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July 17th, 2012, 04:22 AM  #1 
Joined: Jul 2012 Posts: 3 Thanks: 0  The number of soccer balls to build 4 sided pyramid
How to counts the number of soccer balls it would take to build a 4 sided pyramid (a tetrahedron) of soccer balls 6 foot high. The diameter of each ball is 22cm.

July 17th, 2012, 04:38 AM  #2 
Senior Member Joined: Jan 2010 Posts: 205 Thanks: 0  Re: The number of soccer balls to build 4 sided pyramid
Isn't this related to sphere packing and the Kepler conjecture?

July 17th, 2012, 04:54 AM  #3 
Joined: Jul 2012 Posts: 3 Thanks: 0  Re: The number of soccer balls to build 4 sided pyramid
Probably yes, this problem has connection to sphere packing. But how to approach this problem? I can calculate length of base side of a tetrahedron and divide it by sphere's diameter. From that I know how many spheres will be in the base level and then I can calculate next levels. But is this a good approach? 
July 17th, 2012, 12:19 PM  #4 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs  Re: The number of soccer balls to build 4 sided pyramid
Imagine 3 identical spheres of diameter d, all tangent and whose centers lie in the same horizontal plane. Now we place a 4th sphere whose diameter is also d on top so that it is tangent to the other three. If we draw line segments joining the centers of the 4 spheres we form a regular tetrahedron whose side lengths are d. Now, to find the height h of the tetrahedron, we may drop a vertical line segment from the apex to the base and label its length h. Next draw a line segment from the apex down one of its faces bisecting the face, and we find its length from . Now, observing that the vertical line h intersects the base in the middle (the point equidistant from the 3 vertices), we can find the length of the line segment b joining h and l from dividing the equilateral base into 3 congruent isosceles triangles each of which has an area 1/3 that of the equilateral base: Now we have a right triangle whose legs are h and b and whose hypotenuse is l, thus from the Pythagorean theorem, we obtain: From this we may determine how many levels n: We are given and we wish to form a tetrahedral stack high. Hence: Our stack will be just over 6 ft. high. Thus, the total number of balls S is found by adding up the first 10 triangular numbers: For n layers, we would have: 
July 17th, 2012, 01:22 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 12,895 Thanks: 642 
According to the question, d = 22cm, not 22in.

July 17th, 2012, 03:12 PM  #6 
Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,131 Thanks: 431 Math Focus: Calculus/ODEs  Re: The number of soccer balls to build 4 sided pyramid
I corrected that and another mistake.


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