Draw a regular nonagon. If we label the vertices of the polygon sequentially, starting from A, we can draw three line segments: AB, AC, and AE.
Show that AB + AC = AE.
WOLOG let the circumradius of the nonagon be 1.
AE = AB + AC
2*sin(80) = 2*sin(20) + 2*sin(40)
4*sin(40)*cos(40) = 2*sin(20) + 4*sin(20)*cos(20)
8*sin(20)*cos(20)*cos(40) = 2*sin(20) + 4*sin(20)*cos(20)
4*cos(20)*cos(40) = 1 + 2*cos(20)
2*cos(20)*[2*cos(40) 1] = 1
2*cos(20)*[2*{2*cos(20)^2 1}  1] = 1
2*[4*cos(20)^3  3*cos(20)] = 1
2*cos(60) = 1
cos(60) = 1/2
QED
Edited on November 26, 2004, 5:05 pm

Posted by Bractals
on 20041126 16:40:35 