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VII.12

Part of Ukrainian TYM Qualifying - geometry

Problems(1)

\pi/ 3 <= (\alpha a +\beta b + \gamma c)/(a+b+c) < \pi}/2

Source: VII All-Ukrainian Tournament of Young Mathematicians, Qualifying p12

5/25/2021
Let a,ba, b, and cc be the lengths of the sides of an arbitrary triangle, and let α,β\alpha,\beta, and γ\gamma be the radian measures of its corresponding angles. Prove that π3≤αa+βb+γca+b+c<π2. \frac{\pi}{3}\le \frac{\alpha a +\beta b + \gamma c}{a+b+c} < \frac{\pi}{2}. Suggest spatial analogues of this inequality.
geometrygeometric inequalityUkrainian TYMalgebraInequality