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The second largest square inscribed in a quarter circle of side length ${\displaystyle r_{0}}$

The quarter circle as base element. Inscribed is the second largest square.

0) The radius of the quarter circle ${\displaystyle r_{0}}$
1) Side length of the inscribed square: ${\displaystyle a_{1}={\sqrt {\frac {2}{5}}}\cdot r_{0}\quad \approx 0.632\cdot r_{0}\quad }$, see Calculation 1

0) Perimeter of base quarter circle: ${\displaystyle P_{0}=2\cdot r_{0}+{\frac {\pi \cdot r_{0}}{2}}=(2+{\frac {\pi }{2}})\cdot r_{0}\quad \approx 3.571\cdot r_{0}}$
1) Perimeter of the inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot {\sqrt {\frac {2}{5}}}\cdot r_{0}={\sqrt {\frac {32}{5}}}\cdot r_{0}\quad \approx 2.53\cdot r_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=\left(2+{\frac {\pi }{2}}+{\sqrt {\frac {32}{5}}}\right)\cdot r_{0}\quad \approx 6.101\cdot r_{0}}$

0) Area of the base quarter circle: ${\displaystyle A_{0}={\frac {\pi \cdot r_{0}^{2}}{4}}}$
1) Area of the inscribed square: ${\displaystyle A_{1}=a_{1}^{2}=\left({\sqrt {\frac {2}{5}}}\cdot r_{0}\right)^{2}={\frac {2}{5}}\cdot r_{0}^{2}=0.400\cdot r_{0}^{2}\qquad \Rightarrow A_{1}={\frac {8}{5\pi }}\cdot A_{0}\quad \approx 0.509\cdot A_{0}}$

0) Centroid position of the quarter circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed square measured from the centroid of the base shape: ${\displaystyle S_{1}=\left({\frac {1}{\sqrt {5}}}-{\frac {4}{3\pi }}\right)\cdot r_{0}\cdot (1+i)\quad \approx 0.023\cdot (1+i)\cdot r_{0}\quad }$, see calculation 2

W) Weighted centroid: ${\displaystyle S_{w}={\frac {8}{5\pi +8}}\cdot \left({\frac {3\pi -4{\sqrt {5}}}{3\pi {\sqrt {5}}}}\right)\cdot r_{0}\cdot (1+i)\quad \approx 0.007\cdot r_{0}\cdot (1+i)}$, see calculation 3

In the normalised case the area of the base quarter circle is set to 1.
So ${\displaystyle A={\frac {\pi \cdot r_{0}^{2}}{4}}=1\quad \Rightarrow r_{0}^{2}={\frac {4}{\pi }}\quad \Rightarrow r_{0}={\sqrt {\frac {4}{\pi }}}={\frac {2}{\sqrt {\pi }}}}$

0) Radius of the base quarter circle ${\displaystyle r_{0}={\frac {2}{\sqrt {\pi }}}\quad \approx 1.128}$
1) Side length of the inscribed square: ${\displaystyle a_{1}={\sqrt {\frac {2}{5}}}\cdot {\frac {2}{\sqrt {\pi }}}={\sqrt {\frac {8}{5\pi }}}\quad \approx 0.714}$

0) Perimeter of base quarter circle${\displaystyle P_{0}=(2+{\frac {\pi }{2}})\cdot r_{0}=(2+{\frac {\pi }{2}})\cdot {\frac {2}{\sqrt {\pi }}}\quad \approx 4.0292122}$
1) Perimeter of the inscribed square${\displaystyle P_{1}=4\cdot a_{1}=4\cdot {\sqrt {\frac {8}{5\pi }}}\quad \approx 2.8545986}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 6.8838108}$

0) Area of the base quarter circle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed square: ${\displaystyle A_{1}=a_{1}^{2}=\left({\sqrt {\frac {8}{5\pi }}}\right)^{2}={\frac {8}{5\pi }}\quad \approx 0.509}$

0) Centroid position of the quarter circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed square: ${\displaystyle S_{1}=\left({\frac {1}{\sqrt {5}}}-{\frac {4}{3\pi }}\right)\cdot {\frac {2}{\sqrt {\pi }}}\cdot (1+i)\quad \approx 0.026\cdot (1+i)}$

W) Weighted centroid: ${\displaystyle S_{w}={\frac {8}{5\pi +8}}\cdot \left({\frac {1}{\sqrt {5}}}-{\frac {4}{3\pi }}\right)\cdot {\frac {2}{\sqrt {\pi }}}\cdot (1+i)\quad \approx 0.009\cdot (1+i)\quad }$

The distance between the centroid of the base element and the centroid of the quarter circle is:
${\displaystyle d_{01}\approx 0.0363842}$

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.

${\displaystyle decimalpart(6.8838108+0.0363842)=decimalpart(6.9201950)=.9201950}$

So the identifying number is: ${\displaystyle 1.9201950}$

(1) ${\displaystyle |AB|=|AC|=|AF|=r_{0}}$

(2) ${\displaystyle |DG|=|DE|=|EF|=|FG|=a_{1}\quad }$, since ${\displaystyle DEFG}$ is a square

(3) ${\displaystyle |AD|=|AG|\quad }$, since the triangle ADG has a 45°-angle in D

(4) ${\displaystyle |AD|^{2}+|AG|^{2}=|DG|^{2}\quad }$, applying the Pythagorean theorem
${\displaystyle \quad \Leftrightarrow |AD|^{2}+|AD|^{2}=|DG|^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow 2\cdot |AD|^{2}=a_{1}^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow |AD|^{2}={\frac {a_{1}^{2}}{2}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |AD|={\frac {a_{1}}{\sqrt {2}}}\quad }$, rearranging

(5)${\displaystyle |AD|^{2}+|DF|^{2}=|AF|^{2}\quad }$, applying the Pythagorean theorem since ${\displaystyle {\overrightarrow {DF}}}$ is perpendicular to ${\displaystyle {\overrightarrow {AB}}}$
${\displaystyle \quad \Leftrightarrow |AD|^{2}+|DF|^{2}=r_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |AD|^{2}+\left({\sqrt {2}}\cdot |DE|\right)^{2}=r_{0}^{2}\quad }$, since ${\displaystyle DEFG}$ is a square and ${\displaystyle {\overrightarrow {DF}}}$ is the diagonal
${\displaystyle \quad \Leftrightarrow |AD|^{2}+\left({\sqrt {2}}\cdot a_{1}\right)^{2}=r_{0}^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow |AD|^{2}+2\cdot a_{1}^{2}=r_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {a_{1}^{2}}{2}}+2\cdot a_{1}^{2}=r_{0}^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow {\frac {a_{1}^{2}}{2}}+{\frac {4\cdot a_{1}^{2}}{2}}=r_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {5}{2}}\cdot a_{1}^{2}=r_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}^{2}={\frac {2}{5}}\cdot r_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {\frac {2}{5}}}\cdot r_{0}\quad \approx 0.632\cdot r_{0}}$, rearranging
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${\displaystyle \quad S_{1}={\overrightarrow {S_{0}A}}+{\overrightarrow {AS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+|AD|\cdot +|DS_{1}|\cdot i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {a_{1}}{\sqrt {2}}}+|DS_{1}|\cdot i\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {1}{\sqrt {2}}}\cdot {\sqrt {\frac {2}{5}}}\cdot r_{0}+|DS_{1}|\cdot i\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {1}{\sqrt {5}}}\cdot r_{0}+|DS_{1}|\cdot i\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {1}{\sqrt {5}}}\cdot r_{0}+{\frac {\sqrt {2}}{2}}\cdot a_{1}\cdot i\quad }$, since ${\displaystyle {\overrightarrow {DS_{1}}}}$ is half the diagonal of the square
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {1}{\sqrt {5}}}\cdot r_{0}+{\frac {\sqrt {2}}{2}}\cdot {\sqrt {\frac {2}{5}}}\cdot r_{0}\cdot i\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {1}{\sqrt {5}}}\cdot r_{0}+{\frac {1}{\sqrt {5}}}\cdot r_{0}\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {4}{3\pi }}\cdot r_{0}\cdot (1+i)+{\frac {1}{\sqrt {5}}}\cdot r_{0}\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {1}{\sqrt {5}}}-{\frac {4}{3\pi }}\right)\cdot r_{0}\cdot (1+i)\quad }$, rearranging

${\displaystyle S_{w}={\frac {S_{0}\cdot A_{0}+S_{1}\cdot A_{1}}{A_{0}+A_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot A_{1}}{A_{0}+A_{1}}}\quad }$,since ${\displaystyle S_{0}=0}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {A_{1}}{A_{0}+A_{1}}}\cdot S_{1}\quad }$,since ${\displaystyle S_{0}=0}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {8}{5\pi }}\cdot A_{0}}{A_{0}+{\frac {8}{5\pi }}\cdot A_{0}}}\cdot S_{1}\quad }$,since ${\displaystyle A_{1}={\frac {8}{5\pi }}\cdot A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {\frac {8}{5\pi }}{1+{\frac {8}{5\pi }}}}\cdot S_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {\frac {8}{5\pi }}{\frac {5\pi +8}{5\pi }}}\cdot S_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {8}{5\pi +8}}\cdot S_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {8}{5\pi +8}}\cdot \left({\frac {1}{\sqrt {5}}}-{\frac {4}{3\pi }}\right)\cdot r_{0}\cdot (1+i)\quad }$, see Calculation 2

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