Additional interesting factors: 8*b^2-a^2, 12*b^2-a^2. So, a complete list of interesting factors: (3bb-aa), (8bb-aa), (12bb-aa), (15bb-aa), (27bb-aa). It would be interesting to understand why the ratios a/b=sqrt(3), sqrt(8), sqrt(12), sqrt(15), sqrt(27) are so special. At 09:19 AM 11/21/2010, Henry Baker wrote:
I did some more work on embedding an equilateral pentagon (cyclicly, P1,P2,P3,P4,P5; P1=(x1,y1), etc.) into a standard ellipse (x/a)^2+(y/b)^2=1.
Once again, I assumed that (x1,y1)=(a,0).
The problem with the polynomial equations is that they incorporate all sorts of degenerate cases. For example, P4=P2, where the links P2P3 and P3P4 are the same line segment. We then have an equilateral triangle with an extra link hanging off of one corner.
Another degenerate case has P1-P5 all distinct, but the links P2P3 and P4P5 cross each other to the other side.
I don't know how to easily eliminate these degenerate cases from the set of equations.
For this pentagonal case, there are three interesting factors that enter in: (3*b^2-a^2), (15*b^2-a^2), and (27*b^2-a^2). I assume that these again define critical ratios of b/a. For example, as b gets larger relative to a, the degenerate case where P2P3 crosses P4P5 collapses to where P3P4,P2P3,P4P5 all become the same link, and we have just an equilateral triangle. So this behavior can only happen when b/a is small enough.
In order to eliminate these degenerate solutions, we may have to put in additional constraints. For example, in the case of the pentagon with P1=(a,0), the (squared) distances P1P3 and P1P4 are equal. I don't know how many degenerate solutions we can eliminate, but we can add equality constraints for many of the chords based on a presumed symmetry around the x axis.
P1P3 = P1P4, P2P4 = P3P5.
None of these constraints eliminate the case where the pentagon degenerates into a triangle.