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For illustration click here

I have a simple convex irregular polygon (octagon in example image) inside a circle (circle and polygon are not always concentric and never touching or intersecting) and I need to determine the polygon's position relative to the circle.

Unfortunately I only know three basic things...

  1. I know the exact angle between each of the polygon's opposing sides. No sides are parallel.
  2. When extended to the circle, the polygon's edges/sides form inscribed angles. Since one angle accounts for two polygon sides, there are always half as many inscribed angles as polygon sides. In the example there are four inscribed angles because it's an octagon.
  3. The vertices of each inscribed angle are evenly spaced around the circle.

My worry is that this kind of calculation might involve an infinite number of iterations and/or multiple solutions.

If I knew the relative angles (labeled 'Unknown' in drawing) of the inscribed angles, that should be enough to determine everything else, right? Even if I could approximate the 'centre' of the irregular polygon somehow...

*The bottom right edges in the example meet with such an obtuse angle that they appear to be one line, sorry. It is an octagon with each inscribed angle creating two sides of the polygon.

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It is impossible to calculate your figure, because the information you have does not completely determine the figure. Your fear about "multiple solutions" is correct--in fact, infinitely many of them. You need more information to get a unique solution--the "unknown's" in your drawing would do.

Here are two diagrams, both satisfying your conditions and similar to your octagon in your diagram. The only difference between the two is the orientation of the $9.04°$ angle. I hope you can see the differences.

enter image description here

enter image description here

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  • $\begingroup$ Rory, Thank you. I see what you mean. Now as the polygon sides increase, the number of solutions appear to decrease, correct? Approaching a circle, there is only one solution? Is that the only polygon with one solution? I also have area of the polygon to work with, but it's only a very crude measure and using your example it's clear increasing the number of polygons is a better path. Thank you again. $\endgroup$ – JaV Jul 24 '15 at 21:04
  • $\begingroup$ @JaV: There are infinitely many solutions (I just showed two) for any number of sides of the polygon, so the number of solutions does not increase. The more restrictions, such as the number of given angles, the more difficult it is to satisfy them--I had difficulty getting an octagon with the angles you gave, that's why the placement of my octagon differs from yours. Is that difficulty what you mean by "number of solutions"? $\endgroup$ – Rory Daulton Jul 24 '15 at 22:32
  • $\begingroup$ Sorry for the delay. $\endgroup$ – JaV Jul 30 '15 at 14:12
  • $\begingroup$ Thank you again. I now see what you're saying. What led me astray was the apparent shrinking trend I observed. As the number of polygon sides increase, the area section containing all possible solutions decrease, correct? My faulty assumption was that it would reduce to one point, one solution. Unfortunately, even when imperceptibly small, it's still not one. Since this is for an engineering project in which solving for this area section would suffice, is there a way to calculate the extents of this "area of possible solutions"? Have a great day $\endgroup$ – JaV Jul 30 '15 at 14:24

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