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The quarter circle as base element. Inscribed is the largest right triangle.

0) The radius of the quarter circle ${\displaystyle r_{0}}$
1) Side length of the inscribed triangle: ${\displaystyle a_{1}=r_{0}\quad }$, since ${\displaystyle |AB|=r_{0}=a_{1}}$

0) Perimeter of base quarter circle: ${\displaystyle P_{0}=2\cdot r_{0}+{\frac {\pi \cdot r_{0}}{2}}=(2+{\frac {\pi }{2}})\cdot r_{0}\quad \approx 3.571\cdot r_{0}}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}=|AB|+|BC|+|AC|=r_{0}+{\sqrt {2}}\cdot r_{0}+r_{0}=(2+{\sqrt {2}})\cdot r_{0}\quad \approx 3.414\cdot r_{0}}$

S) Sum of perimeters: ${\displaystyle P_{s}=(2+{\frac {\pi }{2}})\cdot r_{0}+(2+{\sqrt {2}})\cdot r_{0}=(4+{\frac {\pi }{2}}+{\sqrt {2}})\cdot r_{0}\quad \approx 6.985\cdot r_{0}}$

0) Area of the base quarter circle ${\displaystyle A_{0}={\frac {\pi }{4}}\cdot r_{0}^{2}\quad \approx 0.785\cdot r_{0}^{2}}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {a_{1}^{2}}{2}}={\frac {1}{2}}\cdot r_{0}^{2}\quad =0.500\cdot r_{0}^{2}\qquad \Rightarrow A_{1}={\frac {2}{\pi }}\cdot A_{0}\approx 0.637\cdot A_{0}}$

0) Centroid position of the quarter circle: ${\displaystyle x_{0}=0\quad y_{0}=0\rightarrow S_{0}=0+0i}$
1) Centroid position of the inscribed triangle: ${\displaystyle S_{1}=\left({\frac {\pi -4}{3\pi }}\right)\cdot (1+i)\cdot r_{0}\quad \approx -0.091\cdot (1+i)\cdot r_{0}\quad }$, see calculation 1

In the normalised case the area of the base quarter circle is set to 1.
So ${\displaystyle A={\frac {\pi \cdot r_{0}^{2}}{4}}=1\quad \Rightarrow r_{0}^{2}={\frac {4}{\pi }}\quad \Rightarrow r_{0}={\sqrt {\frac {4}{\pi }}}={\frac {2}{\sqrt {\pi }}}}$

0) Radius of the base quarter circle ${\displaystyle r_{0}={\frac {2}{\sqrt {\pi }}}\quad \approx 1.128}$
1) Side length of the inscribed triangle: ${\displaystyle a_{1}={\frac {2}{\sqrt {\pi }}}\quad \approx 1.128}$

0) Perimeter of base quarter circle${\displaystyle P_{0}=(2+{\frac {\pi }{2}})\cdot r_{0}=(2+{\frac {\pi }{2}})\cdot {\frac {2}{\sqrt {\pi }}}\quad \approx 4.0292122}$
1) Perimeter of the inscribed triangle${\displaystyle P_{1}=(2+{\sqrt {2}})\cdot {\frac {2}{\sqrt {\pi }}}=\quad \approx 3.8525275}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}=(4+{\frac {\pi }{2}}+{\sqrt {2}})\cdot {\frac {2}{\sqrt {\pi }}}\quad \approx 7.8817396}$

0) Area of the base quarter circle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {1}{2}}\left({\frac {2}{\sqrt {\pi }}}\right)^{2}={\frac {1}{2}}\left({\frac {4}{\pi }}\right)={\frac {2}{\pi }}\quad \approx 0.637}$

0) Centroid position of the quarter circle: ${\displaystyle x_{0}=0\quad y_{0}=0}$

1) Centroid position of the inscribed triangle: ${\displaystyle S_{1}=\left({\frac {\pi -4}{3\pi }}\right)\cdot (1+i)\cdot {\frac {2}{\sqrt {\pi }}}=\left({\frac {2\pi -8}{3\pi {\sqrt {\pi }}}}\right)\cdot (1+i)\quad \approx -0.103\cdot (1+i)}$

The distance between the centroid of the base element and the centroid of the triangle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}\quad \approx 0.1453424}$

Apart of the base element there is one shape allocated. Therefore the integer part of the identifying number is 1.
The decimal part of the identifying number is the decimal part of the sum of the perimeters and the distances of the centroids in the normalised case.

${\displaystyle decimalpart(7.8817396+0.1453424)=decimalpart(8.0270821)=.0270821}$

So the identifying number is: ${\displaystyle 1.0270821}$

Position of the point A: ${\displaystyle x_{A}=-{\frac {4\cdot r_{0}}{3\cdot \pi }}\quad y_{A}=-{\frac {4\cdot r_{0}}{3\cdot \pi }}}$

${\displaystyle \quad S_{1}=|AS_{1}|-|AS_{0}|}$
${\displaystyle \quad \Leftrightarrow x_{1}+y_{1}i=-{\frac {4\cdot r_{0}}{3\cdot \pi }}\cdot (1+i)+{\frac {r_{0}}{3}}\cdot (1+i)}$
${\displaystyle \quad \Leftrightarrow x_{1}+y_{1}i=\left(-{\frac {4}{3\cdot \pi }}+{\frac {1}{3}}\right)\cdot (1+i)\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}+y_{1}i=\left(-{\frac {4}{3\pi }}+{\frac {\pi }{3\pi }}\right)\cdot (1+i)\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}+y_{1}i=\left({\frac {\pi -4}{3\pi }}\right)\cdot (1+i)\cdot r_{0}}$

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