Summary edit

Shows the two second largest quarter circle within a right isosceles triangle that contains the largest quarter circle.

Elements edit

Base is the right isosceles triangle of side length ${\displaystyle a_{0}}$ and centroid ${\displaystyle S_{0}}$
Inscribed is the largest possible quarter circle with radius ${\displaystyle r_{1}}$ and centroid ${\displaystyle S_{1}}$
Inscribed is the second largest circle with radius ${\displaystyle r_{2}}$ and centroid ${\displaystyle S_{2}}$
Inscribed is the next second largest circle with radius ${\displaystyle r_{2}}$ and centroid ${\displaystyle S_{2}}$

Segments in the general case edit

0) The side length of the base right triangle ${\displaystyle |AB|=|AC|=a_{0}}$
1) Radius of the inscribed quarter circle ${\displaystyle r_{1}={\frac {a_{0}}{\sqrt {2}}}\quad }$ (See calculation 1).
2) Radius of the inscribed second largest circle ${\displaystyle r_{2}={\frac {a_{0}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.087\cdot a_{0}\quad }$ ( See Calculation 5).
3) Radius of the other inscribed second largest circle ${\displaystyle r_{3}=r_{2}={\frac {a_{0}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.087\cdot a_{0}\quad }$ ( See Calculation 5).

Perimeters in the general case edit

0) Perimeter of base triangle ${\displaystyle P_{0}=|AB|+|BC|+|AC|=a_{0}+{\sqrt {2}}\cdot a_{0}+a_{0}=(2+{\sqrt {2}})\cdot a_{0}}$
1) Perimeter of the quarter circle ${\displaystyle P_{1}=\left({\frac {4+\pi }{2{\sqrt {2}}}}\right)\cdot a_{0}\quad }$ (See calculation 2 )
2) Perimeter of the inscribed circle: ${\displaystyle P_{2}={\frac {2\pi a_{0}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.548\cdot a_{0}\quad }$
3) Perimeter of the other inscribed circle: ${\displaystyle P_{3}=P_{2}={\frac {2\pi a_{0}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.548\cdot a_{0}\quad }$

Areas in the general case edit

0) Area of the base triangle ${\displaystyle A_{0}={\frac {1}{2}}\cdot a_{0}^{2}}$
1) Area of the inscribed quarter circle ${\displaystyle A_{1}={\frac {\pi }{8}}\cdot a_{0}^{2}\quad }$ (See calculation 3)
2) Area of inscribed circle ${\displaystyle A_{2}=\pi \cdot {\frac {a_{0}^{2}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)^{2}}}\quad \approx 0.024\cdot a_{0}^{2}\approx 0.048\cdot A_{0}}$
3) Area of other inscribed circle ${\displaystyle A_{3}=A_{2}=\pi \cdot {\frac {a_{0}^{2}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)^{2}}}\quad \approx 0.024\cdot a_{0}^{2}\approx 0.048\cdot A_{0}}$

Centroids in the general case edit

Centroid positions are measured from the centroid point of the base shape
0) Centroid position of the base triangle: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed quarter circle: ${\displaystyle S_{1}=\left({\frac {{\sqrt {8}}-\pi }{3\pi }}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0,033-0,033i)\cdot a_{0}}$, (see calculation 4)
2) Centroid position of the inscribed circle: ${\displaystyle S_{2}=\left({\sqrt {\frac {k+2{\sqrt {2}}}{2k}}}-{\frac {1}{3}}+{\frac {3-k}{3k}}i\right)\cdot a_{0}}$ , with ${\displaystyle k=2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\quad \approx (0.456-0.246i)\cdot a_{0}}$, (see calculation 6)
3) Centroid position of the other inscribed circle: ${\displaystyle S_{3}=\left({\frac {3-k}{3k}}+\left({\sqrt {\frac {k+2{\sqrt {2}}}{2k}}}-{\frac {1}{3}}\right)\cdot i\right)\cdot a_{0}}$ , with ${\displaystyle k=2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\quad \approx (-0.246+0.456i)\cdot a_{0}}$, for symmetry reasons

Normalised case edit

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

Segments in the normalised case edit

0) Side length of the base triangle: ${\displaystyle a_{0}={\sqrt {2}}\quad \approx 1.414}$
1) Radius of the inscribed quarter circle: ${\displaystyle r_{1}={\frac {\sqrt {2}}{\sqrt {2}}}=1\quad }$
2) Radius of the inscribed circle: ${\displaystyle r_{2}={\frac {\sqrt {2}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.123}$
3) Radius of the other inscribed circle: ${\displaystyle r_{3}=r_{2}={\frac {\sqrt {2}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.123}$

Perimeters in the normalised case edit

0) Perimeter of base triangle: ${\displaystyle P_{0}=(2+{\sqrt {2}})\cdot {\sqrt {2}}=2{\sqrt {2}}+2\quad \approx 4.8284271}$
1) Perimeter of the inscribed quarter circle: ${\displaystyle P_{1}=\left({\frac {4+\pi }{2{\sqrt {2}}}}\right)\cdot {\sqrt {2}}=\left({\frac {4+\pi }{2}}\right)\quad \approx 3.5707963}$
2) Perimeter of the inscribed circle: ${\displaystyle P_{2}=2\cdot \pi \cdot r_{2}\quad \approx 0.7747710}$
3) Perimeter of the other inscribed circle: ${\displaystyle P_{3}=P_{2}=2\cdot \pi \cdot r_{2}\quad \approx 0.7747710}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}+P_{3}\quad \approx 9.9487655}$

Areas in the normalised case edit

0) Area of the base triangle ${\displaystyle A_{0}=1}$
1) Area of the inscribed quarter circle ${\displaystyle A_{1}={\frac {\pi }{8}}\cdot ({\sqrt {2}})^{2}={\frac {\pi }{8}}\cdot 2={\frac {\pi }{4}}\quad \approx 0.785}$
2) Area of the inscribed circle ${\displaystyle A_{2}=\pi \cdot r_{2}^{2}\quad \approx 0,048}$
3) Area of the other inscribed circle ${\displaystyle A_{3}=A_{2}=\pi \cdot r_{2}^{2}\quad \approx 0,048}$

Centroids in the normalised case edit

Centroid positions are measured from the centroid point of the base shape.
0) Centroid positions of the base triangle: ${\displaystyle S_{0}=0+0i}$
1) Centroid positions of the inscribed quarter circle: ${\displaystyle S_{1}=\left({\frac {{\sqrt {8}}-\pi }{3\pi }}\right)\cdot {\sqrt {2}}\quad \approx (-0.047-0.047i)\quad }$
2) Centroid positions of the inscribed circle: ${\displaystyle S_{2}\quad \approx (0.645-0,348i)\quad }$
3) Centroid positions of the other inscribed circle: ${\displaystyle S_{3}\quad \approx (-0,348+0.645i)\quad }$

Distances of centroids edit

The distance between the centroid of the base element and the centroid of the circle is:
0) ${\displaystyle d_{S_{0}S_{1}}\approx 0.0664558}$
1) ${\displaystyle d_{S_{0}S_{2}}\approx 0.7330382}$
2) ${\displaystyle d_{S_{0}S_{3}}\approx 0.7330382}$
3) ${\displaystyle d_{S_{1}S_{2}}\approx 0.7547689}$
4) ${\displaystyle d_{S_{1}S_{3}}\approx 0.7547689}$
5) ${\displaystyle d_{S_{2}S_{3}}\approx 1.4046132}$
S) ${\displaystyle sum_{d}\approx 4.4466830}$

Identifying number edit

Apart of the base element there are three shapes allocated. Therefore the integer part of the identifying number is 3.
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.

${\displaystyle decimalpart(9.9487655+4.4466830)=decimalpart(14.3954486)=.3954486}$

So the identifying number is: ${\displaystyle 3.3954486}$

Calculations edit

Given elements edit

Because FS_R is a right isosceles triangle the following equations hold:
(1) ${\displaystyle \quad |AB|=|AC|=a_{0}}$
(2) ${\displaystyle \quad |BC|={\sqrt {2}}\cdot a_{0}}$
(3) ${\displaystyle \quad |BD|=|CD|={\frac {{\sqrt {2}}\cdot a_{0}}{2}}={\frac {a_{0}}{\sqrt {2}}}}$
(4) ${\displaystyle \quad r_{1}=|AD|}$
(5) ${\displaystyle \quad |AS_{2}|=r_{1}+r_{2}\quad }$, since both radii are perpendicular to the tangent
(6) ${\displaystyle \quad |FS_{2}|=|ES_{2}|=r_{2}\quad }$, since ${\displaystyle {\overrightarrow {BA}}}$ and ${\displaystyle {\overrightarrow {BC}}}$ are tangent to the circle around ${\displaystyle S_{2}}$ in points E and F respectively
(7) ${\displaystyle \quad |BF|=|BE|\quad }$, since ${\displaystyle {\overrightarrow {BA}}}$ and ${\displaystyle {\overrightarrow {BC}}}$ are tangent to the circle around ${\displaystyle S_{2}}$ in points E and F respectively
(8) ${\displaystyle \quad \alpha ={\frac {45^{\circ }}{2}}=22.5^{\circ }\quad }$, since ${\displaystyle {\overrightarrow {BA}}}$ and ${\displaystyle {\overrightarrow {BC}}}$ are tangent to the circle around ${\displaystyle S_{2}}$ in points E and F respectively and therefore the segment ${\displaystyle {\overline {BS_{2}}}}$ divides the angle in B evenly

Calculation 1 edit

Since ${\displaystyle {\overline {AB}}}$ is tangent to the quarter circle in point ${\displaystyle D}$ the triangle ${\displaystyle \triangle ABD}$ has an right angle in ${\displaystyle D}$. This means:
(5) ${\displaystyle \quad |AD|^{2}+|BD|^{2}=|AB|^{2}}$
${\displaystyle \quad \Leftrightarrow r_{1}^{2}+\left({\frac {a_{0}}{\sqrt {2}}}\right)^{2}=a_{0}^{2}\quad }$, applying equations (1),(3) and (4)
${\displaystyle \quad \Leftrightarrow r_{1}^{2}=a_{0}^{2}-{\frac {a_{0}^{2}}{2}}={\frac {a_{0}^{2}}{2}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {a_{0}}{\sqrt {2}}}\quad }$

Calculation 2 edit

Perimeter of the quarter circle:
${\displaystyle \quad P_{1}=2\cdot r_{1}+{\frac {2\pi r_{1}}{4}}\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=2\cdot r_{1}\cdot \left(1+{\frac {\pi }{4}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=2\cdot r_{1}\cdot \left({\frac {4+\pi }{4}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=r_{1}\cdot \left({\frac {4+\pi }{2}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}={\frac {a_{0}}{\sqrt {2}}}\cdot \left({\frac {4+\pi }{2}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=\left({\frac {4+\pi }{2{\sqrt {2}}}}\right)\cdot a_{0}\quad }$

Calculation 3 edit

Area of the inscribed quarter circle:
${\displaystyle \quad A_{1}={\frac {\pi \cdot r_{1}^{2}}{4}}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{4}}\cdot r_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{4}}\cdot \left({\frac {a_{0}}{\sqrt {2}}}\right)^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{4}}\cdot {\frac {a_{0}^{2}}{2}}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{8}}\cdot a_{0}^{2}}$

Calculation 4 edit

Centroid of the inscribed circle measured from the centroid of the base triangle:
${\displaystyle \quad x_{1}=|Ax_{1}|-|Ax_{0}|}$
${\displaystyle \quad \Leftrightarrow x_{1}=|Ax_{1}|-{\frac {1}{3}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {4}{3\pi }}\cdot r_{1}-{\frac {1}{3}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {4}{3\pi }}\cdot {\frac {a_{0}}{\sqrt {2}}}-{\frac {1}{3}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {4}{3\pi }}\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{3}}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {1}{3\pi }}\cdot {\frac {\sqrt {16}}{\sqrt {2}}}-{\frac {1}{3}}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {\sqrt {8}}{3\pi }}-{\frac {\pi }{3\pi }}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {{\sqrt {8}}-\pi }{3\pi }}\right)\cdot a_{0}}$

Calculation 5 edit

Radius of the inscribed circle around ${\displaystyle S_{2}}$
${\displaystyle \quad |AF|^{2}+|FS_{2}|^{2}=|AS_{2}|^{2}}$ , since ${\displaystyle \triangle {AFS_{2}}}$ is a right triangle
${\displaystyle \quad \Leftrightarrow |AF|^{2}+|FS_{2}|^{2}=(r_{1}+r_{2})^{2}}$ , applying equation (5)
${\displaystyle \quad \Leftrightarrow |AF|^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , applying equation (6)
${\displaystyle \quad \Leftrightarrow \left(|AB|-|BF|\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , since ${\displaystyle {\overrightarrow {AF}}+{\overrightarrow {FB}}={\overrightarrow {AB}}}$
${\displaystyle \quad \Leftrightarrow \left(a_{0}-|BF|\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , applying equation (1)
${\displaystyle \quad \Leftrightarrow \left(a_{0}-{\frac {|FS_{2}|}{\tan {\alpha }}}\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , since ${\displaystyle \triangle {FBS_{2}}}$ is a right triangle
${\displaystyle \quad \Leftrightarrow \left(a_{0}-{\frac {r_{2}}{\tan {\alpha }}}\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , applying equation (6)
${\displaystyle \quad \Leftrightarrow \left(a_{0}-{\frac {r_{2}}{{\sqrt {2}}-1}}\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , applying equation (8)
${\displaystyle \quad \Leftrightarrow \left(a_{0}-{\frac {r_{2}\cdot ({\sqrt {2}}+1)}{({\sqrt {2}}-1)({\sqrt {2}}+1)}}\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , extending
${\displaystyle \quad \Leftrightarrow \left(a_{0}-{\frac {r_{2}\cdot ({\sqrt {2}}+1)}{(2-1)}}\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , applying binomical formula
${\displaystyle \quad \Leftrightarrow \left(a_{0}-r_{2}\cdot ({\sqrt {2}}+1)\right)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , multiplying
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}}$ , multiplying
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}+r_{2}^{2}=({\frac {a_{0}}{\sqrt {2}}}+r_{2})^{2}}$ , applying calculation (1)
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}+r_{2}^{2}={\frac {a_{0}^{2}}{2}}+2r_{2}{\frac {a_{0}}{\sqrt {2}}}+r_{2}^{2}}$ , multiplying
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}={\frac {a_{0}^{2}}{2}}+2r_{2}{\frac {a_{0}}{\sqrt {2}}}\quad }$ , eliminating ${\displaystyle r_{2}^{2}}$ on both sides of the equation
${\displaystyle \quad \Leftrightarrow {\frac {a_{0}^{2}}{2}}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}=2r_{2}{\frac {a_{0}}{\sqrt {2}}}\quad }$ , subtracting ${\displaystyle {\frac {a_{0}^{2}}{2}}}$ on both sides of the equation
${\displaystyle \quad \Leftrightarrow {\frac {a_{0}^{2}}{2}}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}={\sqrt {2}}\cdot a_{0}r_{2}\quad }$ , reducing ${\displaystyle 2r_{2}{\frac {a_{0}}{\sqrt {2}}}}$
${\displaystyle \quad \Leftrightarrow {\frac {a_{0}^{2}}{2}}-2a_{0}\cdot r_{2}\cdot ({\sqrt {2}}+1)-{\sqrt {2}}\cdot a_{0}r_{2}+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}=0\quad }$ , subtracting ${\displaystyle {\sqrt {2}}\cdot a_{0}r_{2}}$ on both sides of the equation
${\displaystyle \quad \Leftrightarrow {\frac {a_{0}^{2}}{2}}-a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}=0\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+2\cdot r_{2}^{2}\cdot ({\sqrt {2}}+1)^{2}=0\quad }$ , multiplying both sides by 2
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot \left({\sqrt {2}}\cdot ({\sqrt {2}}+1)\right)^{2}=0\quad }$ , integration2 into the brackets
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot (2+{\sqrt {2}})^{2}=0\quad }$ , multiplying
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)=-r_{2}^{2}\cdot (2+{\sqrt {2}})^{2}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}=-r_{2}^{2}\cdot (2+{\sqrt {2}})^{2}+r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}\quad }$ , adding ${\displaystyle r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}}$ on both sides
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}=-r_{2}^{2}\cdot (4+2\cdot 2{\sqrt {2}}+2)+r_{2}^{2}\cdot (9\cdot 2+2\cdot 2\cdot 3{\sqrt {2}}+4)\quad }$ , applying binomial formulas
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}=-r_{2}^{2}\cdot (6+4{\sqrt {2}})+r_{2}^{2}\cdot (22+12{\sqrt {2}})\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}=r_{2}^{2}\cdot (16+8{\sqrt {2}})\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow a_{0}^{2}-2\cdot a_{0}\cdot r_{2}\cdot (3{\sqrt {2}}+2)+r_{2}^{2}\cdot (3{\sqrt {2}}+2)^{2}=4\cdot r_{2}^{2}\cdot (4+2{\sqrt {2}})\quad }$ , extracting 4 off the brackets
${\displaystyle \quad \Leftrightarrow \left(a_{0}-r_{2}\cdot (3{\sqrt {2}}+2)\right)^{2}=4\cdot r_{2}^{2}\cdot (4+2{\sqrt {2}})\quad }$ , applying binomial formula
${\displaystyle \quad \Leftrightarrow a_{0}-r_{2}\cdot (3{\sqrt {2}}+2)=2\cdot r_{2}\cdot {\sqrt {(4+2{\sqrt {2}})}}\quad }$ , extracting the root
${\displaystyle \quad \Leftrightarrow a_{0}=2\cdot r_{2}\cdot {\sqrt {(4+2{\sqrt {2}})}}+r_{2}\cdot (3{\sqrt {2}}+2)\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow \left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)\cdot r_{2}=a_{0}\quad }$ , rearranging
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {a_{0}}{\left(2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2\right)}}\quad \approx 0.087\cdot a_{0}}$ , rearranging

Calculation 6 edit

Centroid position of the inscribed circle relative to the centroid of the base triangle:
${\displaystyle S_{2}=S_{0}+{\overrightarrow {S_{0}A}}+{\overrightarrow {AF}}+{\overrightarrow {FS_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}={\overrightarrow {S_{0}A}}+{\overrightarrow {AF}}+{\overrightarrow {FS_{2}}}}$ , since ${\displaystyle S_{0}=0+0i}$
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {1}{3}}\cdot a_{0}\cdot (1+i)+{\overrightarrow {AF}}+{\overrightarrow {FS_{2}}}}$ , definition of centroid of a triangle
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {1}{3}}\cdot a_{0}\cdot (1+i)+{\overrightarrow {AF}}+(0+r_{2}\cdot i)}$ , since applying equation (6)</math>
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}-{\frac {a_{0}}{3}}i+r_{2}i+{\overrightarrow {AF}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\overrightarrow {AF}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+(|AF|+0i)}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {|AF|^{2}}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {|AS_{2}|^{2}-|FS_{2}|^{2}}}}$ , applying the Pythagorean theorem on triangle ${\displaystyle \triangle {AFS_{2}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {|AS_{2}|^{2}-r_{2}^{2}}}}$ , applying equation (6)
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {(r_{1}+r_{2})^{2}-r_{2}^{2}}}}$ , applying equation (5)
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {(r_{1}^{2}+2r_{1}r_{2}+r_{2}^{2})-r_{2}^{2}}}}$ , applying binomial formula
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {r_{1}^{2}+2r_{1}r_{2}}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left(r_{2}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {{\frac {a_{0}^{2}}{2}}+2{\frac {a_{0}}{\sqrt {2}}}r_{2}}}}$ , applying calculation (1)
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left({\frac {a_{0}}{k}}-{\frac {a_{0}}{3}}\right)\cdot i+{\sqrt {{\frac {a_{0}^{2}}{2}}+2{\frac {a_{0}}{\sqrt {2}}}{\frac {a_{0}}{k}}}}}$ , defining ${\displaystyle k=2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2}$ as in Calculation (5)
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left({\frac {3-k}{3k}}\right)\cdot a_{0}\cdot i+{\sqrt {{\frac {a_{0}^{2}}{2}}+2{\frac {a_{0}}{\sqrt {2}}}{\frac {a_{0}}{k}}}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left({\frac {3-k}{3k}}\right)\cdot a_{0}\cdot i+a_{0}{\sqrt {{\frac {1}{2}}+{\frac {2}{k{\sqrt {2}}}}}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left({\frac {3-k}{3k}}\right)\cdot a_{0}\cdot i+a_{0}{\sqrt {{\frac {1}{2}}+{\frac {\sqrt {2}}{k}}}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=-{\frac {a_{0}}{3}}+\left({\frac {3-k}{3k}}\right)\cdot a_{0}\cdot i+a_{0}{\sqrt {\frac {k+2{\sqrt {2}}}{2k}}}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=a_{0}\left({\sqrt {\frac {k+2{\sqrt {2}}}{2k}}}-{\frac {1}{3}}\right)+\left({\frac {3-k}{3k}}\right)\cdot a_{0}\cdot i}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {\frac {k+2{\sqrt {2}}}{2k}}}-{\frac {1}{3}}+{\frac {3-k}{3k}}i\right)\cdot a_{0}}$ , rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=\left({\sqrt {\frac {k+2{\sqrt {2}}}{2k}}}-{\frac {1}{3}}+{\frac {3-k}{3k}}i\right)\cdot a_{0}}$ , with ${\displaystyle k=2{\sqrt {(4+2{\sqrt {2}})}}+3{\sqrt {2}}+2}$
${\displaystyle \quad \Rightarrow S_{2}\approx (0.456165648-0.246140957i)\cdot a_{0}}$

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