Shows the largest circle that can be added to a QH-sangaku.

Base is the square ${\displaystyle [ABCD]}$ of side length s.
Included is a semicircle ${\displaystyle [EKFH]}$ with radius ${\displaystyle r_{1}=(2-{\sqrt {2}})\cdot s}$

In order to find the radius ${\displaystyle r_{2}}$ around point ${\displaystyle S_{2}}$ of the circle within the semicircle the following calculations have to be done:
The line segment ${\displaystyle [KH]}$ is the radius ${\displaystyle r_{1}}$ of the semicircle. It is also the diameter of the circle around point ${\displaystyle S_{2}}$. We get the equation:
${\displaystyle \quad 2\cdot r_{2}=|KH|}$
${\displaystyle \quad \Leftrightarrow 2\cdot r_{2}=r_{1}}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {r_{1}}{2}}}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {2-{\sqrt {2}}}{2}}\cdot s}$

0) The side length of the square: ${\displaystyle s}$
1) Radius of the semicircle ${\displaystyle r_{1}=(2-{\sqrt {2}})\cdot s\quad }$See QH_dia
2) Radius of the additional circle ${\displaystyle r_{2}={\frac {2-{\sqrt {2}}}{2}}\cdot s}$

0) Perimeter of base square ${\displaystyle P_{0}=4\cdot s}$
1) Perimeter of the semicircle ${\displaystyle P_{1}=2\cdot r_{1}+\pi \cdot r_{1}=(2+\pi )\cdot r_{1}=(2+\pi )\cdot (2-{\sqrt {2}})\cdot s}$
2) Perimeter of the additional circle ${\displaystyle P_{2}=2\pi r_{2}=\pi r_{1}=\pi \cdot (2-{\sqrt {2}})\cdot s}$

0) Area of the base square ${\displaystyle A_{0}=s^{2}}$
1) Area of the semicircle ${\displaystyle A_{1}=\pi \cdot (3-2{\sqrt {2}})\cdot s^{2}}$
2) Area of the additional circle ${\displaystyle A_{2}=\pi r_{2}^{2}=\pi \left({\frac {2-{\sqrt {2}}}{2}}\cdot s\right)^{2}=\pi \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot s^{2}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot s^{2}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot A_{0}}$

Centroid positions are measured from centroid point of the base shape.
0) Centroid positions of the base square: ${\displaystyle S_{0}:x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the semicircle: ${\displaystyle S_{1}:x_{1}=\left(({\sqrt {2}}-1)\cdot {\frac {3\pi +4}{3\pi }}-{\frac {1}{2}}\right)\cdot s\quad \quad y_{1}=x_{1}}$
2) Centroid positions of the additional circle: ${\displaystyle S_{2}:}$
${\displaystyle \quad \quad x_{2}=|S_{0}S_{2}|cos(45)}$

${\displaystyle \quad \Leftrightarrow x_{2}=\left(|AS_{2}|-|AS_{0}|\right)\cos(45)}$

${\displaystyle \quad \Leftrightarrow x_{2}=|AS_{2}|\cos(45)-|AS_{0}|\cos(45)}$

${\displaystyle \quad \Leftrightarrow x_{2}=|AS_{2}|\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=(|AH|-|S_{2}H|)\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=(2r_{1}-r_{2})\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=(2\cdot 2r_{2}-r_{2})\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=(4-1)\cdot r_{2}\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=3\cdot \left({\frac {2-{\sqrt {2}}}{2}}\cdot s\right)\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=(6-3{\sqrt {2}})\cdot {\frac {s}{2}}\cos(45)-{\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=\left((6-3{\sqrt {2}})\cdot \cos(45)-1\right)\cdot {\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=\left((6-3{\sqrt {2}})\cdot {\frac {1}{\sqrt {2}}}-{\frac {\sqrt {2}}{\sqrt {2}}}\right)\cdot {\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {6-3{\sqrt {2}}-{\sqrt {2}}}{\sqrt {2}}}\right)\cdot {\frac {s}{2}}}$

${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {6-4{\sqrt {2}}}{2{\sqrt {2}}}}\right)\cdot s}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {3-2{\sqrt {2}}}{\sqrt {2}}}\right)\cdot s}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {3{\sqrt {2}}-4}{2}}\right)\cdot s}$

In the normalised case the area of the base is set to 1.
${\displaystyle ||ABCD||=1\Rightarrow s^{2}=1\Rightarrow s=1}$

0) Segment of the base square ${\displaystyle s=1}$
1) Segment of the semicircle ${\displaystyle r_{1}=2-{\sqrt {2}}\quad \approx 0.585}$
2) Segment of the additional circle ${\displaystyle r_{2}={\frac {2-{\sqrt {2}}}{2}}\quad \approx 0.292}$

0) Perimeter of base square ${\displaystyle P_{0}=4}$
1) Perimeter of the semicircle ${\displaystyle P_{1}=(2+\pi )(2-{\sqrt {2}})=3.0118752...}$
2) Perimeter of the additional circle ${\displaystyle P_{2}=(2-{\sqrt {2}})\cdot \pi =1.8403023...}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}=8.8521776...}$

0) Area of the base square ${\displaystyle A_{0}=1}$
1) Area of the semicircle ${\displaystyle A_{1}=\pi \cdot (3-2{\sqrt {2}})\quad \approx 0.539}$
2) Area of the additional circle: ${\displaystyle A_{2}={\frac {3-2{\sqrt {2}}}{2}}\cdot \pi \quad \approx 0.27}$

Centroid positions are measured from the centroids of the base shape
0) Centroid positions of the base square: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the semicircle:
${\displaystyle \quad x_{1}=({\sqrt {2}}-1)\cdot (1+{\frac {4}{3\cdot \pi }})-{\frac {1}{2}}=0.0900112...}$
${\displaystyle \quad y_{1}=x_{1}=0.0900112...}$

2) Centroid positions of the circle:
${\displaystyle \quad x_{2}={\frac {3{\sqrt {2}}-4}{2}}=0.1213203...}$
${\displaystyle \quad y_{2}=x_{2}=0.1213203...}$

The distances between the centroids of elements are:
01) ${\displaystyle d01={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.127295142...}$
02) ${\displaystyle d02={\sqrt {(x_{0}-x_{2})^{2}+(y_{0}-y_{2})^{2}}}=0.171572875...}$
12) ${\displaystyle d12={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}=0.044277733...}$

S) The sum of the distances is: ${\displaystyle D_{S}=0.343145751...}$

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

${\displaystyle decimalpart(8.8521776...+0,3431457...)=decimalpart(9.1953234...)=.1953234}$

So the identifying number is: ${\displaystyle 2.1953234}$

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