Summary edit

0) The 120°-degree circular sector (fan) as base element.
1) Inscribed is the broadest isosceles triangle.
2) Inscribed is the largest circle.
3) Inscribed is the largest 120°-degree circular sector (fan).
4) Inscribed is the next largest circle.

General case edit

Segments in the general case edit

0) Radius of the base circular sector: ${\displaystyle r_{0}}$
1) side length of the inscribed triangle: ${\displaystyle a_{1}=r_{0}}$, because it is the broadest triangle
2) Radius of the inscribed circle: ${\displaystyle r_{2}={\frac {1}{4}}\cdot r_{0}}$, see calculation (5)
3) Radius of the inscribed circular sector: ${\displaystyle r_{3}=|DM|={\frac {1}{2}}\cdot r_{0}\quad }$, applying calculation (1)
4) Radius of the inscribed circle: ${\displaystyle r_{4}={\frac {3}{16}}\cdot r_{0}\quad }$, see calculation (8)

Perimeters in the general case edit

0) Perimeter of base circular sector: ${\displaystyle P_{0}=|AM|+|MB|+|BCA|=2\cdot r_{0}+{\frac {2\pi }{3}}\cdot r_{0}=2\cdot (1+{\frac {\pi }{3}})\cdot r_{0}\quad \approx 4.094\cdot r_{0}}$
1) Perimeter of inscribed triangle: ${\displaystyle P_{1}=|AM|+|MB|+|BDA|=2\cdot r_{0}+{\sqrt {3}}\cdot r_{0}\quad \approx 3.732\cdot r_{0}}$
2) Perimeter of inscribed circle around ${\displaystyle S_{2}}$: ${\displaystyle P_{2}=2\pi r_{2}=2\pi \cdot {\frac {1}{4}}\cdot r_{0}={\frac {\pi }{2}}\cdot r_{0}\quad \approx 1.571\cdot r_{0}}$
3) Perimeter of inscribed circular sector: ${\displaystyle P_{3}=2\cdot (1+{\frac {\pi }{3}})\cdot r_{3}=2\cdot (1+{\frac {\pi }{3}})\cdot {\frac {1}{2}}\cdot r_{0}=(1+{\frac {\pi }{3}})\cdot r_{0}\quad \approx 2.047\cdot r_{0}}$
4) Perimeter of inscribed circle around ${\displaystyle S_{4}}$: ${\displaystyle P_{4}=2\pi r_{4}=2\pi \cdot {\frac {3}{16}}\cdot r_{0}={\frac {3\pi }{8}}\cdot r_{0}\quad \approx 1.178\cdot r_{0}}$

Areas in the general case edit

0) Area of the base circular sector ${\displaystyle A_{0}={\frac {\pi }{3}}\cdot r_{0}^{2}\quad \approx 1.047\cdot r_{0}^{2}}$
1) Area of the inscribed triangle ${\displaystyle A_{1}={\frac {\sqrt {3}}{4}}\cdot r_{0}^{2}\quad \approx 0.413\cdot r_{0}^{2}\quad }$, see calculation (3)
2) Area of the inscribed circle around ${\displaystyle S_{2}}$: ${\displaystyle A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot \left({\frac {1}{4}}\cdot r_{0}\right)^{2}={\frac {\pi }{16}}\cdot r_{0}^{2}\quad \approx 0.196\cdot r_{0}^{2}}$
3) Area of the inscribed circular sector ${\displaystyle A_{3}={\frac {\pi }{3}}\cdot r_{3}^{2}={\frac {\pi }{3}}\cdot \left({\frac {1}{2}}\cdot r_{0}\right)^{2}={\frac {\pi }{12}}\cdot r_{0}^{2}\quad \approx 0.262\cdot r_{0}^{2}}$
4) Area of the inscribed circle around ${\displaystyle S_{4}}$:${\displaystyle \quad A_{4}=\pi \cdot r_{4}^{2}=\pi \cdot \left({\frac {3}{16}}\cdot r_{0}\right)^{2}={\frac {9\pi }{256}}\cdot r_{0}^{2}\quad \approx 0.11\cdot r_{0}^{2}}$

Centroids in the general case edit

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid of the inscribed triangle relative to the base centroid is: ${\displaystyle S_{1}=\left(0+{\frac {\pi -3{\sqrt {3}}}{3\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.218i)\cdot r_{0}\quad }$, see Calculation (4)
2) The centroid of the inscribed circle relative to the base centroid is: ${\displaystyle S_{2}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0+0.199i)\cdot r_{0}\quad }$, see Calculation (6)
3) The centroid of the inscribed circular sector relative to the base centroid is: ${\displaystyle S_{3}=\left(0-{\frac {\sqrt {3}}{2\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.276i)\cdot r_{0}\quad }$, see Calculation (7)
4) The centroid of the inscribed circle relative to the base centroid is: ${\displaystyle S_{4}=\left({\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0.433+0.136i)\cdot r_{0}\quad }$, see Calculation (9)

Normalised case edit

In the normalised case the area of the base circular sector is set to 1.
So ${\displaystyle A_{0}={\frac {\pi \cdot r_{0}^{2}}{3}}=1\quad \Rightarrow r_{0}^{2}={\frac {3}{\pi }}\quad \Rightarrow r_{0}={\sqrt {\frac {3}{\pi }}}}$

Segments in the normalised case edit

0) Radius of the base circular sector: ${\displaystyle r_{0}={\sqrt {\frac {3}{\pi }}}\quad \approx 0.977}$
1) Side length of the inscribed triangle: ${\displaystyle a_{1}=r_{0}={\sqrt {\frac {3}{\pi }}}\quad \approx 0.977}$
2) Radius of the inscribed circle around ${\displaystyle S_{2}}$ : ${\displaystyle r_{2}={\frac {1}{4}}\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\frac {3}{16\pi }}}\quad \approx 0.244}$
3) Radius of the inscribed circular sector: ${\displaystyle r_{3}={\frac {1}{2}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 0.489}$
4) Radius of the inscribed circle around ${\displaystyle S_{4}}$ : ${\displaystyle r_{4}={\frac {3}{16}}\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\frac {27}{256\pi }}}\quad \approx 0.183}$

Perimeter in the normalised case edit

0) Perimeter of base circular sector: ${\displaystyle P_{0}=2\cdot (1+{\frac {\pi }{3}})\cdot {\sqrt {\frac {3}{\pi }}}\approx 4.0010635}$
1) Perimeter of inscribed triangle: ${\displaystyle P_{1}=(2+{\sqrt {3}})\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 3.6469788}$
2) Perimeter of inscribed circle around ${\displaystyle S_{2}}$: ${\displaystyle P_{2}={\frac {\pi }{2}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 1.5349901}$
3) Perimeter of inscribed circular sector: ${\displaystyle P_{3}=(1+{\frac {\pi }{3}})\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 2.0005317}$
4) Perimeter of inscribed circle around ${\displaystyle S_{4}}$: ${\displaystyle P_{4}={\frac {3\pi }{8}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 1.151242546}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}+P_{3}+P_{4}\quad \approx 12.3348066}$

Area in the normalised case edit

0) Area of the base circular sector is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle ${\displaystyle A_{1}={\frac {\sqrt {3}}{4}}\cdot {\frac {3}{\pi }}\quad \approx 0.413}$
2) Area of the inscribed circle around ${\displaystyle S_{2}}$: ${\displaystyle A_{2}={\frac {\pi }{16}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {3}{16}}\quad \approx 0.188}$
3) Area of the inscribed circular sector ${\displaystyle A_{3}={\frac {\pi }{12}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {3}{12}}=0.250}$
4) Area of the inscribed circle around ${\displaystyle S_{4}}$:${\displaystyle \quad A_{4}={\frac {9\pi }{256}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {27}{256}}\quad \approx 0.105}$

Centroids in the normalised case edit

0) Centroid of the base shape: ${\displaystyle S_{0}=0+0i}$
1) Centroid of the inscribed triangle: ${\displaystyle S_{1}=\left(0+{\frac {\pi -3{\sqrt {3}}}{3\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 0-0.213i}$
2) Centroid of the inscribed circle around ${\displaystyle S_{2}}$: ${\displaystyle S_{2}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx (0+0.194i)\quad }$
3) Centroid of the inscribed circular sector: ${\displaystyle S_{3}=\left(0-{\frac {\sqrt {3}}{2\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}=\left(0-{\frac {3}{2\pi {\sqrt {\pi }}}}\cdot i\right)\quad \approx (0-0.269i)\quad }$
4) The centroid of the inscribed circle around ${\displaystyle S_{4}}$: ${\displaystyle S_{4}=\left({\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}=\left({\sqrt {\frac {9}{16\pi }}}+{\frac {11\pi {\sqrt {3}}-48}{16\pi {\sqrt {\pi }}}}\cdot i\right)\quad \approx (0.423+0.133i)}$

Calculations edit

Given elements edit

(1) ${\displaystyle |AM|=|BM|=|MC|=|MG|=r_{0}=a_{1}}$
(2) Angle in M:${\displaystyle \quad \angle \phi =120^{\circ }={\frac {2\pi }{3}}}$
(3) Angles in A and B of triangle ${\displaystyle \triangle {AMB}}$:${\displaystyle \quad \angle \alpha ={\frac {180^{\circ }-120^{\circ }}{2}}=30^{\circ }={\frac {\pi }{6}}}$, since it is a isosceles triangle
(4) ${\displaystyle |AB|=2\cdot |BD|}$, since the isosceles triangle is symmetric
(5) ${\displaystyle |CD|=|CS_{2}|+|S_{2}D|=2\cdot r_{2}}$
(6) ${\displaystyle |MD|=r_{3}}$
(7) ${\displaystyle |DF|=|ES_{4}|=r_{4}}$, since ${\displaystyle {\overline {FS_{4}}}}$ is perpendicular to ${\displaystyle {\overline {CD}}}$ and therefore parallel to ${\displaystyle {\overline {DE}}}$

Calculation 1 edit

Calculating length of MD
${\displaystyle \sin {30^{\circ }}={\frac {|MD|}{|MB|}}\quad }$, applying equation 3 and the definition of the sinus
${\displaystyle \Leftrightarrow {\frac {1}{2}}={\frac {|MD|}{|MB|}}\quad }$, calculating the sine
${\displaystyle \Leftrightarrow {\frac {1}{2}}={\frac {|MD|}{r_{0}}}\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow {\frac {1}{2}}\cdot r_{0}=|MD|\quad }$, rearranging

Calculation 2 edit

Calculating length of AB
${\displaystyle \cos {30^{\circ }}={\frac {|BD|}{|MB|}}\quad }$, applying equation 3 and the definition of the cosinus
${\displaystyle \Leftrightarrow {\frac {\sqrt {3}}{2}}={\frac {|BD|}{|MB|}}\quad }$, calculating the sine
${\displaystyle \Leftrightarrow {\frac {\sqrt {3}}{2}}={\frac {|BD|}{r_{0}}}\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow {\frac {\sqrt {3}}{2}}\cdot r_{0}=|BD|\quad }$, rearranging
${\displaystyle \Leftrightarrow {\frac {\sqrt {3}}{2}}\cdot r_{0}={\frac {1}{2}}\cdot |AB|\quad }$, applying equation (4)
${\displaystyle \Leftrightarrow |AB|={\sqrt {3}}\cdot r_{0}\quad }$, rearranging

Calculation 3 edit

${\displaystyle A_{1}=|MD|\cdot |BD|\quad }$, calculating the area of triangle ${\displaystyle \triangle {AMB}}$
${\displaystyle \Leftrightarrow A_{1}={\frac {1}{2}}\cdot r_{0}\cdot |BD|\quad }$, applying calculation (1)
${\displaystyle \Leftrightarrow A_{1}={\frac {1}{2}}\cdot r_{0}\cdot {\frac {\sqrt {3}}{2}}\cdot r_{0}\quad }$, applying calculation (2)
${\displaystyle \Leftrightarrow A_{1}={\frac {\sqrt {3}}{4}}\cdot r_{0}^{2}\quad \approx 0.413\cdot r_{0}^{2}}$, rearranging

Calculation 4 edit

${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}D}}+{\overrightarrow {DS_{1}}}\quad }$starting from S_0
${\displaystyle \Leftrightarrow S_{1}=(0+0i)+{\overrightarrow {S_{0}M}}+{\overrightarrow {MD}}+{\overrightarrow {DS_{1}}}\quad }$, extending to M
${\displaystyle \Leftrightarrow S_{1}={\overrightarrow {S_{0}M}}+{\overrightarrow {MD}}+{\overrightarrow {DS_{1}}}\quad }$, since (0+0i)=0
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {2r_{0}\cdot \sin {\frac {\pi }{3}}}{3\cdot {\frac {\pi }{3}}}}\cdot i\right)+{\overrightarrow {MD}}+{\overrightarrow {DS_{1}}}\quad }$, applying the centroid formula
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {2r_{0}\cdot \sin {\frac {\pi }{3}}}{\pi }}\cdot i\right)+{\overrightarrow {MD}}+{\overrightarrow {DS_{1}}}\quad }$, shortening
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {2r_{0}\cdot {\frac {\sqrt {3}}{2}}}{\pi }}\cdot i\right)+{\overrightarrow {MD}}+{\overrightarrow {DS_{1}}}\quad }$, calculating the sine
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+{\overrightarrow {MD}}+{\overrightarrow {DS_{1}}}\quad }$, shortening
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+(0+|MD|\cdot i)+{\overrightarrow {DS_{1}}}\quad }$, expressing the vector as complex number
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+(0+{\frac {1}{2}}r_{0}\cdot i)+{\overrightarrow {DS_{1}}}\quad }$, applying calculation (1)
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+\left(0+{\frac {1}{2}}r_{0}\cdot i\right)+\left(0-|DS_{1}|\cdot i\right)\quad }$, expressing the vector as complex number
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+\left(0+{\frac {1}{2}}r_{0}\cdot i\right)+\left(0-{\frac {1}{3}}\cdot |MD|\cdot i\right)\quad }$, applying the centroid formular for isosceles triangles
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+\left(0+{\frac {1}{2}}r_{0}\cdot i\right)+\left(0-{\frac {1}{3}}\cdot {\frac {1}{2}}r_{0}\cdot i\right)\quad }$, applying calculation (1)
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {3}}\cdot r_{0}}{\pi }}\cdot i\right)+\left(0+{\frac {1}{2}}r_{0}\cdot i\right)+\left(0-{\frac {1}{6}}\cdot r_{0}\cdot i\right)\quad }$, shortening
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {\sqrt {3}}{\pi }}\cdot i\right)\cdot r_{0}+\left(0+{\frac {1}{2}}\cdot i\right)\cdot r_{0}+\left(0-{\frac {1}{6}}\cdot i\right)\cdot r_{0}\quad }$, using distributive property
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {\sqrt {3}}{\pi }}\cdot i+{\frac {1}{2}}\cdot i-{\frac {1}{6}}\cdot i\right)\cdot r_{0}\quad }$, adding complex numbers
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {\sqrt {3}}{\pi }}\cdot i+{\frac {1}{3}}\cdot i\right)\cdot r_{0}\quad }$, adding
${\displaystyle \Leftrightarrow S_{1}=\left(0-{\frac {3{\sqrt {3}}}{3\pi }}\cdot i+{\frac {\pi }{3\pi }}\cdot i\right)\cdot r_{0}\quad }$, adding
${\displaystyle \Leftrightarrow S_{1}=\left(0+{\frac {\pi -3{\sqrt {3}}}{3\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.218i)\cdot r_{0}}$, adding

Calculation 5 edit

Calculating radius ${\displaystyle r_{2}}$
${\displaystyle |MC|=r_{0}\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow |MD|+|CD|=r_{0}\quad }$, applying the construction of the diagram
${\displaystyle \Leftrightarrow {\frac {1}{2}}\cdot r_{0}+|CD|=r_{0}\quad }$, applying calculation (1)
${\displaystyle \Leftrightarrow |CD|={\frac {1}{2}}\cdot r_{0}\quad }$, rearranging
${\displaystyle \Leftrightarrow 2\cdot r_{2}={\frac {1}{2}}\cdot r_{0}\quad }$, applying equation (5)
${\displaystyle \Leftrightarrow r_{2}={\frac {1}{4}}\cdot r_{0}=0.25r_{0}\quad }$, applying equation (5)

Calculation 6 edit

Calculating the centroid ${\displaystyle S_{2}}$ of the circle
${\displaystyle S_{2}=S_{0}+{\overrightarrow {S_{0}M}}+{\overrightarrow {MC}}+{\overrightarrow {CS_{2}}}}$
${\displaystyle \Leftrightarrow S_{2}={\overrightarrow {S_{0}M}}+{\overrightarrow {MC}}+{\overrightarrow {CS_{2}}}\quad }$, since the centroid of the base shape is (0+0i)=0
${\displaystyle \Leftrightarrow S_{2}=-|S_{0}M|\cdot i+|MC|\cdot i-|CS_{2}|\cdot i\quad }$
${\displaystyle \Leftrightarrow S_{2}=-|S_{0}M|\cdot i+|MC|\cdot i-r_{2}\cdot i\quad }$, applying equation (5)
${\displaystyle \Leftrightarrow S_{2}=-|S_{0}M|\cdot i+r_{0}\cdot i-r_{2}\cdot i\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow S_{2}=-|S_{0}M|\cdot i+r_{0}\cdot i-{\frac {1}{4}}\cdot r_{0}\cdot i\quad }$, applying calculation (5)
${\displaystyle \Leftrightarrow S_{2}=-|S_{0}M|\cdot i+{\frac {3}{4}}\cdot r_{0}\cdot i\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{2}=-{\frac {2r_{0}\cdot \sin {\frac {\pi }{3}}}{3\cdot {\frac {\pi }{3}}}}\cdot i+{\frac {3}{4}}\cdot r_{0}\cdot i\quad }$, applying the centroid formula for circular sectors
${\displaystyle \Leftrightarrow S_{2}=-{\frac {2\cdot \sin {\frac {\pi }{3}}}{\pi }}\cdot r_{0}\cdot i+{\frac {3}{4}}\cdot r_{0}\cdot i\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{2}=-{\frac {2\cdot {\frac {\sqrt {3}}{2}}}{\pi }}\cdot r_{0}\cdot i+{\frac {3}{4}}\cdot r_{0}\cdot i\quad }$, calculating the sine
${\displaystyle \Leftrightarrow S_{2}={\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot r_{0}\cdot i=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0+0.199i)\cdot r_{0}\quad }$, rearranging

Calculation 7 edit

Calculating the centroid ${\displaystyle S_{3}}$ of the inscribed circular sector
${\displaystyle S_{3}=S_{0}+{\overrightarrow {S_{0}M}}+{\overrightarrow {MS_{3}}}}$
${\displaystyle \Leftrightarrow S_{3}={\overrightarrow {S_{0}M}}+{\overrightarrow {MS_{3}}}\quad }$, since the centroid of the base shape is (0+0i)=0
${\displaystyle \Leftrightarrow S_{3}=-|S_{0}M|\cdot i+|MS_{3}|\cdot i\quad }$
${\displaystyle \Leftrightarrow S_{3}=-{\frac {2r_{0}\cdot \sin {\frac {\pi }{3}}}{3\cdot {\frac {\pi }{3}}}}\cdot i+|MS_{3}|\cdot i\quad }$, applying the centroid formular for circular sectors
${\displaystyle \Leftrightarrow S_{3}=-{\frac {2\cdot \sin {\frac {\pi }{3}}}{\pi }}\cdot r_{0}\cdot i+|MS_{3}|\cdot i\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{3}=-{\frac {2\cdot {\frac {\sqrt {3}}{2}}}{\pi }}\cdot r_{0}\cdot i+|MS_{3}|\cdot i\quad }$, calculating the sine
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+|MS_{3}|\cdot i\quad }$, shortening
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+{\frac {2r_{3}\cdot \sin {\frac {\pi }{3}}}{3\cdot {\frac {\pi }{3}}}}\cdot i\quad }$, applying the centroid formular for circular sectors
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+{\frac {2\cdot \sin {\frac {\pi }{3}}}{3\cdot {\frac {\pi }{3}}}}\cdot r_{3}\cdot i\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+{\frac {2\cdot \sin {\frac {\pi }{3}}}{\pi }}\cdot r_{3}\cdot i\quad }$, shortening
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+{\frac {2\cdot {\frac {\sqrt {3}}{2}}}{\pi }}\cdot r_{3}\cdot i\quad }$, calculating the sine
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+{\frac {\sqrt {3}}{\pi }}\cdot r_{3}\cdot i\quad }$, shortening
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{\pi }}\cdot r_{0}\cdot i+{\frac {\sqrt {3}}{\pi }}\cdot {\frac {1}{2}}\cdot r_{0}\cdot i\quad }$, applying the relation
${\displaystyle \Leftrightarrow S_{3}=-{\frac {\sqrt {3}}{2\pi }}\cdot r_{0}\cdot i\quad }$, calculating
${\displaystyle \Leftrightarrow S_{3}=(0-{\frac {\sqrt {3}}{2\pi }}\cdot i)\cdot r_{0}\quad \approx (0-0.276i)\cdot r_{0}}$, rearranging

Calculation 8 edit

Calculating the radius ${\displaystyle r_{4}}$ of the inscribed circle around ${\displaystyle S_{4}}$ by making use of the red right triangle ${\displaystyle \triangle {S_{2}FS_{4}}}$ and the green right triangle ${\displaystyle \triangle {FMS_{4}}}$
${\displaystyle |S_{2}S_{4}|^{2}-|S_{2}F|^{2}=|MS_{4}|^{2}-|FM|^{2}\quad }$, since the triangles share the side ${\displaystyle {\overline {FS_{4}}}}$
${\displaystyle \Leftrightarrow (r_{2}+r_{4})^{2}-|S_{2}F|^{2}=|MS_{4}|^{2}-|FM|^{2}\quad }$, since the two circles have the touching point on this line
${\displaystyle \Leftrightarrow (r_{2}+r_{4})^{2}-(|S_{2}D|-|DF|)^{2}=|MS_{4}|^{2}-|FM|^{2}\quad }$, replacing by known lenghts
${\displaystyle \Leftrightarrow (r_{2}+r_{4})^{2}-(r_{2}-|DF|)^{2}=|MS_{4}|^{2}-|FM|^{2}\quad }$, applying equation (5)
${\displaystyle \Leftrightarrow (r_{2}+r_{4})^{2}-(r_{2}-r_{4})^{2}=|MS_{4}|^{2}-|FM|^{2}\quad }$, applying equation (7)
${\displaystyle \Leftrightarrow (r_{2}^{2}+2r_{2}r_{4}+r_{4}^{2})-(r_{2}^{2}-2r_{2}r_{4}+r_{4}^{2})=|MS_{4}|^{2}-|FM|^{2}\quad }$, applying binominal formula
${\displaystyle \Leftrightarrow r_{2}^{2}+2r_{2}r_{4}+r_{4}^{2}-r_{2}^{2}+2r_{2}r_{4}-r_{4}^{2}=|MS_{4}|^{2}-|FM|^{2}\quad }$, rearranging
${\displaystyle \Leftrightarrow 2r_{2}r_{4}+2r_{2}r_{4}=|MS_{4}|^{2}-|FM|^{2}\quad }$, cancelling out
${\displaystyle \Leftrightarrow 4r_{2}r_{4}=|MS_{4}|^{2}-|FM|^{2}\quad }$, rearranging
${\displaystyle \Leftrightarrow 4{\frac {1}{4}}r_{0}r_{4}=|MS_{4}|^{2}-|FM|^{2}\quad }$, applying calculation (5)
${\displaystyle \Leftrightarrow r_{0}r_{4}=(|MG|-|GS_{4}|)^{2}-|FM|^{2}\quad }$, since the circle around ${\displaystyle S_{4}}$ is in point G tangential to the circular sector around M
${\displaystyle \Leftrightarrow r_{0}r_{4}=(r_{0}-r_{4})^{2}-|FM|^{2}\quad }$, applying equation (1)
${\displaystyle \Leftrightarrow r_{0}r_{4}=(r_{0}-r_{4})^{2}-(|MD|+|DF|)^{2}\quad }$,
${\displaystyle \Leftrightarrow r_{0}r_{4}=(r_{0}-r_{4})^{2}-(r_{3}+|DF|)^{2}\quad }$,applying equation (6)
${\displaystyle \Leftrightarrow r_{0}r_{4}=(r_{0}-r_{4})^{2}-(r_{3}+r_{4})^{2}\quad }$,applying equation (7)
${\displaystyle \Leftrightarrow r_{0}r_{4}=(r_{0}-r_{4})^{2}-({\frac {1}{2}}\cdot r_{0}+r_{4})^{2}\quad }$,applying calculation(1)
${\displaystyle \Leftrightarrow r_{0}r_{4}=(r_{0}^{2}-2r_{0}r_{4}+r_{4}^{2})-({\frac {1}{4}}r_{0}^{2}+2{\frac {1}{2}}r_{0}r_{4}+r_{4}^{2})\quad }$, applying binomial formula
${\displaystyle \Leftrightarrow r_{0}r_{4}=r_{0}^{2}-2r_{0}r_{4}+r_{4}^{2}-{\frac {1}{4}}r_{0}^{2}-r_{0}r_{4}-r_{4}^{2}\quad }$, applying binomial formula
${\displaystyle \Leftrightarrow r_{0}r_{4}=r_{0}^{2}-2r_{0}r_{4}-{\frac {1}{4}}r_{0}^{2}-r_{0}r_{4}\quad }$, cancelling out
${\displaystyle \Leftrightarrow r_{0}r_{4}=r_{0}^{2}-2r_{0}r_{4}-{\frac {1}{4}}r_{0}^{2}-r_{0}r_{4}\quad }$, rearranging
${\displaystyle \Leftrightarrow r_{0}r_{4}={\frac {3}{4}}r_{0}^{2}-3r_{0}r_{4}\quad }$, rearranging
${\displaystyle \Leftrightarrow 4r_{0}r_{4}={\frac {3}{4}}r_{0}^{2}\quad }$, rearranging
${\displaystyle \Leftrightarrow 4r_{4}={\frac {3}{4}}r_{0}\quad }$, reducing
${\displaystyle \Leftrightarrow r_{4}={\frac {3}{16}}r_{0}\quad }$, rearranging

Calculation 9 edit

Calculating the centroid ${\displaystyle S_{4}}$ of the inscribed circle
${\displaystyle S_{4}=S_{0}+{\overrightarrow {S_{0}F}}+{\overrightarrow {FS_{4}}}}$
${\displaystyle \Leftrightarrow S_{4}={\overrightarrow {S_{0}F}}+{\overrightarrow {FS_{4}}}\quad }$, since the centroid of the base shape is (0+0i)=0
${\displaystyle \Leftrightarrow S_{4}={\overrightarrow {S_{0}S_{2}}}+{\overrightarrow {S_{2}F}}+{\overrightarrow {FS_{4}}}\quad }$,
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}+{\overrightarrow {S_{2}F}}+{\overrightarrow {FS_{4}}}\quad }$, see calculation (6)
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}-(|S_{2}D|-|DF|)\cdot i+{\overrightarrow {FS_{4}}}\quad }$,
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}-(r_{2}-r_{4})\cdot i+{\overrightarrow {FS_{4}}}\quad }$,
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}-\left({\frac {1}{4}}r_{0}-r_{4}\right)\cdot i+{\overrightarrow {FS_{4}}}\quad }$, see calculation (5)
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}-\left({\frac {1}{4}}r_{0}-{\frac {3}{16}}r_{0}\right)\cdot i+{\overrightarrow {FS_{4}}}\quad }$, see calculation (8)
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}-{\frac {1}{16}}\cdot i\cdot r_{0}+{\overrightarrow {FS_{4}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i-{\frac {1}{16}}\cdot i\right)\cdot r_{0}+{\overrightarrow {FS_{4}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i-{\frac {\pi }{16\pi }}\cdot i\right)\cdot r_{0}+{\overrightarrow {FS_{4}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {12\pi -16{\sqrt {3}}}{16\pi }}\cdot i-{\frac {\pi }{16\pi }}\cdot i\right)\cdot r_{0}+{\overrightarrow {FS_{4}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\overrightarrow {FS_{4}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {|S_{2}S_{4}|^{2}-|FS_{2}|^{2}}}\quad }$, applying the Pythagorean theoreme on ${\displaystyle \triangle FS_{4}S_{2}}$
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {(r_{2}+r_{4})^{2}-(r_{2}-r_{4})^{2}}}\quad }$,
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {\left({\frac {1}{4}}r_{0}+{\frac {3}{16}}r_{0}\right)^{2}-\left({\frac {1}{4}}r_{0}-{\frac {3}{16}}r_{0}\right)^{2}}}\quad }$, applying calculations (5) and (8)
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {\left({\frac {7}{16}}r_{0}\right)^{2}-\left({\frac {1}{16}}r_{0}\right)^{2}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {{\frac {49}{256}}r_{0}^{2}-{\frac {1}{256}}r_{0}^{2}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {{\frac {48}{256}}r_{0}^{2}}}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left(0+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}+{\sqrt {\frac {3}{16}}}\cdot r_{0}\quad }$, rearranging
${\displaystyle \Leftrightarrow S_{4}=\left({\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0.433+0.136i)\cdot r_{0}}$, rearranging