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




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



