File:FS CR dia.png

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

0) The radius of the base circle ${\displaystyle =r_{0}}$
1) Side of the triangle ${\displaystyle a_{1}={\sqrt {2}}\cdot r_{0}\quad }$, see Calculation 1

0) Perimeter of base circle ${\displaystyle P_{0}=2\cdot \pi \cdot r_{0}}$
1) Perimeter of the triangle ${\displaystyle P_{1}=({\sqrt {2}}+1)\cdot 2r_{0}}$

0) Area of the base circle ${\displaystyle A_{0}=\pi \cdot r_{0}^{2}}$
1) Area of the inscribed triangle ${\displaystyle A_{1}={\frac {a_{1}^{2}}{2}}={\frac {({\sqrt {2}}r_{0})^{2}}{2}}={\frac {2r_{0}^{2}}{2}}=r_{0}^{2}}$

Covered surface of the base shape: ${\displaystyle C_{b}={\frac {A_{1}}{A_{0}}}={\frac {r_{0}^{2}}{\pi \cdot r_{0}^{2}}}={\frac {1}{\pi }}\quad \approx 31,8\%}$

0) Centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed triangle: ${\displaystyle S_{1}=-{\frac {1}{3{\sqrt {2}}}}\cdot (1+i)\cdot r_{0}\quad \approx (-0.236-0.236i)\cdot r_{0}\quad }$, see Calculation 3

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the inscribed triangle: ${\displaystyle S_{1}=\left(0-{\frac {1}{3}}\cdot i\right)\cdot r_{0}\quad \approx (0-0.333i)\cdot r_{0}\quad }$

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

0) Radius of the base circle ${\displaystyle r_{0}={\frac {1}{\sqrt {\pi }}}\quad \approx 0.564}$
1) Side length of the inscribed triangle ${\displaystyle a_{1}={\sqrt {2}}\cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\frac {2}{\pi }}}\quad \approx 0.798}$

0) Perimeter of base circle ${\displaystyle P_{0}=2\cdot \pi \cdot r_{0}=2\cdot \pi \cdot {\frac {1}{\sqrt {\pi }}}=2\cdot {\sqrt {\pi }}\quad \approx 3.5449077}$
1) Perimeter of the inscribed triangle ${\displaystyle P_{1}=({\sqrt {2}}+1)\cdot 2\cdot {\frac {1}{\sqrt {\pi }}}\quad \approx 2.7241483}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 6.2690560}$

0) Area of the base square ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle ${\displaystyle A_{1}=r_{0}^{2}=({\frac {1}{\sqrt {\pi }}})^{2}={\frac {1}{\pi }}\quad \approx 0.318}$

0) Centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed triangle: ${\displaystyle S_{1}:\quad S_{1}=-{\frac {1}{3{\sqrt {2}}}}\cdot {\frac {1}{\sqrt {\pi }}}(1+i)=-{\frac {1}{3{\sqrt {2\pi }}}}\quad \approx (-0.133-0.133i)}$

The centroid positions of the following shapes will be expressed orientated so that the first shape n with ${\displaystyle S_{n}\neq S_{0}}$ will be of type ${\displaystyle S_{n}=0-ki}$ with ${\displaystyle k>0}$. This means that the graphical representation will not correspond to the mathematical expression.
0) Orientated centroid position of the base circle: ${\displaystyle S_{0}=0+0i}$
1) Orientated centroid position of the inscribed triangle: ${\displaystyle S_{1}=\left(0-{\frac {1}{3}}\cdot i\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left(0-{\frac {1}{3{\sqrt {\pi }}}}\cdot i\right)\quad \approx (0-0.188i)\quad }$

Since the right isosceles triangle ${\displaystyle \triangle ABC}$ is a right isosceles triangle the following equations hold:
(1) ${\displaystyle |AB|=|AC|=a_{1}}$
(2) ${\displaystyle |AB|^{2}+|AC|^{2}=|BC|^{2}}$
${\displaystyle \quad \Leftrightarrow |BC|^{2}=a_{1}^{2}+a_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow |BC|^{2}=2a_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow |BC|={\sqrt {2}}\cdot a_{1}}$

Since ${\displaystyle \triangle ABC}$ is inscribed in a circle, we have also
(3) ${\displaystyle |BC|=2r_{0}}$
Substituting (2) in (3) gives:
${\displaystyle \quad {\sqrt {2}}\cdot a_{1}=2r_{0}}$
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {2r_{0}}{\sqrt {2}}}}$
${\displaystyle \quad \Leftrightarrow a_{1}={\sqrt {2}}\cdot r_{0}}$

${\displaystyle P_{1}=a_{1}+{\sqrt {2}}a_{1}+a_{1}}$
${\displaystyle \quad \Leftrightarrow P_{1}=(2+{\sqrt {2}})\cdot a_{1}}$
${\displaystyle \quad \Leftrightarrow P_{1}=(2+{\sqrt {2}})\cdot {\sqrt {2}}\cdot r_{0}\quad }$, since ${\displaystyle a_{1}={\sqrt {2}}\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow P_{1}=(2{\sqrt {2}}+2)\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow P_{1}=({\sqrt {2}}+1)\cdot 2r_{0}}$

${\displaystyle x_{1}=|AX_{1}|-|AX_{0}|}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {1}{3}}a_{1}-{\frac {1}{2}}a_{1}}$
${\displaystyle \quad \Leftrightarrow x_{1}=-{\frac {1}{6}}\cdot a_{1}}$
${\displaystyle \quad \Leftrightarrow x_{1}=-{\frac {1}{6}}\cdot {\sqrt {2}}\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=-{\frac {\sqrt {2}}{6}}\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=-{\frac {\sqrt {2}}{3\cdot 2}}\cdot r_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=-{\frac {1}{3{\sqrt {2}}}}\cdot r_{0}}$

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