IMO 1981 SL 19

Origin: YUG

Problem

A finite set of unit circles is given in a plane such that the area of their union U is S. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater that 2S 9 .

Solution

Consider the partition of plane \pi into regular hexagons, each having in- radius 2. Fix one of these hexagons, denoted by \gamma. For any other hexagon x in the partition, there exists a unique translation \taux taking it onto \gamma. Define the mapping ϕ : \pi \to\gamma as follows: If A belongs to the interior of a hexagon x, then ϕ(A) = \taux(A) (if A is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals S, while the area of the hexagon \gamma is 8 \sqrt 3. Thus there exists a point B of \gamma that is covered at least S \sqrt 3 times, i.e., such that ϕ−1(B) consists of at least S \sqrt 3 distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than \pi \sqrt 3S \geq2S/9. Remark. The statement becomes false if the constant 2/9 is replaced by any number greater than 1/4. In that case a counterexample is, for exam- ple, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.