File:FS QH dia.png

Shows the largest semicircle within a square.

0) The side length of the square: ${\displaystyle =a_{0}}$
1) Radius of the semicircle ${\displaystyle r_{1}=(2-{\sqrt {2}})\cdot a_{0}\quad \approx 0.586\cdot a_{0}\quad }$see Calculation 1

0) Perimeter of base square ${\displaystyle P_{0}=4\cdot a_{0}}$
1) Perimeter of the semicircle ${\displaystyle P_{1}=2\cdot r_{1}+\pi \cdot r_{1}=(2+\pi )\cdot r_{1}=(2+\pi )\cdot (2-{\sqrt {2}})\cdot a_{0}\quad \approx 3.012\cdot a_{0}\quad }$

0) Area of the base square ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the semicircle ${\displaystyle A_{1}=\pi \cdot (3-2{\sqrt {2}})\cdot a_{0}^{2}\quad \approx 0.539\cdot a_{0}^{2}\quad }$, see Calculation 2

Centroid positions are measured from the centroid point ${\displaystyle S_{0}}$ of the base shape.
0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the semicircle: ${\displaystyle S_{1}=\left({\frac {12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi }}\right)\cdot a_{0}+0i\quad \approx 0.127a_{0}+0i\quad }$, see Calculation 3

Measured from point ${\displaystyle A}$ the positions are:
0) ${\displaystyle {\overrightarrow {AS_{0}}}={\frac {a_{0}}{2}}\cdot (1+i)\quad =0.5\cdot a_{0}\cdot (1+i)\quad }$, see Calculation 4
1) ${\displaystyle {\overrightarrow {AS_{1}}}=\left({\frac {6\pi +8-3\pi {\sqrt {2}}-4{\sqrt {2}}}{3\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad \approx 0.59\cdot a_{0}\cdot (1+i)\quad }$, see Calculation 5

In the normalised case the area of the base is set to 1.
${\displaystyle ||ABCD||=1\Rightarrow a_{0}^{2}=1\Rightarrow a_{0}=1}$

0) Segment of the base square ${\displaystyle a_{0}=1}$
1) Segment of the semicircle ${\displaystyle r_{1}=(2-{\sqrt {2}})\quad \approx 0.586}$

0) Perimeter of base square ${\displaystyle P_{0}=4}$
1) Perimeter of the semicircle ${\displaystyle P_{1}=(2+\pi )(2-{\sqrt {2}})\quad \approx 3.0118752}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}\approx 7.0118752}$

0) Area of the base square ${\displaystyle A_{0}=1}$
1) Area of the semicircle ${\displaystyle A_{1}=\pi \cdot (3-2{\sqrt {2}})\approx 0.539}$

Centroid positions are measured from the centroids of the base shape
0) Centroid positions of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid positions of the semicircle: ${\displaystyle S_{1}\approx 0.127295142+0i}$

The distance between the centroid of the base element and the centroid of the quarter circle is:
${\displaystyle d=0.127295142}$

Apart of the base element there is only 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.

${\displaystyle decimalpart(7.0118752+0.127295142)=decimalpart(7.1391704)=.1391704}$

So the identifying number is: ${\displaystyle 1.1391704}$

Base is the square ${\displaystyle [ACEJ]}$ of side length ${\displaystyle a_{0}}$.
This means that the following equations hold
(0.1)${\displaystyle \quad |AC|=|CE|=|EJ|=|AJ|=a_{0}}$
(0.2)${\displaystyle \quad |AE|={\sqrt {2}}\cdot a_{0}\quad }$, applying the Pythagorean theorem to the triangle ${\displaystyle \triangle ACE}$
For the semicircle ${\displaystyle [BDHFGK]}$ the following equations hold:
(0.3) ${\displaystyle \quad |KB|=|KD|=|KH|=|KF|=|KG|=r_{1}}$

In order to find the radius ${\displaystyle r_{1}}$ of the semicircle the following calculations have to be done:

Considering the square ${\displaystyle [ACEJ]}$ We get the equation:
(1) ${\displaystyle |AE|=|AK|+|KE|={\sqrt {2}}\cdot a_{0}}$

Since ${\displaystyle |KF|=|KD|=r_{1}}$ the rectangle ${\displaystyle [KDEF]}$ is a square with side length ${\displaystyle r_{1}}$. This leads to the equation:
(2) ${\displaystyle |KE|={\sqrt {2}}\cdot r_{1}}$

The line segment ${\displaystyle [BG]}$ is the diameter of the semicircle and has the length: ${\displaystyle |BG|=2\cdot r_{1}}$. The line segment ${\displaystyle [AB]}$ has length ${\displaystyle b}$. For symmetry reasons the line segment ${\displaystyle [AG]}$ has the same length, so ${\displaystyle |AB|=|AG|=b}$. Using the Pythagorean theorem we get equation:
(3) ${\displaystyle |AB|^{2}+|AG|^{2}=|BG|^{2}}$
${\displaystyle \Rightarrow b^{2}+b^{2}=(2\cdot r_{1})^{2}}$
${\displaystyle \Rightarrow 2\cdot b^{2}=4\cdot r_{1}^{2}}$
${\displaystyle \Rightarrow b^{2}=2\cdot r_{1}^{2}}$
${\displaystyle \Rightarrow b={\sqrt {2}}\cdot r_{1}}$

Applying the Pythagorean theorem to the triangle ${\displaystyle \triangle ABK}$ we get the equation
(4) ${\displaystyle |AK|^{2}+r_{1}^{2}=b^{2}}$
${\displaystyle \quad \Leftrightarrow |AK|^{2}+r_{1}^{2}=\left({\sqrt {2}}\cdot r_{1}\right)^{2}\quad }$applying equation (3)
${\displaystyle \quad \Leftrightarrow |AK|^{2}+r_{1}^{2}=2\cdot r_{1}^{2}\quad }$
${\displaystyle \quad \Leftrightarrow |AK|^{2}=2\cdot r_{1}^{2}-r_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow |AK|^{2}=r_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow |AK|=r_{1}}$

Now we use this result together with equations (1) and (2).
${\displaystyle \quad |AK|+|KE|={\sqrt {2}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow r_{1}+|KE|={\sqrt {2}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow r_{1}+{\sqrt {2}}\cdot r_{1}={\sqrt {2}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow r_{1}\cdot (1+{\sqrt {2}})={\sqrt {2}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{1+{\sqrt {2}}}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{1+{\sqrt {2}}}}\cdot {\frac {1-{\sqrt {2}}}{1-{\sqrt {2}}}}\cdot a_{0}={\frac {{\sqrt {2}}-2}{1-{\sqrt {2}}+{\sqrt {2}}-2}}\cdot s={\frac {{\sqrt {2}}-2}{-1}}\cdot a_{0}}$

${\displaystyle \quad \Leftrightarrow r_{1}=(2-{\sqrt {2}})\cdot a_{0}\quad \approx 0.586\cdot a_{0}}$

${\displaystyle \quad A_{1}={\frac {\pi \cdot r_{1}^{2}}{2}}\quad }$a semicircle has half the area of a circle
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{2}}\cdot \left((2-{\sqrt {2}})\cdot a_{0}\right)^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{2}}\cdot (2-{\sqrt {2}})^{2}\cdot a_{0}^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{2}}\cdot (4+2-2\cdot 2\cdot {\sqrt {2}})\cdot a_{0}^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{2}}\cdot (6-4\cdot {\sqrt {2}})\cdot a_{0}^{2}\quad \approx 0.539\cdot a_{0}^{2}}$

If the center of the radius of the semicircle ${\displaystyle K}$ were positioned on ${\displaystyle 0+0i}$ and the diameter ${\displaystyle {\overrightarrow {BG}}}$ were parallel to the y-axis then the centroid position would be ${\displaystyle |KS_{1}|={\frac {4\cdot r_{1}}{3\cdot \pi }}+0i}$
. ${\displaystyle \quad S_{1}=S_{0}+{\overrightarrow {S_{0}S_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-|KS_{0}|+|KS_{1}|}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-|KS_{0}|+{\frac {4\cdot r_{1}}{3\cdot \pi }}+0i}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-\left(|AS_{0}|-|AK|\right)+{\frac {4\cdot r_{1}}{3\cdot \pi }}+0i}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-|AS_{0}|+|AK|+{\frac {4\cdot r_{1}}{3\cdot \pi }}+0i}$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-|AS_{0}|+r_{1}+{\frac {4\cdot r_{1}}{3\cdot \pi }}+0i\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}+r_{1}+{\frac {4\cdot r_{1}}{3\cdot \pi }}+0i\quad }$, since ${\displaystyle S_{0}}$ is the centroid of the square and ${\displaystyle {\overrightarrow {AE}}}$ its diagonale
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}+r_{1}\left(1+{\frac {4}{3\cdot \pi }}\right)+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}+\left(2-{\sqrt {2}}\right)\cdot a_{0}\cdot \left(1+{\frac {4}{3\cdot \pi }}\right)+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left(\left(2-{\sqrt {2}}\right)\cdot \left(1+{\frac {4}{3\cdot \pi }}\right)-{\frac {\sqrt {2}}{2}}\right)\cdot a_{0}+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left(\left(2-{\sqrt {2}}\right)\cdot \left({\frac {3\pi +4}{3\cdot \pi }}\right)-{\frac {\sqrt {2}}{2}}\right)\cdot a_{0}+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left({\frac {6\pi +8-3\pi {\sqrt {2}}-4{\sqrt {2}}}{3\cdot \pi }}-{\frac {\sqrt {2}}{2}}\right)\cdot a_{0}+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left({\frac {12\pi +16-6\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi }}-{\frac {3\pi {\sqrt {2}}}{6\pi }}\right)\cdot a_{0}+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=S_{0}+\left({\frac {12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi }}\right)\cdot a_{0}+0i\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=0+0i+\left({\frac {12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi }}\right)\cdot a_{0}+0i\quad }$, definition of S_0
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi }}\right)\cdot a_{0}+0i\quad \approx 0.127a_{0}+0i<}$

${\displaystyle \quad {\overrightarrow {AS_{0}}}={\frac {a_{0}}{2}}+{\frac {a_{0}}{2}}i\quad }$,since ${\displaystyle S_{0}}$ is the center point of the square
${\displaystyle \quad \Leftrightarrow {\overrightarrow {AS_{0}}}={\frac {a_{0}}{2}}\cdot (1+i)\quad =0.5\cdot a_{0}\cdot (1+i)}$

${\displaystyle \quad {\overrightarrow {AS_{1}}}={\overrightarrow {AS_{0}}}+{\overrightarrow {S_{0}S_{1}}}\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}={\frac {a_{0}}{2}}\cdot (1+i)+{\overrightarrow {S_{0}S_{1}}}\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}={\frac {a_{0}}{2}}\cdot (1+i)+|S_{0}S_{1}|\cdot \cos {45}+|S_{0}S_{1}|\cdot \sin {45}\cdot i\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}={\frac {a_{0}}{2}}\cdot (1+i)+|S_{0}S_{1}|\cdot {\frac {1}{\sqrt {2}}}+|S_{0}S_{1}|\cdot {\frac {1}{\sqrt {2}}}\cdot i\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}={\frac {a_{0}}{2}}\cdot (1+i)+\left(|S_{0}S_{1}|\cdot {\frac {1}{\sqrt {2}}}\right)\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {a_{0}}{2}}+|S_{0}S_{1}|\cdot {\frac {1}{\sqrt {2}}}\right)\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {a_{0}}{2}}+\left({\frac {12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi }}\right)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\right)\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {1}{2}}+{\frac {12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {6\pi {\sqrt {2}}}{2\cdot 6\pi {\sqrt {2}}}}+{\frac {24\pi +32-18\pi {\sqrt {2}}-16{\sqrt {2}}}{2\cdot 6\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {6\pi {\sqrt {2}}+24\pi +32-18\pi {\sqrt {2}}-16{\sqrt {2}}}{2\cdot 6\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {3\pi {\sqrt {2}}+12\pi +16-9\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {12\pi +16-6\pi {\sqrt {2}}-8{\sqrt {2}}}{6\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad }$
${\displaystyle \quad {\overrightarrow {AS_{1}}}=\left({\frac {6\pi +8-3\pi {\sqrt {2}}-4{\sqrt {2}}}{3\pi {\sqrt {2}}}}\right)\cdot a_{0}\cdot (1+i)\quad \approx 0.59\cdot a_{0}\cdot (1+i)}$

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