Краткое описание

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

General case

Segments in the general case

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

Perimeters in the general case

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

Areas in the general case

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: ${\displaystyle A_{1}=\pi \cdot r_{1}^{2}=\pi \cdot \left({\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\right)^{2}={\frac {\pi }{12}}\cdot a_{0}^{2}\quad \approx 0.262\cdot a_{0}^{2}}$

Covered surface of base shape:${\displaystyle C_{b}={\frac {A_{1}}{A_{0}}}={\frac {{\frac {\pi }{12}}\cdot a_{0}^{2}}{{\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}}}={\frac {\frac {\pi }{12}}{\frac {\sqrt {3}}{4}}}={\frac {\pi }{3{\sqrt {3}}}}\quad \approx 60.5\%}$

Centroids in the general case

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed circle relative to the centroid of the base shape is: ${\displaystyle S_{1}=0+0i\quad }$

Normalised case

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

Segments in the normalised case

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: ${\displaystyle r_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}={\frac {1}{2{\sqrt {3}}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {\frac {1}{3{\sqrt {3}}}}}\quad \approx 0.439\quad }$,

Perimeters in the normalised case

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: ${\displaystyle P_{1}={\frac {\pi }{\sqrt {3}}}\cdot a_{0}={\frac {\pi }{\sqrt {3}}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\frac {2\pi }{\sqrt {3{\sqrt {3}}}}}\quad \approx 2.756}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 7.315}$

Areas in the normalised case

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: ${\displaystyle A_{1}={\frac {\pi }{12}}\cdot \left({\frac {2}{\sqrt {\sqrt {3}}}}\right)^{2}={\frac {4\pi }{12{\sqrt {3}}}}={\frac {\pi }{3{\sqrt {3}}}}\quad \approx 0.605}$

Centroids in the normalised case

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed semicircle relative to the centroid of the base shape is: ${\displaystyle S_{1}=0+0i}$

Calculations

Known elements

(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)${\displaystyle \quad \beta ={\frac {\pi }{6}}}$

Calculation 1

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

Calculation 2

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

Calculation 3

${\displaystyle \quad {\frac {|DS_{0}|}{|BD|}}=\tan {\frac {\pi }{6}}\quad }$
${\displaystyle \quad \Leftrightarrow {\frac {r_{1}}{|BD|}}=\tan {\frac {\pi }{6}}\quad }$, 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
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.289\quad }$, rearranging

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