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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.


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closed as not a real question by Igor Rivin, Gerry Myerson, Andrés E. Caicedo, Will Jagy, S. Carnahan Dec 30 '11 at 11:52

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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. – Igor Rivin Dec 29 '11 at 14:17
if yes, how can we calculate number of circles inside a square? – Neeraja Dec 29 '11 at 14:23
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. – Gerry Myerson Dec 29 '11 at 14:30
Posted now at MSE: – Joseph O'Rourke Dec 29 '11 at 17:55

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 and

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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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As @Robert Israel points out, boundary effects are important, so your second statement is almost certainly false. – Igor Rivin Dec 29 '11 at 19:26

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