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

0) The side length of the equilateral base triangle is: ${\displaystyle a_{0}}$
1) The radius of the inscribed quarter circle is: ${\displaystyle r_{1}={\frac {a_{0}}{\sqrt {3}}}\quad \approx 0.577\cdot a_{0}}$, see calculation 4

0) Perimeter of equilateral base triangle: ${\displaystyle P_{0}=3\cdot a_{0}}$
1) Perimeter of inscribed quarter circle: ${\displaystyle P_{1}=2\cdot r_{1}+{\frac {\pi }{2}}\cdot r_{1}={\frac {4+\pi }{2{\sqrt {3}}}}\cdot a_{0}\quad \approx 2.062\cdot a_{0}}$

0) Area of the equilateral base triangle: ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }$
1) Area of the inscribed quarter circle: ${\displaystyle A_{1}={\frac {\pi \cdot r_{1}^{2}}{4}}={\frac {\pi }{4}}\cdot \left({\frac {a_{0}}{\sqrt {3}}}\right)^{2}={\frac {\pi }{4}}\cdot {\frac {a_{0}}{3}}={\frac {\pi }{12}}\cdot a_{0}^{2}\quad \approx 0.262\cdot a_{0}^{2}}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed quarter circle relative to the centroid of the base shape is: ${\displaystyle S_{1}=\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}+{\frac {8{\sqrt {3}}-3\pi {\sqrt {3}}}{18\pi }}\cdot i\right)\cdot a_{0}\quad \approx (0.078-0.044i)\cdot a_{0}}$, see calculation (5)

In the normalised case the area of the base shape is set to 1.
So ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}={\frac {4}{\sqrt {3}}}\quad \Rightarrow a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}$

0) Side length of the triangle ${\displaystyle a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}$
1) The radius of the inscribed quarter circle is: ${\displaystyle r_{1}={\frac {a_{0}}{\sqrt {3}}}={\frac {2}{{\sqrt {3}}\cdot {\sqrt {\sqrt {3}}}}}\quad \approx 0.877\quad }$,

0) Perimeter of base triangle: ${\displaystyle P_{0}=3\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {12{\sqrt {3}}}}\quad \approx 4.5590141}$
1) Perimeter of inscribed quarter circle: ${\displaystyle P_{1}={\frac {4+\pi }{2{\sqrt {3}}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\frac {4+\pi }{{\sqrt {3}}{\sqrt {\sqrt {3}}}}}\quad \approx 3.1329548}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 7.6919689}$

0) Area of the base triangle is by definition ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}=1}$
1) Area of the inscribed quarter circle: ${\displaystyle A_{1}={\frac {\pi }{12}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}={\frac {\pi }{12}}\cdot {\frac {4}{\sqrt {3}}}={\frac {\pi }{3{\sqrt {3}}}}\quad \approx 0.605}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed quarter circle relative to the centroid of the base shape is: ${\displaystyle S_{1}=\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}+{\frac {8{\sqrt {3}}-3\pi {\sqrt {3}}}{18\pi }}\cdot i\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 0.119-0.066i}$

The distance between the centroid of the base triangle and the centroid of the semicircle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.1363143}$
Sum of distances: ${\displaystyle d_{s}=d_{01}=0.1363143}$

Apart of the base element there is one other 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.

${\displaystyle decimalpart(7.6919689+0.1363143)=decimalpart(7.8282832)=.8282832}$

So the identifying number is: ${\displaystyle 1.8282832}$

(0) Given is the side length of the equilateral triangle: ${\displaystyle a_{0}}$
(1)${\displaystyle \quad |AB|=|BC|=|AC|=a_{0}}$
(2)${\displaystyle \quad |AG|=|BG|={\frac {1}{2}}\cdot a_{0}}$
(3)${\displaystyle \quad |DF|=|DE|=|DH|=r_{1}}$
(4)${\displaystyle \quad |AD|+|DH|+|HB|=|AB|=a_{0}}$
(5)${\displaystyle \quad \angle \alpha =\angle \beta =\angle \gamma ={\frac {\pi }{3}}=60^{\circ }}$, because ${\displaystyle \triangle ABC}$ is equilateral
(6)${\displaystyle \quad \angle \epsilon ={\frac {\pi }{2}}=90^{\circ }}$, because the quarter circle is tangent to ${\displaystyle {\overrightarrow {BC}}}$ so that ${\displaystyle {\overrightarrow {DE}}}$ is perpendicular.

The angle ${\displaystyle \delta }$ is calculated:
${\displaystyle \delta +\beta +\epsilon =180^{\circ }\quad }$,applying the sum of angles in the triangle ${\displaystyle \triangle DBE}$
${\displaystyle \quad \Leftrightarrow \delta =180^{\circ }-\beta -\epsilon \quad }$, rearranging
${\displaystyle \quad \Leftrightarrow \delta =180^{\circ }-\beta -90^{\circ }\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow \delta =180^{\circ }-60^{\circ }-90^{\circ }\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow \delta =30^{\circ }\quad }$

The length of ${\displaystyle {\overrightarrow {AD}}}$ in relation to ${\displaystyle r_{1}}$ is calculated:
${\displaystyle \quad {\frac {|AD|}{|DF|}}=\tan {\alpha }\quad }$, since ${\displaystyle \triangle {ADF}}$ is a right triangle
${\displaystyle \Leftrightarrow \quad {\frac {|DF|}{|AD|}}=\tan {60^{\circ }}\quad }$, applying equation (5)
${\displaystyle \Leftrightarrow \quad {\frac {|DF|}{|AD|}}={\sqrt {3}}\quad }$, applying the tan-function
${\displaystyle \Leftrightarrow \quad |DF|={\sqrt {3}}\cdot |AD|\quad }$, applying the tan-function
${\displaystyle \Leftrightarrow \quad {\frac {|DF|}{\sqrt {3}}}=|AD|\quad }$, rearranging
${\displaystyle \Leftrightarrow \quad {\frac {r_{1}}{\sqrt {3}}}=|AD|\quad }$, applying equation(3)

The length of ${\displaystyle {\overrightarrow {DB}}}$ in relation to ${\displaystyle r_{1}}$ is calculated:
${\displaystyle \quad {\frac {|DE|}{|DB|}}=\cos {\delta }\quad }$, since ${\displaystyle \triangle {DBE}}$ is a right triangle
${\displaystyle \quad \Leftrightarrow {\frac {|DE|}{|DB|}}=\cos {30^{\circ }}\quad }$, applying calculation 1
${\displaystyle \quad \Leftrightarrow {\frac {|DE|}{|DB|}}={\frac {\sqrt {3}}{2}}\quad }$, applying the cos-function
${\displaystyle \quad \Leftrightarrow {\frac {r_{1}}{|DB|}}={\frac {\sqrt {3}}{2}}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow {\frac {2\cdot r_{1}}{\sqrt {3}}}=|DB|\quad }$, rearranging

The length of ${\displaystyle r_{1}}$ in relation to ${\displaystyle a_{0}}$ is calculated:
${\displaystyle \quad |AD|+|DB|=|AB|\quad }$, since ${\displaystyle {\overrightarrow {AB}}={\overrightarrow {AD}}+{\overrightarrow {DB}}}$
${\displaystyle \quad \Leftrightarrow |AD|+|DB|=a_{0}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow |AD|+{\frac {2\cdot r_{1}}{\sqrt {3}}}=a_{0}\quad }$, applying calculation 3
${\displaystyle \quad \Leftrightarrow {\frac {r_{1}}{\sqrt {3}}}+{\frac {2\cdot r_{1}}{\sqrt {3}}}=a_{0}\quad }$, applying calculation 2
${\displaystyle \quad \Leftrightarrow {\frac {3\cdot r_{1}}{\sqrt {3}}}=a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\sqrt {3}}\cdot r_{1}=a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {a_{0}}{\sqrt {3}}}\quad }$, rearranging

The position of the centroid ${\displaystyle S_{1}}$ of the inscribed quarter circle in relation to the centroid ${\displaystyle S_{0}}$ of the equilateral triangle is calculated:
${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}G}}+{\overrightarrow {GD}}+{\overrightarrow {DS_{1}}}\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+\mathbf {|S_{0}G|\cdot (0-i)} +{\overrightarrow {GD}}+{\overrightarrow {DS_{1}}}\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+|S_{0}G|\cdot (0-i)+\mathbf {|GD|\cdot (-1+0i)} +{\overrightarrow {DS_{1}}}\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+|S_{0}G|\cdot (0-i)+|GD|\cdot (-1+0i)+\mathbf {{\frac {4\cdot r_{1}}{3\pi }}\cdot (1+i)} \quad }$, applying the centroid formula for quarter circles with ${\displaystyle \angle \delta =45^{\circ }}$
${\displaystyle \quad \Leftrightarrow S_{1}=0-|S_{0}G|\cdot i-|GD|+{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, simplifying the terms
${\displaystyle \quad \Leftrightarrow S_{1}=0-\mathbf {{\frac {1}{3}}\cdot |GC|\cdot i} -|GD|+{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, applying formula for centroids of triangles
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{3}}\cdot \mathbf {{\frac {\sqrt {3}}{2}}\cdot a_{0}} \cdot i-|GD|+{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, applying formula for height of triangles
${\displaystyle \quad \Leftrightarrow S_{1}=0-\mathbf {{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i} -|GD|+{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, simplifying the term
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-\mathbf {\left(|AG|-|AD|\right)} +{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, reducing to known elements
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-\left(\mathbf {\frac {a_{0}}{2}} -|AD|\right)+{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-\mathbf {{\frac {a_{0}}{2}}+|AD|} +{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, simplifying term
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-{\frac {a_{0}}{2}}+\mathbf {\frac {r_{1}}{\sqrt {3}}} +{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, applying calculation (2)
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-{\frac {a_{0}}{2}}+\mathbf {\frac {\frac {a_{0}}{\sqrt {3}}}{\sqrt {3}}} +{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, applying calculation (4)
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-{\frac {a_{0}}{2}}+\mathbf {\frac {a_{0}}{3}} +{\frac {4\cdot r_{1}}{3\pi }}+{\frac {4\cdot r_{1}}{3\pi }}\cdot i\quad }$, simplifying term
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i-{\frac {a_{0}}{2}}+{\frac {a_{0}}{3}}+\mathbf {{\frac {4\cdot a_{0}}{3\pi {\sqrt {3}}}}+{\frac {4\cdot a_{0}}{3\pi {\sqrt {3}}}}\cdot i} \quad }$, applying calculation (4)
${\displaystyle \quad \Leftrightarrow S_{1}={\frac {4\cdot a_{0}}{3\pi {\sqrt {3}}}}-{\frac {a_{0}}{2}}+{\frac {a_{0}}{3}}+{\frac {4\cdot a_{0}}{3\pi {\sqrt {3}}}}\cdot i-{\frac {a_{0}}{2{\sqrt {3}}}}\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2}}+{\frac {1}{3}}+{\frac {4}{3\pi {\sqrt {3}}}}\cdot i-{\frac {1}{2{\sqrt {3}}}}\cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{6}}\right)+\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2{\sqrt {3}}}}\right)\cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\mathbf {\left({\frac {24}{18\pi {\sqrt {3}}}}-{\frac {3\pi {\sqrt {3}}}{18\pi {\sqrt {3}}}}\right)} +\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2{\sqrt {3}}}}\right)\cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\mathbf {\left({\frac {24-3\pi {\sqrt {3}}}{18\pi {\sqrt {3}}}}\right)} +\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2{\sqrt {3}}}}\right)\cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\mathbf {\left({\frac {8-\pi {\sqrt {3}}}{6\pi {\sqrt {3}}}}\right)} +\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2{\sqrt {3}}}}\right)\cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\mathbf {\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}\right)} +\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2{\sqrt {3}}}}\right)\cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}\right)+\mathbf {\left({\frac {4}{3\pi {\sqrt {3}}}}-{\frac {1}{2{\sqrt {3}}}}\right)} \cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}\right)+\mathbf {\left({\frac {8}{6\pi {\sqrt {3}}}}-{\frac {3\pi }{6\pi {\sqrt {3}}}}\right)} \cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}\right)+\mathbf {\left({\frac {8-3\pi }{6\pi {\sqrt {3}}}}\right)} \cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}\right)+\mathbf {\left({\frac {8{\sqrt {3}}-3\pi {\sqrt {3}}}{18\pi }}\right)} \cdot i\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {8{\sqrt {3}}-3\pi }{18\pi }}+{\frac {8{\sqrt {3}}-3\pi {\sqrt {3}}}{18\pi }}\cdot i\right)\cdot a_{0}\quad \approx (0.078-0.044i)\cdot a_{0}}$, rearranging

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