In article <email@example.com>, Brian Chandler <firstname.lastname@example.org> wrote:
> Michael Press wrote: > > In article <Pine.NEB.email@example.com>, > > William Elliot <firstname.lastname@example.org> 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 > 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)
> > 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.