```Date: Dec 16, 2012 7:15 AM
Author: Brian Chandler
Subject: Re: Missouri State University Problem Corner

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 simplercalculation.Three unit spheres touch; therefore their centres (S1, S2, S3) are atthe vertices of an equilateral triangle of side 2. They also sit onthe flat face of the hemisphere; so the height of each of the centresover the flat face is 1, and symmetry implies that the centre (C) ofthe 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 abase of side 2.Since each sphere touches the curved surface of the hemisphere, thenormal to the point of contact goes through the centre of the sphereand the centre of the flat face. Therefore the radius of thehemisphere is the length of a sloping edge of the pyramid (e) plus theradius of a sphere (1).(Using r() for square root...)Two applications of Pythagoras' theorem give us the distance from thecentre of the triangle to a vertex:d = 2 / r(3)  (30-60-90 triangle; longer right side = 1)And sloping edgee = r( d^2 + 1^2) = r( 4/3 + 1) = r(7/3)So radius of hemisphere is 1 + r(7/3)> 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 ahemisphere...Brian Chandler
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