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 July 17th, 2012, 04:22 AM #1 Newbie   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 Newbie   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 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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 $\sin$$60^{\circ}$$=\frac{l}{d}\:\therefore\:l=\fra c{\sqrt{3}}{2}d$. 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: $\frac{1}{2}db=\frac{1}{3}\cdot\frac{1}{2}\sin$$60^ {\circ}$$d^2$ $b=\frac{1}{3}\cdot\frac{\sqrt{3}}{2}d=\frac{1}{2\s qrt{3}}d$ Now we have a right triangle whose legs are h and b and whose hypotenuse is l, thus from the Pythagorean theorem, we obtain: $h=\sqrt{l^2-b^2}=\sqrt{$$\frac{\sqrt{3}}{2}d$$^2-$$\frac{1}{2\sqrt{3}}d$$^2}$ $h=d\sqrt{\frac{3}{4}-\frac{1}{12}}=d\sqrt{\frac{2}{3}}$ From this we may determine how many levels n: We are given $d=22\text{ cm}$ and we wish to form a tetrahedral stack $72\text{ in}$ high. Hence: $h=22\sqrt{\frac{2}{3}}\text{ cm}$ $h(n-1)+d=72\text{ in}\cdot\frac{2.54\text{ cm}}{1\text{ in}}=\frac{4572}{25}\:\text{cm}$ $n=\frac{\frac{4572}{25}-22}{22\sqrt{\frac{2}{3}}}+1\approx10$ 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: $S_n=\frac{n(n+1)(n+2)}{6}={n+2 \choose 3}$ $S_{10}=220$
 July 17th, 2012, 01:22 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 According to the question, d = 22cm, not 22in.
 July 17th, 2012, 03:12 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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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