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The equilateral triangle as base element.
Inscribed is the smallest equilateral triangle.

0) The side length of the equilateral base triangle is: ${\displaystyle a_{0}}$
1) The radius of the inscribed equilateral triangle is: ${\displaystyle a_{1}={\frac {1}{2}}\cdot a_{0}=0.500\cdot a_{0}\quad }$

0) Perimeter of equilateral base triangle: ${\displaystyle P_{0}=3\cdot a_{0}}$
1) Perimeter of inscribed equilateral triangle: ${\displaystyle P_{1}=3\cdot a_{1}={\frac {3}{2}}\cdot a_{0}=1.500\cdot a_{0}}$

0) Area of the equilateral base triangle: ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }$, see calculation (1)
1) Area of the inscribed equilateral triangle: ${\displaystyle A_{1}={\frac {\sqrt {3}}{4}}\cdot a_{1}^{2}={\frac {\sqrt {3}}{4}}\cdot \left({\frac {1}{2}}\cdot a_{0}\right)^{2}={\frac {\sqrt {3}}{4}}\cdot {\frac {1}{4}}\cdot a_{0}^{2}={\frac {\sqrt {3}}{16}}\cdot a_{0}^{2}\quad \approx 0.108\cdot a_{0}^{2}={\frac {1}{4}}A_{0}}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed circle relative to the centroid of the base shape is: ${\displaystyle S_{1}=S_{0}\quad }$see calculation (2)

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 side length of the inscribed triangle is: ${\displaystyle a_{1}={\frac {1}{2}}\cdot a_{0}={\frac {1}{2}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {\frac {1}{\sqrt {3}}}}\quad \approx 0.76\quad }$

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 triangle: ${\displaystyle P_{1}={\frac {1}{2}}\cdot P_{0}={\sqrt {3{\sqrt {3}}}}\quad \approx 2,2795071}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 6.8385212}$

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 triangle: ${\displaystyle A_{1}={\frac {1}{4}}\cdot A_{0}=0.250}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed triangle relative to the centroid of the base shape is: ${\displaystyle S_{1}=0+0i}$

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

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.

${\displaystyle decimalpart(6.8385212+0)=decimalpart(6.8385212)=.8385212}$

So the identifying number is: ${\displaystyle 1.8385212}$

(0) Given is the side length of the equilateral triangle: ${\displaystyle a_{0}}$
(1)${\displaystyle \quad |AB|=|BC|=|AC|=a_{0}}$
(2)${\displaystyle \quad |AD|=|BD|=|BE|=|CE|={\frac {1}{2}}\cdot a_{0}}$

The height ${\displaystyle |CD|}$ is calculated:
${\displaystyle \quad |AD|^{2}+|CD|^{2}=|AC|^{2}\quad }$,applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle ABC}$
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=|AC|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=a_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{0}^{2}-{\frac {1}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |CD|^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |CD|={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, rearranging

The area of triangle is height multiplied by the half of the side length: ${\displaystyle \quad A_{0}=|AD|\cdot |CD|}$
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot |CD|\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }$

${\displaystyle \quad S_{1}=S_{0}+|S_{0}D|+|DS_{1}|}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+|S_{0}D|+|DS_{1}|}$, since ${\displaystyle S_{0}}$ is in the centre of the figure
${\displaystyle \quad \Leftrightarrow S_{1}=|S_{0}D|+|DS_{1}|}$, since ${\displaystyle (0+0i)=0}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {1}{3}}\cdot |DC|)+|DS_{1}|}$, definition of centroid in an equilateral triangle
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {1}{3}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0})+|DS_{1}|}$, see calculation (1) for height ${\displaystyle |CD|}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+|DS_{1}|}$, multiplying
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+(|DG|-|GS_{1}|)}$, since ${\displaystyle {\overrightarrow {DS_{1}}}={\overrightarrow {DG}}+{\overleftarrow {S_{1}G}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+(0+{\frac {\sqrt {3}}{2}}\cdot a_{1})-(|GS_{1}|)}$, since ${\displaystyle |DG|}$ is the height of the inscribed triangle
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+({\frac {\sqrt {3}}{2}}\cdot {\frac {1}{2}}\cdot a_{0})-(|GS_{1}|)}$, since ${\displaystyle a_{1}={\frac {1}{2}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+({\frac {\sqrt {3}}{4}}\cdot a_{0})-(|GS_{1}|)}$, multiplying
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+({\frac {\sqrt {3}}{4}}\cdot a_{0})-({\frac {1}{3}}\cdot |DG|)}$, definition of centroid in a triangle
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {\sqrt {3}}{6}}\cdot a_{0})+({\frac {\sqrt {3}}{4}}\cdot a_{0})-({\frac {1}{3}}\cdot {\frac {\sqrt {3}}{4}}\cdot a_{0})}$, see line above
${\displaystyle \quad \Leftrightarrow S_{1}=({\frac {\sqrt {3}}{4}}\cdot a_{0})-({\frac {\sqrt {3}}{6}}\cdot a_{0})-({\frac {\sqrt {3}}{12}}\cdot a_{0})}$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=({\frac {3{\sqrt {3}}}{12}}\cdot a_{0})-({\frac {2{\sqrt {3}}}{12}}\cdot a_{0})-({\frac {\sqrt {3}}{12}}\cdot a_{0})=0+0i=S_{0}}$, rearranging
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