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Topic: Missouri State University Problem Corner
Replies: 5   Last Post: Dec 17, 2012 2:23 AM

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 Brian Chandler Posts: 1,899 Registered: 12/6/04
Re: Missouri State University Problem Corner
Posted: Dec 16, 2012 7:15 AM

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.

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

(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.

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...

Brian Chandler

Date Subject Author
12/4/12 Les
12/9/12 William Elliot
12/15/12 Michael Press
12/16/12 Brian Chandler
12/16/12 Michael Press
12/17/12 Brian Chandler