On Tue, 30 Jul 2013, firstname.lastname@example.org wrote:
> If you create a hexagon with the radius of one, and recursively double the > number of vertices of the hexagon(have same radius as side length) > Maybe this is a valid way to calculate Pi? Given a regular n sided polygon, with distance from center to the vertices of r, the length of one side is 2r.sin pi/n and the total length of all the sides is 2rn.sin pi/n =
Now lim(n->oo) 2r.pi.(sin pi/n)/(pi/n) = lim(n->oo) 2r.pi.(sin pi/n)/(pi/n) = 2pi.r
For n = 6, sin pi/6 = 1/2, so 2*6r(1/2) = 6r is a crude first approximation for the circumference of a circle of radius r, giving approximately 3 for pi.