IMO 2002 SL G2

Origin: KOR | Category: Geometry

Problem

Let ABC be a triangle for which there exists an interior point F such that \angleAFB = \angleBFC = \angleCFA. Let the lines BF and CF meet the sides AC and AB at D and E respectively. Prove that AB + AC \geq4DE.

Solution

Construct equilateral triangles ACP and ABQ outside the triangle ABC. Since \angleAPC + \angleAFC = 60◦+ 120◦= 180◦, the points A, C, F, P lie on a circle; hence \angleAFP = \angleACP = 60◦= \angleAFD, so D lies on the segment FP; similarly, E lies on FQ. Further, note that FP FD = 1 + DP FD = 1 + SAPC SAF C \geq4 with equality if F is the midpoint of the smaller arc AC: hence FD \leq1 4FP and FE \leq1 4FQ. Then by the law of cosines, DE =

FD2 + FE2 + FD \cdot FE \leq1

FP 2 + FQ2 + FP \cdot FQ = 1 4PQ \leqAP + AQ = AB + AC. Equality holds if and only if \triangleABC is equilateral.