User:Hans G. Oberlack/Sandkiste

Base is the square of given side length ${\displaystyle a_{0}}$ with centroid at ${\displaystyle S_{0}}$
Inscribed are the largest possible circle and right triangle of the same area.

0) The side length of the base square: ${\displaystyle a_{0}}$
1) The radius of the inscribed circle: ${\displaystyle r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.34\cdot a_{0}\quad }$, see Calculation 1
2) The side length of the inscribed right triangle: ${\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}\quad }$, see Calculation 1

0) Perimeter of base square ${\displaystyle P_{0}=4\cdot a_{0}\quad }$
1) Perimeter of the inscribed circle: ${\displaystyle P_{1}=2\pi \cdot r_{1}={\frac {2\pi {\sqrt {2}}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 2.122\cdot a_{0}\quad }$
2) Perimeter of the inscribed triangle: ${\displaystyle P_{2}=4\cdot a_{2}={\frac {4\cdot 2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 3.387\cdot a_{0}\quad }$

0) Area of the base square ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed circle: ${\displaystyle A_{1}=\pi r_{1}^{2}=\pi \left({\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }$
2) Area of the inscribed triangle: ${\displaystyle A_{2}={\frac {a_{2}^{2}}{2}}={\frac {1}{2}}\left({\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i=0}$
1) Centroid position of the inscribed circle: ${\displaystyle S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.162+0.162i)\cdot a_{0}\quad }$, see Calculation 2
2) Centroid position of the inscribed square: ${\displaystyle S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0.218-0.218i)\quad }$, see Calculation 3

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. The graphical representation does not correspond to the mathematical expression.
0) Orientated centroid position of the base square: ${\displaystyle SO_{0}=0+0i=0}$
1) Orientated centroid position of the inscribed circle: ${\displaystyle SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0-0.229i)\quad }$, see calculation 4
2) Orientated centroid position of the inscribed right triangle: ${\displaystyle SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0+0.308i)}$, see calculation 5

(0) The area of the right triangle ${\displaystyle \triangle AKH}$ has to be the same as the area of the circle with radius ${\displaystyle r_{1}}$ around center point ${\displaystyle S_{1}\quad \Rightarrow A_{1}=A_{2}}$
(1) ${\displaystyle |AB|=|BC|=|CD|=|AD|=a_{0}\quad }$since ${\displaystyle \square ABCD}$ is a square
(2) ${\displaystyle |AC|={\sqrt {2}}\cdot a_{0}\quad }$since ${\displaystyle {\overline {AC}}}$ is the diagonal of square ${\displaystyle \square ABCD}$
(3) ${\displaystyle |AK|=|AH|=a_{2}\quad }$since ${\displaystyle \triangle AKH}$ is a right triangle
(4) ${\displaystyle |S_{1}F|=|FC|=|CG|=|GS_{1}|=r_{1}\quad }$since ${\displaystyle \square S_{1}FCG}$ is a square, because ${\displaystyle F}$ and ${\displaystyle G}$ are tangent points of the circle around ${\displaystyle S_{1}}$ with the sides of square ${\displaystyle \square ABCD}$

The radius ${\displaystyle r_{1}}$ is calculated:

a) First the relation between ${\displaystyle r_{1}}$ and ${\displaystyle a_{2}}$ is determined from equation (0):
${\displaystyle A_{1}=A_{2}\quad }$ applying equation (0)
${\displaystyle \quad \Leftrightarrow {\frac {a_{2}^{2}}{2}}=\pi r_{1}^{2}\quad }$, definitions of areas of right triangles and circles
${\displaystyle \quad \Leftrightarrow a_{2}^{2}=2\pi r_{1}^{2}\quad }$, multiplying both sides by two
${\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad }$, since negative length cannot be applied

b) Then the relation between ${\displaystyle r_{1}}$ and ${\displaystyle a_{0}}$ is determined from the following identity:
${\displaystyle |AC|={\sqrt {2}}\cdot a_{0}\quad }$ since ${\displaystyle \square ABCD}$ is a square
${\displaystyle \quad \Leftrightarrow |AE|+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, because the three segments are forming the diagonal of the square
${\displaystyle \quad \Leftrightarrow {\frac {\sqrt {2}}{2}}\cdot a_{2}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, since ${\displaystyle \triangle AKH}$ is a right isosceles trinagle
${\displaystyle \quad \Leftrightarrow {\frac {\sqrt {2}}{2}}({\sqrt {2\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, applying part a) of the calculation
${\displaystyle \quad \Leftrightarrow {\frac {{\sqrt {2}}{\sqrt {2}}}{2}}({\sqrt {\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, because E is a tangent point on the circle and a diagonal side of the triangle ${\displaystyle \triangle AKH}$
${\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+{\sqrt {2}}\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, because ${\displaystyle \square S_{1}FCG}$ is a square with side length ${\displaystyle r_{1}}$
${\displaystyle \quad \Leftrightarrow r_{1}\cdot ({\sqrt {\pi }}+1+{\sqrt {2}})={\sqrt {2}}\cdot a_{0}\quad }$, extracting ${\displaystyle r_{1}}$ out of the bracket
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.338\cdot a_{0}}$, rearranging

c) Eventually the relation between ${\displaystyle a_{2}}$ and ${\displaystyle a_{0}}$ is determined by using part a) of the calculation:
${\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad }$, from part a) of the calculation
${\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot {\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}}$, using the result from part b) of the calculation
${\displaystyle \quad \Leftrightarrow a_{2}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}}$, rearranging

Calculating the centroid of the inscribed circle as displayed:
${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}S_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+{\overrightarrow {S_{0}C}}+{\overrightarrow {CS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=\left(|S_{0}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }$, since ${\displaystyle {\overrightarrow {S_{0}C}}}$ is collinear to the diagonal of the square
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }$, since ${\displaystyle {\overrightarrow {S_{0}C}}}$ is half the diagonal of the square
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(|S_{1}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left({\sqrt {2}}\cdot r_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, since ${\displaystyle {\overrightarrow {S_{1}C}}}$ is the diagonal of the square ${\displaystyle \square {S_{1}FCG}}$ with side length ${\displaystyle r_{1}}$
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(r_{1}\cdot (1+i)\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-r_{1}\right)\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad }$, applying Calculation 1b
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {1}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {\pi }}+{\sqrt {2}}+1}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}-{\frac {2{\sqrt {2}}}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}\right)\cdot (1+i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {\pi }}-{\sqrt {2}}+1}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}\right)\cdot (1+i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}-{\sqrt {2}}+1}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.162+0.162i)\cdot a_{0}}$, rearranging

Calculating the centroid of the right triangle as displayed:
${\displaystyle S_{2}=S_{0}+{\overrightarrow {S_{0}S_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}E}}+{\overrightarrow {ES_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}A}}+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-|S_{0}A|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad }$, since ${\displaystyle {\overrightarrow {S_{0}A}}}$ is collinear to the diagonal of the square ${\displaystyle \square {ABCD}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad }$, since ${\displaystyle {\overrightarrow {S_{0}A}}}$ is half the diagonal of the square ${\displaystyle \square {ABCD}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left(|AE|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad }$, since ${\displaystyle {\overrightarrow {AE}}}$ is the height of the triangle ${\displaystyle \triangle {AKH}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{\sqrt {2}}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad }$, since ${\displaystyle {\frac {a_{2}}{\sqrt {2}}}}$ is the value of the height of the triangle ${\displaystyle \triangle {AKH}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-|ES_{2}|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, since ${\displaystyle {\overrightarrow {ES_{2}}}}$ is collinear to the diagonal of the square ${\displaystyle \square {ABCD}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}|AE|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, since the centroid of a triangle is a third of the height
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}\cdot {\frac {a_{2}}{\sqrt {2}}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, since the height of a right isoceles triangle with side length a is: ${\displaystyle h={\frac {a}{\sqrt {2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}\cdot {\frac {a_{2}}{2}}\cdot (1+i)\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+a_{2}-{\frac {1}{3}}\cdot a_{2}\right)\cdot (1+i)\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+{\frac {2}{3}}\cdot a_{2}\right)\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+{\frac {2}{3}}\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad }$, applying result of calculation 1c
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-1+{\frac {2}{3}}\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left(-3+2\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}-3\right)\cdot (1+i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}-3({\sqrt {\pi }}+{\sqrt {2}}+1)}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }$, having a common denominator
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}-3{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }$, multiplying into the bracket
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0.218-0.218i)}$, adding

The centroid of the inscribed circle is calculated in orientated expression:
This will be done by rotating the expression as displayed by ${\displaystyle 225^{\circ }}$ which will be done by a ${\displaystyle 45^{\circ }}$ rotation followed by a ${\displaystyle 180^{\circ }}$ rotation
${\displaystyle SO_{1}=S_{1}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (i^{2})}$
${\displaystyle \quad \Leftrightarrow SO_{1}=S_{1}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (-1)\quad }$, since ${\displaystyle i^{2}=-1}$
${\displaystyle \quad \Leftrightarrow SO_{1}=S_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(1+i\right)\cdot (-1)\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow SO_{1}=S_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }$, applying Calculation 2
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot \left(-1-i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (-1-i-i-i^{2})\cdot a_{0}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (-1-i-i+1)\cdot a_{0}\quad }$, since ${\displaystyle i^{2}=-1}$
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-2i)\cdot a_{0}\quad }$, adding up
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {2}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0-0.229i)\quad }$

The centroid of the inscribed triangle is calculated in orientated expression:
This will be done by rotating the expression as displayed by ${\displaystyle 225^{\circ }}$ which will be done by a ${\displaystyle 45^{\circ }}$ rotation followed by a ${\displaystyle 180^{\circ }}$ rotation
${\displaystyle SO_{2}=S_{2}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (i^{2})}$
${\displaystyle \quad \Leftrightarrow SO_{2}=S_{2}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (-1)\quad }$, since ${\displaystyle i^{2}=-1}$
${\displaystyle \quad \Leftrightarrow SO_{2}=S_{2}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(1+i\right)\cdot (-1)\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow SO_{2}=S_{2}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }$, applying Calculation 3
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot \left(-1-i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (-1-i-i-i^{2})\cdot a_{0}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (-1-2i-(-1))\cdot a_{0}\quad }$, since ${\displaystyle i^{2}=-1}$
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-2i)\cdot a_{0}\quad }$, adding
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {2}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0+0.308i)}$, reducing