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Problem 14 (2011 AIME I)

AIME 2011
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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

Let A1A2A3A4A5A6A7A8A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 be a regular octagon. Let M1M_1 , M3M_3 , M5M_5 , and M7M_7 be the midpoints of sides A1A2‾\overline{A_1 A_2} , A3A4‾\overline{A_3 A_4} , A5A6‾\overline{A_5 A_6} , and A7A8‾\overline{A_7 A_8} , respectively. For i=1,3,5,7i = 1, 3, 5, 7 , ray RiR_i is constructed from MiM_i towards the interior of the octagon such that R1⊥R3R_1 \perp R_3 , R3⊥R5R_3 \perp R_5 , R5⊥R7R_5 \perp R_7 , and R7⊥R1R_7 \perp R_1 . Pairs of rays R1R_1 and R3R_3 , R3R_3 and R5R_5 , R5R_5 and R7R_7 , and R7R_7 and R1R_1 meet at B1B_1 , B3B_3 , B5B_5 , B7B_7 respectively. If B1B3=A1A2B_1 B_3 = A_1 A_2 , then cos⁡2∠A3M3B1\cos 2 \angle A_3 M_3 B_1 can be written in the form m−nm - \sqrt{n} , where mm and nn are positive integers. Find m+nm + n . Diagram $

figure

$ All distances are to scale.

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