Circle PackingDate: 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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