IMO 2012 Shortlist G6

Category: Geometry

Problem

Let ABC be a triangle with circumcenter O and incenter I. The points D, E and F on the sides BC, CA and AB respectively are such that BD + BF = CA and CD + CE = AB. The circumcircles of the triangles BFD and CDE intersect at P 6= D. Prove that OP = OI. Solution. By Miquel’s theorem the circles (AEF) = ωA, (BFD) = ωB and (CDE) = ωC have a common point, for arbitrary points D, E and F on BC, CA and AB. So ωA passes through the common point P 6= D of ωB and ωC. Let ωA, ωB and ωC meet the bisectors AI, BI and CI at A 6= A′ , B 6= B′ and C 6= C′ respectively. The key observation is that A′ , B′ and C′ do not depend on the particular choice of D, E and F, provided that BD + BF = CA, CD + CE = AB and AE + AF = BC hold true (the last equality follows from the other two). For a proof we need the following fact. Lemma. Given is an angle with vertex A and measure α. A circle ω through A intersects the angle bisector at L and sides of the angle at X and Y . Then AX + AY = 2ALcos α . Proof. Note that L is the midpoint of arc \ XLY in ω and set XL = Y L = u, XY = v. By Ptolemy’s theorem AX ·Y L+AY ·XL = AL·XY , which rewrites as (AX +AY )u = AL·v. Since ∠LXY = α and ∠XLY = 180◦ − α, we have v = 2cos α u by the law of sines, and the claim follows. X L u u v A Y Apply the lemma to ∠BAC = α and the circle ω = ωA, which intersects AI at A′ . This gives 2AA′ cos α = AE + AF = BC; by symmetry analogous relations hold for BB′ and CC′ . It follows that A′ , B′ and C′ are independent of the choice of D, E and F, as stated. We use the lemma two more times with ∠BAC = α. Let ω be the circle with diameter AI. Then X and Y are the tangency points of the incircle of ABC with AB and AC, and hence AX = AY = 1 (AB + AC − BC). So the lemma yields 2AI cos α = AB + AC − BC. Next, if ω is the circumcircle of ABC and AI intersects ω at M 6= A then {X,Y } = {B,C}, and so 2AM cos α = AB + AC by the lemma. To summarize, 2AA′ cos α = BC, 2AI cos α = AB + AC − BC, 2AM cos α = AB + AC. (*) These equalities imply AA′

AI = AM, hence the segments AM and IA′ have a common midpoint. It follows that I and A′ are equidistant from the circumcenter O. By symmetry OI = OA′ = OB′ = OC′ , so I,A′ ,B′ ,C′ are on a circle centered at O. To prove OP = OI, now it suffices to show that I,A′ ,B′ ,C′ and P are concyclic. Clearly one can assume P 6= I,A′ ,B′ ,C′ . We use oriented angles to avoid heavy case distinction. The oriented angle between the lines l and m is denoted by ∠(l,m). We have ∠(l,m) = −∠(m,l) and ∠(l,m) + ∠(m,n) = ∠(l,n) for arbitrary lines l,m and n. Four distinct non-collinear points U,V,X,Y are concyclic if and only if ∠(UX,V X) = ∠(UY,V Y ).35 M C B′ I O E P C′ A B A′ ωA D F ωB ωC Suppose for the moment that A′ ,B′ ,P,I are distinct and noncollinear; then it is enough to check the equality ∠(A′ P,B′ P) = ∠(A′ I,B′ I). Because A,F,P,A′ are on the circle ωA, we have ∠(A′ P,FP) = ∠(A′ A,FA) = ∠(A′ I,AB). Likewise ∠(B′ P,FP) = ∠(B′ I,AB). Therefore ∠(A′ P,B′ P) = ∠(A′ P,FP) + ∠(FP,B′ P) = ∠(A′ I,AB) − ∠(B′ I,AB) = ∠(A′ I,B′ I). Here we assumed that P 6= F. If P = F then P 6= D,E and the conclusion follows similarly (use ∠(A′ F,B′ F) = ∠(A′ F,EF) + ∠(EF,DF) + ∠(DF,B′ F) and inscribed angles in ωA,ωB,ωC). There is no loss of generality in assuming A′ ,B′ ,P,I distinct and noncollinear. If ABC is an equilateral triangle then the equalities (*) imply that A′ ,B′ ,C′ ,I,O and P coincide, so OP = OI. Otherwise at most one of A′ ,B′ ,C′ coincides with I. If say C′ = I then OI ⊥ CI by the previous reasoning. It follows that A′ ,B′ 6= I and hence A′ 6= B′ . Finally A′ ,B′ and I are noncollinear because I,A′ ,B′ ,C′ are concyclic. Comment. The proposer remarks that the locus γ of the points P is an arc of the circle (A′B′C′I). The reflection I′ of I in O belongs to γ; it is obtained by choosing D, E and F to be the tangency points of the three excircles with their respective sides. The rest of the circle (A′B′C′I), except I, can be included in γ by letting D, E and F vary on the extensions of the sides and assuming signed lengths. For instance if B is between C and D then the length BD must be taken with a negative sign. The incenter I corresponds to the limit case where D tends to infinity.36