Краткое описание

Task

The equilateral triangle as base element.
Inscribed is the largest regular hexagon.

General case

Segments in the general case

0) The side length of the equilateral base triangle is: ${\displaystyle a_{0}}$
1) The side length of the inscribed regular hexagon is: ${\displaystyle a_{1}={\frac {1}{3}}\cdot a_{0}\quad }$, see calculation (2)

Perimeters in the general case

0) Perimeter of equilateral base triangle: ${\displaystyle P_{0}=3\cdot a_{0}}$
1) Perimeter of inscribed regular hexagon: ${\displaystyle P_{1}=6\cdot a_{1}=6\cdot {\frac {1}{3}}\cdot a_{0}=2\cdot a_{0}}$

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 (1)
1) Area of the inscribed regular hexagon: ${\displaystyle A_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}^{2}={\frac {2}{3}}\cdot A_{0}\quad \approx 0.289\cdot a_{0}^{2}\quad }$ see calculation (3)

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 regular hexagon relative to the centroid of the base shape is: ${\displaystyle S_{1}=0+0i}$, because the hexagon is rotationally symmetrical

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 side length of the inscribed regular hexagon is: ${\displaystyle a_{1}={\frac {1}{3}}\cdot a_{0}={\frac {1}{3}}\cdot {\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx \approx 0.507\quad }$

Perimeters in the normalised case

0) Perimeter of base triangle: ${\displaystyle P_{0}=3\cdot {\frac {2}{\sqrt {\sqrt {3}}}}={\sqrt {12{\sqrt {3}}}}\quad \approx 4.5590141}$
1) Perimeter of regular hexagon: ${\displaystyle P_{1}=2\cdot {\frac {2}{\sqrt {\sqrt {3}}}}\quad \approx 3.0393427}$
S) Sum of perimeters: ${\displaystyle P_{S}=P_{0}+P_{1}\quad \approx 7.5983569}$

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 regular hexagon: ${\displaystyle A_{1}={\frac {2}{3}}\cdot A_{0}\approx 0.667}$

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 regular hexagon relative to the centroid of the base shape is: ${\displaystyle S_{1}=0+0i}$

Distances of centroids

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

Identifying number

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.

${\displaystyle decimalpart(7.5983569+0)=decimalpart(7.5983569)=.5983569}$

So the identifying number is: ${\displaystyle 1.5983569}$

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)${\displaystyle \quad |AD|=|BD|={\frac {1}{2}}\cdot a_{0}}$
(3)${\displaystyle \quad |ED|={\frac {1}{2}}\cdot a_{1}}$
(4)${\displaystyle \quad |ES_{0}|=a_{1}\quad }$, since the inscribed hexagon is regular

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 ${\displaystyle \triangle ADC}$
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{0}\right)^{2}+|CD|^{2}=|AC|^{2}\quad }$, 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

The side length ${\displaystyle a_{1}}$ is calculated:
${\displaystyle \quad |ED|^{2}+|DS_{0}|^{2}=|ES_{0}|^{2}\quad }$,applying the Pythagorean theorem on the rectangular triangle ${\displaystyle \triangle EDS_{0}}$
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{1}\right)^{2}+|DS_{0}|^{2}=|ES_{0}|^{2}\quad }$,applying applying equation (3)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{1}\right)^{2}+|DS_{0}|^{2}=a_{1}^{2}\quad }$,applying applying equation (4)
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{1}\right)^{2}+\left({\frac {1}{3}}\cdot |CD|\right)^{2}=a_{1}^{2}\quad }$, since ${\displaystyle S_{0}}$ is the centroid of the equilateral triangle
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2}}\cdot a_{1}\right)^{2}+\left({\frac {1}{3}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{0}\right)^{2}=a_{1}^{2}\quad }$, applying calculation (1)
${\displaystyle \quad \Leftrightarrow {\frac {1}{4}}\cdot a_{1}^{2}+\left({\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\right)^{2}=a_{1}^{2}\quad }$, multiplicating
${\displaystyle \quad \Leftrightarrow \left({\frac {1}{2{\sqrt {3}}}}\cdot a_{0}\right)^{2}={\frac {3}{4}}\cdot a_{1}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\frac {1}{2{\sqrt {3}}}}\cdot a_{0}={\frac {\sqrt {3}}{2}}\cdot a_{1}\quad }$, extracting the roots
${\displaystyle \quad \Leftrightarrow {\frac {1}{\sqrt {3}}}\cdot a_{0}={\sqrt {3}}\cdot a_{1}\quad }$, multiplying by 2
${\displaystyle \quad \Leftrightarrow {\frac {1}{3}}\cdot a_{0}=a_{1}\quad }$, rearranging
-->

Calculation 3

A regular hexagon with side length ${\displaystyle a_{1}}$ can be partitioned into six equilateral triangles with side length ${\displaystyle a_{1}}$. The rectangular triangle ${\displaystyle \triangle EDS_{0}}$ is one half of one of those six equilateral triangles. So:

${\displaystyle \quad A_{1}=6\cdot \left(|ED|\cdot |DS_{0}|\right)}$
${\displaystyle \quad \Leftrightarrow A_{1}=6\cdot \left({\frac {1}{2}}\cdot a_{1}\cdot |DS_{0}|\right)\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot |DS_{0}|\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot {\sqrt {|ES_{0}|^{2}-|ED|^{2}}}\quad }$, applying the Pythagorean theorem on triangle ${\displaystyle \triangle EDS_{0}}$
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot {\sqrt {|ES_{0}|^{2}-\left({\frac {1}{2}}\cdot a_{1}\right)^{2}}}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot {\sqrt {a_{1}^{2}-\left({\frac {1}{2}}\cdot a_{1}\right)^{2}}}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot {\sqrt {a_{1}^{2}-{\frac {1}{4}}\cdot a_{1}^{2}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot {\sqrt {{\frac {3}{4}}\cdot a_{1}^{2}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}=3\cdot a_{1}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{1}\quad }$, extracting the root
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {3{\sqrt {3}}}{2}}\cdot a_{1}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {3{\sqrt {3}}}{2}}\cdot ({\frac {1}{3}}\cdot a_{0})^{2}\quad }$, applying calculation (2)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {3{\sqrt {3}}}{2}}\cdot {\frac {1}{9}}\cdot a_{0}^{2}\quad }$, squaring
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {3{\sqrt {3}}}{2\cdot 9}}\cdot a_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\sqrt {3}}{2\cdot 3}}\cdot a_{0}^{2}\quad }$, shortening
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2{\sqrt {3}}}}\cdot a_{0}^{2}\quad }$, 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