Re: Intersecting Circles: a question
- From: ol3@xxxxxxxxx (Oscar Lanzi III)
- Date: Sun, 3 Jun 2007 16:00:05 +0000 (UTC)
You have to be careful dealing with even the three-circle problem.
There are several possible topoligies with three disks in a plane, and
not all are linked together by the same mathematical relationship. A
situation in which one disk is inside the area of other two does not
yield to the same treatment as one in which each disk has a region
external to the others.
With that in mind, I propose a solution for the case where all three
pairs of disks are known to have partial overlap. In that case the
topological quesation is whether the figure has a hole like a badly
shaped donut, or is filled in the center like a badly shaped danish.
Imagine that the three circles are great circles of spheres. Then the
circles will enclose a common region iff the spheres have two points of
intersection in the three-dimensional space; the enclosed region is the
cross-section of a (non-polyhedral) solid whose vertices are the points
of intersection of the spheres. We can then form a tetrahedron with
vertices A = center of circle 1, B = center of circle 2, C = center of
circle 3, and D = either point of intersection of the spheres. We know
A, B, and C and thus the distances between them; and while we do not
know D (or whether it even exists), we know that AD must be the radius
of circle A, BD must be the radius of circle B, and CD must be the
radius of circle D. This enables one to form the Slater matrix using
the six distances:
0 AB^2 AC^2 AC^2
AB^2 0 BC^2 BD^2
AC^2 BC^2 0 BD^2
AD^2 BC^2 CD^2 0
The deteminant of this matrix is 288 times the volume of the tetrahedron
squared, meaning the tetrahedron can exist only if this determinant is
positive. (We do have the borderline case of a plane figure in the case
where the determinant is exactly zero.) Therefore a positive Slater
determinant is a necessary condition for the disks in the plane to
overlap in a region. With a little geometry and trignonometry, we can
also show that a positive Slater determinant is also sufficient for the
existence of a nondegenerate tetrahedron and therefore a region of
overlap. (A zero determinant means the overlap is reduced to one point
inside the triangle fomremed by the circle centers.)
For your case, plugging in the information for your circles gives at
most a fourth-degree polynomial function of gamma. No inverse cosines
required.
--OL
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