Date: Dec 16, 2012 6:56 PM
Author: Michael Press
Subject: Re: Missouri State University Problem Corner
Brian Chandler <firstname.lastname@example.org> wrote:
> Michael Press wrote:
> > In article <Pine.NEB.email@example.com>,
> > William Elliot <firstname.lastname@example.org> 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
> 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)
> > That looks large. I get a different answer.
> > 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
You labeled it C. I labeled it O. No question here.
Looks like the difference is whether the unit spheres
are internally or externally tangent to the hemisphere.