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Packing 4 Spheres Into a Tetrahedron


Date: 09/03/99 at 16:11:06
From: Daniel Lam
Subject: Packing 4 spheres into a tetrahedron

I have 4 spheres. I need to pack them as snugly as possible, and then 
find the dimensions of the smallest tetrahedron that can serve as a 
container for the spheres, so the spheres are in a tetrahedral 
arrangement (3 at the bottom in a triangular fashion, with one on 
top.) The method I have been trying involves the angle between the 
faces (I believe it's 73.5 degrees) but I can't show conclusively that 
this is correct. I have been advised to construct a tetrahedron with 
vertices at the center of each sphere, so that its edge length is 2r 
(where r is the radius of the sphere), but I don't know where to go 
from there.


Date: 09/04/99 at 17:48:12
From: Doctor Mitteldorf
Subject: Re: Packing 4 spheres into a tetrahedron

Dear Dan,

This is a sophisticated problem, which requires many steps and a good 
ability to visualize in 3 dimensions. The latter has always been 
difficult for me, and requires all the concentration I can muster. 
There are many ways to go about it. I'll suggest one.

The Rosetta stone for a great number of 3-D geometry problems is a 
relation I call the "hinge formula." It can be applied at each of the 
corners of the tetrahedron, and also at the middle. Draw two angles on 
a hinge, with the spine of the hinge for their common edge. What is 
the angle between the two lines that you drew? The hinge formula says

     cos(c) = cos(a)cos(b) + sin(a)sin(b)cos(t)

where angles a and b are painted on the hinge with the axis of the 
hinge for a common side, the hinge is open to a plane angle t, and c 
is the angle formed by the outer limbs of angles a and b.

If the hinge is closed, then the angle c is the difference of a and b, 
and the hinge formula reduces to the formula for the cosine of the 
difference of two angles. If the hinge is open flat, then angle c is 
the sum of a and b, and the hinge formula reduces to the formula for 
the cosine of the sum. (Remember that the cosine of the sum has a 
minus sign in it, and the cosine of the difference has a plus sign.)

But the interesting cases are the ones in between. You can apply the 
hinge formula at any vertex of the tetrahedron, where a, b and c are 
all 60 degrees, and solve for t, which is the angle formed by any two 
faces of the tetrahedron.

Now draw 4 lines from each vertex of the tetrahedron that meet at the 
solid center. Imagine these lines on 4 fins, planes that meet at 120 
degrees. Use the hinge formula again with t = 120 degrees to solve for 
a, b and c, which are all the same in this case. The angle you've 
found is the "tetrahedral angle," the angle formed by any two lines 
from the center of a tetrahedron to two different vertices.

What angle do these lines make when they meet the edges of the 
tetrahedron at a vertex? Again, use the hinge formula, this time 
imagining the hinge open to 120 degrees and the line to the center on 
the spine of the hinge. The angle you want is either a or b (they are 
equal) and the angle t is 120 degrees. The angle between two of the 
edges that meet at the vertex is c, which is 60 degrees.

Now you have all the angles you need. Do a little plane geometry to 
locate the center of one face. Use the angles you have and 
trigonometry definitions to find the vertical height of the 
tetrahedron: the distance between any vertex and the center of the 
opposite face. Also, use the angles you've found to compute the 
distance between the solid center and the center of one face, and 
subtract these two to give the distance between the solid center and 
any vertex.

If you get this far, you've calculated all the dimensions and angles 
of the tetrahedron, and there's not far to go to get your answer. 
Take the center of each ball as a vertex to make a small tetrahedron. 
You know that the edge of that small tetrahedron has length 2r, and 
you can use the above calculation to find the distance between the 
solid center and the 3-ball center, i.e. the distance between the 
midpoint of all 4 balls and the midpoint of any three balls. Imagine 
drawing that line and extending it past the plane of the 
three-ball-centers by an additional distance r. Now it just reaches 
the surface of the large tetrahedron, the one that can enclose all 4 
balls. You know then, the distance between any face of this large 
tetrahedron and its center, and you can use all the calculations you 
did at the beginning to work backwards and find the length of one edge 
of the large tetrahedron in terms of r.

Why don't you get started with this calculation? If you get all the 
way to the end, please send me your answer and I'll check it against 
what I get. If you get stuck, be sure to write again and I'll try to 
fill in some more details of the calculation.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   


Date: 09/11/99 at 09:45:50
From: Daniel Lam
Subject: Re: Packing 4 spheres into a tetrahedron

Dear Dr. Mitteldorf,

Thanks very much for replying to my question. Your reply gave me a lot 
of insight into the problem, and I now have a good idea of what has to 
be done, but I haven't arrived at a correct answer (the balls don't 
fit into the box) The reason could be, in order of likelihood:

     1) I made a mistake in the calculation
     2) The method is incorrect
     3) The solution doesn't apply to reality.

Here are my results for the calculation. Maybe you can tell me where I 
went wrong.

Angle between two faces of a tetrahedron:

     Approx. 70.528 degrees

Tetrahedral Angle:

     Approx. 109.471 degrees

Angle where line from center meets vertex:

     Approx. 35.264 degrees

Center of one face (where e is the length of an edge):

     e/[2 * sqrt(3)] from a perpendicular to an edge

Height of a tetrahedron (where f is the angle between two faces):

     (e tan f)/[2 * sqrt(3)]

Distance from solid center to center of a face (where d is half the 
tetrahedral angle):

     2e / [2 * sqrt(3) * tan d]

Distance between solid center and vertex:

     e(tan f tan d - 2) / [2 * sqrt(3) * tan d]

Distance from 4 ball center to 3 ball center: (e = 2r)

     4r / [2 * sqrt(3) * tan d]

Distance from center of large tetrahedron to face:

     4r / [2 * sqrt(3) * tan d] + r

and

     2e / [2 * sqrt(3) * tan d]

therefore

     4r / [2 * sqrt(3) * tan d] + r = 2e / [2 * sqrt(3) * tan d]

therefore

                                  e = r[2 + sqrt(3) * tan d)

                                    = approx. 4.45r

Measuring the radius of a tennis ball to approx. 3.28 cm, the length 
of an edge of the box would be approx. 14.594 cm. But, I built a box 
of this size and 4 tennis balls definitely don't fit. By trial and 
error, I found that the edge needs to be about 21 cm, so my answer is 
quite a long way off.


Date: 10/07/1999 at 16:51:55
From: Doctor Mitteldorf
Subject: Re: [Re: Packing 4 Spheres into a tetrahedron]

Dear Dan,

You're doing great - you clearly understand the whole process, and 
you've done all the little algebra problems. I think the only reason 
the answer didn't come out right is that you lost a factor of 2 in the 
following place. I get 2r*[1+sqrt(6)] = 6.899 r as an answer.


>Distance from solid center to center of a face (where d is half the 
>tetrahedral angle):
>
>     2e / [2 * sqrt(3) * tan d]  

***************************** Check this *****************************
************* I think you're a factor of 2 too big here **************


One suggestion: you don't need to express any of this in terms of trig 
functions, because all the trig functions derive from geometry, so 
they are all fractions or radicals. For example, tan(d) = sqrt(2) and 
tan(f) = 2sqrt(2).

If you want to stretch a bit further, here's a challenge: I realized 
there's a way to derive the tetrahedral angle without the "hinge 
formula." An edge and an adjacent altitude are part of a right 
triangle whose base is the line between the center of a triangle and 
its vertex, which we know from plane geometry to be e/sqrt(3). This 
gives us a simpler way to calculate the 35.26-degree angle. Then 
consider the triangle formed by two lines center-to-vertex and one 
edge. The fact that it's an isosceles triangle gives us a way to 
calculate the tetrahedral angle in terms of the 35.26-degree angle.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Higher-Dimensional Geometry
College Trigonometry
High School Higher-Dimensional Geometry
High School Trigonometry

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