User:Hans G. Oberlack/FS E

The equilateral triangle as base element.

General case

Segments in the general case

The side of the equilateral triangle ABC $|AB|=|BC|=|AC|=s$
The segment DC is the height h of the triangle. h depends on the length of the side:
$|AD|={\frac {s}{2}}\quad$
Applying the Pythagorean theorem to the triangle ADC leads to:
$|AD|^{2}+|DC|^{2}=|AC|^{2}$
$\Rightarrow h^{2}+({\frac {s}{2}})^{2}=s^{2}$
$\Rightarrow h^{2}=s^{2}-{\frac {s^{2}}{4}}={\frac {3}{4}}\cdot s^{2}$
$\Rightarrow h={\sqrt {(}}{\frac {3}{4}}s^{2})={\frac {\sqrt {3}}{2}}\cdot s$

Perimeter in the general case

Perimeter of base equilateral triangle: $P_{0}=3\cdot s$

Areas in the general case

Area of the base equilateral triangle: $A_{0}=h\cdot {\frac {s}{2}}={\frac {\sqrt {3}}{2}}\cdot s\cdot {\frac {s}{2}}={\frac {\sqrt {3}}{4}}\cdot s^{2}$

Centroids in the general case

By definition the centroid points of a base shape are $x_{0}=0\quad y_{0}=0$. Relatively is the lower left point of the of the base of the triangle at: $x_{L}=-{\frac {s}{2}}\quad y_{L}=-{\frac {h}{3}}=-{\frac {\sqrt {3}}{2\cdot 3}}\cdot s=-{\frac {s}{2\cdot {\sqrt {3}}}}$

Normalised case

Areas in the normalised case

In the normalised case the area of the base equilateral triangle is set to A_0=1.
So $A={\frac {\sqrt {3}}{4}}\cdot s^{2}=1\quad \Rightarrow s^{2}={\frac {4}{\sqrt {3}}}\quad \Rightarrow s={\frac {2}{\sqrt[{4}]{3}}}=1.519671...$

Segments in the normalised case

Segment of the base equilateral triangle $s={\frac {2}{\sqrt[{4}]{3}}}=1.519671...$

Perimeters in the normalised case

Perimeter of base equilateral triangle: $P_{0}=3\cdot s=3\cdot {\frac {2}{\sqrt[{4}]{3}}}={\frac {6}{\sqrt[{4}]{3}}}=4.5590141...$

Centroids in the normalised case

Centroid positions of the base equilateral triangle are $x_{0}=0\quad y_{0}=0$. So the lower left point of the base of the triangle is at:
$\quad x_{L}=-{\frac {s}{2}}=-{\frac {1}{\sqrt[{4}]{3}}}=-0.7598...$
$\quad y_{L}=-{\frac {s}{2\cdot {\sqrt {3}}}}=-s\cdot {\frac {1}{2\cdot {\sqrt {3}}}}=-{\frac {2}{\sqrt[{4}]{3}}}\cdot {\frac {1}{2\cdot {\sqrt {3}}}}=-{\frac {1}{{\sqrt[{4}]{3}}\cdot {\sqrt {3}}}}=-{\frac {1}{\sqrt[{4}]{(3^{3})}}}=-{\frac {1}{\sqrt[{4}]{27}}}=-0.43869...$

Identifying number

Apart of the base element there is no other shape allocated. Therefore the integer part of the identifying number is 0.
The decimal part of the identifying number is the decimal part of the sum of the perimeters and the distances of the centroids in the normalised case.

$decimalpart(4.5590141...+0)=decimalpart(4.5590141...)=.5590141...$

So the identifying number is: $0.5590141$