Here the red circle circumscribes the outer radius and the green circle the inner one.
![enter image source here]()
Let r be the outer radius - that is the radius of the red circle.
Then the vertices of the octagon centred at (0, 0) are at:
(+-r, 0), (0, +-r), (+-r/sqrt(2), +-r/sqrt(2))
The length of one side is the distance between (r, 0) and (r/sqrt(2), r/sqrt(2)):
sqrt((r-r/sqrt(2))^2+(r/sqrt(2))^2)
=r sqrt((1-1/sqrt(2))^2 + 1/2)
=r sqrt(1-2/sqrt(2) + 1/2 + 1/2)
=r sqrt(2-sqrt(2))
So the total perimeter is:
color(red)(8r sqrt(2-sqrt(2)))
So if the outer radius is 20, then the perimeter is:
8*20 sqrt(2-sqrt(2)) = 160 sqrt(2-sqrt(2)) ~~ 122.46
color(white)()
The inner radius will be r_1 = r cos(pi/8) = r/2 (sqrt(2+sqrt(2)))
So r = (2r_1)/(sqrt(2+sqrt(2)))
Then the total perimeter is
8r sqrt(2-sqrt(2)) = 8(2r_1)/(sqrt(2+sqrt(2)))sqrt(2-sqrt(2))
=16r_1 sqrt(2-sqrt(2))/sqrt(2+sqrt(2))
=16r_1 (sqrt(2-sqrt(2))sqrt(2+sqrt(2)))/(2+sqrt(2))
=16r_1 (sqrt((2-sqrt(2))(2+sqrt(2))))/(2+sqrt(2))
=16r_1 sqrt(2)/(2+sqrt(2))
=16r_1 (sqrt(2)(2-sqrt(2)))/((2+sqrt(2))(2-sqrt(2)))
=8r_1(2sqrt(2)-2)
=color(green)(16r_1(sqrt(2)-1))
So if the inner radius is 20, then the perimeter is:
16*20(sqrt(2) - 1) = 320 (sqrt(2) - 1) ~~ 132.55
color(white)()
How good an approximation for pi does this give us?
While we're here, what approximation for pi do we get by averaging the inner and outer radii?
pi ~~ 2(2(sqrt(2) - 1) + sqrt(2-sqrt(2))) ~~ 3.1876
...so not great.
To get as good an approximation as 355/113 ~~ 3.1415929, the Chinese mathematician Zu Chongzhi used a 24576 (= 2^13 xx 3) sided polygon and counting rods.
https://en.wikipedia.org/wiki/Zu_Chongzhi
