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MMO 365 Moscow MO 1957 A'O/A'G + B'O/B'G + C'O/C'G = 3

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March 20, 2021
geometryratioEquilateralCentroid3D geometryspheretetrahedron

Problem Statement

(a) Given a point OO inside an equilateral triangle △ABC\vartriangle ABC. Line OGOG connects OO with the center of mass GG of the triangle and intersects the sides of the triangle, or their extensions, at points A′,B′,C′A', B', C' . Prove that A′OA′G+B′OB′G+C′OC′G=3.\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G} = 3. (b) Point GG is the center of the sphere inscribed in a regular tetrahedron ABCDABCD. Straight line OGOG connecting GG with a point OO inside the tetrahedron intersects the faces at points A′,B′,C′,D′A', B', C', D'. Prove that A′OA′G+B′OB′G+C′OC′G+D′OD′G=4.\frac{A'O}{A'G} + \frac{B'O}{B'G} + \frac{C'O}{C'G}+ \frac{D'O}{D'G} = 4.
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