Quote:
Originally Posted by Dr Sardonicus

Still in high school's math camp we constructed a regular 17gon.
Quote:
Originally Posted by a1call
Similarly for 3*17=51 gon
1/33/17= 8/51 of a circle. Then you can bisect 3 times to get 1/51 of a circle.

There is an elementary way to show that if you can make a regular m and ngon [using straightedge and compass] and gcd(m,n)=1 then you can make a regular m*ngon. Because making a regular kgon is equivalent with constructing a 2*Pi/k angle.
So we can make a 2*Pi/n and 2*Pi/m angle.
We assumed that gcd(m,n)=1 so with extended Euclidean algorithm there exists x and y integers:
n*x+m*y=1 divide this equation by m*n
x/m+y/n=1/(m*n) multiplie by 2*Pi
x*(2*Pi/m)+y*(2*Pi/n)=2*Pi/(m*n), what we needed.