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Circle Packing


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


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, 
plus another radius. Your knowledge of the characteristics of an 
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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