From: "Sir Charles W. Shults III"
References: <3D7B8D10.5070603@BOGUS.earthlink.net> <3D7CFD55.firstname.lastname@example.org> <6p3f9.201882$8aG1.email@example.com> <3D7D6CF1.firstname.lastname@example.org> <3D7E7CB1.9078347A@managesoft.com>
Subject: Re: Complex number quiz
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Date: Wed, 11 Sep 2002 03:31:51 GMT
NNTP-Posting-Date: Tue, 10 Sep 2002 23:31:51 EDT
Organization: RoadRunner - Central Florida
"Clifford Heath" wrote in message
> I liked this one: A cube of side c contains a sphere of radius r.
> What's the largest 2nd sphere that can fit in the cube?
Hmm. My first thought would be how neatly a fractal solution emerges.
Let's call the largest sphere that can occupy the cube "L". Its radius will be
1/2 the length of any edge of the box (called c). For neatness' sake, we will
assume c to be 2 and the radius to be a unit, r. Knowing that, and simple
Pythagorean logic, we also know that the distance from the center of L (which is
also the center of the cube) to and corner will be sqrt(3r) or 1.7320508...
The distance from the corner to the center of the largest sphere now known,
we subtract the radius (1) and end up with the distance from the corner to the
surface of the sphere. These are our boundary conditions. We end up with a
line segment of length 0.7320508... and have the property that one end is
touching the surface of the sphere L and the other end is reaching to the corner
of the box. Now for the fractal part.
Whatever the radius of the smaller sphere (the largest possible sphere to
fit in the corner that is left after subtracting the larger sphere), we know
that the length of the diagonal leftover part will be its radius plus sqrt(3)
times its radius. Therefore, the sphere will be 2/(2.7320508) times 0.7320508
or (oddly enough!) the square root of 3 minus one, squared!
The numeric value I get is 0.535898... for the diameter of the smaller
sphere, or a radius of 0.267949... Since we defined c as being 2, all values
are halved, and the answer if we first create a sphere with diameter c in the
box would be a sphere with diameter 0.267949 or a radius of 0.1339746.
Now for the generalized case.
If I create a sphere of radius r inside a cubic box with side length of c,
packing it in the corner will yield the most space for packing a second
arbitrary sphere. And, we know that for any sphere we create in a corner, its
origin will always be (once again, simple Pythagoras) sqrt(3) from the corner
compared to its radius. This is always true for any sphere in three dimensions.
But this also makes the second sphere we create always be sqrt(3) from its
corner, in relation to its radius. In other words, both spheres will always
have this relationship, regardless of the other sphere.
So, the solution is simple- for any sphere A that we pack in a cubic box,
the largest second sphere we can ever pack in the same box with it (call it B to
be friendly) will result in dividing the diagonal of the box into two line
segments. The two segments will add to c*sqrt(3), and they will be
(1+sqrt(3) )*rA and (1+sqrt(3) )*rB.
And numerically? What is the maximum length of that diagonal that will be
inside the two spheres? Simple. For the largest possible sphere and its
largest bother brother, the sums of their diameters will be 1.267949... And if
we assume two equally large spheres, the solution is 1.267949... also. In fact,
this works out for ANY two sphere cases I have tried. Therefore, for any two
spheres packed as tightly as possible inside a unit box, their radii must add up
to 1.267949 times the edge of the box.
This is 0.7320508 of the diagonal (and once more that value, sqrt(3)-1,
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