File:FS QCE dia.png

Summary[edit]

Task[edit]

The largest equilateral triangle inscribed in the largest circle that is inscribed in a square of side length ${\displaystyle a_{0}}$

General case[edit]

Segments in the general case[edit]

0) The side length of the base square: ${\displaystyle a_{0}}$
1) The radius of the inscribed circle: ${\displaystyle r_{1}={\frac {1}{2}}\cdot a_{0}}$
2) Side of the inscribed triangle: ${\displaystyle a_{2}={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad \approx 0.866\cdot a_{0}}$, see Calculation 1

Perimeters in the general case[edit]

0) Perimeter of the base square: ${\displaystyle P_{0}=4a_{0}}$
1) Perimeter of inscribed circle ${\displaystyle P_{1}=2\cdot \pi \cdot r_{1}=\pi \cdot a_{0}\quad \approx 3.14\cdot a_{0}}$
2) Perimeter of the inscribed triangle ${\displaystyle P_{2}=3a_{2}={\frac {3{\sqrt {3}}}{2}}\cdot a_{0}\quad \approx 2.598\cdot a_{0}}$

Areas in the general case[edit]

0) Area of the base square: ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed circle ${\displaystyle A_{1}=\pi \cdot r_{1}^{2}=\pi \cdot {\frac {1}{4}}\cdot a_{0}^{2}={\frac {\pi }{4}}\cdot A_{0}}$
2) Area of the inscribed triangle ${\displaystyle A_{2}={\frac {3{\sqrt {3}}}{16}}\cdot a_{0}^{2}\quad \approx 0.325\cdot A_{0}\quad }$, see Calculation 2

Centroids in the general case[edit]

Centroid positions are measured from the centroid point of the base shape
0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i=0}$
1) Centroid position of the inscribed circle: ${\displaystyle S_{1}=0+0i=0}$
2) Centroid position of the inscribed triangle: ${\displaystyle S_{2}=0+0i=0}$
W) Weighted centroid: ${\displaystyle S_{w}=0+0i=0}$

Normalised case[edit]

In the normalised case the area of the base is set to 1.
${\displaystyle A_{0}=1\quad \Rightarrow a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}=1}$

Segments in the normalised case[edit]

0) Side length of the base square: ${\displaystyle a_{0}=1}$
1) Radius of the inscribed circle ${\displaystyle r_{1}={\frac {1}{2}}=0.500}$
2) Side length of the inscribed triangle ${\displaystyle a_{2}={\frac {\sqrt {3}}{2}}\quad \approx 0.866}$

Perimeters in the normalised case[edit]

0) Perimeter of the base square: ${\displaystyle P_{0}=4\cdot a_{0}=4.0000000}$
1) Perimeter of inscribed square ${\displaystyle P_{1}=2\cdot \pi \cdot {\frac {1}{2}}=\pi \quad \approx 3.1415927}$
2) Perimeter of the inscribed triangle ${\displaystyle P_{2}=3\cdot {\frac {\sqrt {3}}{2}}\quad \approx 2.5980762}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}\quad \approx 9.7396689}$

Areas in the normalised case[edit]

0) Area of the base square ${\displaystyle A_{0}=1}$
1) Area of the inscribed circle ${\displaystyle A_{1}={\frac {\pi }{4}}\quad \approx 0.785}$
2) Area of the inscribed triangle ${\displaystyle A_{2}={\frac {3{\sqrt {3}}}{16}}\quad \approx 0.325\cdot A_{0}}$

Centroids in the normalised case[edit]

Centroid positions are measured from the centroid point of the base shape.
0) Centroid of the base square: ${\displaystyle S_{0}=0}$
1) Centroid of the inscribed circle: ${\displaystyle S_{1}=0}$
2) Centroid of the inscribed triangle:${\displaystyle S_{2}=0}$
W) Weighted centroid:${\displaystyle S_{w}=0}$

Distances of centroids[edit]

The distance between the centroid of the base element and the centroid of the triangle is:
${\displaystyle d={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0}$

Identifying number[edit]

Apart of the base element there are two shapes allocated. Therefore the integer part of the identifying number is 2.
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.

${\displaystyle decimalpart(9.7396689+0.0000000)=decimalpart(9.7396689)=.7396689}$

So the identifying number is: ${\displaystyle 2.7396689}$

Calculations[edit]

Given elements[edit]

(0) ${\displaystyle |EF|=|FG|=|GH|=|EH|=a_{0}\quad }$, because ${\displaystyle \square {EFGH}}$ is a square
(1) ${\displaystyle |AB|=|AC|=|BC|=a_{2}\quad }$, since the inscribed triangle ${\displaystyle \triangle ABC}$ is an equilateral triangle
(2) ${\displaystyle |AD|=|BD|={\frac {a_{2}}{2}}\quad }$, since the inscribed triangle ${\displaystyle \triangle ABC}$ is an equilateral triangle
(3) ${\displaystyle |DS|+|CS|=|CD|\quad }$, since the inscribed triangle ${\displaystyle \triangle ABC}$ is an equilateral triangle
(4) ${\displaystyle |AS|=|BS|=|CS|=r_{1}\quad }$, since the inscribed triangle ${\displaystyle \triangle ABC}$ is an equilateral triangle
(5) ${\displaystyle r_{1}={\frac {a_{0}}{2}}\quad }$, since the circle around ${\displaystyle S}$ is th largest circle inscribed the square

Calculation 1[edit]

${\displaystyle |CD|^{2}+|BD|^{2}=|BC|^{2}\quad }$, since ${\displaystyle \triangle BCD}$ is a right triangle
${\displaystyle \quad \Leftrightarrow |CD|^{2}+\left({\frac {a_{2}}{2}}\right)^{2}=a_{2}^{2}\quad }$, applying equations (1) and (2)
${\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{2}^{2}-\left({\frac {a_{2}}{2}}\right)^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{2}^{2}-{\frac {a_{2}^{2}}{4}}\quad }$
${\displaystyle \quad \Leftrightarrow |CD|^{2}={\frac {3}{4}}\cdot a_{2}^{2}\quad }$
${\displaystyle \quad \Leftrightarrow |CD|={\frac {\sqrt {3}}{2}}\cdot a_{2}\quad }$
${\displaystyle \quad \Leftrightarrow |DS|+|CS|={\frac {\sqrt {3}}{2}}\cdot a_{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow |DS|+r_{1}={\frac {\sqrt {3}}{2}}\cdot a_{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow {\sqrt {r_{1}^{2}-|BD|^{2}}}+r_{1}={\frac {\sqrt {3}}{2}}\cdot a_{2}\quad }$, since ${\displaystyle \triangle DBS}$ is a right triangle
${\displaystyle \quad \Leftrightarrow {\sqrt {r_{1}^{2}-|BD|^{2}}}={\frac {\sqrt {3}}{2}}\cdot a_{2}-r_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow r_{1}^{2}-|BD|^{2}=\left({\frac {\sqrt {3}}{2}}\cdot a_{2}-r_{1}\right)^{2}\quad }$, squaring both sides
${\displaystyle \quad \Leftrightarrow r_{1}^{2}-\left({\frac {a_{2}}{2}}\right)^{2}=\left({\frac {\sqrt {3}}{2}}\cdot a_{2}-r_{1}\right)^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow r_{1}^{2}-{\frac {a_{2}^{2}}{4}}=\left({\frac {\sqrt {3}}{2}}\cdot a_{2}-r_{1}\right)^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow r_{1}^{2}-{\frac {a_{2}^{2}}{4}}={\frac {3}{4}}\cdot a_{2}^{2}-2\cdot {\frac {\sqrt {3}}{2}}\cdot a_{2}\cdot r_{1}+r_{1}^{2}\quad }$, applying binomial formula
${\displaystyle \quad \Leftrightarrow -{\frac {a_{2}^{2}}{4}}={\frac {3}{4}}\cdot a_{2}^{2}-2\cdot {\frac {\sqrt {3}}{2}}\cdot a_{2}\cdot r_{1}\quad }$, shortening
${\displaystyle \quad \Leftrightarrow 0={\frac {3}{4}}\cdot a_{2}^{2}+{\frac {a_{2}^{2}}{4}}-2\cdot {\frac {\sqrt {3}}{2}}\cdot a_{2}\cdot r_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow 0={\frac {4}{4}}\cdot a_{2}^{2}-{\sqrt {3}}\cdot a_{2}\cdot r_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow 0=a_{2}^{2}-{\sqrt {3}}\cdot a_{2}\cdot r_{1}\quad }$, reducing
${\displaystyle \quad \Leftrightarrow 0=a_{2}-{\sqrt {3}}\cdot r_{1}\quad }$, dividing
${\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {3}}\cdot r_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{2}={\frac {\sqrt {3}}{2}}\cdot a_{0}\quad \approx 0.866\cdot a_{0}}$, applying equation (5)

Calculation 2[edit]

${\displaystyle A_{1}={\frac {1}{2}}\cdot |AD|\cdot |CD|}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2}}\cdot a_{2}\cdot |CD|\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2}}\cdot a_{2}\cdot {\sqrt {|BC|^{2}-|BD|^{2}}}\quad }$, applying the Pythagorean theoreme
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2}}\cdot a_{2}\cdot {\sqrt {a_{2}^{2}-|BD|^{2}}}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2}}\cdot a_{2}\cdot {\sqrt {a_{2}^{2}-({\frac {a_{2}}{2}})^{2}}}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2}}\cdot a_{2}\cdot {\sqrt {{\frac {3}{4}}\cdot a_{2}^{2}}}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {1}{2}}\cdot a_{2}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{2}\quad }$, rooting
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\sqrt {3}}{4}}\cdot a_{2}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\sqrt {3}}{4}}\cdot ({\frac {\sqrt {3}}{2}}\cdot a_{0})^{2}\quad }$, applying calculation 1
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\sqrt {3}}{4}}\cdot {\frac {3}{4}}\cdot a_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {3{\sqrt {3}}}{16}}\cdot a_{0}^{2}\quad \approx 0.325\cdot A_{0}\quad }$, multiplying

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