Sum of angles in regular polygon
By Arun Iyer on Monday, January 21, 2002 -
06:58 pm:
this problem is from some book on complex number and was posed
before me by my friend...the problem really seems out of this
world atleast for me...
the problem.....
Given->
ABCDEFG..... is a regular polygon of n sides,inscribed in a
circle of radius a and center O.
Show that->
the sum of the angles that AP,BP,CP,.... make with OP is,
tan-1([ansinnq]/[ancosnq-rn])
where OP=r and angle AOP=q
love arun
By David Loeffler on Thursday, January
24, 2002 - 01:13 pm:
My reasoning was as follows. Let's suppose we are in the
complex plane and O is the origin, and A the point (a,0). Then the
other vertices are the solutions of the equation xn=an, so we obtain
xn-an=(x-a)(x-b)(x-c)¼
Now, the angle you want is the sum of the arguments of (a-x)-argx,
(b-x)-argx etc where x is the point P.
So this is
arg(a-x)(b-x)¼-argxn=arg(-1)n(x-a)(x-b)¼-argxn = arg(-1)n(1-(a/x)n)
Now put x=r eiq, so xn=rn en iq, and we must find
arg(-1)n(1-an r-n e-n iq)
=tan-1[(ansinnq)/(ancosnq- rn)]
(possibly ± some multiple of p).
David