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May 10th, 2009, 09:33 AM   #1
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How does area of an n-gon contrain the length of its sides?

Hi,
I have been busy working on some ideas for my own education and amusement, and have continuously run into a problem that I really do need to address now (i.e. it is important to me) And that problem is this...

Let's say that on the euclidean plane we have an n-gon with area A. Edges do not need to be the same length (these need not be regular n-gons).

Each of its sides are geodesics (in other words - they are lines connecting the vertices through the shortest possible paths on the euclidean plane) or more simply, they are straight lines.

What would be the constraints and relationships amongst the lengths of the n-gons edges given a set area A. How many edges would be completely free or independent, and how many would be constrained, and what would be their constraints?

Oddly enough, I can't seem to find information on this anywhere. It's not really an optimization problem, per se, but a constraint problem of some sort. I would also be interested in how a volume constrains the areas of its faces.

Thank you very much to anyone who can point me in the right direction on this. I am not lazy, and not against figuring things out on my own, but a little help can make a massive difference.
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May 10th, 2009, 11:13 AM   #2
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Re: How does area of an n-gon contrain the length of its sides?

*Note - I also don't necessarily see any reason that the n-gons should have to be concave or convex, they can be either. I do wish the n-gons to be simple, though. Meaning the area cannot overlap itself.
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May 10th, 2009, 06:02 PM   #3
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Re: How does area of an n-gon contrain the length of its sides?

Quote:
Originally Posted by telltree
Hi,
I have been busy working on some ideas for my own education and amusement, and have continuously run into a problem that I really do need to address now (i.e. it is important to me) And that problem is this...

Let's say that on the euclidean plane we have an n-gon with area A. Edges do not need to be the same length (these need not be regular n-gons).

Each of its sides are geodesics (in other words - they are lines connecting the vertices through the shortest possible paths on the euclidean plane) or more simply, they are straight lines.

What would be the constraints and relationships amongst the lengths of the n-gons edges given a set area A. How many edges would be completely free or independent, and how many would be constrained, and what would be their constraints?

Oddly enough, I can't seem to find information on this anywhere. It's not really an optimization problem, per se, but a constraint problem of some sort. I would also be interested in how a volume constrains the areas of its faces.

Thank you very much to anyone who can point me in the right direction on this. I am not lazy, and not against figuring things out on my own, but a little help can make a massive difference.
I am not sure what you are looking for. However, consider a case where n is even and you center the figure at the origin. Place alternate points near the origin in a tight ring. Place the other on a ring far enough away so that you have the area you want. You will have a star shaped figure with long side lengths. The total area is approximately the sum of n/2 equal sided triangles, where the bases are arbitrarily small. To get the area the sides (which are approximately equal to the altitude) would be arbitrarily long.
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May 11th, 2009, 07:56 AM   #4
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Re: How does area of an n-gon contrain the length of its sides?

Thank you very much for that response, and it is an interesting case that you bring up, and I will consider it more. At the moment I am looking at a more general case where lengths and exterior angles can be anything up until the point that they constrain each other, and am using this equation to work from...

The area A of a simple polygon can also be computed if the lengths of the sides, and the exterior angles,are known. The formula is



The formula was described by Lopshits in 1963, and can be found on the wikipedia page.

I will probably be playing around with this formula to derive what I need.
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May 11th, 2009, 05:04 PM   #5
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Re: How does area of an n-gon contrain the length of its sides?

The simplest example I can describe is a rhombus. Let a and b be the lengths of the diagonals.
Area = ab/2, while perimeter=2(a^2 + b^2)^1/2.
Let a be arbitrarily small, then for fixed area, b becomes arbitrarily large, with the perimeter approx = 2b.

You can also do it with a triangle by making one side arbitrarily small.

What I was trying to describe is a generalization of the rhombus case for an arbitrary (even) number of sides.
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May 12th, 2009, 06:23 AM   #6
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Re: How does area of an n-gon contrain the length of its sides?

Thank you mathman again, that is a great example, and gives me some direction.
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May 12th, 2009, 07:13 AM   #7
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Re: How does area of an n-gon contrain the length of its sides?

Also - I have been away from geometry for so long that I have grown week in it, and I'm reviewing some old geometry books but haven't found this simple bit of information.

Assuming I am dealing with irregular polygons, let's say convex irregular polygons - If I have the length of all of the sides, I assume this is enough information to calculate the interior angles, yes? What is the general formula(s) for determining the interior angles of an irregular convex n-gon? I suppose I could treat them as vectors and go about it this way, but I am wondering if there is an easier way, as I will have to calculate these values repeatedly over the next few days, countless times.

Thank you.
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May 12th, 2009, 05:01 PM   #8
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Re: How does area of an n-gon contrain the length of its sides?

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If I have the length of all of the sides, I assume this is enough information to calculate the interior angles, yes?
No. Only for triangles. For a quadrilateral, consdier a rhombus of fixed side length. The angles are arbitrary as long as they add up to 2pi and opposite angles are equal.
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