55rd Ukrainian National Mathematical Olympiad - Fourth Round

Question

Inside an equilateral triangle ABC point M is chosen. Let points M_1, M_2, M_3 be symmetric corresponding to sides BC, AC, AB of the triangle. Prove that MM_1 + MM_2 + MM_3 is equal to MA + MB + MC. (Tereshin Dmitro) FIGURE_0

Accepted Answer

Let us draw through point M lines that are parallel to sides ABC. Let them intersect AB, BC, AC at C_1, C_2; A_1, A_2; B_1, B_2 (Fig. 30). FIGURE_0 Hence, C_1A_2 AC, A_1B_2 BA, B_1C_2 BC, so C_1A_2, A_1B_2 and B_1C_2 intersect at M. Consider A_1MA_2. Obviously, this triangle is equilateral. Line M_1M contains its altitude, because it is perpendicular to BC. So M_1M is doubled median, hence MA_1 + MA_2 = MM_1, similarly MB_1 + MB_2 = MM_2 and MC_1 + MC_2 = MM_3. On the other hand MC_1AB_2 is a parallelogram, so MA = MC_1 + MB_2. Similarly MB = MC_2 + MA_1 and MC = MA_2 + MB_1. Hence: MM_1 + MM_2 + MM_3 = MA_1 + MA_2 + MB_1 + MB_2 + MC_1 + MC_2 = MA + MB + MC.