Summary edit

The 120°-degree circular sector (fan) as base element.
Inscribed is the broadest isosceles triangle.

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

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: ${\displaystyle P_{1}=|AM|+|MB|+|BDA|=2\cdot r_{0}+{\sqrt {3}}\cdot r_{0}\quad \approx 3.732\cdot r_{0}}$

Areas in the general case edit

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

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)

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: ${\displaystyle a_{1}=r_{0}={\sqrt {\frac {3}{\pi }}}\quad \approx 0.977}$

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: ${\displaystyle P_{1}=(2+{\sqrt {3}})\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 3.6469788}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}\approx 7.6480423}$

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 ${\displaystyle A_{1}={\frac {\sqrt {3}}{4}}\cdot {\frac {3}{\pi }}\approx 0.413}$

Centroids in the normalised case edit

0) Centroid of the base shape: ${\displaystyle S_{0}=0+0i}$
1) Centroid of the inscribed triangle: ${\displaystyle S_{1}=\left(0+{\frac {\pi -3{\sqrt {3}}}{3\pi }}\cdot i\right)\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 0-0.213i}$

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

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

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