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Bay Area Mathematical Olympiad #0kao

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.
Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleAngle chasingDistance chasing

Problem

Problem:

In triangle △ABC\triangle A B C, we have marked points A1A_{1} on side BCB C, B1B_{1} on side ACA C, and C1C_{1} on side ABA B so that AA1A A_{1} is an altitude, BB1B B_{1} is a median, and CC1C C_{1} is an angle bisector. It is known that △A1B1C1\triangle A_{1} B_{1} C_{1} is equilateral. Prove that △ABC\triangle A B C is equilateral too.

(Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median BB1B B_{1} we know that B1B_{1} is the midpoint of side ACA C in △ABC\triangle A B C.)

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