Maths Olympiad Prep

Library / Macedonian Mathematical Competitions /60 of 73

Macedonian Mathematical Competitions

Geometry Difficulty 5.6 AIME, harder Prove it North Macedonia

Let A1B1C1A_1B_1C_1 and A2B2C2A_2B_2C_2 are given triangles. Let T1T_1 and T2T_2 are their centers of mass correspondently. Prove that 3T1T2‾=A1A2‾+B1B2‾+C1C2‾3\overline{T_1T_2} = \overline{A_1A_2} + \overline{B_1B_2} + \overline{C_1C_2}.

Solution

We have A1A2‾=A1T1‾+T1A2‾+T2A2‾\overline{A_1A_2} = \overline{A_1T_1} + \overline{T_1A_2} + \overline{T_2A_2}, B1B2‾=B1T1‾+T1B2‾+T2B2‾\overline{B_1B_2} = \overline{B_1T_1} + \overline{T_1B_2} + \overline{T_2B_2}, C1C2‾=C1T1‾+T1C2‾+T2C2‾\overline{C_1C_2} = \overline{C_1T_1} + \overline{T_1C_2} + \overline{T_2C_2}. If we sum the last three equalities we obtain
A1A2‾+B1B2‾+C1C2‾=3T1T2‾+(A1T1‾+B1T1‾+C1T1‾)+(T2A2‾+T2B2‾+T2C2‾)…(1) \overline{A_1A_2} + \overline{B_1B_2} + \overline{C_1C_2} = 3\overline{T_1T_2} + (\overline{A_1T_1} + \overline{B_1T_1} + \overline{C_1T_1}) + (\overline{T_2A_2} + \overline{T_2B_2} + \overline{T_2C_2}) \dots(1)
From the properties of medians and center of mass we have
A1T1‾=23(A1B1‾+12B1C1‾),B1T1‾=23(B1C1‾+12C1A1‾),C1T1‾=23(C1A1‾+12A1B1‾) \overline{A_1T_1} = \frac{2}{3}(\overline{A_1B_1} + \frac{1}{2}\overline{B_1C_1}), \quad \overline{B_1T_1} = \frac{2}{3}(\overline{B_1C_1} + \frac{1}{2}\overline{C_1A_1}), \quad \overline{C_1T_1} = \frac{2}{3}(\overline{C_1A_1} + \frac{1}{2}\overline{A_1B_1})
So we have
A1T1‾+B1T1‾+C1T1‾=23(A1B1‾+12B1C1‾)+23(B1C1‾+12C1A1‾)+23(C1A1‾+12A1B1‾)==13(A1B1‾+B1C1‾+C1A1‾)+13(B1C1‾+C1A1‾+A1B1‾)==23A1A1‾+13B1B1‾=23o‾+13o‾=o‾+o‾‾=o‾‾. \begin{aligned} \overline{A_1T_1} + \overline{B_1T_1} + \overline{C_1T_1} &= \frac{2}{3}(\overline{A_1B_1} + \frac{1}{2}\overline{B_1C_1}) + \frac{2}{3}(\overline{B_1C_1} + \frac{1}{2}\overline{C_1A_1}) + \frac{2}{3}(\overline{C_1A_1} + \frac{1}{2}\overline{A_1B_1}) = \\ &= \frac{1}{3}(\overline{A_1B_1} + \overline{B_1C_1} + \overline{C_1A_1}) + \frac{1}{3}(\overline{B_1C_1} + \overline{C_1A_1} + \overline{A_1B_1}) = \\ &= \frac{2}{3}\overline{A_1A_1} + \frac{1}{3}\overline{B_1B_1} = \frac{2}{3}\overline{o} + \frac{1}{3}\overline{o} = \overline{o} + \overline{\overline{o}} = \overline{\overline{o}}. \end{aligned}
Similarly we have A2T2‾+B2T2‾+C2T2‾=o‾\overline{A_2T_2} + \overline{B_2T_2} + \overline{C_2T_2} = \overline{o} from where we have that T2A2‾+T2B2‾+T2C2‾=o‾\overline{T_2A_2} + \overline{T_2B_2} + \overline{T_2C_2} = \overline{o}. Finally if we substitute in (1) we get
A1A2‾+B1B2‾+C1C2‾=3T1T2‾+o‾+o‾‾=3T1T2‾. \overline{A_1A_2} + \overline{B_1B_2} + \overline{C_1C_2} = 3\overline{T_1T_2} + \overline{o} + \overline{\overline{o}} = 3\overline{T_1T_2}.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.