Date: Dec 15, 2012 7:21 PM
Author: Michael Press
Subject: Re: Missouri State University Problem Corner
In article <Pine.NEB.4.64.1212090027020.10188@panix1.panix.com>,

William Elliot <marsh@panix.com> wrote:

> > 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

That looks large. I get a different answer.

Let r denote the radius of the hemisphere.

Label the center of the hemisphere O.

The distance from O to the center of a unit sphere is r+1.

Project the centers of the unit spheres onto the

equatorial plane of the hemisphere and label them A,B,C.

2 2 2 2

OA = OB = OC = (r + 1) - 1

Angle AOB = 2.pi/3. By the law of cosines

2 2

4 = 2.((r + 1) - 1 )(1 - cos 2.pi/3) = 2.((r + 1) - 1)(3/2)

2

(r + 1) = 1 + 4/3

r = -1 +|- sqrt{7/3}.

Taking the positive root we get

r = -1 + sqrt{7/3} ~ 0.5275252316.

--

Michael Press