```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 toa hemisphere" to mean "touching a part of the hemisphere", so the bitabout 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 thehemisphere, for example (wherever that is, exactly)...  But OK, I'llcall 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 thesurface of the hemisphere equals the distance to the centre of asphere (=sloping edge of the pyramid (e)) plus the distance (1) fromthis centre to the surface of the hemisphere. (Because they're in astraight 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 ona plane, then there are two hemispherical bubble on the plane whichtouch the spheres, one inside, one outside, and their radii are:e +- 1 QEDBrian Chandler
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