58th Ukrainian National Mathematical Olympiad

Question

In triangle ABC there are points M_1, M_2, M_3 – midpoints of sides BC, AC, AB correspondingly. Point K is symmetrical to M_2 with respect to line BC, AH is the altitude of triangle ABC. Prove that line KM_3 splits the segment HM_1 in half. FIGURE_0

Accepted Answer

From right-angled triangles ABH and ACH we get that M_3H = 12AB, M_2H = 12AC (Fig. 42). Also M_1M_2 = 12AB, M_1M_3 = 12AC as centerlines. From symmetry M_2M_1 = M_1K and M_2H = HK. So M_3H = M_1K and M_3M_1 = HK, so quadrilateral HM_3M_1K is a parallelogram, so line KM_3 splits segment HM_1 in half. FIGURE_0 **Fig. 42**