IMO 2002 SL G6

Origin: UKR | Category: Geometry

Problem

Let n \geq3 be a positive integer. Let C1, C2, C3, . . . , Cn be unit circles in the plane, with centers O1, O2, O3, . . . , On respectively. If no line meets more than two of the circles, prove that  1\leqi<j\leqn OiOj \leq(n −1)\pi .

Solution

Let l(MN) denote the length of the shorter arc MN of a given circle.

Lemma. Let PR, QS be two chords of a circle k of radius r that meet each other at a point X, and let \anglePXQ = \angleRXS = 2\alpha. Then l(PQ) + l(RS) = 4\alphar. Proof. Let O be the center of the circle. Then l(PQ) + l(RS) = \anglePOQ \cdot r + \angleROS \cdot r = 2(\angleQSP + \angleRPS)r = 2\angleQXP \cdot r = 4\alphar. Consider a circle k, sufficiently large, whose interior contains all the given circles. For any two circles Ci, Cj, let their exterior common tangents PR, QS (P, Q, R, S \ink) form an angle 2\alpha. Then OiOj = sin \alpha, so \alpha > sin \alpha = OiOj . By the lemma we have l(PQ) + l(RS) = 4\alphar \geq 8r OiOj , and hence OiOj \leql(PQ) + l(RS) 8r . (1) Now sum all these inequalities for i < j. The result will follow if we show that every point of the circle k belongs to at most n −1 arcs such as PQ, RS. Indeed, that will imply that the sum of all the arcs is at most 2(n −1)\pir, hence from (1) we conclude that  OiOj \leq(n−1)\pi . Consider an arbitrary point T of k. We prove by induction (the basis n = 1 is trivial) that the number of pairs of circles that are simultaneously in- tercepted by a ray from T is at most n −1. Let Tu be a ray touching k at T . If we let this ray rotate around T , it will at some moment intercept a pair of circles for the first time, say C1, C2. At some further moment the interception with one of these circles, say C1, is lost and never estab- lished again. Thus the pair (C1, C2) is the only pair containing C1 that is intercepted by some ray from T . On the other hand, by the inductive hypothesis the number of such pairs not containing C1 does not exceed n −2, justifying our claim.