equilateral triangles around S 0 {\displaystyle S_{0}} we have: | A S 0 | = | B S 0 | = | C S 0 | = | D S 0 | = | E S 0 | = | F S 0 | = a 0 {\displaystyle...
1 = ( 2 − 2 ) ⋅ s {\displaystyle r_{1}=(2-{\sqrt {2}})\cdot s\quad } See QH_dia 2) Radius of the additional circle r 2 = 2 − 2 2 ⋅ s {\displaystyle r_{2}={\frac...
{\displaystyle P_{1}={\frac {4+\pi }{2}}\cdot a\quad } , see details under FS QV dia.png 2) Perimeter of the circle: P 2 = 2 π r 2 = 2 π ⋅ ( 2 − 1 ) ⋅ a {\displaystyle...
r_{0}+{\sqrt {|S_{2}S_{4}|^{2}-|FS_{2}|^{2}}}\quad } , applying the Pythagorean theoreme on △ F S 4 S 2 {\displaystyle \triangle FS_{4}S_{2}} ⇔ S 4 = ( 0 +...
r_{0}+{\sqrt {|S_{2}S_{4}|^{2}-|FS_{2}|^{2}}}\quad } , applying the Pythagorean theoreme on △ F S 4 S 2 {\displaystyle \triangle FS_{4}S_{2}} ⇔ S 4 = ( 0 +...
circle: S 2 : {\displaystyle S_{2}:} x 2 = | S 0 S 2 | c o s ( 45 ) {\displaystyle \quad \quad x_{2}=|S_{0}S_{2}|cos(45)} ⇔ x 2 = ( | A S 2 | − | A S 0 |...
triangle: S 2 = S 0 + S 0 A → + A F → + F S 2 → {\displaystyle S_{2}=S_{0}+{\overrightarrow {S_{0}A}}+{\overrightarrow {AF}}+{\overrightarrow {FS_{2}}}} ⇔ S 2...
{\displaystyle r_{1}} around point A {\displaystyle A} . From the developing of FS_CV we know that r 1 = 2 ⋅ r 0 {\displaystyle r_{1}={\sqrt {2}}\cdot r_{0}}...
{S_{0}S_{1}}}} ⇔ S 1 = S 0 − | K S 0 | + | K S 1 | {\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-|KS_{0}|+|KS_{1}|} ⇔ S 1 = S 0 − | K S 0 | +...
base circle: S 0 = S = 0 + 0 i {\displaystyle S_{0}=S=0+0i} 1) Centroid positions of the square: S 1 = S = 0 + 0 i {\displaystyle S_{1}=S=0+0i} In the...
_dia.png){kind=link}
