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The right isosceles triangle as base element. Inscribed is the largest equilateral triangle.

0) The side length of the right isosceles triangle is: ${\displaystyle a_{0}}$
1) The side length of the equilateral triangle is: ${\displaystyle a_{1}={\sqrt {\frac {2}{3}}}\cdot a_{0}\quad \approx 0.816\cdot a_{0}}$, see calculation 3

0) Perimeter of right isosceles triangle: ${\displaystyle P_{0}=(2+{\sqrt {2}})\cdot a_{0}\quad \approx 3.414\cdot a_{0}}$
1) Perimeter of inscribed equilateral triangle: ${\displaystyle P_{1}=3\cdot a_{1}=3\cdot {\sqrt {\frac {2}{3}}}\cdot a_{0}={\sqrt {6}}\cdot a_{0}\quad \approx 2.449\cdot a_{0}}$

0) Area of the right isosceles base triangle: ${\displaystyle A_{0}={\frac {1}{2}}\cdot a_{0}^{2}=0.500\cdot a_{0}^{2}\quad }$
1) Area of the inscribed equilateral triangle: ${\displaystyle A_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}^{2}\quad \approx 0.289\cdot a_{0}^{2}\quad }$, see calculation 4

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

In the normalised case the area of the base shape is set to 1.
So ${\displaystyle A_{0}={\frac {1}{2}}\cdot a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}=2\quad \Rightarrow a_{0}={\sqrt {2}}\quad \approx 1.414}$

0) Side length of the right isosceles base triangle ${\displaystyle a_{0}={\sqrt {2}}\quad \approx 1.414}$
1) Side length inscribed equilateral triangle: ${\displaystyle a_{1}={\sqrt {\frac {2}{3}}}\cdot {\sqrt {2}}={\frac {2}{\sqrt {3}}}\quad \approx 1.155}$

0) Perimeter of right isosceles base triangle: ${\displaystyle P_{0}=(2+{\sqrt {2}})\cdot {\sqrt {2}}=2\cdot ({\sqrt {2}}+1)\quad \approx 4.8284271}$
1) Perimeter of inscribed equilateral triangle: ${\displaystyle P_{1}={\sqrt {6}}\cdot {\sqrt {2}}\quad \approx 3.4641016}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 8.2925287}$

0) Area of the right isosceles base triangle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed equilateral triangle: ${\displaystyle A_{1}={\frac {1}{2{\sqrt {3}}}}\cdot ({\sqrt {2}})^{2}={\frac {1}{\sqrt {3}}}\quad \approx 0.577}$

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

${\displaystyle decimalpart(8.2925287+0)=decimalpart(8.2925287)=.2925287}$

So the identifying number is: ${\displaystyle 1.2925287}$

(0) Given is the side length of the equilateral triangle: ${\displaystyle a_{0}}$
(1)${\displaystyle \quad |AB|=|AC|=a_{0}}$
(2)${\displaystyle \quad |AE|=|AF|=|EF|=a_{1}}$
(3)${\displaystyle \quad |BD|=|CD|={\frac {|BC|}{2}}}$
(4)${\displaystyle \quad |DE|=|DF|={\frac {a_{1}}{2}}}$

The length ${\displaystyle |BC|}$ is calculated:
${\displaystyle \quad |AB|^{2}+|AC|^{2}=|BC|^{2}\quad }$,applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle ABC}$
${\displaystyle \quad \Leftrightarrow a_{0}^{2}+a_{0}^{2}=|BC|^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow 2\cdot a_{0}^{2}=|BC|^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |BC|={\sqrt {2}}\cdot a_{0}\quad }$, rearranging

The length ${\displaystyle |AD|}$ is calculated:
${\displaystyle \quad |AD|^{2}+|BD|^{2}=|AB|^{2}\quad }$,applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle ABD}$
${\displaystyle \quad \Leftrightarrow |AD|^{2}+\left({\frac {|BC|}{2}}\right)^{2}=|AB|^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow |AD|^{2}+\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\right)^{2}=|AB|^{2}\quad }$, applying the result of calculation (1)
${\displaystyle \quad \Leftrightarrow |AD|^{2}+\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\right)^{2}=a_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |AD|^{2}=a_{0}^{2}-\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\right)^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |AD|^{2}=a_{0}^{2}-{\frac {2a_{0}^{2}}{4}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |AD|^{2}=a_{0}^{2}-{\frac {a_{0}^{2}}{2}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |AD|^{2}={\frac {1}{2}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |AD|={\frac {1}{\sqrt {2}}}\cdot a_{0}\quad }$, rearranging

The length ${\displaystyle a_{1}}$ is calculated:
${\displaystyle \quad |AD|^{2}+|DE|^{2}=|AE|^{2}\quad }$,applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle AED}$
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)^{2}+|DE|^{2}=|AE|^{2}\quad }$, applying result of calculation (2)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=|AE|^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=a_{1}^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}^{2}+{\frac {a_{1}^{2}}{4}}=a_{1}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}^{2}=a_{1}^{2}-{\frac {a_{1}^{2}}{4}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}^{2}={\frac {3}{4}}\cdot a_{1}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {4}{2\cdot 3}}\cdot a_{0}^{2}=a_{1}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}^{2}={\frac {4}{2\cdot 3}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}^{2}={\frac {2}{3}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {\frac {2}{3}}}\cdot a_{0}\quad \approx 0.816\cdot a_{0}}$, rearranging

${\displaystyle \quad A_{1}=|AD|\cdot |DE|\quad }$
${\displaystyle \quad \Leftrightarrow A_{1}=\left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)\cdot |DE|\quad }$, applying result of calculation (2)
${\displaystyle \quad \Leftrightarrow A_{1}=\left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)\cdot \left({\frac {a_{1}}{2}}\right)\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow A_{1}=\left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)\cdot \left({\frac {1}{2}}{\sqrt {\frac {2}{3}}}\cdot a_{0}\right)\quad }$, applying result of calculation (3)
${\displaystyle \quad \Leftrightarrow A_{1}={\sqrt {\frac {1}{2}}}\cdot {\sqrt {\frac {1}{4}}}\cdot {\sqrt {\frac {2}{3}}}\cdot a_{0}^{2}\quad }$, applying result of calculation (3)
${\displaystyle \quad \Leftrightarrow A_{1}={\sqrt {\frac {2}{24}}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}={\sqrt {\frac {1}{12}}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}^{2}\quad \approx 0.289\cdot a_{0}^{2}\quad }$, rearranging

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