MathDB
AO/OK =? <A=60^0, ABCD # - All-Russian MO 2000 Regional (R4) 10.3

Source:

September 26, 2024
ratiogeometryparallelogram

Problem Statement

Given a parallelogram ABCDABCD with angle AA equal to 60o60^o. Point OO is the the center of a circle circumscribed around triangle ABDABD. Line AOAO intersects the bisector of the exterior angle CC at point KK. Find the ratio AO/OKAO/OK.