Team Selection Test Selection Test #0j9y

Team Selection Test Selection TestHard

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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.

Problem

Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$ . Line $l$ is tangent to $\omega_1$ at $P$ and to $\omega_2$ at $Q$ so that $A$ is closer to $l$ than $B$ . Let $X$ and $Y$ be points on major arcs $\widehat{PA}$ (on $\omega_1$ ) and $\widehat{AQ}$ (on $\omega_2$ ), respectively, such that $AX/PX = AY/QY = c$ . Extend segments $PA$ and $QA$ through $A$ to $R$ and $S$ , respectively, such that $AR = AS = c \cdot PQ$ . Given that the circumcenter of triangle $ARS$ lies on line $XY$ , prove that $\angle XPA = \angle AQY$ .

(This problem was suggested by Delong Meng.)