Apollonian Gasket CalculationsFirst, here's the Saudi Equation2*(a^2 + b^2 + c^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;
So then:d + dprime = 2*a + 2*b + 2*c;
d * dprime = 2*(a^2 + b^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);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));
Some things we can are downright meaingless: For instance:e(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));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);But we want simpler looking equations, so let M = (xy+xz+yz)^(1/2):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)))));
All of these are "nice" equations, as expected!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:='z': x='x': y:='y': 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);
(x+y+z+2*(x*y+x*z+y*z)^(1/2), x, y, z);