Apollonian Gasket Calculations

First, we deal with the function w that returns the radius of the inner kissing circle. Maple Code:

# Here is the radius equation.
w := (x,y,z) -> simplify((x*y*z)/( (x*y) + (y*z) + (x*z) + 2 * ((x*y*z*(x+y+z))^(1/2)) ));

# Clusterfy w into a 3-cluster.  
f := (x,y,z) -> (w(x,y,z),y,z);
g := (x,y,z) -> (x,w(x,y,z),z);
h := (x,y,z) -> (x,y,w(x,y,z));


f(1,1,1);
g(f(1,1,1));
f(g(f(1,1,1)));
h(g(f(1,1,1)));
f(f(h(g(f(1,1,1)))));


 

Let's deal with curvature = 1/radius.

 
# Here is the Curvature equation.
solve(2*(a^2 + b^2 + c^2 + d^2) = (a+b+c+d)^2,d);
 
u := (a,b,c) -> simplify(a+b+c + 2*(a*b+a*c+b*c)^(1/2));

# Clusterfy w into a 3-cluster.  
f := (x,y,z) -> (u(x,y,z),y,z);
g := (x,y,z) -> (x,u(x,y,z),z);
h := (x,y,z) -> (x,y,u(x,y,z));

 
f(1,1,1);
g(f(1,1,1));
f(g(f(1,1,1)));
h(g(f(1,1,1)));
f(f(h(g(f(1,1,1)))));
 
#What if one of the lines is staright (r==infty or curvature==0)???
f(0,1,1);
f(f(0,1,1));
h(g(f(0,1,1)));

Apollonian Gasket Calculations

At this point I finally get something really imortant through my head. The Apollonian Gasket makes a Cluster Algebra of rank 4!

Why? Well, given three kissing circles, you can alyways find two other kissing circles! This does not an exchange relation make! Stephen finally expained to me that four kissing circles make a cluster. Now I can apply the methodology suggested by Jim in his 2003-10-11 email. Let's see how this concept simplifies the equations:

2003-10-30-apollonian.mw

2003-10-30-apollonian.html

First, here's the Soddy Equation

2*(a^2 + b^2 + c^2 + d^2) = (a+b+c+d)^2;

2*a^2+2*b^2+2*c^2+2*d^2 = (a+b+c+d)^2

Suppose that for a given (a,b,c), both d and d' are solutions this equation:

2*(a^2 + b^2 + c^2) + 2*d^2 = (a+b+c)^2 + 2*d*(a+b+c) + d^2;
2*(a^2 + b^2 + c^2) + 2*d^2 - (a+b+c)^2 - 2*d*(a+b+c) - d^2 = 0;
d^2 - (2*a + 2*b + 2*c)*d + 2*(a^2 + b^2 + c^2) - (a+b+c)^2 = 0;

2*a^2+2*b^2+2*c^2+2*d^2 = (a+b+c)^2+2*d*(a+b+c)+d^2

2*a^2+2*b^2+2*c^2+d^2-(a+b+c)^2-2*d*(a+b+c) = 0

d^2-(2*a+2*b+2*c)*d+2*a^2+2*b^2+2*c^2-(a+b+c)^2 = 0

So then:

d + dprime = 2*a + 2*b + 2*c;
d * dprime = 2*(a^2 + b^2 + c^2) - (a+b+c)^2;

d+dprime = 2*a+2*b+2*c

d*dprime = 2*a^2+2*b^2+2*c^2-(a+b+c)^2

Next define a function u which takes in four cluster variables and return the exchange for the fourth.

u := (a,b,c,d) -> simplify(2*a + 2*b + 2*c - d);

u := proc (a, b, c, d) options operator, arrow; simplify(2*a+2*b+2*c-d) end proc

Clusterfy u into a 4-cluster.

e := (w,x,y,z) -> (u(x,y,z,w),x,y,z); 
f := (w,x,y,z) -> (w,u(w,y,z,x),y,z);
g := (w,x,y,z) -> (w,x,u(w,x,z,y),z);
h := (w,x,y,z) -> (w,x,y,u(w,x,y,z));

e := proc (w, x, y, z) options operator, arrow; u(x, y, z, w), x, y, z end proc

f := proc (w, x, y, z) options operator, arrow; w, u(w, y, z, x), y, z end proc

g := proc (w, x, y, z) options operator, arrow; w, x, u(w, x, z, y), z end proc

h := proc (w, x, y, z) options operator, arrow; w, x, y, u(w, x, y, z) end proc

Some things are downright meaingless: For instance:

e(x,x,x,x);

5*x, x, x, x

doesn't mean anything. Can you imagine four circles that all have the same radius kissing? No. So we define.

v1 := (a,b,c) -> simplify(a+b+c + 2*(a*b+a*c+b*c)^(1/2));
v2 := (a,b,c) -> simplify(a+b+c - 2*(a*b+a*c+b*c)^(1/2));

v1 := proc (a, b, c) options operator, arrow; simplify(a+b+c+2*(a*b+a*c+b*c)^(1/2)) end proc

v2 := proc (a, b, c) options operator, arrow; simplify(a+b+c-2*(a*b+a*c+b*c)^(1/2)) end proc

So if we had three circles with radius x,y, and z, we can let w be the inner circle for the kiss of x, y, and z.

w := v1(x,y,z);

w := x+y+z+2*(x*y+x*z+y*z)^(1/2)

But we want simpler looking equations, so let M = (xy+xz+yz)^(1/2):

w := x+y+z+2*M;

w := x+y+z+2*M

Then we can look at some of the clusters that are accessible form this cluster.

(w,x,y,z);
e(w,x,y,z);
e(f((w,x,y,z)));
e(f(e(w,x,y,z)));
e(e(f(e(w,x,y,z))));
h(h(e(f(e(w,x,y,z)))));

x+y+z+2*M, x, y, z

x+y+z-2*M, x, y, z

x+9*y+9*z+6*M, x+4*y+4*z+4*M, y, z

x+9*y+9*z-6*M, x+4*y+4*z-4*M, y, z

x+y+z-2*M, x+4*y+4*z-4*M, y, z

x+9*y+9*z-6*M, x+4*y+4*z-4*M, y, z

All of these are “nice” equations, as expected! QUESTION: what do the coefficients count?

But what if we start with 0,1,1?

z:=0;
x:=1;
y:=1;
w:=v1(x,y,z);

(w,x,y,z);
h(w,x,y,z);
e(f((w,x,y,z)));
e(f(e(w,x,y,z)));
e(e(h(e(w,x,y,z))));
h(h(e(f(e(w,x,y,z)))));
h(g(h(h(e(f(e(w,x,y,z)))))));

z := 0

x := 1

y := 1

w := 4

4, 1, 1, 0

4, 1, 1, 12

16, 9, 1, 0

4, 1, 1, 0

0, 1, 1, 4

4, 1, 1, 0

4, 1, 9, 28

z:='z': x='x': y:='y': w:='w';

w := w

x:='x':y:='y':z:='z':
(w,x,y,z);
e(w,x,y,z);
f(w,x,y,z);
g(w,x,y,z);
h(w,x,y,z);

w, x, y, z

2*x+2*y+2*z-w, x, y, z

w, 2*w+2*y+2*z-x, y, z

w, x, 2*w+2*x+2*z-y, z

w, x, y, 2*w+2*x+2*y-z

(x+y+z+2*(x*y+x*z+y*z)^(1/2), x, y, z);

x+y+z+2*(x*y+x*z+y*z)^(1/2), x, y, z


applolonian.mw

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