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Cyclic equalities involving floor.

Source: SMMC 2024 A1

October 12, 2024
algebra

Problem Statement

Let a,b,ca,b,c be real number greater than 1 satisfying ab=bc=ca.\lfloor a\rfloor b = \lfloor b \rfloor c = \lfloor c\rfloor a.Prove that a=b=ca=b=c (Here, x\lfloor x \rfloor denotes the laregst integer that is less than or equal to xx.)