MathDB
Octagon midpoints

Source:

March 18, 2011
trigonometrysymmetryanalytic geometrygraphing linesslopePythagorean Theoremgeometry

Problem Statement

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 R1R3R_1 \perp R_3, R3R5R_3 \perp R_5, R5R7R_5 \perp R_7, and R7R1R_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 cos2A3M3B1\cos 2 \angle A_3 M_3 B_1 can be written in the form mnm - \sqrt{n}, where mm and nn are positive integers. Find m+nm + n.