Hi,
How to calculate number of circles in side a square, if we know the side of the square and the circles all are equal size.
Thanks.
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$\begingroup$ The answer is "probably yes", but (a) who wants to know? and (b) I am fairly sure that the subject is NOT complex geometry. Voting to close. $\endgroup$– Igor RivinCommented Dec 29, 2011 at 14:17
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$\begingroup$ if yes, how can we calculate number of circles inside a square? $\endgroup$– NeerajaCommented Dec 29, 2011 at 14:23
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$\begingroup$ This is a website for questions of interest to research mathematicians - please see the faq. You might try math.stackexchange for your question, although you will also need to work a bit on the phrasing of the question so as to make it less ambiguous. $\endgroup$– Gerry MyersonCommented Dec 29, 2011 at 14:30
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1$\begingroup$ Posted now at MSE: math.stackexchange.com/questions/94941 $\endgroup$– Joseph O'RourkeCommented Dec 29, 2011 at 17:55
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2 Answers
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The right question is: given a positive integer $n$, what is the largest $r$ such that $n$ non-overlapping circles of radius $r$ can fit inside a unit square? It's not simply a matter of hexagonal close-packing, because boundary effects are important. There is no known closed-form formula, and not likely to be one, but the values are known for $n$ up to 30. See http://en.wikipedia.org/wiki/Circle_packing_in_a_square and http://hydra.nat.uni-magdeburg.de/packing/csq/csq.html
answered Dec 29, 2011 at 18:14
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Given a square of side $L$, the area covered by circles, including partial circles, will be $\frac{\pi L^2}{\sqrt{12}}$; since the optimal circle packing density in the plane is $\frac{\pi}{\sqrt{12}}$.
Therefore, given circles of radius $r$, the number of circles which fit in the square should be the greatest integer less than $\frac{1}{\sqrt{12}}\frac{L^2}{r^2}$
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$\begingroup$ As @Robert Israel points out, boundary effects are important, so your second statement is almost certainly false. $\endgroup$ Commented Dec 29, 2011 at 19:26