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

 Messages: [ Previous | Next ]
 Brian Chandler Posts: 1,899 Registered: 12/6/04
Re: Missouri State University Problem Corner
Posted: Dec 17, 2012 2:23 AM

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

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