IMO 1999 SL G8

Origin: RUS | Category: Geometry

Problem

Points A, B, C divide the circumcircle Ωof the triangle ABC into three arcs. Let X be a variable point on the arc AB, and let O1, O2 be the incenters of the triangles CAX and CBX. Prove that the circumcircle of the triangle XO1O2 intersects Ωin a fixed point.

Solution

We first introduce the same lemma as in problem 12 and state it here without proof. Lemma. Let ABC be a triangle and I the center of its incircle. Let M be the center of the arc : BC of the circumcircle not containing A. Then MB = MC = MI. Let the circle XO1O2 intersect the circle Ωagain at point T . Let M and N be respectively the midpoints of arcs : BC and : AC, and let P be the intersection of Ωand the line through C parallel to MN. Then the lemma gives MP = NC = NI = NO1 and NP = MC = MI = MO2. Since O1 and O2 lie on XN and XM respectively, we have \angleNTM = \angleNXM = \angleO1XO2 = \angleO1TO2 and hence \angleNTO1 = \angleMTO2. Moreover, \angleTNO1 = \angleTNX = \angleTMO2, from which it fol- lows that \triangleO1NT ∼\triangleO2MT . Thus NT MP = NT NO1 = MT MO2 = MT NP ⇒ MP \cdot MT = NP \cdot NT ⇒SMPT = SNPT . It follows that TP bisects the segment MN, and hence it passes through I. We conclude that T belongs to the line PI and does not depend on X. Remark. An alternative approach is to apply an inversion at point C. Points O1 and O2 become excenters of \triangleAXC and \triangleBXC, and T be- comes the projection of Ic onto AB.