File:FS EC(3) dia.png

The equilateral triangle as base element.
Inscribed is the largest circle.
Inscribed is the next largest circle.
Inscribed is the next largest circle.

0) The side length of the equilateral base triangle is: ${\displaystyle a_{0}}$
1) The radius of the circle is: ${\displaystyle r_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.289\cdot a_{0}\quad }$, see calculation 3
2) The radius of the circle is: ${\displaystyle r_{2}={\frac {1}{6{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.096\cdot a_{0}\quad }$, see calculation 4
3) The radius of the circle is: ${\displaystyle r_{3}=r_{2}={\frac {1}{6{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.096\cdot a_{0}\quad }$, for symmetry reasons

0) Perimeter of equilateral base triangle: ${\displaystyle P_{0}=3\cdot a_{0}}$
1) Perimeter of inscribed circle: ${\displaystyle P_{1}=2\cdot \pi \cdot r_{1}=2\pi \cdot {\frac {1}{2{\sqrt {3}}}}\cdot a_{0}={\frac {\pi }{\sqrt {3}}}\cdot a_{0}\quad \approx 1.814\cdot a_{0}}$
2) Perimeter of inscribed circle: ${\displaystyle P_{2}=2\cdot \pi \cdot r_{2}=2\pi \cdot {\frac {1}{6{\sqrt {3}}}}\cdot a_{0}={\frac {\pi }{3{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.605\cdot a_{0}}$
3) Perimeter of inscribed circle: ${\displaystyle P_{3}=P_{2}={\frac {\pi }{3{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.605\cdot a_{0}}$

0) Area of the equilateral base triangle: ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }$, see calculation (2)
1) Area of the inscribed circle: ${\displaystyle A_{1}=\pi \cdot r_{1}^{2}=\pi \cdot \left({\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\right)^{2}={\frac {\pi }{12}}\cdot a_{0}^{2}\quad \approx 0.262\cdot a_{0}^{2}}$
2) Area of the inscribed circle: ${\displaystyle A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot \left({\frac {1}{6{\sqrt {3}}}}\cdot a_{0}\right)^{2}={\frac {\pi }{108}}\cdot a_{0}^{2}\quad \approx 0.029\cdot a_{0}^{2}}$
2) Area of the inscribed circle: ${\displaystyle A_{3}=A_{2}=\pi \cdot r_{2}^{2}={\frac {\pi }{108}}\cdot a_{0}^{2}\quad \approx 0.029\cdot a_{0}^{2}}$

Covered surface of base shape:${\displaystyle C_{b}={\frac {A_{1}+A_{2}+A3}{A_{0}}}={\frac {A_{1}+2A_{2}}{A_{0}}}={\frac {{\frac {\pi }{3}}+{\frac {2\pi }{27}}}{\sqrt {3}}}={\frac {11\pi }{27{\sqrt {3}}}}\quad \approx 73,9\%}$

Centroid positions are measured from the centroid point of the base shape
0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed circle: ${\displaystyle S_{1}=(0+0i)\cdot a_{0}\quad }$
2) Centroid of the additional circle: ${\displaystyle S_{2}=S_{0}+r_{1}i+r_{2}i=0+{\frac {a_{0}i}{2{\sqrt {3}}}}+{\frac {a_{0}i}{6{\sqrt {3}}}}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}\quad \approx (0+0.385i)\cdot a_{0}}$
3) Centroid of the additional circle: ${\displaystyle S_{3}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)\cdot a_{0}=\left(-{\frac {1}{3}}-{\frac {1}{3{\sqrt {3}}}}i\right)\cdot a_{0}\quad \approx (-0.333-0.192i)\cdot a_{0}}$, see Calculation (6)

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the inscribed circle: ${\displaystyle S_{2}=(0+0i)\cdot a_{0}}$
2) Orientated centroid of the additional circle: ${\displaystyle S_{2}=\left(0-{\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}\quad \approx (0-0.385i)\cdot a_{0}}$
3) Orientated centroid of the additional circle: ${\displaystyle S_{3}=\left(0-{\frac {2i}{3{\sqrt {3}}}}\right)\cdot \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)\cdot a_{0}=\left({\frac {1}{3}}+{\frac {1}{3{\sqrt {3}}}}i\right)\cdot a_{0}\quad \approx (0.333+0.192i)\cdot a_{0}}$, see Calculation (6)

In the normalised case the area of the base shape is set to 1.
So ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}={\frac {4}{\sqrt {3}}}\quad \Rightarrow a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}$

0) Side length of the triangle ${\displaystyle a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}$
1) The radius of the circle is: ${\displaystyle r_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}={\frac {1}{2{\sqrt {3}}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {\frac {1}{3{\sqrt {3}}}}}\quad \approx 0.439\quad }$,
2) The radius of the additional circle is: ${\displaystyle r_{2}={\frac {1}{6{\sqrt {3}}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {\frac {1}{27{\sqrt {3}}}}}\quad \approx 0.146\quad }$
2) The radius of the additional circle is: ${\displaystyle r_{3}=r_{2}={\sqrt {\frac {1}{27{\sqrt {3}}}}}\quad \approx 0.146\quad }$

0) Perimeter of base triangle: ${\displaystyle P_{0}=3\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {12{\sqrt {3}}}}\quad \approx 4.559}$
1) Perimeter of inscribed circle: ${\displaystyle P_{1}={\frac {\pi }{\sqrt {3}}}\cdot a_{0}={\frac {\pi }{\sqrt {3}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\frac {2\pi }{\sqrt {3{\sqrt {3}}}}}\quad \approx 2.756}$
2) Perimeter of inscribed circle: ${\displaystyle P_{2}={\frac {\pi }{3{\sqrt {3}}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\frac {2\pi }{3{\sqrt {3{\sqrt {3}}}}}}\quad \approx 0.919}$
3) Perimeter of inscribed circle: ${\displaystyle P_{3}=P_{2}={\frac {2\pi }{3{\sqrt {3{\sqrt {3}}}}}}\quad \approx 0.919}$

0) Area of the base triangle is by definition ${\displaystyle A_{0}={\frac {\sqrt {3}}{4}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}=1}$
1) Area of the inscribed circle: ${\displaystyle A_{1}={\frac {\pi }{12}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}={\frac {4\pi }{12{\sqrt {3}}}}={\frac {\pi }{3{\sqrt {3}}}}\quad \approx 0.605}$
2) Area of the inscribed circle: ${\displaystyle A_{2}={\frac {\pi }{108}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}={\frac {4\pi }{108{\sqrt {3}}}}={\frac {\pi }{27{\sqrt {3}}}}\quad \approx 0.067}$
3) Area of the inscribed circle: ${\displaystyle A_{3}=A_{2}={\frac {\pi }{27{\sqrt {3}}}}\quad \approx 0.067}$

Centroid positions are measured from the centroid point of the base shape
0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed circle: ${\displaystyle S_{1}=(0+0i)\quad }$
2) Centroid of the additional circle: ${\displaystyle S_{2}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=\left(0+{\frac {4}{3{\sqrt {3{\sqrt {3}}}}}}\cdot i\right)\quad \approx (0+0.585i)}$
3) Centroid of the second additional circle: ${\displaystyle S_{3}=\left(-{\frac {1}{3}}-{\frac {1}{3{\sqrt {3}}}}i\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=\left(-{\frac {2}{3{\sqrt {\sqrt {3}}}}}-{\frac {2}{3{\sqrt {3{\sqrt {3}}}}}}\cdot i\right)\quad \approx (-0.506-0.292i)}$

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the inscribed circle: ${\displaystyle S_{2}=(0+0i)}$
2) Orientated centroid of the additional circle: ${\displaystyle S_{2}=\left(0-{\frac {4}{3{\sqrt {3{\sqrt {3}}}}}}\cdot i\right)\quad \approx (0-0.585i)}$
3) Centroid of the second additional circle: ${\displaystyle S_{3}=\left({\frac {1}{3}}+{\frac {1}{3{\sqrt {3}}}}i\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=\left({\frac {2}{3{\sqrt {\sqrt {3}}}}}+{\frac {2}{3{\sqrt {3{\sqrt {3}}}}}}\cdot i\right)\quad \approx (0.506+0.292i)}$

(0) Given is the side length of the equilateral triangle: ${\displaystyle a_{0}}$
(1)${\displaystyle \quad |AB|=|BC|=|AC|=a_{0}}$
(2)${\displaystyle \quad |AD|=|BD|=|BE|=|CE|={\frac {1}{2}}\cdot a_{0}}$
(3)${\displaystyle \quad |DS_{0}|=|ES_{0}|=r_{1}}$
(4)${\displaystyle \quad \beta =\gamma ={\frac {\pi }{6}}}$
(5)${\displaystyle \quad |S_{0}F|=|S_{0}G|=r_{2}}$
(6)${\displaystyle \quad |S_{0}S_{2}|=r_{1}+r_{2}}$

The height ${\displaystyle |CD|}$ is calculated:
${\displaystyle \quad |AD|^{2}+|CD|^{2}=|AC|^{2}\quad }$,applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle ABC}$
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=|AC|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=a_{0}^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{0}^{2}-{\frac {1}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |CD|^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |CD|={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, rearranging

${\displaystyle \quad A_{0}=|AD|\cdot |CD|}$
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot |CD|\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, applying result of calculation (2)
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}\quad \approx 0.433\cdot a_{0}^{2}\quad }$

${\displaystyle \quad {\frac {|DS_{0}|}{|BD|}}=\tan {\frac {\pi }{6}}\quad }$
${\displaystyle \quad \Leftrightarrow {\frac {r_{1}}{|BD|}}=\tan {\frac {\pi }{6}}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow {\frac {r_{1}}{{\frac {1}{2}}\cdot a_{0}}}=\tan {\frac {\pi }{6}}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{2}}\cdot a_{0}\cdot \tan {\frac {\pi }{6}}\quad }$rearranging
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{2}}\cdot a_{0}\cdot {\frac {1}{\sqrt {3}}}\quad }$applying the tan-formula
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.289\cdot a_{0}\quad }$, rearranging

To calculate ${\displaystyle r_{2}}$ we use the two right triangles ${\displaystyle \triangle {FCS_{2}}}$ and ${\displaystyle \triangle {ADF}:}$
${\displaystyle \quad |AD|^{2}+|DC|^{2}=|AC|^{2}\quad }$, applying the Pythagorean theoreme on ${\displaystyle \triangle {ADC}}$
${\displaystyle \quad \Leftrightarrow \left(\mathbf {{\frac {1}{2}}\cdot a_{0}} \right)^{2}+|DC|^{2}=|AC|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|DC|^{2}=\mathbf {a_{0}} ^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow |DC|^{2}=a_{0}^{2}-\left({\frac {1}{2}}\cdot a_{0}\right)^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |DC|^{2}=a_{0}^{2}-{\frac {1}{4}}\cdot a_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow |DC|^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow \left(|DS_{0}|+|S_{0}S_{2}|+|S_{2}G|+|GC|\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, breaking into parts
${\displaystyle \quad \Leftrightarrow \left(r_{1}+|S_{0}S_{2}|+|S_{2}G|+|GC|\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow \left(r_{1}+r_{1}+r_{2}+|S_{2}G|+|GC|\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow \left(r_{1}+r_{1}+r_{2}+r_{2}+|GC|\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow \left(2r_{1}+2r_{2}+|GC|\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow \left(2r_{1}+2r_{2}+r_{2}\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, see calculation (5)
${\displaystyle \quad \Leftrightarrow \left(2r_{1}+3r_{2}\right)^{2}={\frac {3}{4}}\cdot a_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow 2r_{1}+3r_{2}={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, extracting the roots
${\displaystyle \quad \Leftrightarrow 3r_{2}={\frac {\sqrt {3}}{2}}\cdot a_{0}-2r_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow 3r_{2}={\frac {\sqrt {3}}{2}}\cdot a_{0}-2\cdot {\frac {1}{2{\sqrt {3}}}}a_{0}\quad }$, applying calculation (3)
${\displaystyle \quad \Leftrightarrow 3r_{2}={\frac {{\sqrt {3}}{\sqrt {3}}}{2{\sqrt {3}}}}\cdot a_{0}-{\frac {2}{2{\sqrt {3}}}}a_{0}\quad }$, expanding
${\displaystyle \quad \Leftrightarrow 3r_{2}=\left({\frac {3}{2{\sqrt {3}}}}-{\frac {2}{2{\sqrt {3}}}}\right)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow 3r_{2}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {1}{6{\sqrt {3}}}}\cdot a_{0}\quad }$, rearranging

In order to find the size of ${\displaystyle |GC|}$ the right triangle ${\displaystyle \triangle {CS_{0}F}}$ is used:
${\displaystyle {\frac {|S_{0}F|}{|GC|+|S_{0}G|}}=\sin {\gamma }}$
${\displaystyle {\frac {r_{2}}{|GC|+|S_{0}G|}}=\sin {\gamma }\quad }$, applying equation (5)
${\displaystyle {\frac {r_{2}}{|GC|+r_{2}}}=\sin {\gamma }=\sin {\frac {\pi }{6}}\quad }$, applying equation (4)
${\displaystyle {\frac {r_{2}}{|GC|+r_{2}}}={\frac {1}{2}}\quad }$, applying the Sinus function
${\displaystyle 2r_{2}=|GC|+r_{2}\quad }$, rearranging
${\displaystyle r_{2}=|GC|\quad }$, rearranging

In order to find the position of centroid ${\displaystyle S_{3}}$ the position of centroid ${\displaystyle S_{2}}$ is rotated by ${\displaystyle 120^{\circ }}$, since the triangle ${\displaystyle \triangle {ABC}}$ is an equilateral triangle. The rotation by ${\displaystyle 120^{\circ }}$ corresponds to a multiplication with ${\displaystyle \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)}$ so this gives:

a) In the general case as displayed:
${\displaystyle S_{3}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}\cdot \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)}$
${\displaystyle \Leftrightarrow S_{3}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left(-{\frac {1}{2}}\cdot {\frac {2i}{3{\sqrt {3}}}}+{\frac {1}{2}}{\sqrt {3}}i\cdot {\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left(-{\frac {i}{3{\sqrt {3}}}}+{\frac {1}{2}}{\sqrt {3}}i\cdot {\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left(-{\frac {i}{3{\sqrt {3}}}}+{\frac {i^{2}}{3}}\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left(-{\frac {i}{3{\sqrt {3}}}}+{\frac {-1}{3}}\right)\cdot a_{0}}$, since ${\displaystyle i^{2}=-1}$
${\displaystyle \Leftrightarrow S_{3}=\left(-{\frac {1}{3}}-{\frac {1}{3{\sqrt {3}}}}i\right)\cdot a_{0}\quad \approx (-0.333-0.192i)\cdot a_{0}}$

b) In the orientated general case:
${\displaystyle S_{3}=\left(0-{\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}\cdot \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)}$
${\displaystyle \Leftrightarrow S_{3}=\left(0-{\frac {2i}{3{\sqrt {3}}}}\right)\cdot \left(-{\frac {1}{2}}+{\frac {1}{2}}{\sqrt {3}}i\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left(-{\frac {1}{2}}\cdot {\frac {-2i}{3{\sqrt {3}}}}+{\frac {1}{2}}{\sqrt {3}}i\cdot {\frac {-2i}{3{\sqrt {3}}}}\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left({\frac {i}{3{\sqrt {3}}}}+{\frac {1}{2}}{\sqrt {3}}i\cdot {\frac {-2i}{3{\sqrt {3}}}}\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left({\frac {i}{3{\sqrt {3}}}}+{\frac {-i^{2}}{3}}\right)\cdot a_{0}}$
${\displaystyle \Leftrightarrow S_{3}=\left({\frac {i}{3{\sqrt {3}}}}+{\frac {1}{3}}\right)\cdot a_{0}}$, since ${\displaystyle -i^{2}=1}$
${\displaystyle \Leftrightarrow S_{3}=\left({\frac {1}{3}}+{\frac {1}{3{\sqrt {3}}}}i\right)\cdot a_{0}\quad \approx (0.333+0.192i)\cdot a_{0}}$

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