File:FS Q(2C) dia.png

The largest pair of equal circles inscribed in a square of side length ${\displaystyle a_{0}}$

0) The side lenght of the base square: ${\displaystyle a_{0}}$
1+2) The radii of the inscribed twin circles: ${\displaystyle r_{1}={\frac {(2-{\sqrt {2}})}{2}}\cdot a_{0}\quad \approx 0.293\cdot a_{0}\quad }$, see Calculation 1

0) Perimeter of base square: ${\displaystyle P_{0}=4a_{0}}$
1+2) Perimeter of the inscribed twin circles: ${\displaystyle P_{1}+P_{2}=2\cdot 2\pi r_{1}=4\pi {\frac {(2-{\sqrt {2}})}{2}}\cdot a_{0}=2\pi \cdot (2-{\sqrt {2}})\cdot a_{0}\quad \approx 3.68\cdot a_{0}\quad }$
S) Sum of perimeters: ${\displaystyle P_{s}=4a_{0}+2\cdot \pi \cdot (2-{\sqrt {2}})\cdot a_{0}=2\cdot \left(2+\pi \cdot (2-{\sqrt {2}})\right)\cdot a_{0}\quad \approx 7.68\cdot a_{0}\quad }$

0) Area of the base square: ${\displaystyle A_{0}=a_{0}^{2}}$
1+2) Area of the inscribed twin circles: ${\displaystyle A_{1}+A_{2}=2\cdot \pi \cdot r_{1}^{2}=\pi \cdot (3-2{\sqrt {2}})\cdot A_{0}\quad \approx 0.539A_{0}}$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the first inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{1}=\left(-{\frac {{\sqrt {2}}-1}{2}}+{\frac {{\sqrt {2}}-1}{2}}i\right)\cdot a_{0}\quad \approx (-0.207+0.207i)\cdot a_{0}}$, see Calculation 3
2) Centroid position of the second inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=-S_{1}=\left({\frac {{\sqrt {2}}-1}{2}}-{\frac {{\sqrt {2}}-1}{2}}i\right)\cdot a_{0}\quad \approx (0.207-0.207i)\cdot a_{0}}$
W) Weighted centroid: ${\displaystyle S_{w}={\frac {A_{0}\cdot S_{0}+A_{1}\cdot S_{1}+A_{2}\cdot S_{2}}{A_{0}+A_{1}+A_{2}}}=0+0i\quad }$, because ${\displaystyle S_{0}=0}$ and ${\displaystyle S_{1}=-S_{2}}$ and ${\displaystyle A_{1}=A_{2}}$

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

0) Side length of the base square ${\displaystyle a_{0}=1}$
1+2) Radii of the inscribed circles: ${\displaystyle r_{1}={\frac {(2-{\sqrt {2}})}{2}}\quad \approx 0.293}$

0) Perimeter of base square: ${\displaystyle P_{0}=4.000000}$
1+2) Perimeter of the inscribed twin circles: ${\displaystyle P_{1}+P_{2}=2\pi \cdot (2-{\sqrt {2}})=3.6806047}$
S) Sum of perimeters: ${\displaystyle P_{s}=2\cdot \left(2+\pi \cdot (2-{\sqrt {2}})\right)=7.6806047\quad }$

0) Area of the base square is by definition ${\displaystyle A_{0}=1}$
1+2) Area of the inscribed circles: ${\displaystyle A_{1}+A_{2}=\pi \cdot (3-2{\sqrt {2}})\quad \approx 0.539A_{0}}$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the first inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{1}=\left(-{\frac {{\sqrt {2}}-1}{2}}+{\frac {{\sqrt {2}}-1}{2}}i\right)\quad \approx (-0.207+0.207i)}$
2) Centroid position of the second inscribed circle measured from the centroid of the base shape: ${\displaystyle S_{2}=-S_{1}=-\left(-{\frac {{\sqrt {2}}-1}{2}}+{\frac {{\sqrt {2}}-1}{2}}i\right)\quad \approx (0.207-0.207i)}$
W) Weighted centroid: ${\displaystyle S_{w}=0-0i}$

The distance between the centroid of the base element and the centroid of the triangle is:
${\displaystyle d_{S_{0}S_{1}}\approx 0.2928932}$
${\displaystyle d_{S_{0}S_{2}}\approx 0.2928932}$
${\displaystyle d_{S_{1}S_{2}}\approx 0.5857864}$
Sum of distances: ${\displaystyle d_{s}\approx 1.1715729}$

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.

${\displaystyle decimalpart(7.6806047+1.1715729)=decimalpart(8.8521776)=.8521776}$

So the identifying number is: ${\displaystyle 2.8521776}$

(1) ${\displaystyle |AB|=|BC|=|CD|=|AD|=a_{0}}$
(2) ${\displaystyle |AE|=|AF|=|S_{1}E|=|S_{1}F|=|S_{2}G|=r_{1}\quad }$, because the circles are tangent to the sides of the square
(3) ${\displaystyle |S_{0}S_{1}|=|S_{0}S_{2}|=r_{1}\quad }$because the circles are tangent to each other and have the same radius
(4) ${\displaystyle |AS_{1}|=|CS_{2}|\quad }$, because the twin circles have the same radius and the figure is symmetric

${\displaystyle \quad |AS_{1}|+|S_{1}S_{0}|+|S_{0}S_{2}|+|S_{2}C|={\sqrt {2}}\cdot a_{0}\quad }$, because ${\displaystyle {\overline {AC}}}$ is the diagonal of the square
${\displaystyle \quad \Leftrightarrow 2\cdot |AS_{1}|+|S_{1}S_{0}|+|S_{0}S_{2}|={\sqrt {2}}\cdot a_{0}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow 2\cdot |AS_{1}|+2\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow 2\cdot {\sqrt {|AE|^{2}+|S_{1}E|^{2}}}+2\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, since ${\displaystyle AES_{1}F}$ form a square
${\displaystyle \quad \Leftrightarrow 2\cdot {\sqrt {r_{1}^{2}+r_{1}^{2}}}+2\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow 2\cdot {\sqrt {2r_{1}^{2}}}+2\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow 2\cdot {\sqrt {2}}\cdot r_{1}+2\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow (2{\sqrt {2}}+2)\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, factoring in
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{(2{\sqrt {2}}+2)}}\cdot a_{0}\quad }$, factoring in
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{2({\sqrt {2}}+1)}}\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{{\sqrt {2}}({\sqrt {2}}+1)}}\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{(2+{\sqrt {2}})}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {(2-{\sqrt {2}})}{(2+{\sqrt {2}})(2-{\sqrt {2}})}}\cdot a_{0}\quad }$, extending
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {(2-{\sqrt {2}})}{(4-2)}}\cdot a_{0}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot a_{0}\quad }$, multiplying

${\displaystyle \quad A_{1}+A_{2}=2\cdot \pi \cdot r_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}+A_{2}=2\cdot \pi \cdot \left({\frac {(2-{\sqrt {2}})}{2}}\cdot a_{0}\right)^{2}\quad }$, applying Calculation 1
${\displaystyle \quad \Leftrightarrow A_{1}+A_{2}=2\cdot \pi \cdot {\frac {(2-{\sqrt {2}})^{2}}{4}}\cdot a_{0}^{2}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow A_{1}+A_{2}=\pi \cdot {\frac {(2-{\sqrt {2}})^{2}}{2}}\cdot a_{0}^{2}\quad }$, reducing
${\displaystyle \quad \Leftrightarrow A_{1}+A_{2}=\pi \cdot {\frac {(4-2\cdot 2{\sqrt {2}}+2)}{2}}\cdot a_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow A_{1}+A_{2}=\pi \cdot {\frac {(6-4{\sqrt {2}})}{2}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}+A_{2}=\pi \cdot (3-2{\sqrt {2}})\cdot a_{0}^{2}=\pi \cdot (3-2{\sqrt {2}})\cdot A_{0}\quad \approx 0.539A_{0}\quad }$reducing

${\displaystyle \quad S_{1}=S_{0}+{\overrightarrow {S_{0}S_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+{\overrightarrow {S_{0}A}}+{\overrightarrow {AS_{1}}}\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\overrightarrow {AS_{0}}}+{\overrightarrow {AS_{1}}}\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=0-({\frac {1}{2}}\cdot a_{0}-{\frac {1}{2}}\cdot a_{0}\cdot i)+{\overrightarrow {AS_{1}}}\quad }$, because ${\displaystyle S_{0}}$ is the centre point of the square
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{2}}\cdot (a_{0}-a_{0}\cdot i)+{\overrightarrow {AS_{1}}}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{2}}\cdot (1-i)\cdot a_{0}+{\overrightarrow {AS_{1}}}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{2}}\cdot (1-i)\cdot a_{0}+(|AE|-|ES_{1}|\cdot i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{2}}\cdot (1-i)\cdot a_{0}+(r_{1}-r_{1}\cdot i)\quad }$, applying equation 2
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{2}}\cdot (1-i)\cdot a_{0}+(1-i)\cdot r_{1}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\frac {1}{2}}\cdot (1-i)\cdot a_{0}+(1-i)\cdot {\frac {2-{\sqrt {2}}}{2}}\cdot a_{0}\quad }$, applying Calculation 1
${\displaystyle \quad \Leftrightarrow S_{1}=0+{\frac {2-{\sqrt {2}}}{2}}\cdot (1-i)\cdot a_{0}-{\frac {1}{2}}\cdot (1-i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {2-{\sqrt {2}}}{2}}-{\frac {1}{2}}\right)\cdot (1-i)\cdot a_{0}\quad }$, factoring out
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {2-{\sqrt {2}}-1}{2}}\right)\cdot (1-i)\cdot a_{0}\quad }$, adding
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {1-{\sqrt {2}}}{2}}\right)\cdot (1-i)\cdot a_{0}\quad }$, adding
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left(-{\frac {{\sqrt {2}}-1}{2}}\right)\cdot (1-i)\cdot a_{0}\quad }$, adding
${\displaystyle \quad \Leftrightarrow S_{1}=\left(-{\frac {{\sqrt {2}}-1}{2}}+{\frac {{\sqrt {2}}-1}{2}}i\right)\cdot a_{0}\quad \approx (-0.207+0.207i)\cdot a_{0}\quad }$, factoring in

I, the copyright holder of this work, hereby publish it under the following license:

This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.

You are free:

• to share – to copy, distribute and transmit the work
• to remix – to adapt the work

Under the following conditions:

• attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
• share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original.

https://creativecommons.org/licenses/by-sa/4.0CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 truetrue