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### A Pyramid of Layered Marbles

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Date: 06/02/99 at 14:45:30
From: Chris
Subject: Pyramid-shaped container, containing layered marbles

A spherical marble has the diameter of a penny. A packaging company is
designing a pyramid-shaped container, with a regular hexagonal base,
to store the marbles. The container is made so that the largest layer
contains 331 marbles, while the number of rows of the next layer of
marbles is one fewer than the number of rows of the bottom layer. Each
subsequent layer has one fewer row than the one beneath it. This
process continues until the last row (containing 1 marble) is reached.
Note that between each row there is a divider with the thickness of a
penny. Find:

(a) the number of layers required
(b) the number of marbles in the container
(c) the height of the container
(d) the volume of the container

If there are 27 new marbles among the marbles in the full container,
calculate the probability that these 27 marbles will be at the top of
the pile of marbles in the container.

This is a very extensive problem, and my difficulty is probably with
the interpretation. It seems like a very practical problem, and I've
tried examining the way marbles fit together, and building a small
model but I'm not sure how to arrive at a mathematical solution for
these questions. Any help would be greatly appreciated. Thank you.
```

```
Date: 06/02/99 at 16:50:44
From: Doctor Rob
Subject: Re: Pyramid-shaped container, containing layered marbles

Very interesting problem!

First, I determined the number n of marbles on each side of the bottom
layer. This equals the number n of layers. A hexagonal arrangement of
marbles would put six equal-sized equilateral triangles of marbles
plus one in the very center. Each equilateral triangle would have n-1
marbles on a side, and (n-1)*n/2 marbles, so that we get the equation

331 = 1 + 6*(n-1)*n/2,

which can easily be solved for n.

Next, we want the sum

n
S = SUM 1 + 6*i*(i-1)/2
i=1

n
= SUM 1*C(i,0) + 6*C(i,2)
i=1

n
= SUM 1*[C(i+1,1)-C(i,1)] + 6*[C(i+1,3)-C(i,3)]
i=1

and these are collapsing or telescoping sums, so

S = C(n+1,1) - C(1,1) + 6*[C(n+1,3) - C(1,3)]
= (n+1) - 1 + 6*[(n+1)*n*(n-1)/6 - 0]
= n + n^3 - n
= n^3

Now comes the tricky geometric part. I let the radius of the marbles
(the radius of a penny) be r and the thickness of the dividers (the
thickness of a penny) be t. I drew a cross-section through the center
of one of the edge marbles in the base and perpendicular to the edge
of the base. I called the distance from the edge of the base to the
tangency points of the marble with the base and the side d. I called
the angle between the base and the side A. I also let the length of
the side of the base be s, the height of the pyramid be h, and the
distance from the top of the top marble to the vertex of the pyramid
be x. Then I started to write down equations relating A, d, h, r, s,
t, and x.

tan(A/2) = r/d
tan(A) = d*x/r^2 = (x+2*r)/d = h/[s*sqrt(3)/2]
h = x + 2*n*r + (n-1)*t
s = d/2 + 2*(n-1)*r + d/2
tan(A) = 2*tan(A/2)/[1-tan^2(A/2)]

The idea is to solve for A, d, h, s, and x in terms of r and t alone.
First

x*(d/r^2-1/d) = 2*r/d
x = 2*r^3/(d^2-r^2)

Then

tan(A) = d*x/r^2 = 2*r*d/(d^2-r^2)
A = arctan[2*r*d/(d^2-r^2)]
h = 2*r^3/(d^2-r^2) + 2*n*r + (n-1)*t
s = d + 2*(n-1)*r

This gives us A, h, s, and x in terms of d, r, and t. We have one
final equation,

h/[s*sqrt(3)/2] = 2*r*d/(d^2-r^2)

which should allow us to solve for d in terms of r and t. I got a
rather messy quadratic equation of which only one root was positive.

At this point, substituting the dimensions of a penny is highly
recommended, and the solution for d should be found numerically. Once
that is obtained, the height h and the area of the base K =
sqrt(3)*s^2/4 can be computed, and the volume V = K*h/3 is easy to
find.

I leave the rest to you.

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

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