Area of fractals - Area of quasifuchsian grandma fractals

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I know that area of fractals yet is more difficult for calculate with math formulas.
Except make IFS surface fractals whick determining with infinity sums of geometric series.
But the quasifuchsian fractals areas it can determining with formulas, whick will coming soon.
Yet I stuck at formula for determining a ray of circle kissing other three circles with rays defined, because seem to need to solve a algebrical equation
kind 8 like sqrt(a*b*x*(a+b+x))+sqrt(a*c*x*(a+c+x))+sqrt(b*c*x*(b+c+x))=sqrt(a*b*c*(a+b+c)) where (a) and (b) and (c) are rays of the three circles, and (x) is the unknown ray of the circle whick touching the three circles and between the three circles. The three circles A,B,C with rays (a) and (b) and (c) have three kissings: AB,BC and AC. The sqrt-s was obtained with Heron formula for areas of triangles. Else this lead to an infinity sum.
Through if area of a triangle with lengths (a+b), (b+c) and (a+c) the area give sqrt(a*b*c*(a+b+c)).

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Finnaly I found a method for solve ray of a circle between other three circles tangents between this, for avoiding formula
sqrt(a*b*c*(a+b+c))=sqrt(a*b*d*(a+b+d)+sqrt(a*c*d*(a+c+d))+sqrt(b*c*d*(b+c+d)) whick lead at 8th degree algebraic equation and is more difficult.
With more geometric renders based on inversion theory and using analytic geometry, I did it formula to this ray:
r=(a*b*c)/(a*b+a*c+b*c+sqrt[a*b*c*(a+b+c)]).
Then for I have rendering 2D quasifuchsian fractal area , I must using an infinity suma based on this formula of ray, but I assert this value give a irrational number whick is not arcsinus or sinus or root or exponential or logarithm or other classic function. If use formula
r=(a*b*c)/(a*b+a*c+b*c+sqrt[a*b*c*(a+b+c)]) in reccurence mode for know rays of all circles in INDRA fractal, the math expressions growth harder, as example (a*b*r)/(a*b+a*r+b*r+sqrt[a*b*r*(a+b+r)]) is expression more difficult and bigger and with 4th degree (contain two
overlaped sqrt-s) like a 4th degree equation solution. So for I composing this infinity sum, I must using reccurence for this formula.
Moment the area give PI*(5*5+3*3+1,34322132*1,34322132+2,902934986*2,902934986+...) if a=3 and b=4 and c=5 and two circles with rays 3 and 5 is inside QUASIFUCHSIAN fractal, yet the render isn't ready
More renders coming soon at this topic.

[hgjf2]

Not. For determining area of all circles in not so difficult because the overlaping square roots it can resolving as perfect squares.
That formula abc/(ab+ac+bc+2S) where S=sqrt (abc(a+b+c)) applied on A=abc/(m11*ab+m12*ac+m13*bc+2*m14*S) ,
B=abc/(m21*ab+m22*ac+m23*bc+2*m24*S) and C=abc/(m31*ab+m32*ac+m33*bc+2*m34*S), so if replace a,b and c with A,B and C the formula give
 abc/(A+B+C+2*sqrt(AB+AC+BC)) where AB+AC+BC allways is perfect square if starting from m13=m22=m31=1 ; m11=m12=m14=m21=m23=m24=
=m32=m33=m34 = 0.

[hgjf2]

Not. For determining area of all circles in not so difficult because the overlaping square roots it can resolving as perfect squares.
That formula abc/(ab+ac+bc+2S) where S=sqrt (abc(a+b+c)) applied on A=abc/(m11*ab+m12*ac+m13*bc+2*m14*S) ,
B=abc/(m21*ab+m22*ac+m23*bc+2*m24*S) and C=abc/(m31*ab+m32*ac+m33*bc+2*m34*S), so if replace a,b and c with A,B and C the formula give
 abc/(A+B+C+2*sqrt(AB+AC+BC)) where AB+AC+BC allways is perfect square if starting from m13=m22=m31=1 ; m11=m12=m14=m21=m23=m24=
=m32=m33=m34 = 0.

The skew:

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Now I try to solve area of this fractal.
So the area formula is under construct yet. I working

[hgjf2]

If after more calculates and research, I founded a 3D equivalent for kissing circles ray formule, than a ray for a sphere neighbour other 4 spheres tangent.
 r = a*b*c*d/(-2*a*b*c-2*a*b*d-2*a*c*d-2*b*c*d+sqrt(3*(a*b*c)^2+3*(a*b*d)^2+3*(a*c*d)^2+3*(b*c*d)^2-6*a*b*c*d*(a*b+a*c+
a*d+b*c+b*d+c*d))).
If in 2D , r=a*b*c/(a*b+a*c+b*c+2*sqrt(a*b*c*(a+b+c))).
Elsewere the infinity sum have same method.

[hgjf2]

If take the ray of the three circles as 1;1;2/3. The infinity sum what generate area of GRANDMA fractal generated, is easy to write:
S[fractal] = 2*PI*(1+1/9+1/49+1/81+1/169+1/225+1/361+1/441+1/625+1/729+1/961+1/961+1/1089+1/1369+1/1849+1/2401+...)
I didn't know yet whick is the order of those number.
The formula it reducing at 1/r1+1/r2 = 2*(1/a+1/b+1/c) a easy formula, where r1=abc/(ab+ac+bc+2*sqrt(abc*(a+b+c))) and
r2=abc/(ab+ac+bc-2*sqrt(abc*(a+b+c))). And can using reccurence method for render and determinating this sum terms.
The approximate result for this sum and for other sum example coming soon. I'm curious how give. I'm trying yet to know and the area of Julia fractal set.

[hgjf2]

The skew:

More GRANDMA QUASIFUCHSIAN type fractals are presented on the site WWW.BRAINJAM.CA by Peter Liepa at chapter "Fractals".