2003-10-30-apollonian.mw

First, here's the Saudi Equation

 > 2*(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!