Date: Dec 9, 2012 3:29 AM
Author: William Elliot
Subject: Re: Missouri State University Problem Corner

> Three unit spheres are mutually tangent to one another and to a
> hemisphere, both along the spherical part of the hemisphere and
> along its equatorial plane. Find the radius of the hemisphere.


Let the spheres be tangent to the xy plain.

Draw lines between the centers to form an equivateral triangle
with side 2. Place the geometric center of the triangle at (0,0,1).

The geometric center is k = 2(sqr 3)/3 from each of the centers
of the spheres. Center one sphere at p = (k,0,1) and take the
xz cross section through p. The large sphere centered at (0,0,0)
has cross section equation of
x^2 + z^2 = r^2
while the small sphere has cross section equation of
(x - k)^2 + (z - 1)^2 = 1

Solving for the cross section points of intersection

x^2 - 2kx + k^2 + z^2 - 2z + 1 = 1
r^2 + k^2 - 2kx - 2z = 0

z = (r^2 + k^2)/2 - kx = c/2 - kx
r^2 = x^2 + z^2
= x^2 + c^2 / 4 - ckx + k^2 x^2
= (1 + k^2)x^2 - ckx + c^2 / 4
4(1 + k^2)x^2 - 4ckx + c^2 - 4r^2 = 0

The discrimanent of that equation, because of tangency,

16c^2 k^2 - 16(1 + k^2)(c^2 - 4r^2) = 0
c^2 k^2 - (1 + k^2)(c^2 - 4r^2) = 0

-c^2 + 4r^2 + 4r^2 k^2 = 0
-(r^4 + 2r^2 k^2 + k^4) + 4r^2 + 4r^2 k^2 = 0
-(r^4 + k^4) + 4r^2 + 2r^2 k^2 = 0

r^4 + k^4 - 4r^2 - 2r^2 k^2 = 0
(r^2 - k^2)^2 = 4r^2; r^2 - k^2 = 2r

r^2 - 2r - k^2 = 0
r = (2 +- sqr(4 + 4k^2))/2
= 1 + sqr(1 + k^2) = 1 + sqr 7/3