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Topic: Missouri State University Problem Corner
Replies: 5   Last Post: Dec 17, 2012 2:23 AM

 Messages: [ Previous | Next ]
 Michael Press Posts: 2,137 Registered: 12/26/06
Re: Missouri State University Problem Corner
Posted: Dec 15, 2012 7:21 PM

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

Date Subject Author
12/4/12 Les
12/9/12 William Elliot
12/15/12 Michael Press
12/16/12 Brian Chandler
12/16/12 Michael Press
12/17/12 Brian Chandler