如图，圆的半径为r; 内接正方形的边长为 $\ell$,由圆心到正方形一边倒垂直距离为 d

${\displaystyle d=\sqrt{r^2-(\frac{\ell }{2}{)}^2}}$

${\displaystyle e=r-d=r-\sqrt{r^2-(\frac{\ell }{2}{)}^2}}$

d 的延长线与圆周相交点将圆周等分为正八边形。

令正八边形的边长为${\displaystyle {\ell }_2}$

${\displaystyle {\ell }_2=\sqrt{(\ell /2{)}^2+e^2}}$

${\displaystyle {\ell }_2=\frac{1}{2}\ast \sqrt{{\ell }^2+4\ast (r-\frac{1}{2}\ast \sqrt{4\ast r^2-{\ell }^2}{)}^2}}$

设${\displaystyle {\ell }_3}$ 为分割圆成正16边形之边长，赵友钦正确地推断${\displaystyle {\ell }_3}$与${\displaystyle {\ell }_2}$的迭代关系：

${\displaystyle {\ell }_3=\frac{1}{2}\ast \sqrt{({\ell }_2{)}^2+4\ast (r-\frac{1}{2}\ast \sqrt{4\ast r^2-({\ell }_2{)}^2}{)}^2}}$

推而广之：

${\displaystyle {\ell }_{n+1}=\frac{1}{2}\ast \sqrt{({\ell }_n{)}^2+4\ast (r-\frac{1}{2}\ast \sqrt{4\ast r^2-({\ell }_n{)}^2}{)}^2}}$

令 r=1;

${\displaystyle {\ell }_1=\sqrt{(}2)}$

${\displaystyle {\ell }_2=\sqrt{2-\sqrt{(}2)}}$

${\displaystyle {\ell }_3=\sqrt{2-\sqrt{2+\sqrt{(}2)}}}$

${\displaystyle {\ell }_4=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{(}2)}}}}$

${\displaystyle {\ell }_5=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{(}2)}}}}}$

……