Olympiad problems

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Found problems: 2

1958 Polish MO Finals, 2

Each side of a convex quadrilateral $ ABCD $ is divided into three equal parts; a straight line is drawn through the dividing points of sides $ AB $ and $ AD $ that lie closer to vertex $ A $, and similarly for vertices $ B $, $ C $, $ D $. Prove that the center of gravity of the quadrilateral formed by the drawn lines coincides with the center of gravity of quadrilateral $ ABCD $.

2015 District Olympiad, 4

Tags: geometry , center of gravity

At the exterior of the square $ ABCD $ it is constructed the isosceles triangle $ ABE $ with $ \angle ABE=120^{\circ} . M $ is the intersection of the bisector line of the angle $ \angle EAB $ with its perpendicular that passes through $ B; N $ is the intersection of the $ AB $ with its perpendicular that passe through $ M; P $ is the intersection of $ CN $ with $ MB. $ If $ G $ is the center of gravity of the triangle $ ABE, $ prove that $ PG $ and $ AE $ are parallel.