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:
$\quad |AH|={\sqrt {2}}\cdot r_{2}+r_{2}={\sqrt {2}}\cdot r_{0}$
$\quad \Leftrightarrow ({\sqrt {2}}+1)\cdot r_{2}={\sqrt {2}}\cdot r_{0}$
$\quad \Leftrightarrow r_{2}={\frac {\sqrt {2}}{{\sqrt {2}}+1}}\cdot r_{0}={\frac {\sqrt {2}}{{\sqrt {2}}+1}}\cdot {\frac {\sqrt {2}}{\sqrt {2}}}\cdot r_{0}={\frac {2}{2+{\sqrt {2}}}}\cdot r_{0}$

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 $r_{2}={\frac {2}{2+{\sqrt {2}}}}\cdot r_{0}$

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
$\quad A_{2}=\pi \cdot r_{2}^{2}=\pi \cdot \left({\frac {2}{2+{\sqrt {2}}}}\cdot r_{0}\right)^{2}=\pi \cdot \left({\frac {2}{2+{\sqrt {2}}}}\right)^{2}\cdot r_{0}^{2}={\frac {4}{(2+{\sqrt {2}})^{2}}}\cdot \pi \cdot r_{0}^{2}={\frac {4}{4+4\cdot {\sqrt {2}}+2}}\cdot A_{0}={\frac {4}{6+4\cdot {\sqrt {2}}}}\cdot A_{0}$
$\quad \Leftrightarrow A_{2}={\frac {2}{3+2\cdot {\sqrt {2}}}}\cdot A_{0}$