User:Hans G. Oberlack/FS CVC1

Shows the 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 $$A$$. From the developing of sangaku FS_CV we know that $$r_1=\sqrt 2 \cdot r_0$$. Inscribed is the largest possible circle in the shape $$[ABC]$$ having radius $$r_2$$ around point $$E$$. This is also the 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 $$[AH] $$ corresponds to the radius $$r_1$$ of the quarter circle around $$A$$. So we have: $$(1) \quad |AH| = |AE| + |EH| = r_1 = \sqrt 2 \cdot r_0$$

The line segments $$[EF],[EG],[EH]$$ correspond to the radius $$r_2$$ of the circle around $$E$$. This means that: $$(2) \quad |EF|=|EG|=|EH|=r_2$$

It also means that the points $$[AGEF]$$ form a square with side length $$r_2$$. So applying the Pythagorean theorem we get: $$(3) \quad |AE|^2=|EF|^2 + |AF|^2 \quad \Leftrightarrow |AE|^2= 2 \cdot r_2^2 \quad \Leftrightarrow |AE|= \sqrt 2 \cdot r_2$$

Using (2) and (3) on (1) gives: $$\quad |AH|=\sqrt 2 \cdot r_2 + r_2 = \sqrt 2 \cdot r_0$$ $$\quad \Leftrightarrow (\sqrt 2 +1) \cdot r_2=\sqrt 2 \cdot r_0$$ $$\quad \Leftrightarrow r_2= \frac {\sqrt 2} {\sqrt 2 +1} \cdot r_0= \frac {\sqrt 2} {\sqrt 2 +1} \cdot \frac {\sqrt 2}{\sqrt 2}\cdot r_0= \frac {2} {2+\sqrt 2} \cdot r_0$$

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 $$r_2 = \frac {2} {2+\sqrt 2} \cdot r_0$$

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 $$P_2 = 2 \cdot \pi \cdot r_2 = 2 \cdot \pi \cdot \frac {2} {2+\sqrt 2} \cdot r_0$$ $$\quad \Leftrightarrow P_2= \frac {2} {2+\sqrt 2} \cdot P_0$$

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 $$\quad A_2= \pi \cdot r_2^2=\pi \cdot \left( \frac {2} {2+\sqrt 2} \cdot r_0 \right)^2=\pi \cdot \left( \frac {2} {2+\sqrt 2}\right)^2 \cdot r_0^2= \frac {4} {(2+\sqrt 2)^2} \cdot \pi \cdot r_0^2= \frac {4} {4+4 \cdot \sqrt 2+2} \cdot A_0= \frac {4} {6+4 \cdot \sqrt 2} \cdot A_0$$ $$\quad \Leftrightarrow A_2= \frac {2} {3+2 \cdot \sqrt 2} \cdot A_0$$

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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: $$\quad x_2=|AF| \cdot cos(45) =(|AG|-|FG|) \cdot cos(45)= (2 \cdot r_0-r_2) \cdot cos(45)= (2 \cdot r_0-\frac {2-\sqrt 2} {2} \cdot r_0) \cdot cos(45)$$ $$\quad \Leftrightarrow x_2= (2 \cdot -\frac {2-\sqrt 2} {2}) \cdot r_0 \cdot cos(45)= (\frac {4-2+\sqrt 2} {2}) \cdot r_0 \cdot cos(45)= (\frac {2+\sqrt 2} {2}) \cdot r_0 \cdot cos(45)$$ $$\quad \Leftrightarrow x_2= ( \frac {2+\sqrt 2} {2}) \cdot r_0 \cdot \frac {1} {\sqrt 2}= \left( \frac {2+\sqrt 2} {2 \cdot \sqrt 2} \right) \cdot r_0 =(\frac {1}{\sqrt 2}+\frac {1}{2}) \cdot r_0$$ $$\quad y_2=x_2$$

Normalised case
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: $$2.9333462$$ -->