Base is the square of given side length
a
0
{\displaystyle a_{0}}
with centroid at
S
0
{\displaystyle S_{0}}
Inscribed are the largest possible circle and right triangle of the same area.
Segments in the general case [ edit ]
0) The side length of the base square:
a
0
{\displaystyle a_{0}}
1) The radius of the inscribed circle:
r
1
=
2
π
+
2
+
1
⋅
a
0
≈
0.34
⋅
a
0
{\displaystyle r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.34\cdot a_{0}\quad }
, see Calculation 1
2) The side length of the inscribed right triangle:
a
2
=
2
π
⋅
r
1
=
2
π
π
+
2
+
1
⋅
a
0
≈
0.847
⋅
a
0
{\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}\quad }
, see Calculation 1
Perimeters in the general case [ edit ]
0) Perimeter of base square
P
0
=
4
⋅
a
0
{\displaystyle P_{0}=4\cdot a_{0}\quad }
1) Perimeter of the inscribed circle:
P
1
=
2
π
⋅
r
1
=
2
π
2
π
+
2
+
1
⋅
a
0
≈
2.122
⋅
a
0
{\displaystyle P_{1}=2\pi \cdot r_{1}={\frac {2\pi {\sqrt {2}}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 2.122\cdot a_{0}\quad }
2) Perimeter of the inscribed triangle:
P
2
=
4
⋅
a
2
=
4
⋅
2
π
π
+
2
+
1
⋅
a
0
≈
3.387
⋅
a
0
{\displaystyle P_{2}=4\cdot a_{2}={\frac {4\cdot 2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 3.387\cdot a_{0}\quad }
Areas in the general case [ edit ]
0) Area of the base square
A
0
=
a
0
2
{\displaystyle A_{0}=a_{0}^{2}}
1) Area of the inscribed circle:
A
1
=
π
r
1
2
=
π
(
2
π
+
2
+
1
)
2
⋅
a
0
2
≈
0.358
⋅
a
0
2
{\displaystyle A_{1}=\pi r_{1}^{2}=\pi \left({\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }
2) Area of the inscribed triangle:
A
2
=
a
2
2
2
=
1
2
(
2
π
π
+
2
+
1
)
2
⋅
a
0
2
≈
0.358
⋅
a
0
2
{\displaystyle A_{2}={\frac {a_{2}^{2}}{2}}={\frac {1}{2}}\left({\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }
Centroids in the general case [ edit ]
Centroids as graphically displayed [ edit ]
0) Centroid position of the base square:
S
0
=
0
+
0
i
=
0
{\displaystyle S_{0}=0+0i=0}
1) Centroid position of the inscribed circle:
S
1
=
1
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
1
+
i
)
⋅
a
0
≈
(
0.162
+
0.162
i
)
⋅
a
0
{\displaystyle S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.162+0.162i)\cdot a_{0}\quad }
, see Calculation 2
2) Centroid position of the inscribed triangle:
S
2
=
1
6
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
≈
(
−
0.218
−
0.218
i
)
⋅
a
0
{\displaystyle S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0.218-0.218i)\cdot a_{0}\quad }
, see Calculation 3
Orientated centroids [ edit ]
The centroid positions of the following shapes will be expressed orientated so that the first shape n with
S
n
≠
S
0
{\displaystyle S_{n}\neq S_{0}}
will be of type
S
n
=
0
−
k
i
{\displaystyle S_{n}=0-ki}
with
k
>
0
{\displaystyle k>0}
. The graphical representation does not correspond to the mathematical expression.
0) Orientated centroid position of the base square:
S
O
0
=
0
+
0
i
=
0
{\displaystyle SO_{0}=0+0i=0}
1) Orientated centroid position of the inscribed circle:
S
O
1
=
1
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
0
−
i
)
⋅
a
0
≈
(
0
−
0.229
i
)
⋅
a
0
{\displaystyle SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0-0.229i)\cdot a_{0}\quad }
, see calculation 4
2) Orientated centroid position of the inscribed right triangle:
S
O
2
=
1
3
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
0
−
i
)
⋅
a
0
≈
(
0
+
0.308
i
)
⋅
a
0
{\displaystyle SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0+0.308i)\cdot a_{0}}
, see calculation 5
In the normalised case the area of the base is set to 1.
A
0
=
1
⇒
a
0
2
=
1
⇒
a
0
2
=
1
{\displaystyle A_{0}=1\quad \Rightarrow a_{0}^{2}=1\quad \Rightarrow a_{0}^{2}=1}
Segments in the normalised case [ edit ]
0) Side length of the base square:
a
0
=
1
{\displaystyle a_{0}=1}
1) The radius of the inscribed circle:
r
1
=
2
π
+
2
+
1
≈
0.340
{\displaystyle r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\quad \approx 0.340\quad }
2) The side length of the inscribed right triangle:
a
2
=
2
π
⋅
r
1
=
2
π
π
+
2
+
1
≈
0.847
{\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\quad \approx 0.847\quad }
Perimeters in the normalised case [ edit ]
0) Perimeter of base square
P
0
=
4
{\displaystyle P_{0}=4\quad }
1) Perimeter of the inscribed circle:
P
1
=
2
π
⋅
r
1
=
2
π
2
π
+
2
+
1
≈
2.122
{\displaystyle P_{1}=2\pi \cdot r_{1}={\frac {2\pi {\sqrt {2}}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\quad \approx 2.122\quad }
2) Perimeter of the inscribed triangle:
P
2
=
4
⋅
a
2
=
4
⋅
2
π
π
+
2
+
1
≈
3.387
⋅
a
0
{\displaystyle P_{2}=4\cdot a_{2}={\frac {4\cdot 2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\quad \approx 3.387\cdot a_{0}\quad }
Areas in the normalised case [ edit ]
0) Area of the base square
A
0
=
1
{\displaystyle A_{0}=1}
1) Area of the inscribed circle:
A
1
=
π
r
1
2
=
π
(
2
π
+
2
+
1
)
2
≈
0.358
{\displaystyle A_{1}=\pi r_{1}^{2}=\pi \left({\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\quad \approx 0.358\quad }
2) Area of the inscribed triangle:
A
2
=
a
2
2
2
=
1
2
(
2
π
π
+
2
+
1
)
2
≈
0.358
{\displaystyle A_{2}={\frac {a_{2}^{2}}{2}}={\frac {1}{2}}\left({\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\quad \approx 0.358\quad }
Covered surface of the base shape [ edit ]
C
b
=
A
1
+
A
2
A
0
≈
0.716
≈
72
%
{\displaystyle C_{b}={\frac {A_{1}+A_{2}}{A_{0}}}\approx 0.716\quad \approx 72\%}
Centroids in the normalised case [ edit ]
Normalized centroids as graphically displayed [ edit ]
0) Centroid position of the base square:
S
0
=
0
+
0
i
=
0
{\displaystyle S_{0}=0+0i=0}
1) Centroid position of the inscribed circle:
S
1
=
1
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
1
+
i
)
≈
(
0.162
+
0.162
i
)
{\displaystyle S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\quad \approx (0.162+0.162i)\quad }
2) Centroid position of the inscribed square:
S
2
=
1
6
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
1
+
i
)
≈
(
−
0.218
−
0.218
i
)
{\displaystyle S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\quad \approx (-0.218-0.218i)\quad }
Normalized centroids orientated [ edit ]
The centroid positions of the following shapes will be expressed orientated so that the first shape n with
S
n
≠
S
0
{\displaystyle S_{n}\neq S_{0}}
will be of type
S
n
=
0
−
k
i
{\displaystyle S_{n}=0-ki}
with
k
>
0
{\displaystyle k>0}
. The graphical representation does not correspond to the mathematical expression.
0) Orientated centroid position of the base square:
S
O
0
=
0
+
0
i
=
0
{\displaystyle SO_{0}=0+0i=0}
1) Orientated centroid position of the inscribed circle:
S
O
1
=
1
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
0
−
i
)
≈
(
0
−
0.229
i
)
{\displaystyle SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\quad \approx (0-0.229i)\quad }
2) Orientated centroid position of the inscribed right triangle:
S
O
2
=
1
3
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
0
−
i
)
≈
(
0
+
0.308
i
)
{\displaystyle SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\quad \approx (0+0.308i)}
Equations of given elements and relations [ edit ]
(0) The area of the right triangle
△
A
K
H
{\displaystyle \triangle AKH}
has to be the same as the area of the circle with radius
r
1
{\displaystyle r_{1}}
around center point
S
1
⇒
A
1
=
A
2
{\displaystyle S_{1}\quad \Rightarrow A_{1}=A_{2}}
(1)
|
A
B
|
=
|
B
C
|
=
|
C
D
|
=
|
A
D
|
=
a
0
{\displaystyle |AB|=|BC|=|CD|=|AD|=a_{0}\quad }
since
◻
A
B
C
D
{\displaystyle \square ABCD}
is a square
(2)
|
A
C
|
=
2
⋅
a
0
{\displaystyle |AC|={\sqrt {2}}\cdot a_{0}\quad }
since
A
C
¯
{\displaystyle {\overline {AC}}}
is the diagonal of square
◻
A
B
C
D
{\displaystyle \square ABCD}
(3)
|
A
K
|
=
|
A
H
|
=
a
2
{\displaystyle |AK|=|AH|=a_{2}\quad }
since
△
A
K
H
{\displaystyle \triangle AKH}
is a right triangle
(4)
|
S
1
F
|
=
|
F
C
|
=
|
C
G
|
=
|
G
S
1
|
=
r
1
{\displaystyle |S_{1}F|=|FC|=|CG|=|GS_{1}|=r_{1}\quad }
since
◻
S
1
F
C
G
{\displaystyle \square S_{1}FCG}
is a square, because
F
{\displaystyle F}
and
G
{\displaystyle G}
are tangent points of the circle around
S
1
{\displaystyle S_{1}}
with the sides of square
◻
A
B
C
D
{\displaystyle \square ABCD}
The radius
r
1
{\displaystyle r_{1}}
is calculated:
a) First the relation between
r
1
{\displaystyle r_{1}}
and
a
2
{\displaystyle a_{2}}
is determined from equation (0):
A
1
=
A
2
{\displaystyle A_{1}=A_{2}\quad }
applying equation (0)
⇔
a
2
2
2
=
π
r
1
2
{\displaystyle \quad \Leftrightarrow {\frac {a_{2}^{2}}{2}}=\pi r_{1}^{2}\quad }
, definitions of areas of right triangles and circles
⇔
a
2
2
=
2
π
r
1
2
{\displaystyle \quad \Leftrightarrow a_{2}^{2}=2\pi r_{1}^{2}\quad }
, multiplying both sides by two
⇔
a
2
=
2
π
⋅
r
1
{\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad }
, since negative length cannot be applied
b) Then the relation between
r
1
{\displaystyle r_{1}}
and
a
0
{\displaystyle a_{0}}
is determined from the following identity:
|
A
C
|
=
2
⋅
a
0
{\displaystyle |AC|={\sqrt {2}}\cdot a_{0}\quad }
since
◻
A
B
C
D
{\displaystyle \square ABCD}
is a square
⇔
|
A
E
|
+
|
E
S
1
|
+
|
S
1
C
|
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow |AE|+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }
, because the three segments are forming the diagonal of the square
⇔
2
2
⋅
a
2
+
|
E
S
1
|
+
|
S
1
C
|
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow {\frac {\sqrt {2}}{2}}\cdot a_{2}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }
, since
△
A
K
H
{\displaystyle \triangle AKH}
is a right isosceles trinagle
⇔
2
2
(
2
π
⋅
r
1
)
+
|
E
S
1
|
+
|
S
1
C
|
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow {\frac {\sqrt {2}}{2}}({\sqrt {2\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }
, applying part a) of the calculation
⇔
2
2
2
(
π
⋅
r
1
)
+
|
E
S
1
|
+
|
S
1
C
|
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow {\frac {{\sqrt {2}}{\sqrt {2}}}{2}}({\sqrt {\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }
, rearranging
⇔
π
⋅
r
1
+
|
E
S
1
|
+
|
S
1
C
|
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }
, rearranging
⇔
π
⋅
r
1
+
r
1
+
|
S
1
C
|
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }
, because E is a tangent point on the circle and a diagonal side of the triangle
△
A
K
H
{\displaystyle \triangle AKH}
⇔
π
⋅
r
1
+
r
1
+
2
⋅
r
1
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+{\sqrt {2}}\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }
, because
◻
S
1
F
C
G
{\displaystyle \square S_{1}FCG}
is a square with side length
r
1
{\displaystyle r_{1}}
⇔
r
1
⋅
(
π
+
1
+
2
)
=
2
⋅
a
0
{\displaystyle \quad \Leftrightarrow r_{1}\cdot ({\sqrt {\pi }}+1+{\sqrt {2}})={\sqrt {2}}\cdot a_{0}\quad }
, extracting
r
1
{\displaystyle r_{1}}
out of the bracket
⇔
r
1
=
2
π
+
2
+
1
⋅
a
0
≈
0.338
⋅
a
0
{\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.338\cdot a_{0}}
, rearranging
c) Eventually the relation between
a
2
{\displaystyle a_{2}}
and
a
0
{\displaystyle a_{0}}
is determined by using part a) of the calculation:
a
2
=
2
π
⋅
r
1
{\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad }
, from part a) of the calculation
⇔
a
2
=
2
π
⋅
2
π
+
2
+
1
⋅
a
0
{\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot {\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}}
, using the result from part b) of the calculation
⇔
a
2
=
2
π
π
+
2
+
1
⋅
a
0
≈
0.847
⋅
a
0
{\displaystyle \quad \Leftrightarrow a_{2}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}}
, rearranging
Calculating the centroid of the inscribed circle as displayed:
S
1
=
S
0
+
S
0
S
1
→
{\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}S_{1}}}}
⇔
S
1
=
(
0
+
0
i
)
+
S
0
C
→
+
C
S
1
→
{\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+{\overrightarrow {S_{0}C}}+{\overrightarrow {CS_{1}}}}
⇔
S
1
=
(
|
S
0
C
|
⋅
1
2
⋅
(
1
+
i
)
)
+
C
S
1
→
{\displaystyle \quad \Leftrightarrow S_{1}=\left(|S_{0}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }
, since
S
0
C
→
{\displaystyle {\overrightarrow {S_{0}C}}}
is collinear to the diagonal of the square
⇔
S
1
=
(
2
⋅
a
0
2
⋅
1
2
⋅
(
1
+
i
)
)
+
C
S
1
→
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }
, since
S
0
C
→
{\displaystyle {\overrightarrow {S_{0}C}}}
is half the diagonal of the square
⇔
S
1
=
(
a
0
2
⋅
(
1
+
i
)
)
+
C
S
1
→
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }
, rearranging
⇔
S
1
=
(
a
0
2
⋅
(
1
+
i
)
)
−
(
|
S
1
C
|
⋅
1
2
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(|S_{1}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }
, rearranging
⇔
S
1
=
(
a
0
2
⋅
(
1
+
i
)
)
−
(
2
⋅
r
1
⋅
1
2
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left({\sqrt {2}}\cdot r_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }
, since
S
1
C
→
{\displaystyle {\overrightarrow {S_{1}C}}}
is the diagonal of the square
◻
S
1
F
C
G
{\displaystyle \square {S_{1}FCG}}
with side length
r
1
{\displaystyle r_{1}}
⇔
S
1
=
(
a
0
2
⋅
(
1
+
i
)
)
−
(
r
1
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(r_{1}\cdot (1+i)\right)\quad }
, rearranging
⇔
S
1
=
(
a
0
2
−
r
1
)
⋅
(
1
+
i
)
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-r_{1}\right)\cdot (1+i)\quad }
, rearranging
⇔
S
1
=
(
a
0
2
−
2
π
+
2
+
1
⋅
a
0
)
⋅
(
1
+
i
)
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad }
, applying Calculation 1b
⇔
S
1
=
(
1
2
−
2
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {1}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }
, rearranging
⇔
S
1
=
(
π
+
2
+
1
2
(
π
+
2
+
1
)
−
2
2
2
(
π
+
2
+
1
)
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {\pi }}+{\sqrt {2}}+1}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}-{\frac {2{\sqrt {2}}}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}\right)\cdot (1+i)\cdot a_{0}\quad }
, rearranging
⇔
S
1
=
(
π
−
2
+
1
2
(
π
+
2
+
1
)
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {\pi }}-{\sqrt {2}}+1}{2({\sqrt {\pi }}+{\sqrt {2}}+1)}}\right)\cdot (1+i)\cdot a_{0}\quad }
, rearranging
⇔
S
1
=
1
2
⋅
(
π
−
2
+
1
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
≈
(
0.162
+
0.162
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}-{\sqrt {2}}+1}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (0.162+0.162i)\cdot a_{0}}
, rearranging
Calculating the centroid of the right triangle as displayed:
S
2
=
S
0
+
S
0
S
2
→
{\displaystyle S_{2}=S_{0}+{\overrightarrow {S_{0}S_{2}}}}
⇔
S
2
=
(
0
+
0
i
)
+
S
0
E
→
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}E}}+{\overrightarrow {ES_{2}}}}
⇔
S
2
=
(
0
+
0
i
)
+
S
0
A
→
+
A
E
→
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}A}}+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}}
⇔
S
2
=
(
−
|
S
0
A
|
⋅
1
2
⋅
(
1
+
i
)
)
+
A
E
→
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-|S_{0}A|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad }
, since
S
0
A
→
{\displaystyle {\overrightarrow {S_{0}A}}}
is collinear to the diagonal of the square
◻
A
B
C
D
{\displaystyle \square {ABCD}}
⇔
S
2
=
(
−
2
⋅
a
0
2
⋅
1
2
⋅
(
1
+
i
)
)
+
A
E
→
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad }
, since
S
0
A
→
{\displaystyle {\overrightarrow {S_{0}A}}}
is half the diagonal of the square
◻
A
B
C
D
{\displaystyle \square {ABCD}}
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
A
E
→
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {AE}}+{\overrightarrow {ES_{2}}}\quad }
, rearranging
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
|
A
E
|
⋅
1
2
⋅
(
1
+
i
)
)
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left(|AE|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad }
, since
A
E
→
{\displaystyle {\overrightarrow {AE}}}
is the height of the triangle
△
A
K
H
{\displaystyle \triangle {AKH}}
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
a
2
2
⋅
1
2
⋅
(
1
+
i
)
)
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{\sqrt {2}}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad }
, since
a
2
2
{\displaystyle {\frac {a_{2}}{\sqrt {2}}}}
is the value of the height of the triangle
△
A
K
H
{\displaystyle \triangle {AKH}}
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
a
2
2
⋅
(
1
+
i
)
)
+
E
S
2
→
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+{\overrightarrow {ES_{2}}}\quad }
, rearranging
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
a
2
2
⋅
(
1
+
i
)
)
+
(
−
|
E
S
2
|
⋅
1
2
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-|ES_{2}|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }
, since
E
S
2
→
{\displaystyle {\overrightarrow {ES_{2}}}}
is collinear to the diagonal of the square
◻
A
B
C
D
{\displaystyle \square {ABCD}}
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
a
2
2
⋅
(
1
+
i
)
)
+
(
−
1
3
|
A
E
|
⋅
1
2
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}|AE|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }
, since the centroid of a triangle is a third of the height
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
a
2
2
⋅
(
1
+
i
)
)
+
(
−
1
3
⋅
a
2
2
⋅
1
2
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}\cdot {\frac {a_{2}}{\sqrt {2}}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }
, since the height of a right isoceles triangle with side length a is:
h
=
a
2
{\displaystyle h={\frac {a}{\sqrt {2}}}}
⇔
S
2
=
(
−
a
0
2
⋅
(
1
+
i
)
)
+
(
a
2
2
⋅
(
1
+
i
)
)
+
(
−
1
3
⋅
a
2
2
⋅
(
1
+
i
)
)
{\displaystyle \quad \Leftrightarrow S_{2}=\left(-{\frac {a_{0}}{2}}\cdot (1+i)\right)+\left({\frac {a_{2}}{2}}\cdot (1+i)\right)+\left(-{\frac {1}{3}}\cdot {\frac {a_{2}}{2}}\cdot (1+i)\right)\quad }
, rearranging
⇔
S
2
=
1
2
⋅
(
−
a
0
+
a
2
−
1
3
⋅
a
2
)
⋅
(
1
+
i
)
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+a_{2}-{\frac {1}{3}}\cdot a_{2}\right)\cdot (1+i)\quad }
, factoring out
⇔
S
2
=
1
2
⋅
(
−
a
0
+
2
3
⋅
a
2
)
⋅
(
1
+
i
)
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+{\frac {2}{3}}\cdot a_{2}\right)\cdot (1+i)\quad }
, rearranging
⇔
S
2
=
1
2
⋅
(
−
a
0
+
2
3
⋅
2
π
π
+
2
+
1
⋅
a
0
)
⋅
(
1
+
i
)
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-a_{0}+{\frac {2}{3}}\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad }
, applying result of calculation 1c
⇔
S
2
=
1
2
⋅
(
−
1
+
2
3
⋅
2
π
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{2}}\cdot \left(-1+{\frac {2}{3}}\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }
, factoring out
⇔
S
2
=
1
6
⋅
(
−
3
+
2
⋅
2
π
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left(-3+2\cdot {\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }
, factoring out
⇔
S
2
=
1
6
⋅
(
4
π
π
+
2
+
1
−
3
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}-3\right)\cdot (1+i)\cdot a_{0}\quad }
, rearranging
⇔
S
2
=
1
6
⋅
(
4
π
−
3
(
π
+
2
+
1
)
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}-3({\sqrt {\pi }}+{\sqrt {2}}+1)}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }
, having a common denominator
⇔
S
2
=
1
6
⋅
(
4
π
−
3
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {4{\sqrt {\pi }}-3{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad }
, multiplying into the bracket
⇔
S
2
=
1
6
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
≈
(
−
0.218
−
0.218
i
)
{\displaystyle \quad \Leftrightarrow S_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\quad \approx (-0.218-0.218i)}
, adding
The centroid of the inscribed circle is calculated in orientated expression:
This will be done by rotating the expression as displayed by
225
∘
{\displaystyle 225^{\circ }}
which will be done by a
45
∘
{\displaystyle 45^{\circ }}
rotation followed by a
180
∘
{\displaystyle 180^{\circ }}
rotation
S
O
1
=
S
1
⋅
(
1
2
+
i
2
)
⋅
(
i
2
)
{\displaystyle SO_{1}=S_{1}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (i^{2})}
⇔
S
O
1
=
S
1
⋅
(
1
2
+
i
2
)
⋅
(
−
1
)
{\displaystyle \quad \Leftrightarrow SO_{1}=S_{1}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (-1)\quad }
, since
i
2
=
−
1
{\displaystyle i^{2}=-1}
⇔
S
O
1
=
S
1
⋅
1
2
⋅
(
1
+
i
)
⋅
(
−
1
)
{\displaystyle \quad \Leftrightarrow SO_{1}=S_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(1+i\right)\cdot (-1)\quad }
, factoring out
⇔
S
O
1
=
S
1
⋅
1
2
⋅
(
−
1
−
i
)
{\displaystyle \quad \Leftrightarrow SO_{1}=S_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }
, multiplying
⇔
S
O
1
=
1
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
1
+
i
)
⋅
a
0
⋅
1
2
⋅
(
−
1
−
i
)
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }
, applying Calculation 2
⇔
S
O
1
=
1
2
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
1
+
i
)
⋅
(
−
1
−
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (1+i)\cdot \left(-1-i\right)\cdot a_{0}\quad }
, rearranging
⇔
S
O
1
=
1
2
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
−
1
−
i
−
i
−
i
2
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (-1-i-i-i^{2})\cdot a_{0}\quad }
, multiplying
⇔
S
O
1
=
1
2
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
−
1
−
i
−
i
+
1
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (-1-i-i+1)\cdot a_{0}\quad }
, since
i
2
=
−
1
{\displaystyle i^{2}=-1}
⇔
S
O
1
=
1
2
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
0
−
2
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-2i)\cdot a_{0}\quad }
, adding up
⇔
S
O
1
=
2
2
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
0
−
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {2}{2{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad }
, factoring out
⇔
S
O
1
=
1
2
⋅
(
π
+
1
−
2
π
+
1
+
2
)
⋅
(
0
−
i
)
⋅
a
0
≈
(
0
−
0.229
i
)
{\displaystyle \quad \Leftrightarrow SO_{1}={\frac {1}{\sqrt {2}}}\cdot \left({\frac {{\sqrt {\pi }}+1-{\sqrt {2}}}{{\sqrt {\pi }}+1+{\sqrt {2}}}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0-0.229i)\quad }
The centroid of the inscribed triangle is calculated in orientated expression:
This will be done by rotating the expression as displayed by
225
∘
{\displaystyle 225^{\circ }}
which will be done by a
45
∘
{\displaystyle 45^{\circ }}
rotation followed by a
180
∘
{\displaystyle 180^{\circ }}
rotation
S
O
2
=
S
2
⋅
(
1
2
+
i
2
)
⋅
(
i
2
)
{\displaystyle SO_{2}=S_{2}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (i^{2})}
⇔
S
O
2
=
S
2
⋅
(
1
2
+
i
2
)
⋅
(
−
1
)
{\displaystyle \quad \Leftrightarrow SO_{2}=S_{2}\cdot \left({\frac {1}{\sqrt {2}}}+{\frac {i}{\sqrt {2}}}\right)\cdot (-1)\quad }
, since
i
2
=
−
1
{\displaystyle i^{2}=-1}
⇔
S
O
2
=
S
2
⋅
1
2
⋅
(
1
+
i
)
⋅
(
−
1
)
{\displaystyle \quad \Leftrightarrow SO_{2}=S_{2}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(1+i\right)\cdot (-1)\quad }
, factoring out
⇔
S
O
2
=
S
2
⋅
1
2
⋅
(
−
1
−
i
)
{\displaystyle \quad \Leftrightarrow SO_{2}=S_{2}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }
, multiplying
⇔
S
O
2
=
1
6
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
a
0
⋅
1
2
⋅
(
−
1
−
i
)
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot a_{0}\cdot {\frac {1}{\sqrt {2}}}\cdot \left(-1-i\right)\quad }
, applying Calculation 3
⇔
S
O
2
=
1
6
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
1
+
i
)
⋅
(
−
1
−
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (1+i)\cdot \left(-1-i\right)\cdot a_{0}\quad }
, rearranging
⇔
S
O
2
=
1
6
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
−
1
−
i
−
i
−
i
2
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (-1-i-i-i^{2})\cdot a_{0}\quad }
, multiplying
⇔
S
O
2
=
1
6
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
−
1
−
2
i
−
(
−
1
)
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (-1-2i-(-1))\cdot a_{0}\quad }
, since
i
2
=
−
1
{\displaystyle i^{2}=-1}
⇔
S
O
2
=
1
6
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
0
−
2
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-2i)\cdot a_{0}\quad }
, adding
⇔
S
O
2
=
2
6
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
0
−
i
)
⋅
a
0
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {2}{6{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad }
, factoring out
⇔
S
O
2
=
1
3
2
⋅
(
π
−
3
2
−
3
π
+
2
+
1
)
⋅
(
0
−
i
)
⋅
a
0
≈
(
0
+
0.308
i
)
{\displaystyle \quad \Leftrightarrow SO_{2}={\frac {1}{3{\sqrt {2}}}}\cdot \left({\frac {{\sqrt {\pi }}-3{\sqrt {2}}-3}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)\cdot (0-i)\cdot a_{0}\quad \approx (0+0.308i)}
, reducing