File:FS HCC dia.png

The semicircle as base element. And the largest inscribed circle. And the next largest inscribed circle.

0) The radius of the semicircle around ${\displaystyle D:r_{0}}$
1) The radius of the inscribed circle around ${\displaystyle S_{1}:r_{1}={\frac {1}{2}}r_{0}}$
2) The radius of the next inscribed circle around ${\displaystyle S_{2}:r_{2}={\frac {1}{4}}r_{0}\quad }$, see calculation 1

0) Perimeter of base semicircle: ${\displaystyle P_{0}=2\cdot r_{0}+\pi \cdot r_{0}=(2+\pi )\cdot r_{0}\quad \approx 5.142\cdot r_{0}}$
1) Perimeter of inscribed circle: ${\displaystyle P_{1}=2\cdot \pi \cdot r_{1}=\pi \cdot r_{0}\quad \approx 3.142\cdot r_{0}}$
2) Perimeter of next inscribed circle: ${\displaystyle P_{2}=2\cdot \pi \cdot r_{2}={\frac {\pi }{2}}\cdot r_{0}\quad \approx 1.571\cdot r_{0}}$

0) Area of the base semicircle ${\displaystyle A_{0}={\frac {\pi \cdot r_{0}^{2}}{2}}}$
1) Area of the inscribed circle ${\displaystyle A_{1}=\pi \cdot r_{1}^{2}={\frac {\pi \cdot r_{0}^{2}}{4}}={\frac {A_{0}}{2}}}$
2) Area of the next inscribed circle ${\displaystyle A_{2}=\pi \cdot r_{2}^{2}={\frac {\pi \cdot r_{0}^{2}}{16}}={\frac {A_{0}}{8}}}$

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid of the inscribed circle relative to the base centroid is: ${\displaystyle S_{1}=0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i}$, see Calculation 2
2) The centroid of the next inscribed circle relative to the base centroid is: ${\displaystyle S_{2}=\left({\sqrt {\frac {1}{2}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot i\right)\cdot r_{0}\quad }$, see Calculation 3

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

0) Radius of the base semicircle ${\displaystyle r_{0}={\sqrt {\frac {2}{\pi }}}\quad \approx 0.798}$
1) Radius of the inscribed circle ${\displaystyle r_{1}={\frac {1}{2}}{\sqrt {\frac {2}{\pi }}}\quad \approx 0.399}$
2) Radius of the next inscribed circle ${\displaystyle r_{2}={\frac {1}{4}}{\sqrt {\frac {2}{\pi }}}\quad \approx 0.199}$

0) Perimeter of base semicircle: ${\displaystyle P_{0}=(2+\pi )\cdot r_{0}=(2+\pi )\cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 4.1023974}$
1) Perimeter of inscribed circle: ${\displaystyle P_{1}=\pi \cdot {\sqrt {\frac {2}{\pi }}}\quad \approx 2.5066283}$
2) Perimeter of next inscribed circle: ${\displaystyle P_{2}={\frac {\pi }{2}}\cdot {\sqrt {\frac {2}{\pi }}}={\sqrt {\frac {2\pi ^{2}}{4\pi }}}={\sqrt {\frac {\pi }{2}}}\quad \approx 1.2533141}$

S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}\quad \approx 7.8623398}$

0) Area of the base semicircle is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed circle ${\displaystyle A_{1}={\frac {A_{0}}{2}}=0.500}$
2) Area of the next inscribed circle ${\displaystyle A_{2}={\frac {A_{0}}{8}}=0.125}$

0) ${\displaystyle S_{0}=0+0i}$
1) ${\displaystyle S_{1}=0+{\frac {3\pi -8}{6\pi }}\cdot {\sqrt {\frac {2}{\pi }}}\cdot i\quad \approx 0+0.060i}$
2) ${\displaystyle S_{2}=\left({\sqrt {\frac {1}{2}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot i\right)\cdot {\sqrt {\frac {2}{\pi }}}={\sqrt {\frac {1}{\pi }}}+\left({\frac {3\pi -16}{12\pi }}\cdot {\sqrt {\frac {2}{\pi }}}\right)\cdot i\quad \approx 0.564-0.139i}$

The distance between the centroid of the base semicircle and the centroid of the circle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}\quad \approx 0.0603096}$
${\displaystyle d_{02}={\sqrt {(x_{0}-x_{2})^{2}+(y_{0}-y_{2})^{2}}}\quad \approx 0.5810988}$
${\displaystyle d_{12}={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}\quad \approx 0.5984134}$
Sum of distances: ${\displaystyle d_{s}=d_{01}+d_{02}+d_{12}\quad \approx 1.2398218}$

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.

${\displaystyle decimalpart(7.8623398+1.2398218)=decimalpart(9.1021616)=.1021616}$

So the identifying number is: ${\displaystyle 2.1021616}$

(0) Segments of given length:
${\displaystyle \quad |DB|=|DF|=|DC|=|DS_{1}|+|CS_{1}|=r_{0}}$
${\displaystyle \quad |DS_{1}|=|HS_{1}|=|CS_{1}|=r_{1}={\frac {1}{2}}r_{0}}$
${\displaystyle \quad |ES_{2}|=|HS_{2}|=|FS_{2}|=r_{2}}$

(1) ${\displaystyle \quad {\overline {DHS_{0}DS_{1}}}}$ is perpendicular on ${\displaystyle {\overline {AB}}}$ since the circle around ${\displaystyle S_{1}}$ is tangent to ${\displaystyle {\overline {AB}}}$
and ${\displaystyle \quad {\overline {ES_{2}}}}$ is perpendicular on ${\displaystyle {\overline {AB}}}$ since the circle around ${\displaystyle S_{2}}$ is tangent to ${\displaystyle {\overline {AB}}}$
${\displaystyle \quad \rightarrow |DE|=|HS_{2}|\quad }$and ${\displaystyle |ES_{2}|=|DH|=r_{2}}$

(2) ${\displaystyle \quad {\overline {DS_{2}F}}}$ is perpendicular in ${\displaystyle F}$ since the circle around ${\displaystyle S_{2}}$ is tangent to the semicircle around ${\displaystyle D}$ with ${\displaystyle \quad |DF|=r_{0}\quad }$ and ${\displaystyle \quad |FS_{2}|=r_{2}}$
${\displaystyle \quad \rightarrow |DS_{2}|=r_{0}-r_{2}\quad }$

(3) ${\displaystyle \quad {\overline {S_{1}GS_{2}}}}$ is perpendicular in ${\displaystyle G}$ since the circle around ${\displaystyle S_{2}}$ is tangent to the circle around ${\displaystyle S_{1}}$ with ${\displaystyle \quad |GS_{1}|=r_{1}\quad }$ and ${\displaystyle \quad |GS_{2}|=r_{2}}$
${\displaystyle \quad \rightarrow |S_{1}S_{2}|=r_{1}+r_{2}\quad }$

(4) ${\displaystyle \quad |DHS_{2}|=r_{1}}$
${\displaystyle \quad \Leftrightarrow \quad |DH|+|HS_{1}|=r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow \quad r_{2}+|HS_{1}|=r_{1}\quad }$, applying eqaution (1)
${\displaystyle \quad \Leftrightarrow \quad |HS_{1}|=r_{1}-r_{2}\quad }$

(5) ${\displaystyle \quad |DE|^{2}+|ES_{2}|^{2}=|DS_{2}|^{2}}$, applying the Pythagorean theorem
${\displaystyle \quad \Leftrightarrow |DE|^{2}+r_{2}^{2}=|ES_{2}|^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |DE|^{2}+r_{2}^{2}=(r_{0}-r_{2})^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow |DE|^{2}+r_{2}^{2}=r_{0}^{2}+r_{2}^{2}-2r_{0}r_{2}\quad }$, applying binomial theorem
${\displaystyle \quad \Leftrightarrow |DE|^{2}=r_{0}^{2}-2r_{0}r_{2}\quad }$, eliminating ${\displaystyle r_{2}^{2}}$ on both sides

(6) ${\displaystyle \quad |HS_{2}|^{2}+|HS_{1}|^{2}=|S_{1}S_{2}|^{2}}$, applying the Pythagorean theorem
${\displaystyle \quad \Leftrightarrow |DE|^{2}+|HS_{1}|^{2}=|S_{1}S_{2}|^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}+|HS_{1}|^{2}=|S_{1}S_{2}|^{2}\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}+(r_{1}-r_{2})^{2}=|S_{1}S_{2}|^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}+(r_{1}-r_{2})^{2}=(r_{1}+r_{2})^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}+(r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2})=(r_{1}+r_{2})^{2}\quad }$, applying binomial theorem
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}+r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}=r_{1}^{2}+r_{2}^{2}+2r_{1}r_{2}\quad }$, applying binomial theorem
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}-2r_{1}r_{2}=2r_{1}r_{2}\quad }$, eliminating ${\displaystyle r_{1}^{2}+r_{2}^{2}}$ on both sides
${\displaystyle \quad \Leftrightarrow r_{0}^{2}-2r_{0}r_{2}=4r_{1}r_{2}\quad }$, adding ${\displaystyle 2r_{1}r_{2}}$ on both sides
${\displaystyle \quad \Leftrightarrow r_{0}^{2}=4r_{1}r_{2}+2r_{0}r_{2}\quad }$, adding ${\displaystyle 2r_{0}r_{2}}$ on both sides
${\displaystyle \quad \Leftrightarrow r_{0}^{2}=(4r_{1}+2r_{0})\cdot r_{2}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {r_{0}^{2}}{4r_{1}+2r_{0}}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {r_{0}^{2}}{4{\frac {1}{2}}r_{0}+2r_{0}}}\quad }$, since ${\displaystyle r_{1}={\frac {1}{2}}r_{0}}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {r_{0}^{2}}{2r_{0}+2r_{0}}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {r_{0}^{2}}{4r_{0}}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {1}{4}}\cdot r_{0}\quad }$

${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}D}}+{\overrightarrow {DS_{1}}}}$
${\displaystyle \Leftrightarrow S_{1}=(0+0i)-\left(0+\left({\frac {4\cdot r_{0}}{3\cdot \pi }}\right)\cdot i\right)+(0+r_{1}\cdot i)}$
${\displaystyle \Leftrightarrow S_{1}=0-\left({\frac {4}{3\cdot \pi }}\right)\cdot r_{0}\cdot i+r_{1}\cdot i}$
${\displaystyle \Leftrightarrow S_{1}=0+\left({\frac {1}{2}}-{\frac {4}{3\cdot \pi }}\right)\cdot r_{0}\cdot i}$
${\displaystyle \Leftrightarrow S_{1}=0+\left({\frac {3\pi }{6\pi }}-{\frac {8}{6\pi }}\right)\cdot r_{0}\cdot i}$
${\displaystyle \Leftrightarrow S_{1}=0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i}$

${\displaystyle S_{2}=S_{1}+{\overrightarrow {S_{1}H}}+{\overrightarrow {HS_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i\right)+{\overrightarrow {S_{1}H}}+{\overrightarrow {HS_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i\right)-\left(0+(r_{1}-r_{2})\cdot i\right)+{\overrightarrow {HS_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i\right)-\left(0+(r_{1}-r_{2})\cdot i\right)+\left(|HS_{2}|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i\right)-\left(0+(r_{1}-r_{2})\cdot i\right)+\left(|DE|+0i\right)\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i\right)-\left(0+({\frac {1}{2}}r_{0}-{\frac {1}{4}}r_{0})\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i\right)-\left(0+{\frac {1}{4}}\cdot r_{0}\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+{\frac {3\pi -8}{6\pi }}\cdot r_{0}\cdot i-{\frac {1}{4}}\cdot r_{0}\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+\left({\frac {3\pi -8}{6\pi }}-{\frac {1}{4}}\right)\cdot r_{0}\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+\left({\frac {12\pi -32-6\pi }{24\pi }}\right)\cdot r_{0}\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+\left({\frac {6\pi -32}{24\pi }}\right)\cdot r_{0}\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(0+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)+\left(|DE|+0i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left(|DE|+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)}$
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {r_{0}^{2}-2r_{0}r_{2}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)\quad }$,applying equation (5)
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {r_{0}^{2}-2r_{0}{\frac {1}{4}}r_{0}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {r_{0}^{2}-{\frac {1}{2}}r_{0}^{2}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)\quad }$
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {{\frac {1}{2}}r_{0}^{2}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)\quad }$
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {\frac {1}{2}}}\cdot r_{0}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot r_{0}\cdot i\right)\quad }$
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {\frac {1}{2}}}+\left({\frac {3\pi -16}{12\pi }}\right)\cdot i\right)\cdot r_{0}\quad }$

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