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HMMT February 2025 #0l7b

HMMT February 2025Hard

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 circleHomothetyCyclic quadrilateralsTangentsSimson linePolar triangles, harmonic conjugatesAngle chasing

Problem

Problem:

Let △ABC\triangle ABC be a triangle with incenter II. The incircle of triangle △ABC\triangle ABC touches BC‾\overline{BC} at DD. Let MM be the midpoint of BC‾\overline{BC}, and let line AIAI meet the circumcircle of triangle △ABC\triangle ABC again at L≠AL \neq A. Let ω\omega be the circle centered at LL tangent to ABAB and ACAC. If ω\omega intersects segment AD‾\overline{AD} at point PP, prove that ∠IPM=90∘\angle IPM = 90^{\circ}.

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