Краткое описание

Display the largest equilateral triangle within a circle

Elements

Base is the circle of given radius ${\displaystyle r_{0}}$ around point ${\displaystyle S_{0}}$
Inscribed is the largest possible equilateral triangle.

General case

Segments in the general case

0) The radius of the base circle: ${\displaystyle r_{0}}$
1) Side of the inscribed triangle: ${\displaystyle a_{1}={\sqrt {3}}\cdot r_{0}\quad \approx 1.732\cdot r_{0}}$, see Calculation 1

Perimeters in the general case

0) Perimeter of base circle ${\displaystyle P_{0}=2\cdot \pi \cdot r_{0}}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}=3a_{1}=3{\sqrt {3}}\cdot r_{0}}$

Areas in the general case

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

Centroids in the general case

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

Normalised case

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

Segments in the normalised case

0) Radius of the base circle: ${\displaystyle r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564}$
1) Side length of the inscribed triangle: ${\displaystyle a_{1}={\sqrt {3}}\cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\frac {3}{\pi }}}\quad \approx 0.977}$

Perimeters in the normalised case

0) Perimeter of base square: ${\displaystyle P_{0}=2\cdot \pi \cdot r_{0}=2\cdot \pi \cdot {\frac {1}{\sqrt {\pi }}}=2\cdot {\sqrt {\pi }}\quad \approx 3.5449077}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}=3\cdot {\sqrt {\frac {3}{\pi }}}\quad \approx 2.9316151}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 6.4765228}$

Areas in the normalised case

0) Area of the base circle: ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {3{\sqrt {3}}}{4}}\cdot \left({\sqrt {\frac {1}{\pi }}}\right)^{2}={\frac {3{\sqrt {3}}}{4}}\cdot {\frac {1}{\pi }}={\frac {3{\sqrt {3}}}{4\pi }}\quad \approx 0.413}$

Covered surface of the base shape

${\displaystyle C_{b}={\frac {A_{1}}{A_{0}}}={\frac {3{\sqrt {3}}}{4\pi }}\quad \approx 41\%}$

Centroids in the normalised case

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

Calculation 1

Since the inscribed triangle ${\displaystyle \triangle ABC}$ is a equilateral triangle the following equations hold:
(1) ${\displaystyle |AB|=|AC|=|BC|=a_{1}}$
(2) ${\displaystyle |BD|={\frac {a_{1}}{2}}}$
(3) ${\displaystyle |DS_{0}|+|CS_{0}|=|CD|}$
(4) ${\displaystyle |BS|=|CS_{0}|=r_{0}}$
(5) ${\displaystyle |CD|^{2}+|BD|^{2}=|BC|^{2}}$
${\displaystyle \quad \Leftrightarrow |CD|^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=a_{1}^{2}\quad }$, applying equations (1) and (2)
${\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{1}^{2}-\left({\frac {a_{1}}{2}}\right)^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow |CD|^{2}=a_{1}^{2}-{\frac {a_{1}^{2}}{4}}\quad }$
${\displaystyle \quad \Leftrightarrow |CD|^{2}={\frac {3}{4}}\cdot a_{1}^{2}\quad }$
${\displaystyle \quad \Leftrightarrow |CD|={\frac {\sqrt {3}}{2}}\cdot a_{1}\quad }$

Further we have:
(6) ${\displaystyle |DS_{0}|^{2}+|BD|^{2}=|BS_{0}|^{2}}$
${\displaystyle \quad \Leftrightarrow \left(|CD|-|CS_{0}|\right)^{2}+|BD|^{2}=|BS_{0}|^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow \left(|CD|-|CS_{0}|\right)^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=|BS_{0}|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow \left(|CD|-|CS_{0}|\right)^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=r_{0}^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow \left(|CD|-r_{0}\right)^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=r_{0}^{2}\quad }$, applying equation (4)
${\displaystyle \quad \Leftrightarrow \left({\frac {\sqrt {3}}{2}}\cdot a_{1}-r_{0}\right)^{2}+\left({\frac {a_{1}}{2}}\right)^{2}=r_{0}^{2}\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow \left({\frac {3}{4}}\cdot a_{1}^{2}+r_{0}^{2}-2{\frac {\sqrt {3}}{2}}\cdot a_{1}\cdot r_{0}\right)+{\frac {a_{1}^{2}}{4}}=r_{0}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow a_{1}^{2}+r_{0}^{2}-2{\frac {\sqrt {3}}{2}}\cdot a_{1}\cdot r_{0}=r_{0}^{2}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}^{2}-2{\frac {\sqrt {3}}{2}}\cdot a_{1}\cdot r_{0}=0\quad }$, subtracting ${\displaystyle r_{0}^{2}}$ on each side
${\displaystyle \quad \Leftrightarrow a_{1}^{2}-{\sqrt {3}}\cdot a_{1}\cdot r_{0}=0\quad }$, shortening
${\displaystyle \quad \Leftrightarrow a_{1}-{\sqrt {3}}\cdot r_{0}=0\quad }$, dividing by ${\displaystyle a_{1}}$
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {3}}\cdot r_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {3}}\cdot r_{0}\quad }$, rearranging

Calculation 2

${\displaystyle A_{1}=|BD|\cdot |CD|}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {a_{1}}{2}}\cdot |CD|\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {a_{1}}{2}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{1}\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {a_{1}}{2}}\cdot {\frac {\sqrt {3}}{2}}\cdot a_{1}\quad }$, applying equation (5)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\sqrt {3}}{4}}\cdot a_{1}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\sqrt {3}}{4}}\cdot \left({\sqrt {3}}\cdot r_{0}\right)^{2}\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {3{\sqrt {3}}}{4}}\cdot r_{0}^{2}\quad }$, multiplying

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