Problem

Source: Romanian TST 4 2007, Problem 2, BMO 2007 Shortlist

Tags: geometry, parallelogram, trapezoid, geometry proposed



Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that \[ [A_{1}A_{2}A_{3}] = [A_{2}A_{3}A_{4}] = [A_{3}A_{4}A_{5}] = [A_{4}A_{5}A_{1}] = [A_{5}A_{1}A_{2}].\] Prove that there exists a point $ M$ in the plane of the pentagon such that \[ [A_{1}MA_{2}] = [A_{2}MA_{3}] = [A_{3}MA_{4}] = [A_{4}MA_{5}] = [A_{5}MA_{1}].\] Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.