IMO 2012 Shortlist G3

Category: Geometry

Problem

In an acute triangle ABC the points D, E and F are the feet of the altitudes through A, B and C respectively. The incenters of the triangles AEF and BDF are I1 and I2 respectively; the circumcenters of the triangles ACI1 and BCI2 are O1 and O2 respectively. Prove that I1I2 and O1O2 are parallel. Solution. Let ∠CAB = α, ∠ABC = β, ∠BCA = γ. We start by showing that A,B,I1 and I2 are concyclic. Since AI1 and BI2 bisect ∠CAB and ∠ABC, their extensions beyond I1 and I2 meet at the incenter I of the triangle. The points E and F are on the circle with diameter BC, so ∠AEF = ∠ABC and ∠AFE = ∠ACB. Hence the triangles AEF and ABC are similar with ratio of similitude AE AB = cosα. Because I1 and I are their incenters, we obtain I1A = IAcosα and II1 = IA − I1A = 2IAsin2 α . By symmetry II2 = 2IB sin2 β . The law of sines in the triangle ABI gives IAsin α = IB sin β . Hence II1 · IA = 2 IAsin α 2 = 2 IB sin β 2 = II2 · IB. Therefore A,B,I1 and I2 are concyclic, as claimed. O2 O1 C A F B I2 Q I E D I3 I1 In addition II1 · IA = II2 · IB implies that I has the same power with respect to the circles (ACI1), (BCI2) and (ABI1I2). Then CI is the radical axis of (ACI1) and (BCI2); in particular CI is perpendicular to the line of centers O1O2. Now it suffices to prove that CI ⊥ I1I2. Let CI meet I1I2 at Q, then it is enough to check that ∠II1Q + ∠I1IQ = 90◦ . Since ∠I1IQ is external for the triangle ACI, we have ∠II1Q + ∠I1IQ = ∠II1Q + (∠ACI + ∠CAI) = ∠II1I2 + ∠ACI + ∠CAI. It remains to note that ∠II1I2 = β from the cyclic quadrilateral ABI1I2, and ∠ACI = γ , ∠CAI = α . Therefore ∠II1Q + ∠I1IQ = α

β
γ = 90◦ , completing the proof. Comment. It follows from the first part of the solution that the common point I3 6= C of the circles (ACI1) and (BCI2) is the incenter of the triangle CDE.32