Problem:
Let A1A2A3A4A5A6A7A8 be a regular octagon. Let M1,M3,M5, and M7 be the midpoints of sides A1A2,A3A4,A5A6, and A7A8, respectively. For i=1,3,5,7, ray Ri is constructed from Mi towards the interior of the octagon such that R1⊥R3,R3⊥R5,R5⊥R7, and R7⊥R1. Pairs of rays R1 and R3,R3 and R5,R5 and R7, and R7 and R1 meet at B1,B3,B5, and B7, respectively. If B1B3=A1A2, then cos2∠A3M3B1 can be written in the form m−n, where m and n are positive integers. Find m+n.
Solution:
Because this configuration can be scaled without affecting the angles, assume that A1A2=2. Then M1A2=M3A3=1.
In this regular octagon AiAi+1⊥Ai+2Ai+3 (where Ak+8=Ak ). Consider the 90∘ counterclockwise rotation centered at the center of the octagon. Under this rotation, Ri+2 goes to Ri. Hence by symmetry, B1B3B5B7 is a square and M1B1=M3B3=M5B5=M7B7. Set M1B1=a and M3B1=b. Then
b−a=M3B1−M1B1=M3B1−M3B3=B1B3=A1A2=2
Let C denote the intersection of lines A1A2 and A3A4. The properties of the regular octagon show that ∠A1A2A3=∠A2A3A4=135∘, ∠M1CM3=90∘,A2C=A3C=2, and M1C=M3C=1+2. In particular, in the right triangles M1M3C and M1M3B1,
The octagon is dissected into 4 congruent pentagons and one square. These five pieces can be reassembled to form a square. Because A1A2=B1B3, this square and the octagon have the same area, from which it follows that (a+b)2=(2+22)2−4.
