Join the USAMO Guide Discord Server. Regular updates are posted there!

Let me Join!
Dismiss

Harvard-MIT Mathematics Tournament #0jcr

Harvard-MIT Mathematics TournamentHard

From module Angle Chasing Mastery

Community difficultyNo ratings yetBe the first to rate this problem.
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.
Cyclic quadrilateralsTriangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleTangentsAngle chasing

Problem

Problem:

Let ABCABC be a triangle with AB<ACAB < AC. Let MM be the midpoint of BCBC. Line ll is drawn through MM so that it is perpendicular to AMAM, and intersects line ABAB at point XX and line ACAC at point YY. Prove that ∠BAC=90∘\angle BAC = 90^{\circ} if and only if quadrilateral XBYCXBYC is cyclic.

← Back to all problems