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In the interior of a unit square, there are $n(n\in \mathbb{N}^*)$ circles whose sum of areas is greater than $n-1$. Prove that the circles has at least a common point of intersection

I really don't know where to start. Thanks

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What about the obvious solutions with $n=1$ or with concentric circles as long as $n\le 4$? –  Hagen von Eitzen Nov 12 '12 at 16:10

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up vote 2 down vote accepted

EDIT after reinterpretation of the problem statement.

If all circles contain the center of the square, we are done. If a circle does not contain the center, then its diameter is $<\frac{\sqrt 2}2$ and its area is $<\frac\pi8$. For all other circles, we have that the diameter is $\le 1$ and area $\le\frac \pi 4$. Thus the total area $A$ of the circles is $$n-1<A\le (n-1)\frac\pi 4+\frac\pi 8.$$ By solving for $n-1$, we find $$(n-1)<\frac{\frac\pi 8}{1-\frac\pi4}=1.829\ldots,$$ i.e. $n\le 2$. The case $n=1$ is trivial. If $n=2$, the total area of the disks is $>1$ hence greater than the area of the unit square, hence they must intersect.

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I think the question was to prove that there exists a point in the unit square which is contained in all the circles –  Vincent Nivoliers Nov 12 '12 at 16:19
    
@VincentNivoliers But that variant is also quite trivial as soon as $n-1\ge1$. Oh, wait - maybe it's not. –  Hagen von Eitzen Nov 12 '12 at 16:22
    
What I find trivial is to prove that any two circles intersect, however proving that one point is contained in every circle seems less trivial. I may be mistaken. –  Vincent Nivoliers Nov 12 '12 at 16:24
    
I was writing the same proof, you're faster ! –  Vincent Nivoliers Nov 12 '12 at 16:46

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