```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 7/3That 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 theequatorial 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
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