# Calculating the area of an irregular polygon

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?

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What is "the arae of the maximum area"? –  Pedro Tamaroff Mar 1 '12 at 18:33

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. –  Kobojunkie Mar 1 '12 at 18:28
For some reason, I missed that you wanted to compute the area. I could only find this paper: 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. –  aelguindy Mar 1 '12 at 18:33
There are apparently also formulas for pentagons and hexagons: springerlink.com/content/37266l8v11r0l512 –  aelguindy Mar 1 '12 at 18:34
Can every set of side-lengths be made into a cyclic polygon? –  BlueRaja - Danny Pflughoeft Mar 1 '12 at 22:06
@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. –  joriki Mar 2 '12 at 7:20

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? –  Kobojunkie Mar 1 '12 at 20:55
$L_k$ are the lengths of the sides, $R$ is the radius of the circle. –  Robert Israel Mar 1 '12 at 21:10
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. –  Robert Israel Mar 1 '12 at 21:25
one more silly question . . . where do I get R from? :) –  Kobojunkie Mar 1 '12 at 22:00
@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 $R$. –  joriki Mar 1 '12 at 22:07

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. –  Kobojunkie Mar 2 '12 at 4:53

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. –  Daan Michiels Feb 6 '13 at 14:46