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Task

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

General case

Segments in the general case

0) The side length of the equilateral base triangle is: ${\displaystyle a_{0}}$
1) The side length of the inscribed right triangle is: ${\displaystyle a_{1}={\frac {1}{\sqrt {2}}}\cdot a_{0}\approx 0.707\cdot a_{0}\quad }$, see calculation (2)

Perimeters in the general case

0) Perimeter of equilateral base triangle: ${\displaystyle P_{0}=3\cdot a_{0}}$
1) Perimeter of inscribed right triangle: ${\displaystyle P_{1}=2\cdot a_{1}+a_{0}=\left({\frac {2}{\sqrt {2}}}+1\right)\cdot a_{0}\approx 2.414\cdot a_{0}}$

Areas in the general case

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 right triangle: ${\displaystyle A_{1}={\frac {a_{1}^{2}}{2}}={\frac {1}{2}}({\frac {1}{\sqrt {2}}}\cdot a_{0})^{2}={\frac {1}{2}}\cdot {\frac {1}{2}}\cdot a_{0}^{2}={\frac {1}{4}}a_{0}^{2}=0.250\cdot a_{0}^{2}={\frac {1}{\sqrt {3}}}\cdot A_{0}}$

Centroids in the general case

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed right triangle relative to the centroid of the base shape is: ${\displaystyle S_{1}=\left(0+{\frac {1-{\sqrt {3}}}{6}}i\right)\cdot a_{0}\quad \approx (0-0.122i)\cdot a_{0}\quad }$see calculation (3)

Normalised case

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

Segments in the normalised case

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: ${\displaystyle a_{1}={\frac {1}{\sqrt {2}}}\cdot a_{0}={\frac {1}{\sqrt {2}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\frac {\sqrt {2}}{\sqrt {\sqrt {3}}}}={\sqrt {\frac {2}{\sqrt {3}}}}\quad \approx 1.075\quad }$

Perimeters in the normalised case

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: ${\displaystyle P_{1}=\left({\frac {2}{\sqrt {2}}}+1\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 3.6688112}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 8.2278253}$

Areas in the normalised case

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: ${\displaystyle A_{1}={\frac {1}{\sqrt {3}}}\cdot A_{0}={\frac {1}{\sqrt {3}}}\approx 0.577}$

Centroids in the normalised case

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: ${\displaystyle S_{1}=\left(0+{\frac {1-{\sqrt {3}}}{6}}i\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=\left(0+{\frac {1-{\sqrt {3}}}{3{\sqrt {\sqrt {3}}}}}i\right)\quad \approx (0-0.1854128i)}$

Distances of centroids

The distance between the centroid of the base triangle and the centroid of the inscribed triangle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.1854128}$
Sum of distances: ${\displaystyle d_{s}=d_{01}=0.1854128}$

Identifying number

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.

${\displaystyle decimalpart(8.2278253+0.1854128)=decimalpart(8.4132381)=.4132381}$

So the identifying number is: ${\displaystyle 1.4132381}$

Calculations

Known elements

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

Calculation 1

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)
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }$

Calculation 2

The side length ${\displaystyle a_{1}}$ is calculated:
${\displaystyle \quad |AE|^{2}+|BE|^{2}=|AB|^{2}\quad }$,applying the Pythagorean theorem on the rectangular, isosceles triangle ${\displaystyle \triangle ABE}$
${\displaystyle \quad \Leftrightarrow a_{1}^{2}+a_{1}^{2}=|AB|^{2}\quad }$,applying equation (3)
${\displaystyle \quad \Leftrightarrow 2\cdot a_{1}^{2}=a_{0}^{2}\quad }$,applying equation (1)
${\displaystyle \quad \Leftrightarrow a_{1}^{2}={\frac {1}{2}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {{\frac {1}{2}}\cdot a_{0}^{2}}}\quad }$, drawing the root
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {1}{\sqrt {2}}}\cdot a_{0}\quad }$, rearranging

With this result the height ${\displaystyle |DE|}$ of triangle ${\displaystyle \triangle ABE}$ can be calculated:
${\displaystyle \quad |DE|^{2}+|DB|^{2}=|BE|^{2}\quad }$, applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle DBE}$
${\displaystyle \quad \Leftrightarrow |DE|^{2}+\left({\frac {1}{2}}\cdot a_{0}\right)^{2}=|BE|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow |DE|^{2}+{\frac {1}{4}}\cdot a_{0}^{2}=a_{1}^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow |DE|^{2}+{\frac {1}{4}}\cdot a_{0}^{2}=\left({\frac {1}{\sqrt {2}}}\cdot a_{0}\right)^{2}\quad }$, applying the first part of calculation (2)
${\displaystyle \quad \Leftrightarrow |DE|^{2}+{\frac {1}{4}}\cdot a_{0}^{2}={\frac {1}{2}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |DE|^{2}={\frac {1}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |DE|={\frac {1}{2}}\cdot a_{0}\quad }$, extracting the root

Calculation 3

${\displaystyle \quad S_{1}=S_{0}+{\overrightarrow {S_{0}D}}+{\overrightarrow {DS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-|S_{0}D|i+|DS_{1}|i}$, since ${\displaystyle S_{0}\quad }$ is in the centre of the figure
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {1}{3}}\cdot |DC|i+|DS_{1}|i\quad }$, definition of centroid in an equilateral triangle
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {1}{3}}\cdot ({\frac {\sqrt {3}}{2}}\cdot a_{0})i+|DS_{1}|i\quad }$, see calculation (1) for height ${\displaystyle |CD|}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {\sqrt {3}}{6}}\cdot a_{0}i+|DS_{1}|i\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {\sqrt {3}}{6}}\cdot a_{0}i+{\frac {1}{3}}\cdot |DE|i\quad }$, definition of centroid in an isosceles triangle
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {\sqrt {3}}{6}}\cdot a_{0}i+{\frac {1}{3}}\cdot {\frac {1}{2}}\cdot a_{0}i\quad }$, see second part of calculation (2)
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left({\frac {1}{6}}-{\frac {\sqrt {3}}{6}}\right)\cdot a_{0}i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left({\frac {1-{\sqrt {3}}}{6}}\right)\cdot a_{0}i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+\left({\frac {1-{\sqrt {3}}}{6}}\right)\cdot a_{0}i\quad }$, definition of centroid of the base shape
${\displaystyle \quad \Leftrightarrow S_{1}=\left(0+{\frac {1-{\sqrt {3}}}{6}}i\right)\cdot a_{0}\quad \approx (0-0.122i)\cdot a_{0}}$, rearranging