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.

General case edit

Segments in the general case edit

0) The radius of the base circular sector: ${\displaystyle r_{0}}$
1) The side length of the inscribed triangle: ${\displaystyle a_{1}=r_{0}}$, because it is the broadest triangle
2) The radius of the inscribed circle: ${\displaystyle r_{2}={\frac {1}{4}}\cdot r_{0}}$, see calculation (5)

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 ${\displaystyle S_{2}}$: ${\displaystyle P_{2}=2\pi r_{2}=2\pi \cdot {\frac {1}{4}}\cdot r_{0}={\frac {\pi }{2}}\cdot r_{0}\quad \approx 1.571r_{0}}$

Areas in the general case edit

0) Area of the base circular sector ${\displaystyle A_{0}={\frac {\pi \cdot r_{0}^{2}}{3}}\quad \approx 1.047\cdot r_{0}^{2}}$
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 ${\displaystyle A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot ({\frac {1}{4}}\cdot r_{0})^{2}={\frac {\pi }{16}}\cdot r_{0}^{2}\quad \approx 0.196\cdot r_{0}^{2}}$

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)

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) The radius of the inscribed circle: ${\displaystyle r_{2}={\frac {1}{4}}\cdot {\sqrt {\frac {3}{\pi }}}={\sqrt {\frac {3}{16\pi }}}\quad \approx 0.244}$

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 ${\displaystyle S_{2}}$: ${\displaystyle P_{2}={\frac {\pi }{2}}\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 1.5349901}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}\approx 9.1830323}$

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 ${\displaystyle A_{2}={\frac {\pi }{16}}\cdot \left({\sqrt {\frac {3}{\pi }}}\right)^{2}\quad \approx 0.188}$

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: ${\displaystyle S_{2}=\left(0+{\frac {3\pi -4{\sqrt {3}}}{4\pi }}\cdot i\right){\sqrt {\frac {3}{\pi }}}\quad \approx (0+0.194i)\quad }$

Calculations edit

Given elements edit

(1) ${\displaystyle |AM|=|BM|=|MC|=r_{0}=a_{1}}$
(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) ${\displaystyle |CD|=2\cdot r_{2}}$

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

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