File:FS EQ dia.png

Summary[edit]

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

General case[edit]

Segments in the general case[edit]

0) The side length of the equilateral base triangle is: ${\displaystyle a_{0}}$
1) The side length of the inscribed square is: ${\displaystyle a_{1}=(2{\sqrt {3}}-3)\cdot a_{0}\quad \approx 0.464\cdot a_{0}\quad }$, see calculation 3

Perimeters in the general case[edit]

0) Perimeter of equilateral base triangle: ${\displaystyle P_{0}=3\cdot a_{0}}$
1) Perimeter of inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot (2{\sqrt {3}}-3)\cdot a_{0}\quad \approx 1.864\cdot a_{0}}$

Areas in the general case[edit]

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 square: ${\displaystyle A_{1}=\left((2{\sqrt {3}}-3)\cdot a_{0}\right)^{2}=(21-12{\sqrt {3}})\cdot a_{0}^{2}\quad \approx 0.215\cdot a_{0}^{2}}$

Centroids in the general case[edit]

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed square relative to the centroid of the base shape is: ${\displaystyle S_{1}=\left(0-{\frac {9-5{\sqrt {3}}}{6}}\cdot i\right)\cdot a_{0}\quad \approx (0-0.057i)\cdot a_{0}\quad }$, see calculation (4)

Normalised case[edit]

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[edit]

0) Side length of the triangle ${\displaystyle a_{0}={\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 1.52}$
1) The side length of the square is: ${\displaystyle a_{1}=(2{\sqrt {3}}-3)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 0.705\quad }$,

Perimeters in the normalised case[edit]

0) Perimeter of base triangle: ${\displaystyle P_{0}=3\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {12{\sqrt {3}}}}\quad \approx 4.5590141}$
1) Perimeter of inscribed square: ${\displaystyle P_{1}=4\cdot a_{1}=4\cdot (2{\sqrt {3}}-3)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=8\cdot {\frac {2{\sqrt {3}}-3}{\sqrt {\sqrt {3}}}}\quad \approx 2.8211278}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 7.3801419}$

Areas in the normalised case[edit]

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 square: ${\displaystyle A_{1}=(21-12{\sqrt {3}})\cdot ({\frac {2}{\sqrt {\sqrt {3}}}})^{2}=3\cdot (7-4{\sqrt {3}})\cdot {\frac {4}{\sqrt {3}}}=4{\sqrt {3}}\cdot (7-4{\sqrt {3}})=4\cdot (7{\sqrt {3}}-12)\quad \approx 0.497}$

Centroids in the normalised case[edit]

0) By definition the centroid point of a base shape is ${\displaystyle S_{0}=0+0i}$
1) The centroid point of the inscribed square relative to the centroid of the base shape is: ${\displaystyle S_{1}=\left(0-{\frac {9-5{\sqrt {3}}}{6}}\cdot i\right)\cdot {\frac {2}{\sqrt {\sqrt {3}}}}=\left(0-{\frac {9-5{\sqrt {3}}}{3{\sqrt {\sqrt {3}}}}}\cdot i\right)\quad \approx 0-0.086i}$

Distances of centroids[edit]

The distance between the centroid of the base triangle and the centroid of the semicircle is:
${\displaystyle d_{01}={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.0860504}$
Sum of distances: ${\displaystyle d_{s}=d_{01}=0.0860504}$

Identifying number[edit]

Apart of the base element there is one other shape allocated. Therefore the integer part of the identifying number is 1.
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.

${\displaystyle decimalpart(7.3801419+0.0860504)=decimalpart(7.4661922)=.4661922}$

So the identifying number is: ${\displaystyle 1.4661922}$

Calculations[edit]

Known elements[edit]

(0) Given is the side length of the equilateral triangle: ${\displaystyle a_{0}}$
(1)${\displaystyle \quad |AB|=|BC|=|AC|=a_{0}}$
(2)${\displaystyle \quad |AD|=|BD|={\frac {1}{2}}\cdot a_{0}}$
(3)${\displaystyle \quad |FG|=|EG|=a_{1}}$
(4)${\displaystyle \quad |FD|=|DG|={\frac {1}{2}}\cdot a_{1}}$
(5)${\displaystyle \quad \beta ={\frac {\pi }{3}}}$, because ${\displaystyle \triangle ABC}$ is equilateral

Calculation 1[edit]

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[edit]

${\displaystyle \quad A_{0}=|AD|\cdot |CD|}$
${\displaystyle \quad \Leftrightarrow A_{0}={\frac {1}{2}}\cdot a_{0}\cdot |CD|\quad }$, 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[edit]

${\displaystyle \quad {\frac {|EG|}{|BG|}}=\tan {\beta }\quad }$
${\displaystyle \quad \Leftrightarrow {\frac {|EG|}{|BG|}}=\tan {\frac {\pi }{3}}\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow {\frac {a_{1}}{|BG|}}=\tan {\frac {\pi }{3}}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow a_{1}=\tan {\frac {\pi }{3}}\cdot |BG|\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {3}}\cdot |BG|\quad }$, applying the tan-formula
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {3}}\cdot \left(|BD|-|DG|\right)\quad }$, applying the tan-formula
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {3}}\cdot \left({\frac {1}{2}}\cdot a_{0}-|DG|\right)\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {3}}\cdot \left({\frac {1}{2}}\cdot a_{0}-{\frac {1}{2}}\cdot a_{1}\right)\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {\sqrt {3}}{2}}\cdot a_{0}-{\frac {\sqrt {3}}{2}}\cdot a_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}+{\frac {\sqrt {3}}{2}}\cdot a_{1}={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {2+{\sqrt {3}}}{2}}\cdot a_{1}={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {2}{2+{\sqrt {3}}}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {\sqrt {3}}{2+{\sqrt {3}}}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {\sqrt {3}}{2+{\sqrt {3}}}}\cdot {\frac {2-{\sqrt {3}}}{2-{\sqrt {3}}}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {2{\sqrt {3}}-3}{2^{2}-({\sqrt {3}})^{2}}}\cdot a_{0}={\frac {2{\sqrt {3}}-3}{4-3}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}=(2{\sqrt {3}}-3)\cdot a_{0}\quad \approx 0.464\cdot a_{0}}$

Calculation 4[edit]

${\displaystyle \quad S_{1}=S_{0}+{\overrightarrow {S_{0}D}}+{\overrightarrow {DS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)-(0+|DS_{0}|\cdot i)+(0+|DS_{1}|\cdot i)\quad }$
${\displaystyle \quad \Leftrightarrow S_{1}=0-|DS_{0}|\cdot i+|DS_{1}|\cdot i\quad }$, adding the real and the complex terms
${\displaystyle \quad \Leftrightarrow S_{1}=0+|DS_{1}|\cdot i-|DS_{0}|\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left(|DS_{1}|-|DS_{0}|\right)\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {1}{2}}a_{1}-|DS_{0}|\right)\cdot i\quad }$, centroid of square
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {1}{2}}a_{1}-{\frac {1}{3}}\cdot |CD|\right)\cdot i\quad }$, centroid of triangle
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {1}{2}}a_{1}-{\frac {1}{3}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\right)\cdot i\quad }$, applying result of calculation (1)
${\displaystyle \quad \Leftrightarrow S_{1}=0+\left({\frac {1}{2}}\cdot (2{\sqrt {3}}-3)\cdot a_{0}-{\frac {1}{3}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\right)\cdot i\quad }$, applying result of calculation (3)
${\displaystyle \quad \Leftrightarrow S_{1}=0+{\frac {1}{2}}\cdot \left((2{\sqrt {3}}-3)-{\frac {\sqrt {3}}{3}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0+{\frac {1}{2}}\cdot \left({\frac {(6{\sqrt {3}}-9)-{\sqrt {3}}}{3}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0+{\frac {1}{2}}\cdot \left({\frac {5{\sqrt {3}}-9}{3}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(0+{\frac {5{\sqrt {3}}-9}{6}}\cdot i\right)\cdot a_{0}\quad \approx (0-0.057i)\cdot a_{0}}$, rearranging

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