IMO 2012 Shortlist G4

Category: Geometry

Problem

Let ABC be a triangle with AB 6= AC and circumcenter O. The bisector of ∠BAC intersects BC at D. Let E be the reflection of D with respect to the midpoint of BC. The lines through D and E perpendicular to BC intersect the lines AO and AD at X and Y respectively. Prove that the quadrilateral BXCY is cyclic. Solution. The bisector of ∠BAC and the perpendicular bisector of BC meet at P, the midpoint of the minor arc d BC (they are different lines as AB 6= AC). In particular OP is perpendicular to BC and intersects it at M, the midpoint of BC. Denote by Y ′ the reflexion of Y with respect to OP. Since ∠BY C = ∠BY ′ C, it suffices to prove that BXCY ′ is cyclic. A D Y ′ Y B C X M O E P We have ∠XAP = ∠OPA = ∠EY P. The first equality holds because OA = OP, and the second one because EY and OP are both perpendicular to BC and hence parallel. But {Y,Y ′ } and {E,D} are pairs of symmetric points with respect to OP, it follows that ∠EY P = ∠DY ′ P and hence ∠XAP = ∠DY ′ P = ∠XY ′ P. The last equation implies that XAY ′ P is cyclic. By the powers of D with respect to the circles (XAY ′ P) and (ABPC) we obtain XD · DY ′ = AD · DP = BD · DC. It follows that BXCY ′ is cyclic, as desired.33