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\sqrt2 KM > AB, midpoint of broken BAC (2019 Kyiv City MO 11.2)

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September 16, 2020
geometrymidpointgeometric inequality

Problem Statement

In an acute-angled triangle ABCABC, in which AB<ACAB<AC, the point MM is the midpoint of the side BC,KBC, K is the midpoint of the broken line segment BACBAC . Prove that 2KM>AB\sqrt2 KM > AB.
(George Naumenko)
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