call it there is the minkowski sum $P+P^{t}$. At IMO, I didn't see the thing with the minkowski sum but my solution is quite similar to the second official solution - I do some things to get vertices at extreme positions and cancel parallel sides.

Since solutions of the german team will be available at some time in the internet and I had to write down the solution to number 6, I have a .pdf-file of my solution. I'm attaching it in German and English. Peter

Here are 2 problems looking very useful to P6. Both are known in Russia. 1. (SPbMO-88) Let $a_{1}, a_{2},\ldots , a_{n}$ be sides and also side-lengths of a convex polygon $P$. Let $d_{1}, d_{2},\ldots , d_{n}$ be lengths of projections of the polygon to $a_{1}, a_{2},\ldots , a_{n}$ respectively. Prove that \[\sum_{i=1}^{n}\frac{a_{i}}{d_{i}}\le 4.\] 2. (SPb math circles) Let $ABCD$ be a convex quadrilateral with a point $O$ in the interior and points $P,Q,R,S$ on the sides $AB,BC,CD,DA$ respectively such that $APOS\text{ and }CROQ$ are parallelogramms. Prove that $\sqrt{S_{BQOP}}+\sqrt{S_{DSOR}}\le \sqrt{S_{ABCD}}.$ 1 contains an idea to reduce an arbitrary polygon to central-symmetric. 2 is used to explain Brunno-Minkowski inequality. So we (team Russia) could solve it much better.

I started to read Scholze's solution to the problem 6, and there is something I don't understand. In the case 1 there was said that "for every side $b$ of $P$ the line $\bar b$ which is parallel to $b$, meets $P$ in some point and has greatest distance to $P$". But in metric topology, I have been taught that if a line $l$ meets a set $A$, then $d(l,A)=0$. Is therefore $\bar b$ an arbitrary line which meets $P$?

Sorry, just a stupid typo: replace the last $P$ by $b$.

On the excellent Russian site http://www.problems.ru this problem is recognized as the middle-level problem (with the mark 5+, while the maximal complexity is 10). However, the solution posted there is pretty unclear (at least, for me). Look http://www.problems.ru/view_problem_details_new.php?id=110751 What do you think?

It's also unclear for me. I'll post here the Russian solution, and my translation (it isn't good, and words aren't translated correctly!) Воспользуемся соображением: 1) если двигать одну из вершин (трех)(много)угольника с постоянной скоростью, то его площадь меняется тоже с постоянной скоростью. 2) линейная комбинация линейных функций тоже линейна. 3) линейная функция достигает максимума (минимума) на границе отрезка. Пусть P - вершина M, и число сторон M не меньше 4, R, T - соседние с P вершины. Будем двигать P параллельно [RT]. Тогда при движении в любом из двух направлений вершина M выйдет на продолжение стороны P. Применив эти соображения, сведем задачу к случаю, когда P лежит на продолжении одной из сторон M, т.е. один из углов M равен π. Но в этом случае дело сводится к многоугольнику с меньшим числом вершин и завершается индукционным спуском. We use the following: 1) When moving from one vertex to another with constant velocity, the area(???) is also changing at constant velocity 2) A linear combination of linear functions is also linear 3) A linear function reaches its maximum (minimum) on the boundaries of its line Let P be a vertex of M, with the number of sides of M not less than 4, and R and T are neighboring vertexes with P. Move P parallel to RT. Then, when moving in any direction, vertex of M will be the extension of sides of P (???) Applying 1),2),3) we reduce the problem to the one when P lies on the extension of one of the sides of M, that is when one of the angles is $ \pi$. Then it reduces to a problem with less vertexes, which ends in an inductive descent. OK, there are things I don't understand. What are M and P (later in the solution, it seems like they aren't vertex and polygon). 1)? And the second ??? This is a very nice solution (partially understood), but can anyone explain the remaining for me?

I came up with a beautiful proof, but then noticed it had a gaping gap! Fortunately Kai Neergard engineered a bridge. See http://camoo.freeshell.org/IMO_2006_6.pdf I also included a short proof based on the Minkowski sum. Laura

The very nice and simple solution in larkl's preceding post is based on a beautiful idea which to my knowledge was never advanced before in relation to this problem. This idea is to consider a linearly parametrised family of polygons XX(t) which contains several polygons relevant to the problem: XX(0) ~= XX(1) ~= PP, where PP is the polygon of the IMO problem, XX(1/2) ~= 1/2 QQ, where QQ is the 'associate' in the language of the shortlist's Solution 2 (which according to Peter Scholze is also Charlydif's F), and in the limits t --> +- oo asymptotically XX(t) ~= |t| RR traversed twice, where RR's side vectors are those of PP in the order of increasing line direction modulo pi when we assume they are traversed in PP in the order of increasing vector direction modulo 2 pi so that PP has positive oriented area [PP]. The factors in the congruences are scale factors. Since [XX(t)] is a quadratic in t, the areas [PP], [QQ] and [RR] are related. In particular [RR] < 0 implies [QQ] >= 4 [PP], which is eqivalent to the IMO problem's assertion. I supplied the proof that indeed [RR] < 0. larkl cites an analytic version of it. I prefer, myself, the following geometric one. Note that the areas [TT_i] introduced below are the terms in the final sum in the analytic calculation. Let RR = A_0...A_{n-1} and A_n = A_0. For i = 0, ..., n - 1, let m_i be the line A_iA_{i+1}, and let l be a line through A_0 within the least positive angle from m_{n-1} to m_0. Since this angle and the least positive angles from m_i to m_{i+1} for i = 1, ..., n - 1 sum to pi, the direction of l is outside any of the angles mentioned most recently. For i = 0, ..., n - 1, let B_i be the point of intersection of l and m_i. We notice B_0 = B_{n-1} = A_0. The area [RR] is the sum of [TT_i] over i = 0, ..., n - 2, where TT_i = B_iA_{i+1}B_{i+1}. For i = 0, ..., n - 2, since the direction of l is outside the least positive angle from m_i to m_{i+1}, the oriented angle \angle B_iA_{i+1}B_{i+1} is congruent to this angle modulo 2 pi, so [TT_i] < 0. Therefore [RR] < 0. The relation between [PP], [QQ] and [RR] is 2[PP] = [QQ]/2 + [RR]. Instead of considering the quadratic [XX(t)], one may derive this relation using vector algebra. To understand the following, notice the relation between PP and RR, which is elucidated very nicely in larkl's text: The polygon RR is composed of the sequences of sides of PP with a common most distant vertex V in PP. These sequences appear in RR in the order so that, in PP, each sequence begins at the V of the preceding sequence and ends at that of the next one. Therefore, RR is got from PP by a parallel displacement of each of its sides with the vector OV, where O is an arbitrary constant point and V its most distant vertex in PP. Also note that [QQ]/2 is the sum of the maximal triangular areas of the IMO problem. Therefore, if P is a point on the side with the vector s, we have [RR] = 1/2 sum ( OP + OV ) x s = 1/2 sum ( 2 OP - VP ) x s = 2 [PP] - [QQ]/2 . Alternatively, let s_i = A_iA_{i+1} in the notation above. Then 2 [RR] = sum_{i<j} s_i x s_j . Let S and T be the sets of such indices i that if u is a vector on l, the angle from u to s_i is in the interval (-pi,0) and (0,pi), respectively. If, in the previous notation, P and V are associated with the side with index j, then if j is in S, the vector VP is the sum of the vectors s_i with i in T and i > j or i in S and i < j. Simlarly if j is in T with S and T interchanged. Thus 2 ([QQ]/2) = sum' s_i x s_j , where the prime indicates that the sum is restricted to the pairs (i,j) just described. Collecting the terms, one finds that 2 ( [RR] + [QQ]/2 ) is the sum of s_i x s_j over the following (i,j). (i) i,j are both in S or both in T, and i < j. These (i,j) occur twice. (ii) i is in S , j is in T or i is in T , j is in S. The terms with (i,j) of the type (ii) sum to U x (-U) + (-U) x U = 0, where U = sum_{i in S} s_i = - sum_{i in T} s_i. The area 2 [PP] is the sum of s_i x s_j over such (i,j) that i,j are both in S or both in T, and i < j, or i is in S and j in T. The terms with i in S and j in T sum to U x (-U) = 0. Thus [RR]+ [QQ]/2 = 2 [PP] .

Can somebody rigorously explain fedja's solution please?

fmasroor wrote: Can somebody rigorously explain fedja's solution please? I finally take the time to write it up properly myself, after having been told the solution during Taiwan IMO camp in 2014. We say a polygon in almost convex if all its angles are at most $180^{\circ}$. Note that given any convex or almost convex polygon, we can take any side $b$ and add another vertex on it, and the sum of the labels doesn't change (since the label of a side is the length of the side times the distance of the farthest point). Lemma: Let $N$ be an even integer. Then any almost convex $N$-gon with area $S$ should have an inscribed triangle with area at least $2S/N$. The main work is the proof of the lemma. Label the polygon $P_0 P_1 \dots P_{N-1}$. Consider the $N/2$ major diagonals of the almost convex $N$-gon, $P_0 P_{N/2}$, $P_1 P_{N/2+1}$, et cetera. A butterfly refers to a self-intersecting quadrilateral $P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$. An example of a butterfly is shown below for $N=8$. [asy][asy] pair P_0 = (-1,-3); pair P_1 = ( 0,-3); pair P_2 = (2.5,-3); pair P_3 = (3,0); pair P_4 = (1.8,1.2); pair P_5 = (1.1,1.9); pair P_6 = (0.3,2.7); pair P_7 = (-2,0.4); filldraw(P_0--P_2--P_3--P_6--P_7--cycle, invisible, blue); filldraw(P_0--P_4--P_5--P_1--cycle, palered, red); draw(P_1--P_5, red); draw(P_2--P_6, red); draw(P_3--P_7, red); dot("$P_0$", P_0, dir(P_0)); dot("$P_1$", P_1, dir(P_1)); dot("$P_2$", P_2, dir(P_2)); dot("$P_3$", P_3, dir(P_3)); dot("$P_4$", P_4, dir(P_4)); dot("$P_5$", P_5, dir(P_5)); dot("$P_6$", P_6, dir(P_6)); dot("$P_7$", P_7, dir(P_7)); /* TSQ Source: P_0 = (-1,-3) P_1 = ( 0,-3) P_2 = (2.5,-3) P_3 = (3,0) P_4 = (1.8,1.2) P_5 = (1.1,1.9) P_6 = (0.3,2.7) P_7 = (-2,0.4) P_0--P_2--P_3--P_6--P_7--cycle 0.1 lightcyan / blue P_0--P_4--P_5--P_1--cycle 0.2 lightred / red P_1--P_5 red P_2--P_6 red P_3--P_7 red */ [/asy][/asy] Claim: Every point $X$ in the polygon is contained in the wingspan of some butterfly. Proof. Consider a windmill-like process which starts from some oriented red line $P_0 P_{N/2}$, oriented to face $P_0 P_{N/2}$ rotates through $P_0 P_{N/2} \cap P_1 P_{N/2+1}$ to get line $P_1 P_{N/2+1}$, rotates through $P_1 P_{N/2+1} \cap P_2 P_{N/2+2}$ to get line $P_2 P_{N/2+2}$, \dots et cetera, until returning to line $P_{N/2} P_0$, but in the reverse orientation. At the end of the process, every point in the plane has switched sides with our moving line. The moment that $X$ crosses the moving red line, we get it contained in a butterfly, as needed. $\blacksquare$ Claim: If $ABDC = P_i P_{i+1} P_{i+1+N/2} P_{i+N/2}$ is a butterfly, one of the triangles $ABC$, $BCD$, $CDA$, $DAB$ has area at least that of the butterfly. Proof. Let the diagonals of the butterfly meet at $O$, and let $a = AO$, $b = BO$, $c = CO$, $d = DO$. If we assume WLOG $d = \min(a,b,c,d)$ then it follows $[ABC] = [AOB] + [BOC] \ge [AOB] + [COD]$, as needed. $\blacksquare$ Now, since the $N/2$ butterflies cover an area of $S$, it follows that one of the butterflies has area at least $S / (N/2) = 2S/N$, and so that butterfly gives a triangle with area at least $2S/N$, completing the proof of the lemma. \bigskip Main proof: Let $a_1$, \dots, $a_n$ be the numbers assigned to the sides. Assume for contradiction $a_1 + \dots + a_n < 2S$. We pick even integers $m_1$, $m_2$, \dots, $m_n$ such that \begin{align*} \frac{a_1}{S} &< \frac{2m_1}{m_1 + \dots + m_n} \\ \frac{a_2}{S} &< \frac{2m_2}{m_1 + \dots + m_n} \\ &\vdotswithin\le \\ \frac{a_n}{S} &< \frac{2m_n}{m_1 + \dots + m_n}. \end{align*}which is possible by rational approximation, since the right-hand sides sum to $2$ and the left-hand sides sum to strictly less than $2$. Now we break every side of $P$ into $m_i$ equal parts to get an almost convex $N$-gon, where $N = m_1 + \dots + m_n$. The main lemma then gives us a triangle $\Delta$ of the almost convex $N$-gon which has area at least $\frac{2S}{N}$. If $\Delta$ used the $i$th side then it then follows the label $a_i$ on that side should be at least $m_i \cdot \frac{2S}{N}$, contradiction.

To each side $a$ of $P$, let $f(a)$ be the unique line parallel to $a$, on the same side of $a$ as $P$, and such that the intersection of $f(a)$ with $P$ has zero area (is either a point or a segment). Let $s(P)$ denote the set of edges of $P$, and let $\ell(a)$ denote the length of side $a\in s(P)$. Then, we see that the sum of the areas assigned to each side is \[\beta(P):=\frac{1}{2}\sum_{a\in s(P)}\ell(a)\cdot d(a,f(a)),\]where $d(x,y)$ is the distance between two parallel lines $x$ and $y$. The first step is to reduce the problem to the case where $P$ has opposite sides that are parallel. Indeed, call a side $a\in s(P)$ good if there is some other $b\in s(P)$ such that $a\parallel b$. We proceed by induction on the number of bad sides. We'll solve the case where all sides are good later, so we may assume for this induction that the base case of $0$ bad sides is given. Suppose we have $P$ with at least one bad side. Let $Q$ be the polygon formed by all the lines $f(a)$ for $a\in s(P)$. Lemma: There is some vertex of $Q$ that is outside of $P$. Proof:

Suppose to the contrary that all vertices of $Q$ were inside $P$ (here $P$ includes its boundary). If any vertex of $Q$ is not a vertex of $P$, then some side of $Q$ intersects $P$ in the interior, which is bad since the sides of $Q$ are segments on the lines $f(a)$. Thus, all vertices of $Q$ are vertices of $P$. However, if any vertex of $P$ is not a vertex of $Q$, then it is easy to see that again, one of the sides of $Q$ intersects the interior of $P$, which is bad. Thus, the vertices of $P$ and $Q$ are identical, so $P=Q$. This implies that every side of $P$ is good, so we have our desired contradiction. $\blacksquare$ Say $X$ is a vertex of $Q$ that is outside $P$. Let $a$ and $b$ be sides of $P$ such that the edges of $Q$ incident to $X$ are from $f(a)$ and $f(b)$. Let $A$ be the vertex of $P$ that is on $f(a)$ and let $B$ be the vertex of $P$ that is on $f(b)$ (note that these may not be unique if one of $a$ or $b$ is good, and we will deal with this issue later). Let $I_1,\ldots,I_m$ be the vertices of $P$ that are on the same side of $AB$ as $X$ is such that $I_0=A,I_1,\ldots,I_m,B=I_{m+1}$ are consecutive vertices in $P$. If there were two possible choices for $A$, then choose $A$ such that $A,I_1,\ldots,I_m,B$ are indeed consecutive. If there was any edge $c$ between $a$ and $b$ (on the other side as $A,I_1,\ldots,I_m,B$), then $f(c)$ would intersect the interior of the convex region $\triangle ABX$, which would imply that $X$ was not a vertex of $Q$, which is a contradiction. Thus, $a$ and $b$ are consecutive in $P$, and suppose they are both incident to $U$. We have the following key claim. Claim: We have $U\in f(I_jI_{j+1})$ for all $0\le j\le m$, and $f(c)\not\in\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$ for $c\in s(P)\setminus\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$. Proof:

We first prove that $U\in f(I_jI_{j+1})$ for all $0\le j\le m$. For some $j$, let $\ell$ be the line through $U$ parallel to $I_jI_{j+1}$. It suffices to show that $\ell$ doesn't intersect the interior of $P$. This is obvious by some directed angles. Now, suppose we had $f(c)\in\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$ for some $c\in s(P)\setminus\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$. Then, $f(c)$ would intersect the interior of the convex region $\triangle ABX$, which is absurd as we saw before. This completes the proof of the claim. $\blacksquare$ The key to the induction is to take $P'$ to be the polygon that is the convex hull of $P$ and $X$. Note that the vertices of $P'$ are the vertices of $P$ with the $I_j$ removed and $X$ added. We have the following claim which shows that reducing $P$ to $P'$ is valid under the inductive monovariant. Claim: The number of bad edges of $P'$ is less than the number of bad edges of $P$. Proof: The number of vertices of $P'$ is the number of vertices of $P$ plus $1-m$. If both $a$ and $b$ are bad, then clearly $m\ge 0$, if just one of them is bad, then $m\ge 1$, and if both are good, then $m\ge 2$. We see that both $a$ and $b$ become good when we replace $P$ with $P'$, and the goodness of all edges in $s(P)\setminus\{a,b,I_0I_1,\ldots,I_mI_{m+1}\}$ is unaffected. We lose the bad edges $\{I_0I_1,\ldots,I_mI_{m+1}\}$ entirely, which are replaced with the two good edges $AX$ and $BX$. Furthermore, the edges $a$ and $b$ become good if they weren't before. All this implies that the number of bad edges goes down (this is just casework on the goodness of $a$ and $b$), so we're done. $\blacksquare$ Thus, we may assume that the problem is true for $P'$. By construction, the areas written down for the edges in $s(P)\setminus\{I_0I_1,\ldots,I_mI_{m+1}\}$ remain unchanged. The area written on the edge $I_jI_{j+1}$ is $[I_jI_{j+1}U]$ (for $P$), and the areas written on the edges $AX$ and $BX$ (for $P'$) are $[AXU]$ and $[BXU]$. Thus, \[\beta(P')-\beta(P)=[AXU]+[BXU]-[AI_1\cdots I_mBU]=[AI_1\cdots I_mBX]=\alpha(P')-\alpha(P),\]where $\alpha(R)$ is the area of $R$. Thus, \[[\beta(P)-2\alpha(P)] - [\beta(P')-2\alpha(P')]= \alpha(P')-\alpha(P)\ge 0,\]so since $\beta(P')-2\alpha(P')$, we must have $\beta(P)-2\alpha(P)\ge 0$, as desired. This completes the induction part of our proof. Now we have to show the base case, i.e. the case where all the edges of $P$ are good. Let the vertices of $P$ be $A_1A_2\cdots A_{2n}$ with indices taken mod $2n$. Let $a_i$ be $A_iA_{i+1}$, and let $d_i=d(a_i,a_{i+n})$. Note that \[\beta(P) = \sum_{i=1}^n [\ell(a_i)+\ell(a_{i+n})]\cdot\frac{d_i}{2}.\]Let $m_i$ be the line halfway between $a_i$ and $a_{i+n}$. The key idea is to move the $m_i$ (while preserving the relative distances between $a_i,m_i,a_{i+n}$). As long as the polygon remains convex, the value of $\beta(P)$ does not change under this movement. However, the value of $\alpha(P)$ may change. We have the following key lemma. Lemma: Suppose we have oriented non-parallel lines $m_1,\ldots,m_n$ that can move while maintaining their direction. We have fixed $r_1,\ldots,r_n$, and define $a_i$ to be the line exactly $r_i$ to the left of $m_i$ (this makes sense as the lines are oriented) and $a_{i+n}$ to be the line exactly $r_i$ to the right of $m_i$. Let $A_i=a_i\cap a_{i-1}$ for $1\le i\le 2n$. For an appropriate global area orientation, the signed area $[A_1A_2\cdots A_{2n}]$ attains a maximum value when all the $m_i$ are concurrent. Proof: The proof proceeds by induction on $n$. The case $n=2$ is trivial, as the area $[A_1A_2A_3A_4]$ is actually always fixed. So suppose the result is true for some $n-1\ge 2$. We'll show it's true for $n$. Here is a diagram for $n=4$.

Let $X=a_{n-1}\cap a_{n+1}$ and $Y=a_{2n-1}\cap a_1$. Fix $m_1,\ldots,m_{n-1}$, while varying $m_n$. Then, we have the signed area equality \[[A_1A_2\cdots A_{2n}] = [YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}] - [YA_1A_{2n}] - [XA_{n+1}A_n].\]In the following diagram, the highlighted parallelogram is fixed as $m_n$ varies, and furthermore, its dimensions and orientation is only dependent on the fixed directions of $m_1,\ldots,m_{2n}$ and the fixed distances $r_1,\ldots,r_n$.

As $m_n$ varies, we see that \[[YA_1A_{2n}] + [XA_{n+1}A_n] = k(p^2+q^2)\]where $p$ and $q$ represent the signed distances $A_1A_{2n}$ and $A_nA_{n+1}$, so $p+q:=s$ is fixed, and $k$ is some fixed proportionality constant only dependent on the fixed directions of $m_1,\ldots,m_{2n}$ and the fixed distances $r_1,\ldots,r_n$. Thus, $[YA_1A_{2n}] + [XA_{n+1}A_n]$ attains the same minimum value $T=ks^2/2$ as $m_n$ varies, no matter where the $m_1,\ldots,m_{n-1}$ are. The area \[[YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}]\]is exactly the area from the $n-1$ case of the lemma with lines $m_1,\ldots,m_{n-1}$, so the area \[[A_1A_2\cdots A_{2n}] = [YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}]-T\]is maximized when $m_1,\ldots,m_{n-1}$ are in optimal position, which is concurrent by the base case. In the case that $m_1,\ldots,m_{n-1}$ are concurrent (so the polygon $YA_2\cdots A_{n-1}XA_{n+2}\cdots A_{2n-1}$ is centrally symmetric) , $m_n$ is placed optimally when $A_1A_{2n}=A_nA_{n+1}$. It's easy to see that this implies that $m_n$ also passes through the same center, so $m_n$ is concurrent with $m_1,\ldots,m_{n-1}$, as desired. $\blacksquare$ Thus, the area $\alpha(P)$ is at most \[\sum_{i=1}^{2n}\frac{\ell(a_i)+\ell(a_{i+n})}{2}\cdot d_i = \sum_{i=1}^{n}[\ell(a_i)+\ell(a_{i+n})]\cdot d_i=2\beta(P),\]as desired. This is because if we translate the $m_i$s to make $P$ centrally symmetric, the (signed) area becomes \[\sum_{i=1}^{2n}\frac{\ell(a_i)+\ell(a_{i+n})}{2}\cdot d_i,\]and this is an upper bound for $\alpha(P)$ by the lemma.

. The boundary of this shape can be parametrized with a function $f \colon \mathbb{R}/2\pi\mathbb{Z} \to \mathbb{C}$, such that $f(\theta)$ is the point on the boundary where $f'(\theta)$ is a positive multiple of $e^{i\theta}$. Now, by Shoelace the area is given by \[A = \int_0^{2\pi} \frac{f'(\theta) \overline{f(\theta)} - \overline{f'(\theta)} f(\theta)}{4i}\,d\theta,\]while the sum of areas in the problem is well-approximated by \[B = \int_0^{2\pi} \frac{f'(\theta) \overline{f(\theta)-f(\theta+\pi)} - \overline{f'(\theta)} (f(\theta)-f(\theta+\pi))}{4i}\,d\theta,\]so it suffices to evaluate these quantities and compare them. To do this, we write $f$ as a Fourier series. First, note that $e^{-i\theta} f'(\theta)$ is real, so it can be written as $\textstyle \sum_{k \in \mathbb{Z}} c_k e^{ik\theta}$ with $c_{-k} = \overline{c_k}$. Also note that $c_1 = 0$, since $f'(\theta)$ has a vanishing zeroth Fourier coefficient. Furthermore, we have \[f(\theta) = \sum_{k\in\mathbb{Z}\setminus\{0\}} \frac{c_{k-1}}{ik} e^{ik\theta},\]modulo some constant that we can WLOG assume is $0$ (we have no convergence issues anywhere since smoothness implies the coefficients decay superpolynomially). Using the fact that \[\int_{0}^{2\pi} \left(\sum_{k\in\mathbb{Z}} a_k e^{ik\theta} \right) \overline{\left(\sum_{k\in\mathbb{Z}} b_k e^{ik\theta} \right)}\,d\theta = 2\pi \sum_{k \in \mathbb{Z}} a_k \overline{b_k},\]the integrals above can be evaluated to be \[A = \pi \sum_{k \in \mathbb{Z}\setminus\{-1\}} \frac{\lvert c_k \rvert^2}{k+1} \text{ and } B = 2\pi \sum_{k \in 2\mathbb{Z}} \frac{\lvert c_k \rvert^2}{k+1},\]respectively. Hence \[B - 2A = 2\pi \sum_{k \in (2\mathbb{Z} + 1)\setminus\{-1\}} -\frac{\lvert c_k \rvert^2}{k+1} = 2\pi \sum_{\text{odd }k \geq 3}\lvert c_k \rvert^2 \left(\frac{1}{k-1} - \frac{1}{k+1}\right) \geq 0,\]which finishes upon taking better and better approximations.