Краткое описание

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

Perimeters in the general case

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

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

Covered surface of base shape:${\displaystyle C_{b}={\frac {A_{1}+A_{2}}{A_{0}}}={\frac {{\frac {\pi }{12}}\cdot a_{0}^{2}+{\frac {\pi }{108}}\cdot a_{0}^{2}}{{\frac {\sqrt {3}}{4}}\cdot a_{0}^{2}}}={\frac {{\frac {\pi }{3}}+{\frac {\pi }{27}}}{\sqrt {3}}}={\frac {10\pi }{27{\sqrt {3}}}}\quad \approx 67.1\%}$

Centroids in the general case

1) Centroids as graphically displayed

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) Centroids of the additional circle: ${\displaystyle S_{2}=S_{0}+r_{1}i+r_{2}i=0+{\frac {a_{0}i}{2{\sqrt {3}}}}+{\frac {a_{0}i}{6{\sqrt {3}}}}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}\quad \approx (0+0.385i)\cdot a_{0}}$

2) Orientated centroids

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: ${\displaystyle S_{2}=\left(0-{\frac {2i}{3{\sqrt {3}}}}\right)\cdot a_{0}\quad \approx (0-0.385i)\cdot a_{0}}$

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

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

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

Centroids in the general case

1) Centroids as graphically displayed

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) Centroids of the additional circle: ${\displaystyle S_{2}=\left(0+{\frac {2i}{3{\sqrt {3}}}}\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=\left(0+{\frac {4}{3{\sqrt {3{\sqrt {3}}}}}}\cdot i\right)\quad \approx (0+0.585i)}$

2) Orientated centroids

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: ${\displaystyle S_{2}=\left(0-{\frac {4}{3{\sqrt {3{\sqrt {3}}}}}}\cdot i\right)\quad \approx (0-0.585i)}$

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)${\displaystyle \quad \beta =\gamma ={\frac {\pi }{6}}}$
(5)${\displaystyle \quad |S_{0}F|=|S_{0}G|=r_{2}}$
(6)${\displaystyle \quad |S_{0}S_{2}|=r_{1}+r_{2}}$

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
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\quad \approx 0.289\cdot a_{0}\quad }$, rearranging

Calculation 4

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

Calculation 5

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

Лицензирование

Я, владелец авторских прав на это произведение, добровольно публикую его на условиях следующей лицензии:

Этот файл доступен по лицензии Creative Commons Attribution-Share Alike 4.0 International

Вы можете свободно:

• делиться произведением – копировать, распространять и передавать данное произведение
• создавать производные – переделывать данное произведение

При соблюдении следующих условий:

• атрибуция – Вы должны указать авторство, предоставить ссылку на лицензию и указать, внёс ли автор какие-либо изменения. Это можно сделать любым разумным способом, но не создавая впечатление, что лицензиат поддерживает вас или использование вами данного произведения.
• распространение на тех же условиях – Если вы изменяете, преобразуете или создаёте иное произведение на основе данного, то обязаны использовать лицензию исходного произведения или лицензию, совместимую с исходной.

https://creativecommons.org/licenses/by-sa/4.0CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 truetrue