# AMO 2009 Problem 7

Problem 7 of the Australian Mathematical Olympiad 2009

Let I be the incentre of a triangle ABC in which AC ≠ BC. Let Γ be the circle passing through A, I and B. Suppose Γ intersects the line AC at A and X and intersects the line BC at B and Y . Show that AX = BY .

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to Pat Scholl, Ph.D.) Since BI is the bisector of ∠ABC, we have ∠ABI = ∠CBI, but ∠CBI = ∠CYI + ∠YIB subtends arcs YB + BI = arc YI. Meanwhile ∠ABI subtends arc AI. Therefore,

AI = YI and ∠ABI = ∠YXI (*)

We also have ∠AXY = ∠ABY (subtend larger arc AY) or ∠ABC = ∠YXC or ∠ABI + ∠CBI = ∠YXI + CXI.

Along with (*), we have ∠CBI = ∠CXI and the two triangles BDI and XEI being congruent, or XE = BD.

We also have EC = DC and thus CX = CB. From point C outside circle г, we have CX·CA = CB·CY or CA = CY. Hence, AX = CA  CX = CY  CB = BY

(There is a different solution that is accompanied by a dynamic illustration and another even more transparent.)

There are five solutions in all: