User:Hans G. Oberlack/FS E



The equilateral triangle as base element.

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:  $$x_L=-\frac {s}{2} \quad   y_L= - \frac{h}{3} = - \frac{\sqrt 3} {2 \cdot 3} \cdot s= - \frac{s}{2 \cdot \sqrt 3}$$

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: $$0.5590141$$