Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Missouri State University Problem Corner
Replies:
5
Last Post:
Dec 17, 2012 2:23 AM




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) (306090 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



