File:FS RCC dia.png

Shows the largest circle within RC sangaku.

Base is the right isosceles triangle of side length ${\displaystyle a}$ and centroid ${\displaystyle S_{0}}$
Inscribed is the largest possible circle with radius ${\displaystyle r_{1}}$ around point ${\displaystyle S_{1}}$
Inscribed is the largest possible circle outside the first circle with radius ${\displaystyle r_{2}}$ around point ${\displaystyle S_{2}}$

0) The side length of the base right triangle ${\displaystyle |AB|=|AC|=a}$
1) Radius of the first circle ${\displaystyle r_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot a\quad \approx 0,293\cdot a\quad }$ (See FS RC).
2) Radius of the second circle ${\displaystyle r_{2}=\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)\cdot r_{1}=\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot a\quad \approx 0.131\cdot a\quad }$, (see Calculation 1)

0) Perimeter of base triangle ${\displaystyle P_{0}=|AB|+|BC|+|AC|=a+{\sqrt {2}}\cdot a+a=(2+{\sqrt {2}})\cdot a}$
1) Perimeter of the first circle ${\displaystyle P_{1}=\pi \cdot (2-{\sqrt {2}})\cdot a\quad }$ (See FS RC )
2) Perimeter of the second circle ${\displaystyle P_{2}=2\pi r_{2}=2\pi \left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot a}$

0) Area of the base triangle ${\displaystyle A_{0}={\frac {a^{2}}{2}}}$
1) Area of the inscribed circle ${\displaystyle A_{1}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot a^{2}\quad }$ (See FS RC)
2) Area of the second circle ${\displaystyle A_{2}=\pi r_{2}^{2}=\pi \left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{4}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{4}\cdot a^{2}}$

Centroid positions are measured from the centroid point of the base shape
0) Centroid positions of the base triangle: ${\displaystyle x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed circle: ${\displaystyle x_{1}=y_{1}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a\quad \approx -0.04\cdot a\quad }$
2) Centroid positions of the second circle:
${\displaystyle \quad x_{2}=\left({\frac {14}{3}}-{\frac {7}{2}}{\sqrt {2}}+(6-4{\sqrt {2}}){\sqrt {2+{\sqrt {2}}}}\right)\cdot a\quad \approx 0.351\cdot a\quad }$, (see Calculation 2)
${\displaystyle \quad y_{2}=\left(\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}-{\frac {1}{3}}\right)\cdot a\quad \approx -0.203\cdot a}$, (see Calculation 3)

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

0) Side length of the base triangle ${\displaystyle a={\sqrt {2}}\quad \approx 1.414}$
1) Radius of the inscribed circle ${\displaystyle r_{1}={\frac {2-{\sqrt {2}}}{2}}\cdot {\sqrt {2}}={\frac {2{\sqrt {2}}-2}{2}}={\sqrt {2}}-1\quad \approx 0.414}$
2) Radius of the second inscribed circle ${\displaystyle r_{2}=\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot {\sqrt {2}}\quad \approx 0.185}$

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 circle ${\displaystyle P_{1}=\pi \cdot (2-{\sqrt {2}})\cdot {\sqrt {2}}=\pi \cdot (2{\sqrt {2}}-2)\quad \approx 2.6025805}$
2) Perimeter of the second circle ${\displaystyle P_{2}=2\pi r_{2}=2\pi \left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot {\sqrt {2}}\quad \approx 1.1619551}$

S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}+P_{2}\approx 8.5929628}$

0) Area of the base triangle ${\displaystyle A_{0}=1}$
1) Area of the inscribed circle ${\displaystyle A_{1}=\pi \left({\frac {3-2{\sqrt {2}}}{2}}\right)\cdot \left({\sqrt {2}}\right)^{2}=\pi (3-2{\sqrt {2}})\quad \approx 0.539}$
2) Area of the second inscribed circle ${\displaystyle A_{2}=\pi \left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{4}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{4}\cdot ({\sqrt {2}})^{2}\quad \approx 0.107}$

Centroid positions are measured from the centroid point of the base shape.
0) Centroid positions of the base triangle: ${\displaystyle S_{0}:\quad x_{0}=0\quad y_{0}=0}$
1) Centroid positions of the inscribed circle: ${\displaystyle S_{1}:x_{1}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot {\sqrt {2}}={\frac {4{\sqrt {2}}-6}{6}}={\frac {2{\sqrt {2}}-3}{3}}=-0.057\quad y_{1}=x_{1}==-0.057}$
2) Centroid positions of the second circle:
${\displaystyle \quad x_{2}=\left({\frac {14}{3}}-{\frac {7}{2}}{\sqrt {2}}+(6-4{\sqrt {2}}){\sqrt {2+{\sqrt {2}}}}\right)\cdot {\sqrt {2}}=\left({\frac {14{\sqrt {2}}}{3}}-{\frac {7}{2}}\cdot 2+(6{\sqrt {2}}-4\cdot 2){\sqrt {2+{\sqrt {2}}}}\right)=\left({\frac {14{\sqrt {2}}}{3}}-7+2\cdot (3{\sqrt {2}}-4){\sqrt {2+{\sqrt {2}}}}\right)\quad \approx 0.496}$
${\displaystyle \quad y_{2}=\left(\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}-{\frac {1}{3}}\right){\sqrt {2}}\quad \approx -0.286}$

01) The distance between the centroid of the base element and the centroid of the circle is:${\displaystyle d={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}=0.0808802...}$
02) The distance between the centroid of the base element and the centroid of the second circle is:${\displaystyle d={\sqrt {(x_{0}-x_{2})^{2}+(y_{0}-y_{2})^{2}}}=0.5730854...}$
12) The distance between the centroids of the circles is:${\displaystyle d={\sqrt {(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}=0.5991445...}$

The sum of the distances is: ${\displaystyle D=1.2531101...}$

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 the distances of the centroids in the normalised case.

${\displaystyle decimalpart(8.5929628...+1.2531101...)=decimalpart(9.8460729...)=.8460729}$

So the identifying number is: ${\displaystyle 2.8460729}$

The following equations hold:
(1)${\displaystyle \quad |AB|=|AC|=a}$
(2)${\displaystyle \quad |AX_{1}|=|AY_{1}|=|S_{1}X_{1}|=|S_{1}Y_{1}|=|DS_{1}|=r_{1}}$
(3)${\displaystyle \quad |AY_{2}|=|S_{2}X_{2}|=|DS_{2}|=|EX_{1}|=r_{2}}$
(4)${\displaystyle \quad |AX_{1}|+|X_{1}X_{2}|+|X_{2}B|=|AB|=a}$
(5)${\displaystyle \quad |X_{1}X_{2}|=|ES_{2}|}$

The triangle ${\displaystyle \triangle BS_{1}X_{1}}$ is similar to triangle ${\displaystyle \triangle S_{2}S_{1}E}$. So the angle ${\displaystyle \theta }$ in ${\displaystyle B}$ is the same as the angle ${\displaystyle \theta }$ in ${\displaystyle S_{2}}$.
The line segments ${\displaystyle AB}$ and ${\displaystyle BC}$ are tangent to the circle around ${\displaystyle S_{1}}$ with radius ${\displaystyle r_{1}}$ and the circle around ${\displaystyle S_{2}}$ with radius ${\displaystyle r_{2}}$. So the angle in ${\displaystyle B}$ in ${\displaystyle \triangle BS_{1}X_{1}}$ bisects the angle in ${\displaystyle \triangle BCA}$. Since the angle in ${\displaystyle \triangle BCA}$ is 45° we get:
(6)${\displaystyle \quad sin{(\theta )}=sin{({\frac {45}{2}})}={\frac {1}{2}}{\sqrt {2-{\sqrt {2}}}}}$
${\displaystyle \quad \Leftrightarrow {\frac {1}{2}}{\sqrt {2-{\sqrt {2}}}}=t}$, by putting ${\displaystyle sin{(\theta )}=t}$

Looking at ${\displaystyle \triangle S_{2}S_{1}E}$, we get:
(7) ${\displaystyle \quad {\frac {|ES_{1}|}{|S_{1}S_{2}|}}=t}$, since ${\displaystyle {\overline {ES_{1}}}}$ is the opposite and ${\displaystyle {\overline {S_{1}S_{2}}}}$ is the hypothenuse of ${\displaystyle \triangle S_{2}S_{1}E}$
${\displaystyle \quad \Leftrightarrow {\frac {r_{1}-r_{2}}{r_{1}+r_{2}}}=t}$, applying equations (2) and (3)
${\displaystyle \quad \Leftrightarrow r_{1}-r_{2}=(r_{1}+r_{2})\cdot t}$
${\displaystyle \quad \Leftrightarrow r_{1}-r_{2}=tr_{1}+tr_{2}+r_{2}}$
${\displaystyle \quad \Leftrightarrow r_{1}=tr_{1}+tr_{2}+r_{2}}$
${\displaystyle \quad \Leftrightarrow r_{1}\cdot (1-t)=r_{2}\cdot (1+t)}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {1-t}{1+t}}\cdot r_{1}}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {1-t}{1+t}}\cdot {\frac {1-t}{1-t}}\cdot r_{1}}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {(1-t)^{2}}{1-t^{2}}}\cdot r_{1}}$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {(1-t)^{2}}{1-\left({\frac {1}{2}}{\sqrt {2-{\sqrt {2}}}}\right)^{2}}}\cdot r_{1}\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {(1-t)^{2}}{1-{\frac {1}{4}}(2-{\sqrt {2}})}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {4\cdot (1-t)^{2}}{4-(2-{\sqrt {2}})}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {4\cdot (1-t)^{2}}{2+{\sqrt {2}}}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {4\cdot (1-t)^{2}}{2+{\sqrt {2}}}}\cdot {\frac {2-{\sqrt {2}}}{2-{\sqrt {2}}}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {4\cdot (1-t)^{2}\cdot (2-{\sqrt {2}})}{(2+{\sqrt {2}})(2-{\sqrt {2}})}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {4\cdot (1-t)^{2}\cdot (2-{\sqrt {2}})}{(2^{2}-({\sqrt {2}})^{2})}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}={\frac {4\cdot (1-t)^{2}\cdot (2-{\sqrt {2}})}{(4-2)}}\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}=2\cdot (1-t)^{2}\cdot (2-{\sqrt {2}})\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}=2\cdot \left(1-{\frac {1}{2}}{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot (2-{\sqrt {2}})\cdot r_{1}\quad }$, applying equation (6)
${\displaystyle \quad \Leftrightarrow r_{2}=2\cdot \left({\frac {2-{\sqrt {2-{\sqrt {2}}}}}{2}}\right)^{2}\cdot (2-{\sqrt {2}})\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}=\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)\cdot r_{1}\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}=\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)\cdot \left({\frac {2-{\sqrt {2}}}{2}}\cdot a\right)\quad }$, applying the equation for r_1
${\displaystyle \quad \Leftrightarrow r_{2}=\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot a\quad }$
${\displaystyle \quad \Leftrightarrow r_{2}=\left(11-{\frac {15}{2}}{\sqrt {2}}-6{\sqrt {2-{\sqrt {2}}}}+4{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}\right)\cdot a\quad }$

${\displaystyle \quad x_{2}=|X_{0}X_{2}|=x_{1}+[X_{1}X_{2}|=x_{1}+|ES_{2}|}$
${\displaystyle \quad \Leftrightarrow x_{2}=x_{1}+|S_{1}S_{2}|\cdot cos{\frac {45}{2}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=x_{1}+|S_{1}S_{2}|\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=x_{1}+(r_{1}+r_{2})\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+(r_{1}+r_{2})\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left({\frac {2-{\sqrt {2}}}{2}}\cdot a+r_{2}\right)\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left({\frac {2-{\sqrt {2}}}{2}}\cdot a+\left(11-{\frac {15}{2}}{\sqrt {2}}-6{\sqrt {2-{\sqrt {2}}}}+4{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}\right)\cdot a\right)\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left({\frac {2-{\sqrt {2}}}{2}}+11-{\frac {15}{2}}{\sqrt {2}}-6{\sqrt {2-{\sqrt {2}}}}+4{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}\right)\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(1-{\frac {1}{2}}{\sqrt {2}}+11-{\frac {15}{2}}{\sqrt {2}}-6{\sqrt {2-{\sqrt {2}}}}+4{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}\right)\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(12-{\frac {16}{2}}{\sqrt {2}}-6{\sqrt {2-{\sqrt {2}}}}+4{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}\right)\cdot {\frac {1}{2}}{\sqrt {2+{\sqrt {2}}}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(6-4{\sqrt {2}}-3{\sqrt {2-{\sqrt {2}}}}+2{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}\right)\cdot {\sqrt {2+{\sqrt {2}}}}\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {2-{\sqrt {2}}}}{\sqrt {2+{\sqrt {2}}}}+2{\sqrt {2}}{\sqrt {2-{\sqrt {2}}}}{\sqrt {2+{\sqrt {2}}}}\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {(2-{\sqrt {2}})(2+{\sqrt {2}})}}+2{\sqrt {2}}{\sqrt {(2-{\sqrt {2}})(2+{\sqrt {2}})}}\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {(4-2)}}+2{\sqrt {2}}{\sqrt {(4-2)}}\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {2}}+2{\sqrt {2}}{\sqrt {2}}\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4-3{\sqrt {2}}}{6}}\right)\cdot a+\left(6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {2}}+2\cdot 2\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {4}{6}}-{\frac {3{\sqrt {2}}}{6}}\right)\cdot a+\left(6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {2}}+4\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {2}{3}}-{\frac {\sqrt {2}}{2}}+6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}-3{\sqrt {2}}+4\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {2+12}{3}}-{\frac {1+6}{2}}{\sqrt {2}}+6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {14}{3}}-{\frac {7}{2}}{\sqrt {2}}+6{\sqrt {2+{\sqrt {2}}}}-4{\sqrt {2}}{\sqrt {2+{\sqrt {2}}}}\right)\cdot a}$
${\displaystyle \quad \Leftrightarrow x_{2}=\left({\frac {14}{3}}-{\frac {7}{2}}{\sqrt {2}}+(6-4{\sqrt {2}}){\sqrt {2+{\sqrt {2}}}}\right)\cdot a\approx 0.351\cdot a}$

${\displaystyle \quad y_{2}=|S_{0}X_{0}|+r_{2}}$
${\displaystyle \quad y_{2}=-{\frac {1}{3}}\cdot a+r_{2}}$
${\displaystyle \quad y_{2}=-{\frac {1}{3}}\cdot a+\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}\cdot a}$
${\displaystyle \quad y_{2}=\left(\left(2-{\sqrt {2-{\sqrt {2}}}}\right)^{2}\cdot \left({\frac {2-{\sqrt {2}}}{2}}\right)^{2}-{\frac {1}{3}}\right)\cdot a}$

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