Three balls in a bowl.

Without loss of generality, we may place the (centres of the ) balls A, B and C on the x-y plane such that

A = {a1, 0, 0}

B= {b1, b2, 0}

C= {c1, c2, 0}

D= { 0, 0, h}

We next add constraints so that the balls "fit" in D:

(1) (d - a)^2 = a1^2 +h^2

(2) (d - b)^2 = b1^2+b2^2+h^2

(3) (d - c)^2 = c1^2+c2^2+h^2

and so that they touch each other

(4) (a+b)^2 = (a1-b1)^2 + b2^2

(5) (b+c)^2 = (b1-c1)^2 + (b2 - c2)^2

(6) (a+c)^2 = (a1 -c1)^2+c2^2

The six equations would allow us to eliminate five variables {a1, b1, b2, c1, c2} and express h in terms of a,b,c,d. We note that h is fixed for a given {a,b,c,d}. However they are useful for computing the centre of gravity,

We denote the centre of gravity by G and since it is proportional to the volume of the spheres we have

G (a^3+b^3+c^3) = A a^3 + B b^3 + C c^3. We note that G lies on the x-y plane.

Then g= |OG| = Sqrt[((a^3*a1 + b^3*b1 + c^3*c1)^2 + (b^3*b2 + c^3*c2)^2)/(a^3 + b^3 + c^3)^2].

Now it is a simple matter to calculate t = tan(alpha) = g/h. I will post separately an expression for h but note that there seem to be possibly 4 different intervals where other expressions apply.