IMO 2003 SL C2

Origin: GEO | Category: Combinatorics

Problem

Let D1, . . . , Dn be closed disks in the plane. (A closed disk is a region bounded by a circle, taken jointly with this circle.) Suppose that every point in the plane is contained in at most 2003 disks Di. Prove that there exists disk Dk that intersects at most 7 \cdot 2003 −1 other disks Di.

Solution

Let S be the disk with the smallest radius, say s, and O the center of that disk. Divide the plane into 7 regions: one bounded by disk s and 6 regions T1, . . . , T6 shown in the figure. Any of the disks different from S, say Dk, has its center in one of the seven regions. If its center is inside S then Dk contains point O. Hence the number of disks different from S having their centers in S is at most 2002. Consider a disk Dk that intersects S and whose center is in the re- gion Ti. Let Pi be the point such that OPi bisects the region Ti and OPi = s \sqrt 3. O P1 P2 P6 P3 P4 P5 T1 T2 T6 T3 T4 T5 K L U6 V6 We claim that Dk contains Pi. Divide the region Ti by a line li through Pi perpendicular to OPi into two regions Ui and Vi, where O and Ui are on the same side of li. Let K be the center of Dk. Consider two cases: (i) K \inUi. Since the disk with the center Pi and radius s contains Ui, we see that KPi \leqs. Hence Dk contains Pi. (ii) K \inVi. Denote by L the intersection point of the segment KO with the circle s. We want to prove that KL > KPi. It is enough to prove that \angleKPiL > \angleKLPi. However, it is obvious that \angleLPiO \leq30◦and \angleLOPi \leq30◦, hence \angleKLPi \leq60◦, while \angleNPiL = 90◦−\angleLPiO \geq 60◦. This implies that \angleKPiL \geq\angleNPiL \geq60◦\geq\angleKLPi (N is the point on the edge of Ti as shown in the figure). Our claim is thus proved. Now we see that the number of disks with centers in Ti that intersect S is less than or equal to 2003, and the total number of disks that intersect S is not greater than 2002 + 6 \cdot 2003 = 7 \cdot 2003 −1.