Formulas: Width, Side Length of Octagon

Date: 03/24/97 at 22:45:29
From: cash perkins
Subject: Width, Side Length of Octagon

Hi,

I trying to find some simple geometric formulas:

1. Given the width of a regular octagon, find the side lengths.
2. Given the side length, find the width of the octagon.

Thanks in advance,

Cash Perkins

Date: 03/25/97 at 07:29:54
From: Doctor Anthony
Subject: Re: Width, Side Length of Octagon

If you draw a circle of radius r and construct the octagon to lie 
inside the circle with the vertices on the circumference of the 
circle, then drop a perpendicular from the centre of the circle to the 
mid-point of one of the sides, it is easy to see that the length of 
this perpendicular is 

 r.cos(22.5), so width of octagon is 2r.cos(22.5)

Also, half the length of the side of the octagon is r.sin(22.5), so 
the length of a side of the octagon is 2r.sin(22.5)

  ratio   side/width = 2rsin(22.5)/2rcos(22.5)
               
                     = tan(22.5) = 0.414213

You could stop here if you like, but if you would prefer the answer in 
surd form, we can use the double angle formula for tan(x)

      tan(45) = 2tan(22.5)/[1-tan^2(22.5)]   

 If tan(22.5) = x  we get
    
            1 = 2x/(1-x^2)    
     so 1-x^2 = 2x
 x^2 + 2x - 1 = 0
            x = [-2 +or- sqrt(4+4)]/2  
            x = [-2 +or- sqrt(8)]/2

we can ignore the negative option since x must be positive, so

            x = [-2 + 2sqrt(2)]/2
              = -1 + sqrt(2)

So ratio   side/width  =  sqrt(2)-1   =  0.414213  as before.

Knowing the ratio, then if given the width the side =(sqrt(2)-1)width. 
Given the side, then  width = side/[sqrt(2)-1] 

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/