Draw a polygon. the sum of its angles is pi*(n-2). The sum of pieces of circles is R*(n*pi-(n-2)*pi)=2*R*pi (you can see it easily if u draw)

Thanks, so nice.

yep, its simple

Beautiful. Solved it by calculating angles between vectors of consecutive nails.

How is the sum of its angles pi*(n-2)?

OP means the sum of angles in a n-polygon is pi*(n-2). This can be attained inductively:
1. The sum of angles in a triangle is pi.
2. for any n-polygon, it can be divided into a (n-1)-polygon and a triangle.

thanks !