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Given are three unit disks on the plane. Let $A$ be the area of the plane covered by exactly $1$ disk. Let $B$ be the area of the plane covered by exactly $2$ disks. Prove that $A\geq B$.

Intuitively, this statement seems unlikely. One might think that if one draws three disks as in the Venn diagram, with a lot of overlapping, then one can get $A<B$. But apparently a large area is covered by all three disks, which is not counted toward $A$ or $B$.

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  • $\begingroup$ B depends on the overlap of the two disks. $\endgroup$ – TKM Nov 13 '14 at 2:33
  • $\begingroup$ Well the most will be when the two discs overlap completely. Which is then the area of one disc. So it makes sense. Just had to say this as it took me a while to understand B is the area of intersection. $\endgroup$ – Eoin Nov 13 '14 at 2:35

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