Problem:
Let A1βA2βA3βA4βA5βA6βA7βA8β be a regular octagon. Let M1β,M3β,M5β, and M7β be the midpoints of sides A1βA2ββ,A3βA4ββ,A5βA6ββ, and A7βA8ββ, respectively. For i=1,3,5,7, ray Riβ is constructed from Miβ towards the interior of the octagon such that R1ββ₯R3β,R3ββ₯R5β,R5ββ₯R7β, and R7ββ₯R1β. Pairs of rays R1β and R3β,R3β and R5β,R5β and R7β, and R7β and R1β meet at B1β,B3β,B5β, and B7β, respectively. If B1βB3β=A1βA2β, then cos2β A3βM3βB1β can be written in the form mβnβ, where m and n are positive integers. Find m+n.
Solution:
Because this configuration can be scaled without affecting the angles, assume that A1βA2β=2. Then M1βA2β=M3βA3β=1.
In this regular octagon AiβAi+1ββ₯Ai+2βAi+3β (where Ak+8β=Akβ ). Consider the 90β counterclockwise rotation centered at the center of the octagon. Under this rotation, Ri+2β goes to Riβ. Hence by symmetry, B1βB3βB5βB7β is a square and M1βB1β=M3βB3β=M5βB5β=M7βB7β. Set M1βB1β=a and M3βB1β=b. Then
Let C denote the intersection of lines A1βA2β and A3βA4β. The properties of the regular octagon show that β A1βA2βA3β=β A2βA3βA4β=135β, β M1βCM3β=90β,A2βC=A3βC=2β, and M1βC=M3βC=1+2β. In particular, in the right triangles M1βM3βC and M1βM3βB1β,
The octagon is dissected into 4 congruent pentagons and one square. These five pieces can be reassembled to form a square. Because A1βA2β=B1βB3β, this square and the octagon have the same area, from which it follows that (a+b)2=(2+22β)2β4.
