olymmath

RU-ALLRUSSIAN 1966 Problem 2

    (a) B and C are on the segment AD with AB=CD\mathrm{AB} = \mathrm{CD} . Prove that for any point P in the plane: PA+PD≥PB+PC\mathrm{PA} + \mathrm{PD} \geq \mathrm{PB} + \mathrm{PC} .

    (b) Given four points A, B, C, D on the plane such that for any point P on the plane we have PA+PD≥PB+PC\mathrm{PA} + \mathrm{PD} \geq \mathrm{PB} + \mathrm{PC} . Prove that B and C are on the segment AD with AB=CD\mathrm{AB} = \mathrm{CD} .

    Dataset text NuminaMath olympiads. Quoted with source.

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