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RU-ALLRUSSIAN 1981 Problem 15

    ABC is a triangle. A' lies on the side BC with BA/BC=1/4\mathrm{BA} / \mathrm{BC} = 1 / 4 . Similarly, B' lies on the side CA with CB/CA=1/4\mathrm{CB} / \mathrm{CA} = 1 / 4 , and C' lies on the side AB with AC/AB=1/4\mathrm{AC} / \mathrm{AB} = 1 / 4 . Show that the perimeter of A'B'C' is between 1/2 and 3/4 of the perimeter of ABC.

    Dataset text NuminaMath olympiads. Quoted with source.

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