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### Circle Packing

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Date: 01/22/2001 at 23:38:51
From: Li
Subject: Circle packing

Here is my question:

One way to pack a 100 by 100 square with 10,000 circles, each of
diameter 1, is to put them in 100 rows with 100 circles in each row.
If the circles are repacked so that the centers of any three tangent
circles form an equilateral triangle, what is the maximum number of
additional circles that can be packed?

To solve this question, I first calculated the total area of the
square: 10000. Then, 10000 - 7850, which is not occupied by the
circles, thus, 2150/4PI, so the answer must be the maximum one, 1443.
However, my math teacher told me that my method was wrong. He said
that the circles could not be inserted simply by sqeezing them in. On
the other hand, he hasn't figured out a way to solve the problem yet.

Yours truly,
A sincere visitor to your Web site
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Date: 01/23/2001 at 16:20:36
From: Doctor Greenie
Subject: Re: Circle packing

Hello, Li -

I see the correct answer (at least the answer I got: 1443) in what you
have written.  But I don't see from the work you tried to show where
the answer comes from.  If I follow the words you use to describe how
you tried to solve the problem, then you just divided the area not
occupied by circles (2150 sq cm) by the area of one circle (pi/4)...
but this calculation gives an answer of 2738.

Here is how I came up with the answer of 1443....

When the circles are packed so that the centers of any three tangent
circles form an equilateral triangle, then alternate rows of circles
will contain 100 or 99 circles. We need to determine how many rows of
100 and how many rows of 99 we can fit into the 100-cm width if we
pack them in this way.

With one row of circles, the amount of the width that is used is just
the diameter of each circle, which is 1 cm.  If you draw a rough
picture of how the circles look when they are packed as described, you
will be able to see that the amount of width used by two rows of
circles is equal to the radius of one circle, plus the height of one
of the equilateral triangles formed by the centers of the circles,
equilateral triangle should show you that the height of the
equilateral triangle is equal to one-half the square root of 3. So
the amount of the width used by two rows of circles is (1 + one-half
the square root of 3).

It can then be seen from a sketch of the arrangement that the width
used by n rows of circles is

(n-1) * square root of 3
1 + --------------------------
2

The number of rows that can fit in the 100X100 container can be found
by determining the largest value of n for which the expression above
is less than 100. The answer is n = 115 rows.

The container then holds 58 rows of 100 circles each and 57 rows of 99
circles each, for a total of

58*100 + 57*99 = (115*100) - 57 = 11443

So the number of additional circles that can fit is

11443-10000 = 1443

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