Round Shapes

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:

a b c x y s t r
440 385 330 77 616 -1 112 840
880 770 660 154 1232 -2 224 1680
1218 1015 957 290 1740 -7 420 2310
1560 728 715 1092 1365 -8 210 2310
1755 945 910 1083 1729 -9 273 2730
1760 1540 1320 308 2464 -4 448 3360
2200 1925 1650 385 3080 -5 560 4200
2436 2030 1914 580 3480 -14 840 4620
2511 1953 1736 775 3255 -9 567 4536
2856 1768 1547 1452 3003 -4 273 4641
3080 2695 2310 539 4312 -7 784 5880
3120 1456 1430 2184 2730 -16 420 4620
3245 1947 1888 1770 3540 -22 660 5280

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.

Originally posted by joe shmo
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?

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 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 iamatiger
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

I 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).

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.

Originally posted by joe shmo
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.

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.

Originally posted by iamatiger
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.

Ah 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.

Sorry about that, not quite how the original question was stated!

Hmmm, it looks like the centre of the containing circle may always lie on a line joining the intersection of circles A&B to the intersection of circles B&C...

I will look into that

Originally posted by iamatiger
Hmmm, it looks like the centre of the containing circle may always lie on a line joining the intersection of circles A&B to the intersection of circles B&C...

I will look into that

Nope, that is not true