## The number of soccer balls to build 4 sided pyramid

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### 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.
simonbinxs
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### Re: The number of soccer balls to build 4 sided pyramid

Isn't this related to sphere packing and the Kepler conjecture?
daigo
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### 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?
simonbinxs
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### 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=\frac{\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$

$b=\frac{1}{3}\cdot\frac{\sqrt{3}}{2}d=\frac{1}{2\sqrt{3}}d$

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:

$S_n=\frac{n(n+1)(n+2)}{6}={n+2 \choose 3}$

$S_{10}=220$
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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According to the question, d = 22cm, not 22in.
skipjack
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### Re: The number of soccer balls to build 4 sided pyramid

I corrected that and another mistake.
Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

MarkFL
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