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Problem 13 (2019 AIME II)

AIME 2019
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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.

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Problem

Regular octagon A1A2A3A4A5A6A7A8A_1A_2A_3A_4A_5A_6A_7A_8 is inscribed in a circle of area 1.1. Point PP lies inside the circle so that the region bounded by PA1‾,PA2‾,\overline{PA_1},\overline{PA_2}, and the minor arc A1A2^\widehat{A_1A_2} of the circle has area 17,\tfrac{1}{7}, while the region bounded by PA3‾,PA4‾,\overline{PA_3},\overline{PA_4}, and the minor arc A3A4^\widehat{A_3A_4} of the circle has area 19.\tfrac{1}{9}. There is a positive integer nn such that the area of the region bounded by PA6‾,PA7‾,\overline{PA_6},\overline{PA_7}, and the minor arc A6A7^\widehat{A_6A_7} of the circle is equal to 18−2n.\tfrac{1}{8}-\tfrac{\sqrt2}{n}. Find n.n.

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