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13th Bay Area Mathematical Olympiad #0j81

13th Bay Area Mathematical OlympiadHard

From module Angle Chasing Mastery

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Understand the scale
Scale 0.5: Easiest math competition problems, often solvable even for early middle school students without experience. 1: Problems strictly for beginners, often at a moderate middle school level. 1.5: Problems for stronger beginner students, on the level of the middling problems in most middle school contests. 2: For motivated beginners, harder questions from the previous categories. 2.5: More advanced beginner problems, hardest questions from previous categories. 3: Early intermediate problems that require more creative thinking. 3.5: Tougher early intermediate problems that consistently stretch into higher-level creative thinking or conceptual knowledge. 4: Intermediate-level problems. 4.5: Upper intermediate problems approaching the upper bound for non-invitational sprint math competitions. 5: More difficult AIME problems or simple proof-based Olympiad-style problems. 6: High-level AIME-styled questions or introductory Olympiad-level questions. 7: Tougher Olympiad-level questions, may require more technical knowledge. 8: High-level Olympiad-level questions. 9: Expert Olympiad-level questions. 9.5: The hardest problems appearing on Olympiads which the strongest students could reasonably solve. 10: Historically notorious problems, generally unsuitable for even very hard competitions because they are exceedingly tedious, long, and difficult.
InversionAngle chasingDistance chasing

Problem

Problem:

Three circles k1k_{1}, k2k_{2}, and k3k_{3} intersect in point OO. Let AA, BB, and CC be the second intersection points (other than OO) of k2k_{2} and k3k_{3}, k1k_{1} and k3k_{3}, and k1k_{1} and k2k_{2}, respectively. Assume that OO lies inside of the triangle ABCABC. Let lines AOAO, BOBO, and COCO intersect circles k1k_{1}, k2k_{2}, and k3k_{3} for a second time at points A′A', B′B', and C′C', respectively. If ∣XY∣|XY| denotes the length of segment XYXY, prove that

∣AO∣∣AA′∣+∣BO∣∣BB′∣+∣CO∣∣CC′∣=1\frac{|AO|}{|AA'|} + \frac{|BO|}{|BB'|} + \frac{|CO|}{|CC'|} = 1

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