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 spheres

are internally or externally tangent to the hemisphere.

--

Michael Press