MathDB
<A_1A_2A_3 +< A_2A_3A_4 +... + < A_{2021}A_1A_2 = 360^o

Source: Iranian Geometry Olympiad 2021 IGO Elementary p5

January 25, 2022
geometryangles

Problem Statement

Let A1,A2,...,A2021A_1, A_2, . . . , A_{2021} be 20212021 points on the plane, no three collinear and ∠A1A2A3+∠A2A3A4+...+∠A2021A1A2=360o,\angle A_1A_2A_3 + \angle A_2A_3A_4 +... + \angle A_{2021}A_1A_2 = 360^o, in which by the angle ∠Ai−1AiAi+1\angle A_{i-1}A_iA_{i+1} we mean the one which is less than 180o180^o (assume that A2022=A1A_{2022} =A_1 and A0=A2021A_0 = A_{2021}). Prove that some of these angles will add up to 90o90^o.
Proposed by Morteza Saghafian - Iran
Back to ProblemsView on AoPS