IMO 1997 SL 9

Origin: USA

Problem

Let A1A2A3 be a nonisosceles triangle with incenter I. Let Ci, i = 1, 2, 3, be the smaller circle through I tangent to AiAi+1 and AiAi+2 (the addition of indices being mod 3). Let Bi, i = 1, 2, 3, be the second point of intersection of Ci+1 and Ci+2. Prove that the circumcenters of the triangles A1B1I, A2B2I, A3B3I are collinear.

Solution

For i = 1, 2, 3 (all indices in this problem will be modulo 3) we denote by Oi the center of Ci and by Mi the midpoint of the arc Ai+1Ai+2 that does not contain Ai. First we have that Oi+1Oi+2 is the perpendicular bisector of IBi, and thus it contains the circumcenter Ri of AiBiI. Ad- ditionally, it is easy to show that Ti+1Ai

Ti+1I and Ti+2Ai

Ti+2I, which implies that Ri lies on the line Ti+1Ti+2. Therefore Ri = Oi+1Oi+2 \capTi+1Ti+2. A1 A2 A3 I B1 B2 B3 R3 R1 Now, the lines T1O1, T2O2, T3O3 are concurrent at I. By Desargues’s the- orem, the points of intersection of Oi+1Oi+2 and Ti+1Ti+2, i.e., the Ri’s, lie on a line for i = 1, 2, 3. Second solution. The centers of three circles passing through the same point I and not touching each other are collinear if and only if they have another common point. Hence it is enough to show that the circles AiBiI have a common point other than I. Now apply inversion at center I and with an arbitrary power. We shall denote by X′ the image of X under this inversion. In our case, the image of the circle Ci is the line B′ i+1B′ i+2 while the image of the line Ai+1Ai+2 is the circle IA′ i+1A′ i+2 that is tangent to B′ iB′ i+2, and B′ iB′ i+2. These three circles have equal radii, so their centers P1, P2, P3 form a triangle also homothetic to \triangleB′ 1B′ 2B′ 3. Consequently, points A′ 1, A′ 2, A′ 3, that are the reflections of I across the sides of P1P2P3, are vertices of a triangle also homothetic to B′ 1B′ 2B′ 3. It follows that A′ 1B′ 1, A′ 2B′ 2, A′ 3B′ 3 are concurrent at some point J′, i.e., that the circles AiBiI all pass through J.