The largest square inscribe in a right isosceles triangle of side length ${\displaystyle a_{0}}$

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

0) Perimeter of base triangle: ${\displaystyle P_{0}=2\cdot a_{0}+{\sqrt {2}}\cdot a_{0}=(2+{\sqrt {2}})\cdot a_{0}\quad \approx 3.414\cdot a_{0}}$
1) Perimeter of inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot {\frac {1}{2}}\cdot a_{0}\quad =2.000\cdot a_{0}}$

0) Area of the base triangle: ${\displaystyle A_{0}={\frac {1}{2}}\cdot a_{0}^{2}\quad =0.500\cdot a_{0}^{2}\quad }$
1) Area of the inscribed square: ${\displaystyle A_{1}=a_{1}^{2}=\left({\frac {1}{2}}\cdot a_{0}\right)^{2}={\frac {1}{4}}\cdot a_{0}^{2}\quad =0.250\cdot a_{0}^{2}}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed square relative to the centroid of the base shape is: ${\displaystyle S_{1}=-{\frac {1}{12}}\cdot a_{0}\cdot (1+i)\quad \approx -0.083\cdot a_{0}\cdot (1+i)}$
, see calculation (1)

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 triangle ${\displaystyle a_{0}={\sqrt {2}}\quad \approx 1.414}$
1) The side length of the square is: ${\displaystyle a_{1}={\frac {\sqrt {2}}{2}}\quad \approx 0.707\quad }$,

0) Perimeter of base triangle: ${\displaystyle P_{0}=(2+{\sqrt {2}})\cdot {\sqrt {2}}\quad \approx 4.8284271}$
1) Perimeter of inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot {\frac {1}{2}}\cdot {\sqrt {2}}\quad \approx 2.8284271}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 7.6568542}$

0) Area of the base triangle is by definition ${\displaystyle A_{0}={\frac {1}{2}}\cdot ({\sqrt {2}})^{2}=1}$
1) Area of the inscribed square: ${\displaystyle A_{1}=a_{1}^{2}=\left({\frac {1}{2}}\cdot a_{0}\right)^{2}={\frac {1}{4}}\cdot ({\sqrt {2}})^{2}={\frac {1}{2}}\quad =0.500}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed square relative to the centroid of the base shape is: ${\displaystyle S_{1}=-{\frac {1}{12}}\cdot {\sqrt {2}}\cdot (1+i)=-{\frac {1}{6{\sqrt {2}}}}\cdot (1+i)\quad \approx -0.118\cdot (1+i)}$

The distance between the centroid of the base triangle and the centroid of the semicircle is:
${\displaystyle d_{01}=0.1666667}$
Sum of distances: ${\displaystyle d_{s}=d_{01}=0.1666667}$

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.

${\displaystyle decimalpart(7.6568542+0.1666667)=decimalpart(7.8235209)=.8235209}$

So the identifying number is: ${\displaystyle 1.8235209}$

(0) Given is the side length of the equilateral triangle: ${\displaystyle a_{0}}$
(1)${\displaystyle \quad |AB|=|AC|=a_{0}}$

The centroid of the square relative to the centroid of the triangle is:
${\displaystyle \quad S_{1}=S_{0}+{\overrightarrow {S_{0}A}}+{\overrightarrow {AS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)-{\frac {|AS_{0}|}{\sqrt {2}}}\cdot (1+i)+{\frac {|AS_{1}|}{\sqrt {2}}}\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {{\frac {2}{3}}\cdot |AD|}{\sqrt {2}}}\cdot (1+i)+{\frac {{\frac {1}{2}}\cdot |AD|}{\sqrt {2}}}\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {{\frac {2}{3}}\cdot {\sqrt {2}}\cdot a_{1}}{\sqrt {2}}}\cdot (1+i)+{\frac {{\frac {1}{2}}\cdot {\sqrt {2}}\cdot a_{1}}{\sqrt {2}}}\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=-\left({\frac {2}{3}}\cdot a_{1}\right)\cdot (1+i)+\left({\frac {1}{2}}\cdot a_{1}\right)\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {1}{2}}-{\frac {2}{3}}\right)\cdot a_{1}\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {3}{6}}-{\frac {4}{6}}\right)\cdot a_{1}\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {3}{6}}-{\frac {4}{6}}\right)\cdot {\frac {1}{2}}\cdot a_{0}\cdot (1+i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{12}}\cdot a_{0}\cdot (1+i)\quad \approx -0.083\cdot a_{0}\cdot (1+i)}$

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