IMO 1972 SL 11

Origin: NET

Problem

Consider a sequence of circles K1, K2, K3, K4, . . . of radii r1, r2, r3, r4, . . . , respectively, situated inside a triangle ABC. The circle K1 is tangent to AB and AC; K2 is tangent to K1, BA, and BC; K3 is tangent to K2, CA, and CB; K4 is tangent to K3, AB, and AC; etc. (a) Prove the relation r1 cot 1 2A + 2\sqrtr1r2 + r2 cot 1 2B = r  cot 1 2A + cot 1 2B  , where r is the radius of the incircle of the triangle ABC. Deduce the existence of a t1 such that r1 = r cot 1 2B cot 1 2C sin2 t1. (b) Prove that the sequence of circles K1, K2, . . . is periodic.

Solution

Let \angleA = 2x, \angleB = 2y, \angleC = 2z. (a) Denote by Mi the center of Ki, i = 1, 2, . . . . If N1, N2 are the projec- tions of M1, M2 onto AB, we have AN1 = r1 cot x, N2B = r2 cot y, and N1N2 =

(r1 + r2)2 −(r1 −r2)2 = 2\sqrtr1r2. The required rela- tion between r1, r2 follows from AB = AN1 + N1N2 + N2B. If this relation is further considered as a quadratic equation in \sqrtr2, then its discriminant, which equals ∆= 4 (r(cot x + cot y) cot y −r1(cot x cot y −1)) , must be nonnegative, and therefore r1 \leqr cot y cot z. Then t1, t2, . . . exist, and we can assume that ti \in[0, \pi/2]. (b) Substituting r1 = r cot y cot z sin2 t1, r2 = r cot z cot x sin2 t2 in the relation of (a) we obtain that sin2 t1 +sin2 t2 +k2 +2k sin t1 sin t2 = 1, where we set k = \sqrttan x tan y. It follows that (k + sin t1 sin t2)2 = (1 −sin2 t1)(1 −sin2 t2) = cos2 t1 cos2 t2, and hence cos(t1 + t2) = cos t1 cos t2 −sin t1 sin t2 = k = \sqrttan x tan y, which is constant. Writing the analogous relations for each ti, ti+1 we conclude that t1 + t2 = t4 + t5, t2 + t3 = t5 + t6, and t3 + t4 = t6 + t7. It follows that t1 = t7, i.e., K1 = K7.