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The second largest circle inscribed in a square of side length ${\displaystyle a_{0}}$, that contains the largest non-right isosceles triangle

0) The side length of the base square: ${\displaystyle a_{0}}$
1) Side length of the inscribed isosceles triangle: ${\displaystyle a_{1}={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad \approx 1.118\cdot a_{0}\quad }$, see Calculation 1
2) Radius of the inscribed square: ${\displaystyle r_{2}={\frac {3-{\sqrt {5}}}{4}}\cdot a_{0}\quad \approx 0.191\cdot a_{0}}$, see Calculation 4

0) Perimeter of base square: ${\displaystyle P_{0}=4\cdot a_{0}}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}=2\cdot a_{1}+a_{0}=2\cdot {\frac {\sqrt {5}}{2}}\cdot a_{0}+a_{0}=({\sqrt {5}}+1)\cdot a_{0}\quad \approx 3.236\cdot a_{0}}$
2) Perimeter of the inscribed circle: ${\displaystyle P_{2}=2\pi r_{2}=\pi \cdot {\frac {3-{\sqrt {5}}}{2}}\cdot a_{0}\quad \approx 1.2\cdot a_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=4\cdot a_{0}+({\sqrt {5}}+1)\cdot a_{0}+\pi \cdot {\frac {3-{\sqrt {5}}}{2}}\cdot a_{0}=\left({\sqrt {5}}+5+\pi \cdot {\frac {3-{\sqrt {5}}}{2}}\right)\cdot a_{0}\quad \approx 8.436\cdot a_{0}}$

0) Area of the base square: ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {a_{0}}{2}}\cdot a_{0}={\frac {1}{2}}\cdot a_{0}^{2}={\frac {1}{2}}\cdot A_{0}}$
2) Area of the inscribed circle: ${\displaystyle A_{2}=\pi r_{2}^{2}=\pi \cdot \left({\frac {3-{\sqrt {5}}}{4}}\cdot a_{0}\right)^{2}=\pi \cdot {\frac {9-2\cdot 3{\sqrt {5}}+5}{16}}\cdot a_{0}^{2}=\pi \cdot {\frac {14-6{\sqrt {5}}}{16}}\cdot a_{0}^{2}=\pi \cdot {\frac {7-3{\sqrt {5}}}{8}}\cdot a_{0}^{2}\quad \approx 0.115\cdot A_{0}}$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed triangle measured from the centroid of the base shape: ${\displaystyle S_{1}=(0-{\frac {1}{6}}\cdot i)\cdot a_{0}\quad \approx (0-0.167i)\cdot a_{0}}$, see calculation 2
2) Centroid position of the inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=\left({\frac {{\sqrt {5}}-1}{4}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.309+0.309i)\cdot a_{0}\quad }$, see Calculation (5)
W) Weighted centroid: ${\displaystyle S_{w}={\frac {-4i+3\pi \cdot (5{\sqrt {5}}-11)\cdot (1+i)}{72+6\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad \approx (0.022-0.030i)\cdot a_{0}}$, see calculation 3

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

0) Side lenght of the base square ${\displaystyle a_{0}=1}$
1) Side length of the inscribed triangle: ${\displaystyle a_{1}={\frac {\sqrt {5}}{2}}\cdot 1\quad \approx 1.118\quad }$
2) Radius of the inscribed square: ${\displaystyle r_{2}={\frac {3-{\sqrt {5}}}{4}}\cdot 1\quad \approx 0.191}$

0) Perimeter of base square${\displaystyle P_{0}=4\cdot a_{0}=4}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}={\sqrt {5}}+1\quad \approx 3.2360680}$
2) Perimeter of the inscribed circle: ${\displaystyle P_{2}=\pi \cdot {\frac {3-{\sqrt {5}}}{2}}\cdot 1\quad \approx 1.1999816}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}\quad \approx 8.4360496}$

0) Area of the base square is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {1}{2}}\cdot A_{0}=0.500\quad }$
2) Area of the inscribed circle: ${\displaystyle A_{2}=\pi \cdot {\frac {7-3{\sqrt {5}}}{8}}\cdot 1\quad \approx 0.115}$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed triangle: ${\displaystyle S_{1}=(0-{\frac {1}{6}}\cdot i)\quad \approx (0-0.167i)}$
2) Centroid position of the inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=\left({\frac {{\sqrt {5}}-1}{4}}\right)\cdot (1+i)\quad \approx (0.309+0.309i)\quad }$
W) Weighted centroid: ${\displaystyle S_{w}=\left(3\pi \cdot (5{\sqrt {5}}-11)+\left(3\pi \cdot (5{\sqrt {5}}-11)-4\right)i\right)\cdot {\frac {1}{72+6\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad \approx (0.022-0.030i)\cdot a_{0}}$

The distance between the centroid of the base element and the centroid of the triangle is:
${\displaystyle d_{S_{0}S_{1}}\approx 0.1666667}$
${\displaystyle d_{S_{0}S_{2}}\approx 0.4370160}$
${\displaystyle d_{S_{1}S_{2}}\approx 0.5672446}$
Sum of distances: ${\displaystyle d_{s}\approx 1.1709273}$

Apart of the base element there are two shapes allocated. Therefore the integer part of the identifying number is 2.
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.

${\displaystyle decimalpart(8.4360496+1.1709273)=decimalpart(9.6069769)=.6069769}$

So the identifying number is: ${\displaystyle 2.6069769}$

(1) ${\displaystyle |AB|=|BC|=|CD|=|AD|=|EF|=a_{0}}$, since ${\displaystyle AEBCFD}$ is a square
(2) ${\displaystyle |AE|=|BE|=|CF|=|DF|=|ES_{0}|=|FS_{0}|={\frac {1}{2}}\cdot a_{0}}$
(3) ${\displaystyle |AF|=|BF|=a_{1}\quad }$
(4) ${\displaystyle |CH|=|CG|=|HS_{2}|=|GS_{2}|=|JS_{2}|=r_{2}}$

${\displaystyle \quad |AE|^{2}+|EF|^{2}=|AF|^{2}\quad }$, applying the Pythagorean theorem on triangle ${\displaystyle \triangle {AEF}}$
${\displaystyle \quad \Leftrightarrow ({\frac {1}{2}}\cdot a_{0})^{2}+|EF|^{2}=|AF|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow ({\frac {1}{2}}\cdot a_{0})^{2}+a_{0}^{2}=|AF|^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow ({\frac {1}{2}}\cdot a_{0})^{2}+a_{0}^{2}=a_{1}^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow {\frac {1}{4}}\cdot a_{0}^{2}+a_{0}^{2}=a_{1}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow {\frac {5}{4}}\cdot a_{0}^{2}=a_{1}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad \approx 1.118\cdot a_{0}}$,

${\displaystyle \quad S_{1}=S_{0}+{\overrightarrow {S_{0}E}}+{\overrightarrow {ES_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=-|ES_{0}|\cdot i+|ES_{1}|\cdot i}$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{2}}\cdot a_{0}\cdot i+|ES_{1}|\cdot i\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{2}}\cdot a_{0}\cdot i+{\frac {1}{3}}\cdot |EF|\cdot i\quad }$, applying the formula for centroids of triangles
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{2}}\cdot a_{0}\cdot i+{\frac {1}{3}}\cdot a_{0}\cdot i\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {1}{3}}-{\frac {1}{2}}\right)\cdot a_{0}\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{6}}\cdot a_{0}\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {1}{6}}\cdot i)\cdot a_{0}\quad \approx (0-0.167i)\cdot a_{0}}$, rearranging

${\displaystyle \quad S_{w}={\frac {S_{0}\cdot A_{0}+S_{1}\cdot A_{1}+S_{2}\cdot A_{2}}{A_{0}+A_{1}+A_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot A_{1}+S_{2}\cdot A_{2}}{A_{0}+A_{1}+A_{2}}}\quad }$, since ${\displaystyle S_{0}=0+0i=0}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot {\frac {1}{2}}\cdot A_{0}+S_{2}\cdot A_{2}}{A_{0}+{\frac {1}{2}}\cdot A_{0}+A_{2}}}\quad }$, since ${\displaystyle A_{1}={\frac {1}{2}}\cdot A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot {\frac {1}{2}}\cdot A_{0}+S_{2}\cdot \pi \cdot {\frac {7-3{\sqrt {5}}}{8}}\cdot A_{0}}{A_{0}+{\frac {1}{2}}\cdot A_{0}+\pi \cdot {\frac {7-3{\sqrt {5}}}{8}}\cdot A_{0}}}\quad }$, since ${\displaystyle A_{2}=\pi \cdot {\frac {7-3{\sqrt {5}}}{8}}\cdot A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot {\frac {1}{2}}+S_{2}\cdot \pi \cdot {\frac {7-3{\sqrt {5}}}{8}}}{1+{\frac {1}{2}}+\pi \cdot {\frac {7-3{\sqrt {5}}}{8}}}}\quad }$, eliminating ${\displaystyle A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot {\frac {4}{8}}+S_{2}\cdot \pi \cdot {\frac {7-3{\sqrt {5}}}{8}}}{{\frac {8}{8}}+{\frac {4}{8}}+\pi \cdot {\frac {7-3{\sqrt {5}}}{8}}}}\quad }$, rearranging</math>
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot 4+S_{2}\cdot \pi \cdot (7-3{\sqrt {5}})}{8+4+\pi \cdot (7-3{\sqrt {5}})}}\quad }$, eliminating 8
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {4\cdot S_{1}+\pi \cdot (7-3{\sqrt {5}})\cdot S_{2}}{12+\pi \cdot (7-3{\sqrt {5}})}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {4\cdot (-{\frac {i}{6}}\cdot a_{0})+\pi \cdot (7-3{\sqrt {5}})\cdot S_{2}}{12+\pi \cdot (7-3{\sqrt {5}})}}\quad }$, since ${\displaystyle S_{1}=-{\frac {i}{6}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {-{\frac {2i}{3}}\cdot a_{0}+\pi \cdot (7-3{\sqrt {5}})\cdot \left({\frac {{\sqrt {5}}-1}{4}}\right)\cdot (1+i)\cdot a_{0}}{12+\pi \cdot (7-3{\sqrt {5}})}}\quad }$, since ${\displaystyle S_{2}=\left({\frac {{\sqrt {5}}-1}{4}}\right)\cdot (1+i)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {-{\frac {2i}{3}}+\pi \cdot (7-3{\sqrt {5}})\cdot \left({\frac {{\sqrt {5}}-1}{4}}\right)\cdot (1+i)}{12+\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {-{\frac {8i}{12}}+3\pi \cdot (7-3{\sqrt {5}})\cdot \left({\frac {{\sqrt {5}}-1}{12}}\right)\cdot (1+i)}{12+\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {-8i+3\pi \cdot (7-3{\sqrt {5}})\cdot \left({\sqrt {5}}-1\right)\cdot (1+i)}{144+12\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {-8i+3\pi \cdot (10{\sqrt {5}}-22)\cdot (1+i)}{144+12\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {-4i+3\pi \cdot (5{\sqrt {5}}-11)\cdot (1+i)}{72+6\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}=(3\pi \cdot (5{\sqrt {5}}-11)\cdot (1+i)-4i)\cdot {\frac {1}{72+6\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}=\left(3\pi \cdot (5{\sqrt {5}}-11)+\left(3\pi \cdot (5{\sqrt {5}}-11)-4\right)i\right)\cdot {\frac {1}{72+6\pi \cdot (7-3{\sqrt {5}})}}\cdot a_{0}\quad \approx (0.022-0.030i)\cdot a_{0}}$

${\displaystyle \quad |FJ|+|BJ|=|BF|=a_{1}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow |FJ|+|BJ|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, applying the result of calculation (1)
${\displaystyle \quad \Leftrightarrow |FG|+|BJ|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, since ${\displaystyle {\overrightarrow {FJ}}}$ and ${\displaystyle {\overrightarrow {FG}}}$ emanate from point ${\displaystyle F}$ and are tangent to the circle around ${\displaystyle S_{2}}$ with radius ${\displaystyle r_{2}}$
${\displaystyle \quad \Leftrightarrow |FC|-|GC|+|BJ|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$,since ${\displaystyle {\overrightarrow {FG}}+{\overrightarrow {GC}}={\overrightarrow {FC}}}$
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}-|GC|+|BJ|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}-r_{2}+|BJ|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}-r_{2}+|BH|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$,since ${\displaystyle {\overrightarrow {BJ}}}$ and ${\displaystyle {\overrightarrow {BH}}}$ emanate from point ${\displaystyle B}$ and are tangent to the circle around ${\displaystyle S_{2}}$ with radius ${\displaystyle r_{2}}$
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}-r_{2}+|BC|-|CH|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, since ${\displaystyle {\overrightarrow {BH}}+{\overrightarrow {HC}}={\overrightarrow {BC}}}$
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}-r_{2}+a_{0}-|CH|={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}\cdot a_{0}-r_{2}+a_{0}-r_{2}={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow {\frac {3}{2}}\cdot a_{0}-2\cdot r_{2}={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {3}{2}}\cdot a_{0}-{\frac {\sqrt {5}}{2}}\cdot a_{0}=2\cdot r_{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {3-{\sqrt {5}}}{2}}\cdot a_{0}=2\cdot r_{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {3-{\sqrt {5}}}{4}}\cdot a_{0}\quad }$, rearranging

${\displaystyle \quad S_{2}=S_{0}+{\overrightarrow {S_{0}F}}+{\overrightarrow {FG}}+{\overrightarrow {GS_{2}}}=0+{\overrightarrow {S_{0}F}}+{\overrightarrow {FG}}+{\overrightarrow {GS_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=|S_{0}F|\cdot i+|FG|-|GS_{2}|\cdot i}$
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {a_{0}}{2}}\cdot i+|FG|-|GS_{2}|\cdot i\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {a_{0}}{2}}\cdot i+|FC|-|CG|-|GS_{2}|\cdot i\quad }$
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {a_{0}}{2}}\cdot i+{\frac {a_{0}}{2}}-|CG|-|GS_{2}|\cdot i\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {a_{0}}{2}}\cdot i+{\frac {a_{0}}{2}}-r_{2}-|GS_{2}|\cdot i\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {a_{0}}{2}}\cdot i+{\frac {a_{0}}{2}}-r_{2}-r_{2}\cdot i\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow S_{2}={\frac {a_{0}}{2}}\cdot (1+i)-r_{2}\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\frac {a_{0}}{2}}-r_{2}\right)\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\frac {a_{0}}{2}}-{\frac {3-{\sqrt {5}}}{4}}\cdot a_{0}\right)\cdot (1+i)\quad }$, applying the result of Calculation (4)
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\frac {2}{4}}-{\frac {3-{\sqrt {5}}}{4}}\right)\cdot a_{0}\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\frac {2-3+{\sqrt {5}}}{4}}\right)\cdot (1+i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\frac {{\sqrt {5}}-1}{4}}\right)\cdot (1+i)\cdot a_{0}\quad \approx 0.309\cdot (1+i)\cdot a_{0}\quad }$, rearranging

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