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centers of mass, and a geometric double inequality

Source: Romanian District Olympiad 2002, Grade IX, Problem 3

October 7, 2018
inequalitiesgeometrycenter of mass

Problem Statement

Let G G be the center of mass of a triangle ABC, ABC, and the points M,N,P M,N,P on the segments AB,BC, AB,BC, respectively, CA CA (excluding the extremities) such that AMMB=BNNC=CPPA. \frac{AM}{MB} =\frac{BN}{NC} =\frac{CP}{PA} . G1,G2,G3 G_1,G_2,G_3 are the centers of mass of the triangles AMP,BMN, AMP, BMN, respectively, CNP. CNP. Pove that:
a) The centers of mas of ABC ABC and G1G2G3 G_1G_2G_3 are the same. b) For any planar point D, D, the inequality 3⋅DG<DG1+DG2+DG3<DA+DB+DC 3\cdot DG< DG_1+DG_2+DG_3<DA+DB+DC holds.
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