# Circles on a plane

$n$ circles with total area A have been drawn on the plane (overlapping circles are not counted multiple times). Prove that we can select a disjoint union of circles that has area greater than $\frac{A}{9}$.

I'm thinking that if there is a circle that has an area that is greater than 1/9 then we're basically done. But if not AND the statement is false then we must be full of overlaps. More specifically a point has been covered by at least 9 circles.

• Funny problem; I like it. Where does it come from? Why do you think it's true? What have you tried? – user228113 Feb 1 '16 at 3:25
• @G.Sassatelli This is the same as being able to cover a table with 10 dots on it by 10 circles. See here very instructive. – Shailesh Feb 1 '16 at 3:54
• Thank you for the reference, @Shailesh. – user228113 Feb 1 '16 at 3:57
• @Shailesh Why is that the same? – Michael Biro Feb 1 '16 at 4:08
• Updated the question – The Math Penguin Feb 1 '16 at 4:08

Now, if we accept a disk of radius $r$, we will disallow any disk that intersects it (which must have radius $\leq r$), all of which are completely covered by a disk of radius $3r$.
Therefore, we cover $\pi r^2$ with our chosen disk, while disallowing at most $\pi (3r)^2$ in area from $A$. In other words, we cumulatively cover a total area of at least $\frac{A}{9}$.