File:FS FJCFC2(3) dia.png

Size of this preview: 800 × 529 pixels. Other resolutions: 320 × 211 pixels | 640 × 423 pixels | 1,024 × 677 pixels | 1,280 × 846 pixels | 2,412 × 1,594 pixels.

Original file ‎(2,412 × 1,594 pixels, file size: 159 KB, MIME type: image/png)

Captions

English

Sangaku with circular sectors, isoscles triangle and circles

Summary edit

DescriptionFS FJCFC2(3) dia.png	English: Second largest circle in a 120° circular sector (fan) that contains the broadest isosceles triangle and the following largest circle, the following largest 120° circular sector (fan), a further following second largest circle and a another following second largest circle - Details: FJCFC2(3) dia.png Deutsch: Zweitgrößter Kreis in einem Drittelkreis (Fächer), der bereits das breitestes gleichschenkliges Dreieck und den nächstgrößten Kreis, den nächstmöglichen größten Drittelkreis, einen nächstmöglichen zweitgrößten Kreis und einen weiteren nächstmöglichen zweitgrößten Kreis enthält - Details: FJCFC2(3) dia.png
Date	26 November 2023
Source	Own work
Author	Hans G. Oberlack

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 second largest circle.
5) Inscribed is the next second largest circle.
6) Inscribed is the next second largest circle.

General case edit

Segments in the general case edit

0) Radius of the base circular sector: $r_{0}$
1) Side length of the inscribed triangle: $a_{1}=r_{0}$ , because it is the broadest triangle
2) Radius of the inscribed circle: $r_{2}={\frac {1}{4}}\cdot r_{0}$ , see calculation (5)
3) Radius of the inscribed circular sector: $r_{3}=|DM|={\frac {1}{2}}\cdot r_{0}\quad$ , applying calculation (1)
4) Radius of the inscribed circle around $S_{4}$ : $r_{4}={\frac {3}{16}}\cdot r_{0}\quad$ , see calculation (8)
5) Radius of the inscribed circle around $S_{5}$ : $r_{5}=r_{4}={\frac {3}{16}}\cdot r_{0}\quad$ , for symmetry reasons
6) Radius of the inscribed circle around $S_{6}$ : $r_{6}=\left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\approx 0.108\cdot r_{0}\quad$ , see calculation (10)

Perimeters in the general case edit

0) Perimeter of base circular sector: $P_{0}=|AM|+|MB|+|BCA|=2\cdot r_{0}+{\frac {2\pi }{3}}\cdot r_{0}=2\cdot (1+{\frac {\pi }{3}})\cdot r_{0}\quad \approx 4.094\cdot r_{0}$
1) Perimeter of inscribed triangle: $P_{1}=|AM|+|MB|+|BDA|=2\cdot r_{0}+{\sqrt {3}}\cdot r_{0}\quad \approx 3.732\cdot r_{0}$
2) Perimeter of inscribed circle around $S_{2}$ : $P_{2}=2\pi r_{2}=2\pi \cdot {\frac {1}{4}}\cdot r_{0}={\frac {\pi }{2}}\cdot r_{0}\quad \approx 1.571\cdot r_{0}$
3) Perimeter of inscribed circular sector: $P_{3}=2\cdot (1+{\frac {\pi }{3}})\cdot r_{3}=2\cdot (1+{\frac {\pi }{3}})\cdot {\frac {1}{2}}\cdot r_{0}=(1+{\frac {\pi }{3}})\cdot r_{0}\quad \approx 2.047\cdot r_{0}$
4) Perimeter of inscribed circle around $S_{4}$ : $P_{4}=2\pi r_{4}=2\pi \cdot {\frac {3}{16}}\cdot r_{0}={\frac {3\pi }{8}}\cdot r_{0}\quad \approx 1.178\cdot r_{0}$
5) Perimeter of inscribed circle around $S_{5}$ : $P_{5}=P_{4}=2\pi r_{4}=2\pi \cdot {\frac {3}{16}}\cdot r_{0}={\frac {3\pi }{8}}\cdot r_{0}\quad \approx 1.178\cdot r_{0}$
6) Perimeter of inscribed circle around $S_{6}$ : $P_{6}=2\pi r_{6}=2\pi \cdot \left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\quad \approx 0.677\cdot r_{0}$

Areas in the general case edit

0) Area of the base circular sector $A_{0}={\frac {\pi }{3}}\cdot r_{0}^{2}\quad \approx 1.047\cdot r_{0}^{2}$
1) Area of the inscribed triangle $A_{1}={\frac {\sqrt {3}}{4}}\cdot r_{0}^{2}\quad \approx 0.413\cdot r_{0}^{2}\quad$ , see calculation (3)
2) Area of the inscribed circle around $S_{2}$ : $A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot \left({\frac {1}{4}}\cdot r_{0}\right)^{2}={\frac {\pi }{16}}\cdot r_{0}^{2}\quad \approx 0.196\cdot r_{0}^{2}$
3) Area of the inscribed circular sector $A_{3}={\frac {\pi }{3}}\cdot r_{3}^{2}={\frac {\pi }{3}}\cdot \left({\frac {1}{2}}\cdot r_{0}\right)^{2}={\frac {\pi }{12}}\cdot r_{0}^{2}\quad \approx 0.262\cdot r_{0}^{2}$
4) Area of the inscribed circle around $S_{4}$ : $\quad A_{4}=\pi \cdot r_{4}^{2}=\pi \cdot \left({\frac {3}{16}}\cdot r_{0}\right)^{2}={\frac {9\pi }{256}}\cdot r_{0}^{2}\quad \approx 0.11\cdot r_{0}^{2}$
5) Area of the inscribed circle around $S_{5}$ : $\quad A_{5}=A_{4}=\pi \cdot r_{5}^{2}=\pi \cdot \left({\frac {3}{16}}\cdot r_{0}\right)^{2}={\frac {9\pi }{256}}\cdot r_{0}^{2}\quad \approx 0.11\cdot r_{0}^{2}$
6) Area of the inscribed circle around $S_{6}$ : $\quad A_{6}=\pi \cdot r_{6}^{2}=\pi \cdot \left({\frac {21-12{\sqrt {3}}}{2}}\right)^{2}\cdot r_{0}^{2}\quad \approx 0.036\cdot r_{0}^{2}$

Centroids in the general case edit

0) By definition the centroid point of a base shape is $S_{0}=0+0i$
1) The centroid of the inscribed triangle relative to the base centroid is: $S_{1}=\left(0+{\frac {\pi -3{\sqrt {3}}}{3\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.218i)\cdot r_{0}\quad$ , see Calculation (4)
2) The centroid of the inscribed circle relative to the base centroid is: $S_{2}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0+0.199i)\cdot r_{0}\quad$ , see Calculation (6)
3) The centroid of the inscribed circular sector relative to the base centroid is: $S_{3}=\left(0-{\frac {\sqrt {3}}{2\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0-0.276i)\cdot r_{0}\quad$ , see Calculation (7)
4) The centroid of the inscribed circle relative to the base centroid is: $S_{4}=\left({\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}\quad \approx (0.433+0.136i)\cdot r_{0}\quad$ , see Calculation (9)
5) The centroid of the inscribed circle relative to the base centroid is: $S_{5}=\left(-{\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot r_{0}\quad \approx (-0.433+0.136i)\cdot r_{0}\quad$ , for symmetry reasons
6) The centroid of the inscribed circle relative to the base centroid is: $S_{6}=\left((2{\sqrt {3}}-3)+\left({\frac {6\pi {\sqrt {3}}-{\sqrt {3}}-10\pi }{\pi }}\right)\cdot i\right)\cdot r_{0}\quad \approx (0.464-0.159i)\cdot r_{0}\quad$ , see Calculation (11)

Normalised case edit

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

Segments in the normalised case edit

0) Radius of the base circular sector: $r_{0}={\sqrt {\frac {3}{\pi }}}\quad \approx 0.977$
1) Side length of the inscribed triangle: $a_{1}=r_{0}={\sqrt {\frac {3}{\pi }}}\quad \approx 0.977$
2) Radius of the inscribed circle around $S_{2}$ : $r_{2}={\frac {1}{4}}\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\frac {3}{16\pi }}}\quad \approx 0.244$
3) Radius of the inscribed circular sector: $r_{3}={\frac {1}{2}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 0.489$
4) Radius of the inscribed circle around $S_{4}$ : $r_{4}={\frac {3}{16}}\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\frac {27}{256\pi }}}\quad \approx 0.183$
5) Radius of the inscribed circle around $S_{5}$ : $r_{5}=r_{4}={\frac {3}{16}}\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\frac {27}{256\pi }}}\quad \approx 0.183$
6) Radius of the inscribed circle around $S_{6}$ : $r_{6}=\left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot {\sqrt {\frac {3}{\pi }}}=\left({\frac {21{\sqrt {3}}-36}{2{\sqrt {\pi }}}}\right)\quad \approx 0.105$

Perimeter in the normalised case edit

0) Perimeter of base circular sector: $P_{0}=2\cdot (1+{\frac {\pi }{3}})\cdot {\sqrt {\frac {3}{\pi }}}\approx 4.0010635$
1) Perimeter of inscribed triangle: $P_{1}=(2+{\sqrt {3}})\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 3.6469788$
2) Perimeter of inscribed circle around $S_{2}$ : $P_{2}={\frac {\pi }{2}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 1.5349901$
3) Perimeter of inscribed circular sector: $P_{3}=(1+{\frac {\pi }{3}})\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 2.0005317$
4) Perimeter of inscribed circle around $S_{4}$ : $P_{4}={\frac {3\pi }{8}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 1.151242546$
5) Perimeter of inscribed circle around $S_{5}$ : $P_{5}=P_{4}={\frac {3\pi }{8}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 1.151242546$
6) Perimeter of inscribed circle around $S_{6}$ : $P_{6}=2\pi r_{6}=2\pi \cdot \left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\pi }}\cdot \left(21{\sqrt {3}}-36\right)\quad \approx 0.6612440$

Area in the normalised case edit

0) Area of the base circular sector is by definition $A_{0}=1$
1) Area of the inscribed triangle $A_{1}={\frac {\sqrt {3}}{4}}\cdot {\frac {3}{\pi }}\quad \approx 0.413$
2) Area of the inscribed circle around $S_{2}$ : $A_{2}={\frac {\pi }{16}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {3}{16}}\quad \approx 0.188$
3) Area of the inscribed circular sector $A_{3}={\frac {\pi }{12}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {3}{12}}=0.250$
4) Area of the inscribed circle around $S_{4}$ : $\quad A_{4}={\frac {9\pi }{256}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {27}{256}}\quad \approx 0.105$
5) Area of the inscribed circle around $S_{5}$ : $\quad A_{5}=A_{4}={\frac {9\pi }{256}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {27}{256}}\quad \approx 0.105$
6) Area of the inscribed circle around $S_{6}$ : $\quad A_{6}=\pi \cdot r_{6}^{2}=\pi \cdot \left({\frac {21-12{\sqrt {3}}}{2}}\right)^{2}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}={\frac {27}{4}}\cdot \left(97-56{\sqrt {3}}\right)\quad \approx 0.038$

Centroids in the normalised case edit

0) Centroid of the base shape: $S_{0}=0+0i$
1) Centroid of the inscribed triangle: $S_{1}=\left(0+{\frac {\pi -3{\sqrt {3}}}{3\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 0-0.213i$
2) Centroid of the inscribed circle around $S_{2}$ : $S_{2}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx (0+0.194i)\quad$
3) Centroid of the inscribed circular sector: $S_{3}=\left(0-{\frac {\sqrt {3}}{2\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}=\left(0-{\frac {3}{2\pi {\sqrt {\pi }}}}\cdot i\right)\quad \approx (0-0.269i)\quad$
4) The centroid of the inscribed circle around $S_{4}$ : $S_{4}=\left({\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}=\left({\sqrt {\frac {9}{16\pi }}}+{\frac {11\pi {\sqrt {3}}-48}{16\pi {\sqrt {\pi }}}}\cdot i\right)\quad \approx (0.423+0.133i)$
5) The centroid of the inscribed circle around $S_{5}$ : $S_{5}=\left(-{\sqrt {\frac {3}{16}}}+{\frac {11\pi -16{\sqrt {3}}}{16\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}=\left(-{\sqrt {\frac {9}{16\pi }}}+{\frac {11\pi {\sqrt {3}}-48}{16\pi {\sqrt {\pi }}}}\cdot i\right)\quad \approx (-0.423+0.133i)$
6) The centroid of the inscribed circle relative to the base centroid is: $S_{6}=\left((2{\sqrt {3}}-3)+\left({\frac {6\pi {\sqrt {3}}-{\sqrt {3}}-10\pi }{\pi }}\right)\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx (0.454-0.155i)\quad$

Calculations edit

Given elements edit

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

Calculation 1 edit

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

Calculation 2 edit

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

Calculation 3 edit

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

Calculation 4 edit

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

Calculation 5 edit

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

Calculation 6 edit

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

Calculation 7 edit

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

Calculation 8 edit

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

Calculation 9 edit

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

Calculation 10 edit

Calculating the radius $r_{6}$ of the inscribed circle around $S_{6}$ by making use of the red right triangle $\triangle {BHS_{6}}$ and the blue right triangle $\triangle {HMS_{6}}$
${\frac {|HS_{6}|}{|BH|}}=\tan {15^{\circ }}\quad$ , since ${\overrightarrow {BS_{6}}}$ divides the angle in B in two equal halves because ${\overrightarrow {BK}}$ and ${\overrightarrow {BH}}$ are tangent to the circle around $S_{6}$
$\Leftrightarrow {\frac {r_{6}}{|BH|}}=\tan {15^{\circ }}\quad$ , see equation (8)
$\Leftrightarrow r_{6}=|BH|\cdot c\quad$ , rearranging and setting $c=\tan {15^{\circ }}$
$\Leftrightarrow r_{6}=\left(|BM|-|HM|\right)\cdot c\quad$ , see diagram
$\Leftrightarrow r_{6}=\left(r_{0}-|HM|\right)\cdot c\quad$ , see equation (1)
$\Leftrightarrow {\frac {r_{6}}{c}}=r_{0}-|HM|\quad$ , rearranging
$\Leftrightarrow |HM|=r_{0}-{\frac {r_{6}}{c}}\quad$ , rearranging
$\Leftrightarrow |HM|^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , squaring
$\Leftrightarrow |MS_{6}|^{2}-|HS_{6}|^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying the Pythagorean theorem on the right triangle $\triangle {HMS_{6}}$ with $|HM|^{2}+|HS_{6}|^{2}=|MS_{6}|^{2}\quad$
$\Leftrightarrow |MS_{6}|^{2}-r_{6}^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying equation (8)
$\Leftrightarrow \left(|MJ|+|JS_{6}|\right)^{2}-r_{6}^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying equation (9)
$\Leftrightarrow \left(r_{3}+|JS_{6}|\right)^{2}-r_{6}^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying equation (6)
$\Leftrightarrow \left(r_{3}+r_{6}\right)^{2}-r_{6}^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying equation (8)
$\Leftrightarrow \left({\frac {r_{0}}{2}}+r_{6}\right)^{2}-r_{6}^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying calculation (1)
$\Leftrightarrow \left({\frac {r_{0}^{2}}{4}}+2\cdot {\frac {r_{0}}{2}}\cdot r_{6}+r_{6}^{2}\right)-r_{6}^{2}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , applying binomial formula
$\Leftrightarrow {\frac {r_{0}^{2}}{4}}+r_{0}\cdot r_{6}=\left(r_{0}-{\frac {r_{6}}{c}}\right)^{2}\quad$ , eliminating
$\Leftrightarrow {\frac {r_{0}^{2}}{4}}+r_{0}\cdot r_{6}=\left(r_{0}^{2}-2\cdot r_{0}\cdot {\frac {r_{6}}{c}}+{\frac {r_{6}^{2}}{c^{2}}}\right)\quad$ , applying binomial formula
$\Leftrightarrow -{\frac {3}{4}}\cdot r_{0}^{2}+r_{0}\cdot r_{6}={\frac {r_{6}^{2}}{c^{2}}}-2\cdot r_{0}\cdot {\frac {r_{6}}{c}}\quad$ , rearranging
$\Leftrightarrow -{\frac {3}{4}}\cdot r_{0}^{2}={\frac {r_{6}^{2}}{c^{2}}}-2\cdot r_{0}\cdot {\frac {r_{6}}{c}}-r_{0}\cdot r_{6}\quad$ , rearranging
$\Leftrightarrow -{\frac {3}{4}}\cdot r_{0}^{2}={\frac {r_{6}^{2}}{c^{2}}}-2r_{0}r_{6}\cdot {\frac {1}{c}}-2r_{0}r_{6}\cdot {\frac {1}{2}}\quad$ , rearranging
$\Leftrightarrow -{\frac {3}{4}}\cdot r_{0}^{2}={\frac {r_{6}^{2}}{c^{2}}}-2r_{0}r_{6}\cdot \left({\frac {1}{c}}+{\frac {1}{2}}\right)\quad$ , putting outside brackets
$\Leftrightarrow -{\frac {3}{4}}\cdot r_{0}^{2}={\frac {r_{6}^{2}}{c^{2}}}-2r_{0}r_{6}\cdot \left({\frac {2+c}{2c}}\right)\quad$ , rearranging
$\Leftrightarrow -{\frac {3}{4}}\cdot r_{0}^{2}={\frac {r_{6}^{2}}{c^{2}}}-2r_{0}{\frac {r_{6}}{c}}\cdot \left({\frac {2+c}{2}}\right)\quad$ , rearranging
$\Leftrightarrow r_{0}^{2}\cdot \left({\frac {2+c}{2}}\right)^{2}-{\frac {3}{4}}\cdot r_{0}^{2}={\frac {r_{6}^{2}}{c^{2}}}-2r_{0}{\frac {r_{6}}{c}}\cdot \left({\frac {2+c}{2}}\right)+r_{0}^{2}\cdot \left({\frac {2+c}{2}}\right)^{2}\quad$ , adding $r_{0}^{2}\cdot \left({\frac {2+c}{2}}\right)^{2}$ on both sides
$\Leftrightarrow r_{0}^{2}\cdot \left({\frac {2+c}{2}}\right)^{2}-{\frac {3}{4}}\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , applying binomial formula
$\Leftrightarrow r_{0}^{2}\cdot \left({\frac {4+4c+c^{2}}{4}}\right)-{\frac {3}{4}}\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , applying binomial formula
$\Leftrightarrow \left({\frac {4+4c+c^{2}}{4}}-{\frac {3}{4}}\right)\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , putting outside the brackets
$\Leftrightarrow \left({\frac {4+4c+c^{2}-3}{4}}\right)\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow \left({\frac {1+4c+c^{2}}{4}}\right)\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow \left(1+4c+c^{2}\right)\cdot \left({\frac {r_{0}}{2}}\right)^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow \left(1+4\cdot (2-{\sqrt {3}})+(2-{\sqrt {3}})^{2}\right)\cdot \left({\frac {r_{0}}{2}}\right)^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , using $c=\tan {15^{\circ }}=2-{\sqrt {3}}$
$\Leftrightarrow \left(1+(8-4{\sqrt {3}})+(4-4{\sqrt {3}}+3)\right)\cdot \left({\frac {r_{0}}{2}}\right)^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging and applying binomial formula
$\Leftrightarrow (16-8{\sqrt {3}})\cdot \left({\frac {r_{0}}{2}}\right)^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow 4(4-2{\sqrt {3}})\cdot \left({\frac {r_{0}}{2}}\right)^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (4-2{\sqrt {3}})\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2+c}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (4-2{\sqrt {3}})\cdot r_{0}^{2}=\left({\frac {r_{6}}{c}}-r_{0}\cdot \left({\frac {2c+c^{2}}{2c}}\right)\right)^{2}\quad$ , expanding the fraction
$\Leftrightarrow (4-2{\sqrt {3}})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {2c+c^{2}}{2}}\right)\right)^{2}\cdot {\frac {1}{c^{2}}}\quad$ , putting c outside the brackets
$\Leftrightarrow (4-2{\sqrt {3}})\cdot r_{0}^{2}\cdot c^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {2c+c^{2}}{2}}\right)\right)^{2}\quad$ , multiplying by $c^{2}$
$\Leftrightarrow (4-2{\sqrt {3}})\cdot r_{0}^{2}\cdot (2-{\sqrt {3}})^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {2c+c^{2}}{2}}\right)\right)^{2}\quad$ , using $c=\tan {15^{\circ }}=2-{\sqrt {3}}$
$\Leftrightarrow (4-2{\sqrt {3}})\cdot (7-4{\sqrt {3}})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {2c+c^{2}}{2}}\right)\right)^{2}\quad$ , applying binomial formula
$\Leftrightarrow (28-16{\sqrt {3}}-14{\sqrt {3}}+24)\cdot (7-4{\sqrt {3}})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {2c+c^{2}}{2}}\right)\right)^{2}\quad$ , multiplying out
$\Leftrightarrow (52-30{\sqrt {3}})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {2c+c^{2}}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (52-30{\sqrt {3}})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {(4-2{\sqrt {3}})+(7-4{\sqrt {3}})}{2}}\right)\right)^{2}\quad$ , using $c=\tan {15^{\circ }}=2-{\sqrt {3}}$
$\Leftrightarrow (52-30{\sqrt {3}})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (25-30{\sqrt {3}}+27)\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (5^{2}-30{\sqrt {3}}+9*3)\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (5^{2}-2\cdot 5\cdot 3{\sqrt {3}}+(3{\sqrt {3}})^{2})\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow (5-3{\sqrt {3}})^{2}\cdot r_{0}^{2}=\left(r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\right)^{2}\quad$ , rearranging
$\Leftrightarrow -(5-3{\sqrt {3}})\cdot r_{0}=r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\quad$ or $\quad (5-3{\sqrt {3}})\cdot r_{0}=r_{6}-r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)\quad$ , extracting the root
$\Leftrightarrow r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)-(5-3{\sqrt {3}})\cdot r_{0}=r_{6}\quad$ or $\quad r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)+(5-3{\sqrt {3}})\cdot r_{0}=r_{6}\quad$ , rearranging
$\Leftrightarrow r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)-{\frac {2\cdot (5-3{\sqrt {3}})}{2}}\cdot r_{0}=r_{6}\quad$ or $\quad r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)+{\frac {2\cdot (5-3{\sqrt {3}})}{2}}\cdot r_{0}=r_{6}\quad$ , rearranging
$\Leftrightarrow r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)-{\frac {2\cdot (5-3{\sqrt {3}})}{2}}\cdot r_{0}=r_{6}\quad$ or $\quad r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)+{\frac {2\cdot (5-3{\sqrt {3}})}{2}}\cdot r_{0}=r_{6}\quad$ , rearranging
$\Leftrightarrow r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)-{\frac {(10-6{\sqrt {3}})}{2}}\cdot r_{0}=r_{6}\quad$ or $\quad r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)+{\frac {10-6{\sqrt {3}}}{2}}\cdot r_{0}=r_{6}\quad$ , rearranging
$\Leftrightarrow \left({\frac {11-6{\sqrt {3}}-10+6{\sqrt {3}}}{2}}\right)\cdot r_{0}=r_{6}\quad$ or $\quad r_{0}\cdot \left({\frac {11-6{\sqrt {3}}}{2}}\right)+{\frac {10-6{\sqrt {3}}}{2}}\cdot r_{0}=r_{6}\quad$ , rearranging
$\Leftrightarrow \left({\frac {11-6{\sqrt {3}}-10+6{\sqrt {3}}}{2}}\right)\cdot r_{0}=r_{6}\quad$ or $\quad \left({\frac {11-6{\sqrt {3}}+10-6{\sqrt {3}}}{2}}\right)\cdot r_{0}=r_{6}\quad$ , rearranging
$\Leftrightarrow r_{6}=\left({\frac {1}{2}}\right)\cdot r_{0}=0.5\cdot r_{0}\quad$ or $\quad r_{6}=\left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\quad \approx 0.108\cdot r_{0}$ , rearranging
$\rightarrow r_{6}=\left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\approx 0.108\cdot r_{0}\quad$ , since $r_{6}<0.5\cdot r_{0}$

Calculation 11 edit

Calculating the centroid $S_{6}$ of the inscribed circle of radius $r_{6}$
$S_{6}=S_{0}+{\overrightarrow {S_{0}D}}+{\overrightarrow {DK}}+{\overrightarrow {KS_{6}}}$
$\Leftrightarrow S_{6}=S_{0}+({\overrightarrow {S_{0}M}}+{\overrightarrow {MD}})+{\overrightarrow {DK}}+{\overrightarrow {KS_{6}}}$
$\Leftrightarrow S_{6}=S_{0}+({\overrightarrow {S_{0}M}}+{\overrightarrow {MD}})+({\overrightarrow {DB}}+{\overrightarrow {BK}})+{\overrightarrow {KS_{6}}}$
$\Leftrightarrow S_{6}=(0+0i)-|MS_{0}|i+|DM|i+|DB|-|BK|-|KS_{6}|i$
$\Leftrightarrow S_{6}=(0+0i)-|MS_{0}|i+|DM|i+|DB|-|BK|-r_{6}i\quad$ , applying equation (8)
$\Leftrightarrow S_{6}=(0+0i)-|MS_{0}|i+|DM|i+{\frac {|DB|}{2}}-|BK|-r_{6}i\quad$ , applying calculation(2)
$\Leftrightarrow S_{6}=(0+0i)-|MS_{0}|i+|DM|i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-|BK|-r_{6}i\quad$ , applying calculation(2)
$\Leftrightarrow S_{6}=(0+0i)-\left({\frac {2r_{0}\cdot \sin {\frac {\pi }{3}}}{3\cdot {\frac {\pi }{3}}}}\right)\cdot i+|DM|i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-|BK|-r_{6}i\quad$ , applying the centroid formula for circular sectors
$\Leftrightarrow S_{6}=(0+0i)-\left({\frac {2r_{0}\cdot {\frac {\sqrt {3}}{2}}}{\pi }}\right)\cdot i+|DM|i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-|BK|-r_{6}i\quad$ , applying sine function
$\Leftrightarrow S_{6}=(0+0i)-\left({\frac {r_{0}\cdot {\sqrt {3}}}{\pi }}\right)\cdot i+|DM|i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-|BK|-r_{6}i\quad$ , reducing fraction
$\Leftrightarrow S_{6}=(0+0i)-\left({\frac {r_{0}\cdot {\sqrt {3}}}{\pi }}\right)\cdot i+{\frac {1}{2}}r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-|BK|-r_{6}i\quad$ , applying calculation(1)
$\Leftrightarrow S_{6}=(0+0i)-\left({\frac {r_{0}\cdot {\sqrt {3}}}{\pi }}\right)\cdot i+{\frac {1}{2}}r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-|BH|-r_{6}i\quad$ , since ${\overline {BD}}$ and ${\overline {BM}}$ are tangent to the circle around $S_{6}$
$\Leftrightarrow S_{6}=(0+0i)-\left({\frac {r_{0}\cdot {\sqrt {3}}}{\pi }}\right)\cdot i+{\frac {1}{2}}r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}}{\tan {15^{\circ }}}}-r_{6}i\quad$ , applying the tan-formula to the right triangle $\triangle {BHS_{6}}$
$\Leftrightarrow S_{6}=(0+0i)+{\frac {1}{2}}r_{0}\cdot i-\left({\frac {r_{0}\cdot {\sqrt {3}}}{\pi }}\right)\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}}{\tan {15^{\circ }}}}-r_{6}i\quad$ , rearranging
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {1}{2}}-{\frac {\sqrt {3}}{\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}}{\tan {15^{\circ }}}}-r_{6}i\quad$ , rearranging
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}}{\tan {15^{\circ }}}}-r_{6}i\quad$ , rearranging
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}}{2-{\sqrt {3}}}}-r_{6}i\quad$ , applying tangent-formula
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}\cdot (2+{\sqrt {3}})}{(2-{\sqrt {3}})(2+{\sqrt {3}})}}-r_{6}i\quad$ , expanding
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-{\frac {r_{6}\cdot (2+{\sqrt {3}})}{(4-3)}}-r_{6}i\quad$ , applying binomial formula
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-\left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\cdot {\frac {(2+{\sqrt {3}})}{(1)}}-r_{6}i\quad$ , applying calculation (10)
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+{\frac {{\sqrt {3}}\cdot r_{0}}{2}}-\left({\frac {6-3{\sqrt {3}}}{2}}\right)\cdot r_{0}-r_{6}i\quad$ , multiplying
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+\left({\frac {\sqrt {3}}{2}}-{\frac {6-3{\sqrt {3}}}{2}}\right)\cdot r_{0}-r_{6}i\quad$ , rearranging
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+\left({\frac {4{\sqrt {3}}-6}{2}}\right)\cdot r_{0}-r_{6}i\quad$ , rearranging
$\Leftrightarrow S_{6}=(0+0i)+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i+\left(2{\sqrt {3}}-3\right)\cdot r_{0}-r_{6}i\quad$ , reducing fraction
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+0i\right)\cdot r_{0}+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i-r_{6}i\quad$ , rearranging
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+0i\right)\cdot r_{0}+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i-\left({\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\cdot i\quad$ , applying calculation (10)
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+0i\right)\cdot r_{0}+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}-{\frac {21-12{\sqrt {3}}}{2}}\right)\cdot r_{0}\cdot i\quad$ , rearranging
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+0i\right)\cdot r_{0}+\left({\frac {\pi -2{\sqrt {3}}}{2\pi }}-{\frac {21\pi -12\pi {\sqrt {3}}}{2\pi }}\right)\cdot r_{0}\cdot i\quad$ , expanding fraction
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+0i\right)\cdot r_{0}+\left({\frac {12\pi {\sqrt {3}}-2{\sqrt {3}}-20\pi }{2\pi }}\right)\cdot r_{0}\cdot i\quad$ , adding
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+0i\right)\cdot r_{0}+\left({\frac {6\pi {\sqrt {3}}-{\sqrt {3}}-10\pi }{\pi }}\right)\cdot r_{0}\cdot i\quad$ , reducing fraction
$\Leftrightarrow S_{6}=\left((2{\sqrt {3}}-3)+\left({\frac {6\pi {\sqrt {3}}-{\sqrt {3}}-10\pi }{\pi }}\right)\cdot i\right)\cdot r_{0}\quad \approx (0.464-0.159i)\cdot r_{0}\quad$ , rearranging

Licensing edit

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

This file is licensed under the Creative Commons Attribution 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.

File history

Click on a date/time to view the file as it appeared at that time.

	Date/Time	Thumbnail	Dimensions	User	Comment
current	18:41, 26 November 2023		2,412 × 1,594 (159 KB)	Hans G. Oberlack (talk \| contribs)	Points corrected
	15:50, 26 November 2023		2,412 × 1,594 (180 KB)	Hans G. Oberlack (talk \| contribs)	Uploaded own work with UploadWizard

You cannot overwrite this file.

File usage on Commons

The following 2 pages use this file:

Metadata

This file contains additional information such as Exif metadata which may have been added by the digital camera, scanner, or software program used to create or digitize it. If the file has been modified from its original state, some details such as the timestamp may not fully reflect those of the original file. The timestamp is only as accurate as the clock in the camera, and it may be completely wrong.

Horizontal resolution	129.92 dpc
Vertical resolution	129.92 dpc