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Using a perl program testing for integer solutions to the derived equations I have found the following answers, where a,b,c are the initial radii, x,y are the centre of c, s,t are the centre of the containing circle and r is its radius:
abcxystr
44038533077616-1112840
8807706601541232-22241680
121810159572901740-74202310
156072871510921365-82102310
175594591010831729-92732730
1760154013203082464-44483360
2200192516503853080-55604200
2436203019145803480-148404620
2511195317367753255-95674536
28561768154714523003-42734641
3080269523105394312-77845880
31201456143021842730-164204620
32451947188817703540-226605280
These have all been verified against the original equations and confirm to the requirement that the length of a line from the centre of the containing circle, through the centre and to the end of a diameter of each other circle, is of length r.
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Originally posted by joe shmoHi Joe
Next bisect all of angles using the point of intersection from the bisectors as the one vertex for a new triangle that has side length "b" (assuming r_1>r_2>r_3) and solve for the remaining legs from the half angles and the law of sines, taking the larger of the legs to be r'
Once you have the legs, draw a circle of radius "R" = ( r_1 + r' ) centered on ...[text shortened]... ctor intersection of the original triangle.
Like I said, seems messy, is it a valid approach?
Can you explain, what the "remaining legs" are in this context, and what the orientation of the "new triangle" is - where are its corners?
I would like to test whether your method gives the same results as mine (if circle 3 is large enough) but I do not understand your algorithm description well enough, sorry if I am being dense.
However with the tool geogebra, I have drawn one of my listed integer solutions (which proves that is definitely correct), and for the one I tested the centre of the containing circle does NOT lie on the lines bisecting the angles of your triangle, so I suspect your method is not valid.
I am trying to find a valid geometrical solution (using geogebra) but it seems very difficult, as might be expected given the complexity of the algebraic solution
Originally posted by iamatigerI intended to use the "centroid" of the triangle that joins the centers of 3 the circles to find the larger circle. It worked swimmingly for 3 circles all of radius "r", so I thought ( rather prematurely, and without good reason ) that extending this method to 3 circles of radius a,b,c, where a>b>c would just be an extension into a little more trig ( I was wrong).
Hi Joe
Can you explain, what the "remaining legs" are in this context, and what the orientation of the "new triangle" is - where are its corners?
I would like to test whether your method gives the same results as mine (if circle 3 is large enough) but I do not understand your algorithm description well enough, sorry if I am being dense.
However with t ...[text shortened]... but it seems very difficult, as might be expected given the complexity of the algebraic solution
Since then I have tried using the circumcenter, orthocenter, and fermat point of the respective triangle. Nothing seems to bear any fruit.
For instance, for circles with radius "R"
we have an equilateral of side length 2R
from its known interior angles, find the length of the line that connects any vertex to the centroid.
the line has a length:
2R*sin(120)/sin(30).
The raduis of the large circle tangent to all three is then:
R + 2R*sin(120)/sin(30)
But as I said, it doesn't generalize to the posed problem.
Its also becoming apparent to me that for there to be a solution where all three minor circles are tangent to each other and all simultaniously tangent to the major circle that encompasses them all that the dimension and size of the smallest minor, and the major circle is entirely dependent on the size of the two largest minor circles in the group.
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Originally posted by joe shmoI don't think so.
Its also becoming apparent to me that for there to be a solution where all three minor circles are tangent to each other and all simultaniously tangent to the major circle that encompasses them all that the dimension and size of the smallest minor, and the major circle is entirely dependent on the size of the two largest minor circles in the group.
for a given largest, and second largest circle, A and B, there is certainly a threshold smallest size for the smallest circle C, below which an encompassing circle that is tangent to them all does not exist. However there is a solution for a tangent, encompassing circle for all sizes C in between that threshold and B.
Originally posted by iamatigerAh Ha, your right. I was confusing myself. I was thinking that the "threshold" was the solution, because it would give the "smallest" possible radius for the Major circle tangent to all in a family of circles A,B,C where A>B>C.
I don't think so.
for a given largest, and second largest circle, A and B, there is certainly a threshold smallest size for the smallest circle C, below which an encompassing circle that is tangent to them all does not exist. However there is a solution for a tangent, encompassing circle for all sizes C in between that threshold and B.
Sorry about that, not quite how the original question was stated!