```Date: Dec 16, 2012 6:56 PM
Author: Michael Press
Subject: Re: Missouri State University Problem Corner

In article <ee9686cf-d6f4-4b14-aacd-f490073b1054@6g2000pbh.googlegroups.com>, Brian Chandler <imaginatorium@despammed.com> wrote:> Michael Press wrote:> > 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.> > > All those equations and things made my head spin. Here's a simpler> calculation.Heh!> Three unit spheres touch; therefore their centres (S1, S2, S3) are at> the vertices of an equilateral triangle of side 2. They also sit on> the flat face of the hemisphere; so the height of each of the centres> over the flat face is 1, and symmetry implies that the centre (C) of> the hemisphere is under the centre (T) of the equilateral triangle,> and the centres S1, S2, S3, and C form a pyramid  of height 1 on a> base of side 2.> > Since each sphere touches the curved surface of the hemisphere, the> normal to the point of contact goes through the centre of the sphere> and the centre of the flat face. Therefore the radius of the> hemisphere is the length of a sloping edge of the pyramid (e) plus the> radius of a sphere (1).> > (Using r() for square root...)> Two applications of Pythagoras' theorem give us the distance from the> centre of the triangle to a vertex:> > d = 2 / r(3)  (30-60-90 triangle; longer right side = 1)> > And sloping edge> > e = r( d^2 + 1^2) = r( 4/3 + 1) = r(7/3)> > So radius of hemisphere is 1 + r(7/3)Why? > > That looks large. I get a different answer.> > Hmm.> > > Let r denote the radius of the hemisphere.> > Label the center of the hemisphere O.> > There's a real question: what position _is_ the centre of a> hemisphere...You labeled it C. I labeled it O. No question here. Looks like the difference is whether the unit spheresare internally or externally tangent to the hemisphere.-- Michael Press
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