In article <Pine.NEB.firstname.lastname@example.org>, William Elliot <email@example.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.