Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.