User:Hans G. Oberlack/FS CVC2

Shows the second largest circle within a forced sudoku of CV-type.

Elements

Base is the circle of given radius $r_{0}$ around point $D$ and the resulting quarter circle $[ABC]$ of radius $r_{1}$ around point $A$

Inscribed is the largest possible circle in the shape $[CBG]$ having radius $r_{2}$ around point $F$. This is the second largest circle in a CV type sangaku.

In order to find radius $r_{2}$ of the circle, the following reasoning is used:
The line segment $[AG]$ is the diameter of the circle around $D$ with radius $r_{0}$. The line segment $[AE]$ is radius $r_{1}$ of the quarter circle around $A$. The line segment $[EG]$ is the diameter of the circle around $F$ with radius $r_{2}$.

$|AG|=|AE|+|EG|=2\cdot r_{0}$
$\Rightarrow r_{1}+2\cdot r_{2}=2\cdot r_{0}$
$\Leftrightarrow 2\cdot r_{2}=2\cdot r_{0}-r_{1}$

From the construction of the quarter circle (see FS_CV) we know that $r_{1}={\sqrt {2}}\cdot r_{0}$. So:
$\quad 2\cdot r_{2}=2\cdot r_{0}-{\sqrt {2}}\cdot r_{0}$
$\Leftrightarrow 2\cdot r_{2}=(2-{\sqrt {2}})\cdot r_{0}$
$\Leftrightarrow r_{2}={\frac {2-{\sqrt {2}}}{2}}\cdot r_{0}$

General case

Segments in the general case

0) The radius of the base circle $=r_{0}$
1) Radius of the quarter circle $r_{1}={\sqrt {2}}\cdot r_{0}$
2) Radius of the additional circle $r_{2}={\frac {2-{\sqrt {2}}}{2}}\cdot r_{0}$

Perimeters in the general case

0) Perimeter of base circle $P_{0}=2\cdot \pi \cdot r_{0}$
1) Perimeter of the quarter circle $P_{1}={\frac {4+\pi }{\sqrt {2}}}\cdot r_{0}$
2) Perimeter of additional circle $P_{2}=2\cdot \pi \cdot r_{2}=2\cdot \pi \cdot {\frac {2-{\sqrt {2}}}{2}}\cdot r_{0}=(2-{\sqrt {2}})\cdot \pi \cdot r_{0}=(1-{\frac {1}{\sqrt {2}}})\cdot 2\cdot \pi \cdot r_{0}$
$\quad \Leftrightarrow P_{2}=(1-{\frac {1}{\sqrt {2}}})\cdot P_{0}$

Areas in the general case

0) Area of the base circle $A_{0}=\pi \cdot r_{0}^{2}$
1) Area of the inscribed quarter circle $A_{1}={\frac {1}{2}}\cdot A_{0}$
2) Area of the additional circle
$\quad A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot ((1-{\frac {1}{\sqrt {2}}})\cdot r_{0})^{2}=(1-{\frac {1}{\sqrt {2}}})^{2}\cdot \pi \cdot r_{0}^{2}=({\frac {\sqrt {2}}{\sqrt {2}}}-{\frac {1}{\sqrt {2}}})^{2}\cdot \pi \cdot r_{0}^{2}=({\frac {{\sqrt {2}}-1}{\sqrt {2}}})^{2}\cdot \pi \cdot r_{0}^{2}={\frac {({\sqrt {2}}-1)^{2}}{2}}\cdot \pi \cdot r_{0}^{2}$
$\quad \Leftrightarrow A_{2}={\frac {2-2\cdot {\sqrt {2}}+1}{2}}\cdot \pi \cdot r_{0}^{2}={\frac {3-2\cdot {\sqrt {2}}}{2}}\cdot \pi \cdot r_{0}^{2}=({\frac {3}{2}}-{\sqrt {2}})\cdot \pi \cdot r_{0}^{2}$
$\quad \Leftrightarrow A_{2}=({\frac {3}{2}}-{\sqrt {2}})\cdot A_{0}$

Centroids in the general case

Centroid positions are measured from the lower left point of the surrounding square.
0) Centroid positions of the base square: $x_{0}=r_{0}\quad y_{0}=r_{0}$
1) Centroid positions of the inscribed quarter circle:
$\quad x_{1}=({\frac {\sqrt {32}}{3\cdot \pi }}+{\frac {{\sqrt {2}}-1}{\sqrt {2}}})\cdot r_{0}$
$\quad y_{1}=x_{1}$

2) Centroids of the additional circle:
$\quad x_{2}=|AF|\cdot cos(45)=(|AG|-|FG|)\cdot cos(45)=(2\cdot r_{0}-r_{2})\cdot cos(45)=(2\cdot r_{0}-{\frac {2-{\sqrt {2}}}{2}}\cdot r_{0})\cdot cos(45)$
$\quad \Leftrightarrow x_{2}=(2\cdot -{\frac {2-{\sqrt {2}}}{2}})\cdot r_{0}\cdot cos(45)=({\frac {4-2+{\sqrt {2}}}{2}})\cdot r_{0}\cdot cos(45)=({\frac {2+{\sqrt {2}}}{2}})\cdot r_{0}\cdot cos(45)$
$\quad \Leftrightarrow x_{2}=({\frac {2+{\sqrt {2}}}{2}})\cdot r_{0}\cdot {\frac {1}{\sqrt {2}}}=\left({\frac {{\sqrt {2}}+1}{2}}\right)\cdot r_{0}\cdot {\frac {\sqrt {2}}{\sqrt {2}}}={\frac {{\sqrt {2}}+1}{2}}\cdot r_{0}\cdot$
$\quad y_{2}=x_{2}$

Normalised case

Black-and-White version

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

Segments in the normalised case

0) Radius of the base circle $r_{0}={\frac {1}{\sqrt {\pi }}}=0.56...$
1) Radius of the inscribed quarter circle $r_{1}={\sqrt {\frac {2}{\pi }}}=0.79...$
2) Radius of the additional circle
$\quad r_{2}={\frac {2-{\sqrt {2}}}{2}}\cdot r_{0}={\frac {2-{\sqrt {2}}}{2}}\cdot {\frac {1}{\sqrt {\pi }}}={\frac {2-{\sqrt {2}}}{2\cdot {\sqrt {\pi }}}}=0.165...$

Perimeters in the normalised case

0) Perimeter of base square $P_{0}=2\cdot {\sqrt {\pi }}=3.5449077...$
1) Perimeter of the inscribed quarter circle $P_{1}={\frac {4+\pi }{\sqrt {2\cdot \pi }}}=2.8490832...$
2) Perimeter of additional circle $P_{2}=(1-{\frac {1}{\sqrt {2}}})\cdot 2\cdot {\sqrt {\pi }}=1.0382794...$
S) Sum of perimeters $P_{s}=P_{0}+P_{1}+P_{2}=7.4322703...$

Areas in the normalised case

0) Area of the base square $A_{0}=1$
1) Area of the inscribed quarter circle $A_{1}=0.5$
2) Area of the additional circle $A_{2}=({\frac {3}{2}}-{\sqrt {2}})\cdot A_{0}=({\frac {3}{2}}-{\sqrt {2}})=0.086...$

Centroids in the normalised case

Centroid positions are measured from the lower left point of the surrounding square.
0) Centroid positions of the base square: $x_{0}=r_{0}=0.5641895...\quad y_{0}=r_{0}=0.5641895...$
1) Centroid positions of the inscribed quarter circle: $x_{1}=0.503880...\quad y_{1}=x_{1}=0.503880...$
2) Centroids of the additional circle:
$\quad x_{2}=\left({\frac {2+{\sqrt {2}}}{2\cdot {\sqrt {2}}}}\right)\cdot r_{0}=\left({\frac {2+{\sqrt {2}}}{2\cdot {\sqrt {2}}}}\right)\cdot {\frac {1}{\sqrt {\pi }}}=\left({\frac {2+{\sqrt {2}}}{2\cdot {\sqrt {2\pi }}}}\right)=0.68103707...$
$\quad y_{2}=x_{2}=0.68103707...$

Distances of centroids

The distance between the centroid of the base element and the centroid of the quarter circle is:
$d01={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.0852905...$
$d02={\sqrt {(x_{0}-x_{2})^{2}+(y_{0}-y_{2})^{2}}}=0.1652473...$
$d12={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}=0.2505378...$

The sum of the distances is $D_{s}=0.5010757...$

Identifying number

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 sum of the distances of the centroids in the normalised case.

$decimalpart(7.4322704...+0.5010758...)=decimalpart(7.9333462...)=.9333462$

So the identifying number is: $2.9333462$