Date: Dec 17, 2012 2:23 AM Author: Brian Chandler Subject: Re: Missouri State University Problem Corner Michael Press wrote: (but not necessarily in this order)

> Brian Chandler <imaginatorium@despammed.com> wrote:

> > Michael Press wrote:

> > > 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.

Well, the question is slightly vaguely worded, but I take "tangent to

a hemisphere" to mean "touching a part of the hemisphere", so the bit

about the equatorial plane needs to be inside the "equator".

> > There's a real question: what position _is_ the centre of a

> > hemisphere...

>

> You labeled it C. I labeled it O. No question here.

Just a nitpick: is the "centre" of a hemisphere the centre of its

"equatorial plane"? It could be the centre of gravity of the

hemisphere, for example (wherever that is, exactly)... But OK, I'll

call it O too.

> > 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?

Because the distance from the centre (O) of the flat face to the

surface of the hemisphere equals the distance to the centre of a

sphere (=sloping edge of the pyramid (e)) plus the distance (1) from

this centre to the surface of the hemisphere. (Because they're in a

straight line)

>

> > > That looks large. I get a different answer.

> >

> > Hmm.

> >

> > > Let r denote the radius of the hemisphere.

> > > Label the center of the hemisphere O.

> >

> Looks like the difference is whether the unit spheres

> are internally or externally tangent to the hemisphere.

Right. It you phrase the question as three touching spheres resting on

a plane, then there are two hemispherical bubble on the plane which

touch the spheres, one inside, one outside, and their radii are:

e +- 1 QED

Brian Chandler