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Given the length of the sides of an irregular polygon (no coordinates provided) how do you compute the area of the maximum area of the polygon?

Thanks in advance

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    What is "the arae of the maximum area"?2012-03-01

5 Answers 5

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The polygon must be cyclic. For details, check this out.

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    Yes, the irregular polygon is cyclic. But I am stuck on how to compute the area of the irregular cyclic polygon or the area of the contained triangles, given that for each of the (n-2) triangles in my polygon, I am only aware of the length of just two sides.2012-03-01
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    For some reason, I missed that you wanted to compute the area. I could only find this paper: http://arxiv.org/pdf/math/0408104v1.pdf which (if I understand correctly) implies that an algebraic relation between the sides and the area is not known.2012-03-01
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    There are apparently also formulas for pentagons and hexagons: http://www.springerlink.com/content/37266l8v11r0l512/2012-03-01
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    Can every set of side-lengths be made into a cyclic polygon?2012-03-01
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    @BlueRaja-DannyPflughoeft: No. The diameter of the circle must be at least the length of the longest side, and then the angles formed by the remaining sides at that radius must add up to at least $\pi$. On the other hand, if they do, then the intermediate value theorem applied to the left-hand side of the first equation in my answer says that there is a suitable radius.2012-03-02
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    @joriki: Er, so this answer doesn't answer the question, at least not for the case that the polygon can't be made cyclic...2012-03-02
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    @BlueRaja-DannyPflughoeft: Sorry, I was wrong there. The whole polygon could be on one side of the centre, and then the angles of the remaining sides only have to add up to the angle of the one side. As the radius goes to infinity, that turns into the condition that the remaining sides have to add up to at least the one side, which is a generalized form of the triangle equality that has to hold for any polygon, not just a cyclic one. So while not every set of side lengths can be made into a cyclic polygon, every set of side lengths that can be made into a polygon can be made into a cyclic one.2012-03-02
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Since the very interesting treatments that aelguindy linked to appear to indicate that no formulas are known beyond hexagons, it seems you'll need to determine the area numerically. One efficient approach to doing this might be to solve the condition for the angles to add up to a full circle,

$$\sum_k\arcsin\frac{L_k}{2R}=\pi\;,$$

for $R$ numerically (e.g. using Newton's method). Then the area is

$$ \begin{eqnarray} A &=& \frac{R^2}2\sum_k\sin\left(2\arcsin\frac{L_k}{2R}\right) \\ &=& \frac{R}2\sum_kL_k\sqrt{1-\left(\frac{L_k}{2R}\right)^2}\;. \end{eqnarray} $$

[Edit as requested:]

Here's how you could solve for $R$ using Newton's method. The general prescription for using Newton's method to solve the equation $f(x)=0$ is

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\;.$$

This is where the tangent to the graph at $x_n$ intersects the $x$ axis. In the present case, we want to solve

$$f(R)=\sum_k\arcsin\frac{L_k}{2R}-\pi=0\;,$$

so we form

$$f'(R)=-\sum_k\frac{L_k}{2R\sqrt{(2R)^2-L_k^2}}$$

and iterate using

$$R_{n+1}=R_n+\frac{\sum_k\arcsin\frac{L_k}{2R_n}-\pi}{\sum_k\frac{L_k}{2R_n\sqrt{(2R_n)^2-L_k^2}}}\;.$$

Now we just need a suitable initial value $R_0$. This must be somewhere between the minimal radius $L_\max/2$ (where $L_\max$ is the maximal side length) and infinity; a reasonable choice might be $L_\max$.

Note that Newton's method isn't guaranteed to converge in general; I think it should work in this case, but I haven't tried it. If it doesn't work, you may have to use a more robust root-finding algorithm.

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    I am not certain I follow. I don't know what R is and what does L(k) stand for?2012-03-01
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    $L_k$ are the lengths of the sides, $R$ is the radius of the circle.2012-03-01
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    You can also do it "algebraically", as the condition $\sum_k \arcsin(L_k/(2R)) = \pi$ is equivalent to $\prod_k \left( \sqrt{1 - (L_k/(2R))^2} + i L_k/(2R)\right) = -1$. This will imply a polynomial equation in $R$ -- but it's going to be horrendously complicated I think.2012-03-01
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    one more silly question . . . where do I get R from? :)2012-03-01
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    @Kobojunkie: Sorry, I don't think I understand the question. I suggested to solve the condition for the angles for $R$ numerically; "solving for $R$" means finding a value of $R$ such that the condition is satisfied, so this step would yield a value for $R$.2012-03-01
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    @Kobojunkie: a) If your $\TeX$ input doesn't work (as above), you can edit a comment for up to five minutes to fix it. If you can't edit it anymore, you can delete and repost it. Please don't leave it in such a state; it's ugly and hard to read. b) I don't understand how you arrived at that equation, but you can guess from the fact that lots of papers have been written about this that it can't be that easy. If you could express $R$ like that, you could simply calculate the area according to the formula in my answer. As I wrote, you'll need to determine $R$ numerically.2012-03-02
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    P.S.: $\sin\pi=0$, so that expression actually isn't defined since you're dividing by $0$.2012-03-02
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    Thanks. What I was trying to do is figure out a formula for computing R from the first equation you have there in your answer. I figure if I can compute R by using the edges I have, I can then compute Area using the second approach as well.2012-03-02
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    @Kobojunkie: Yes, that was the idea, just that this can't be done in such a simple closed form. As Robert pointed out, you can try doing it algebraically, but you'll get very complicated expressions. If you want to try to deal with those for theoretical reasons, best of luck; but if you just want a practical way of finding the area, applying Newton's method to the formula as I stated it is going to be a lot easier.2012-03-02
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    @joriki, I am writing a program to do this(not much of a mathematician yet) but I hope that by coding some of the problems i have picked, I can get a better idea of what is out there. So, far, I have a solution for polygons <=4. I need a better explaination please of your Newton idea, since all I know is that I need to find R before I can find Area.2012-03-02
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    @Kobojunkie: I added a part on how to apply Newton's method to find $R$.2012-03-02
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    @joriki, I am still having some problem here understanding the Newton's method please. I get what you are trying to do but I still see R on both sides of the final equation that I am still missing. For instance, from what I see, you have initial value $R_0$, as well as $R_n$. What values are these for me?2012-03-05
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    @Kobojunkie: I suggest you read about [Newton's method](http://en.wikipedia.org/wiki/Newton%27s_method) at Wikipedia. The article has a worked out example. I already gave a suggestion for the value of $R_0$ in my answer. All other values are determined iterally according to the iteration rule, which expresses the next value $R_{n+1}$ in terms of the previous value $R_n$.2012-03-06
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The polygon should be cyclic, meaning that its vertices will lie on the circumference of a circle(assuming the polygon is convex). If the side lengths are given as $x_1, x_2, x_3, x_4,\ldots$ and $s=\frac{p}{2}$ where $p$ is the perimeter then the area is given by something akin to Brahmgupta's formula for $4$ sides or Heron's formula for a triangle. Not sure what the area is given by when the number of sides exceeds $4$.

You can usually find the areas of irregular convex polygons by dividing them into triangles wisely, though.

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    Thanks. I did go ahead and implement the Brahmgupta solution to handle 3/4 sided polygons, but I still would prefer a solution that handles all edge counts.2012-03-02
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Following on @aelguindy's comment, this was an area of research by one Prof. David Robbins (who passed away prematurely). Here is a typical paper from that research effort: he was able to find explicit expressions for the areas of cyclic heptagons and octagons. Anyone who browses through that paper will understand that this problem is extraordinarily difficult in general.

For amusement, I compared relatively simple results I got from a Putnam exam problem with his result and got agreement, happily. Here is the problem:

Find the area of a convex octagon inscribed in a circle that has 4 consecutive sides of length 3 units and 4 consecutive sides of length 2 units. Give your answer in the form $r+s \sqrt{t}$, where $r$, $s$ and $t$ are positive integers.

I'll leave it as an exercise for the reader.

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because we know the sides So the angles formed by these sides at the center of the circle will be in the same ratio as the sides for example if the sides of the polygon are a,b,c,d and e then the sum of all the angles at the center will be = a*x+b*x+c*x+d*x+e*x and this sum will be equal to 360.Sonow we know all the angles made by the sides at the centre. Now we can find the area of each triangle and sum of these areas is the area of the polygon

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    Could you explain how this answers the question? Also, your statement "the angles will be in the same ratio as the sides" is incorrect. Maybe you meant to write "the angles will be in the same ratio as the corresponding arcs", but it is not the length of the arcs that are given.2013-02-06