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Smallest square inscribed (touching all sides) in a square of side length ${\displaystyle a_{0}}$

The square as base element. Inscribed is the smallest square that touches all sides of the base element

0) Side length of the base square ${\displaystyle a_{0}}$
1) Side length of the inscribed square: ${\displaystyle a_{1}={\frac {a_{0}}{\sqrt {2}}}\quad \approx 0.707\cdot a_{0}\quad }$, see Calculation 1

0) Perimeter of base square: ${\displaystyle P_{0}=4\cdot a_{0}}$
1) Perimeter of the inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot {\frac {a_{0}}{\sqrt {2}}}=2{\sqrt {2}}\cdot a_{0}\quad \approx 2.828\cdot a_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=4\cdot a_{0}+2{\sqrt {2}}\cdot a_{0}=2\cdot ({\sqrt {2}}+2)\cdot a_{0}\quad \approx 6.828\cdot a_{0}}$

0) Area of the base square: ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed square: ${\displaystyle A_{1}=a_{1}^{2}=({\frac {a_{0}}{\sqrt {2}}})^{2}={\frac {a_{0}^{2}}{2}}=0.500\cdot A_{0}}$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed square: ${\displaystyle S_{1}=S_{0}=0+0i}$
W) Weighted centroid: ${\displaystyle S_{w}=0+0i}$

In the normalised case the area of the base square is set to 1.
So ${\displaystyle A_{0}=1\quad \Rightarrow a_{0}^{2}=1\quad \Rightarrow a_{0}=1}$

0) Side length of the base square: ${\displaystyle a_{0}=1}$
1) Side length of the inscribed square: ${\displaystyle a_{1}={\frac {1}{\sqrt {2}}}\quad \approx 0.707}$

0) Perimeter of base square: ${\displaystyle P_{0}=4\cdot a_{0}=4}$
1) Perimeter of the inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot {\frac {1}{\sqrt {2}}}=2{\sqrt {2}}\quad \approx 2.8284271}$
S) Sum of perimeters: ${\displaystyle P_{s}=4+2{\sqrt {2}}=2\cdot ({\sqrt {2}}+2)\quad \approx 6.8284271}$

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

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed square: ${\displaystyle S_{1}=S_{0}=0+0i}$
W) Weighted centroid: ${\displaystyle S_{w}=0+0i}$

The distance between the centroid of the base element and the centroid of the inscribed square is:
${\displaystyle d_{01}=0}$

Apart of the base element there is one 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.

${\displaystyle decimalpart(6.8284271+0)=decimalpart(6.8284271)=.8284271}$

So the identifying number is: ${\displaystyle 1.8284271}$

(1) ${\displaystyle |AB|=|AD|=|BC|=|CD|=|EG|=|FH|=a_{0}}$

(2) ${\displaystyle |EF|=|EH|=|FG|=|GH|=a_{1}\quad }$, since ${\displaystyle EFGH}$ is the inscribed square

(3) ${\displaystyle |ES_{0}|=|FS_{0}|={\frac {|EG|}{2}}={\frac {|FH|}{2}}={\frac {1}{2}}\cdot a_{0}\quad }$, applying equation (1)

(4) ${\displaystyle |EF|^{2}=|ES_{0}|^{2}+|FS_{0}|^{2}\quad }$, applying the Pythagorean theorem
${\displaystyle \quad \Leftrightarrow a_{1}^{2}=|ES_{0}|^{2}+|FS_{0}|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow a_{1}^{2}=({\frac {1}{2}}\cdot a_{0})^{2}+({\frac {1}{2}}\cdot a_{0})^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow a_{1}^{2}=2\cdot ({\frac {1}{2}}\cdot a_{0})^{2}\quad }$, adding the two terms
${\displaystyle \quad \Leftrightarrow a_{1}^{2}=2\cdot {\frac {1}{4}}\cdot a_{0}^{2}\quad }$, squaring the bracket term
${\displaystyle \quad \Leftrightarrow a_{1}^{2}={\frac {1}{2}}\cdot a_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {1}{\sqrt {2}}}\cdot a_{0}\quad }$, applying the root

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