User:Hans G. Oberlack/Sandkiste: Difference between revisions

Base is the square of given side length ${\displaystyle a_{0}}$ with centroid at ${\displaystyle S_{0}}$
Inscribed are the largest possible circle and right triangle of the same area.

0) The side length of the base square: ${\displaystyle a_{0}}$
1) The radius of the inscribed circle: ${\displaystyle r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.34\cdot a_{0}\quad }$, see Calculation 1
2) The side length of the inscribed right triangle: ${\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}\quad }$, see Calculation 1

0) Perimeter of base square ${\displaystyle P_{0}=4\cdot a_{0}\quad }$
1) Perimeter of the inscribed circle: ${\displaystyle P_{1}=2\pi \cdot r_{1}={\frac {2\pi {\sqrt {2}}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 2.122\cdot a_{0}\quad }$
2) Perimeter of the inscribed triangle: ${\displaystyle P_{2}=4\cdot a_{2}={\frac {4\cdot 2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 3.387\cdot a_{0}\quad }$

0) Area of the base square ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed circle: ${\displaystyle A_{1}=\pi r_{1}^{2}=\pi \left({\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }$
2) Area of the inscribed triangle: ${\displaystyle A_{2}={\frac {a_{2}^{2}}{2}}={\frac {1}{2}}\left({\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\right)^{2}\cdot a_{0}^{2}\quad \approx 0.358\cdot a_{0}^{2}\quad }$

(0) The area of the right triangle ${\displaystyle \triangle AKH}$ has to be the same as the area of the circle with radius ${\displaystyle r_{1}}$ around center point ${\displaystyle S_{1}\quad \Rightarrow A_{1}=A_{2}}$
(1) ${\displaystyle |AB|=|BC|=|CD|=|AD|=a_{0}\quad }$since ${\displaystyle \square ABCD}$ is a square
(2) ${\displaystyle |AC|={\sqrt {2}}\cdot a_{0}\quad }$since ${\displaystyle {\overline {AC}}}$ is the diagonal of square ${\displaystyle \square ABCD}$
(3) ${\displaystyle |AK|=|AH|=a_{2}\quad }$since ${\displaystyle \triangle AKH}$ is a right triangle
(4) ${\displaystyle |S_{1}F|=|FC|=|CG|=|GS_{1}|=r_{1}\quad }$since ${\displaystyle \square S_{1}FCG}$ is a square, because ${\displaystyle F}$ and ${\displaystyle G}$ are tangent points of the circle around ${\displaystyle S_{1}}$ with the sides of square ${\displaystyle \square ABCD}$

The radius ${\displaystyle r_{1}}$ is calculated:

a) First the relation between ${\displaystyle r_{1}}$ and ${\displaystyle a_{2}}$ is determined from equation (0):
${\displaystyle A_{1}=A_{2}\quad }$ applying equation (0)
${\displaystyle \quad \Leftrightarrow {\frac {a_{2}^{2}}{2}}=\pi r_{1}^{2}\quad }$, definitions of areas of right triangles and circles
${\displaystyle \quad \Leftrightarrow a_{2}^{2}=2\pi r_{1}^{2}\quad }$, multiplying both sides by two
${\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad }$, since negative length cannot be applied

b) Then the relation between ${\displaystyle r_{1}}$ and ${\displaystyle a_{0}}$ is determined from the following identity:
${\displaystyle |AC|={\sqrt {2}}\cdot a_{0}\quad }$ since ${\displaystyle \square ABCD}$ is a square
${\displaystyle \quad \Leftrightarrow |AE|+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, because the three segments are forming the diagonal of the square
${\displaystyle \quad \Leftrightarrow {\frac {\sqrt {2}}{2}}\cdot a_{2}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, since ${\displaystyle \triangle AKH}$ is a right isosceles trinagle
${\displaystyle \quad \Leftrightarrow {\frac {\sqrt {2}}{2}}({\sqrt {2\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, applying part a) of the calculation
${\displaystyle \quad \Leftrightarrow {\frac {{\sqrt {2}}{\sqrt {2}}}{2}}({\sqrt {\pi }}\cdot r_{1})+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+|ES_{1}|+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+|S_{1}C|={\sqrt {2}}\cdot a_{0}\quad }$, because E is a tangent point on the circle and a diagonal side of the triangle ${\displaystyle \triangle AKH}$
${\displaystyle \quad \Leftrightarrow {\sqrt {\pi }}\cdot r_{1}+r_{1}+{\sqrt {2}}\cdot r_{1}={\sqrt {2}}\cdot a_{0}\quad }$, because ${\displaystyle \square S_{1}FCG}$ is a square with side length ${\displaystyle r_{1}}$
${\displaystyle \quad \Leftrightarrow r_{1}\cdot ({\sqrt {\pi }}+1+{\sqrt {2}})={\sqrt {2}}\cdot a_{0}\quad }$, extracting ${\displaystyle r_{1}}$ out of the bracket
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.338\cdot a_{0}}$, rearranging

c) Eventually the relation between ${\displaystyle a_{2}}$ and ${\displaystyle a_{0}}$ is determined by using part a) of the calculation:
${\displaystyle a_{2}={\sqrt {2\pi }}\cdot r_{1}\quad }$, from part a) of the calculation
${\displaystyle \quad \Leftrightarrow a_{2}={\sqrt {2\pi }}\cdot {\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}}$, using the result from part b) of the calculation
${\displaystyle \quad \Leftrightarrow a_{2}={\frac {2{\sqrt {\pi }}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\quad \approx 0.847\cdot a_{0}}$, rearranging

The centroid of the inscribed circle is calculated in orientated expression:
${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}C}}+{\overleftarrow {CS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+{\overrightarrow {S_{0}C}}+{\overleftarrow {CS_{1}}}}$, applying definition of ${\displaystyle S_{0}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+(0-|S_{0}C|i)+{\overleftarrow {CS_{1}}}}$, applying the orientation principle
${\displaystyle \quad \Leftrightarrow S_{1}=(0-|S_{0}C|i)+{\overleftarrow {CS_{1}}}}$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=(0-|S_{0}C|i)+(0+|CS_{1}|i)}$, since ${\displaystyle {\overline {S_{0}C}}}$ and ${\displaystyle {\overline {CS_{1}}}}$ are collinear
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot i)+(0+|CS_{1}|i)\quad }$, since ${\displaystyle {\overline {S_{0}C}}}$ is half the diagonal of square ${\displaystyle \square {ABCD}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot i)+(0+{\sqrt {2}}\cdot r_{1}\cdot i)\quad }$, since ${\displaystyle {\overline {CS_{1}}}}$ is the diagonal of square ${\displaystyle \square {S_{1}FCG}}$ with side length ${\displaystyle r_{1}}$
${\displaystyle \quad \Leftrightarrow S_{1}=0-\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}-{\sqrt {2}}\cdot r_{1}\right)\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\sqrt {2}}\cdot \left({\frac {a_{0}}{2}}-r_{1}\right)\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\sqrt {2}}\cdot \left({\frac {a_{0}}{2}}-{\frac {\sqrt {2}}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot i\quad }$, applying result from Calculation 1
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\sqrt {2}}\cdot \left({\frac {1}{2}}-{\frac {\sqrt {2}}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i\cdot a_{0}\quad }$, extracting ${\displaystyle a_{0}}$
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\sqrt {2}}\cdot \left({\frac {{\sqrt {2\pi }}+{\sqrt {2}}+1}{2\cdot ({\sqrt {2\pi }}+{\sqrt {2}}+1)}}-{\frac {2{\sqrt {2}}}{2\cdot ({\sqrt {2\pi }}+{\sqrt {2}}+1)}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\sqrt {2}}\cdot \left({\frac {{\sqrt {2\pi }}+{\sqrt {2}}+1-2{\sqrt {2}}}{2\cdot ({\sqrt {2\pi }}+{\sqrt {2}}+1)}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0-{\sqrt {2}}\cdot \left({\frac {{\sqrt {2\pi }}-{\sqrt {2}}+1}{2\cdot ({\sqrt {2\pi }}+{\sqrt {2}}+1)}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=0-\left({\frac {{\sqrt {2\pi }}-{\sqrt {2}}+1}{{\sqrt {2}}\cdot ({\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left(0-{\frac {{\sqrt {2\pi }}-{\sqrt {2}}+1}{2{\sqrt {\pi }}+2+{\sqrt {2}}}}\cdot i\right)\cdot a_{0}\quad \approx (0-0.3i)\cdot a_{0}\quad }$, rearranging

The centroid of the inscribed square is calculated in orientated expression:
${\displaystyle S_{2}=S_{0}+{\overrightarrow {S_{0}A}}+{\overrightarrow {AS_{2}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+0i)+{\overrightarrow {S_{0}A}}+{\overrightarrow {AS_{2}}}}$, applying definition of ${\displaystyle S_{0}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+0i)+(0+|S_{0}A|i)+{\overrightarrow {AS_{2}}}}$, applying the orientation
${\displaystyle \quad \Leftrightarrow S_{2}=(0+|S_{0}A|i)+{\overrightarrow {AS_{2}}}}$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+|S_{0}A|i)+(0-|AS_{2}|i)}$, applying the orientation
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot i)+(0-|AS_{2}|i)}$, since ${\displaystyle {\overline {S_{0}A}}}$ is half the diagonal of square ${\displaystyle \square {ABCD}}$ with side length ${\displaystyle a_{0}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot i)+(0-{\frac {{\sqrt {2}}\cdot a_{2}}{2}}\cdot i)}$, since ${\displaystyle {\overline {AS_{2}}}}$ is half the diagonal of square ${\displaystyle \square {AKEH}}$ with side length ${\displaystyle a_{2}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot i-{\frac {{\sqrt {2}}\cdot a_{2}}{2}}\cdot i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\cdot i\cdot a_{0}-{\frac {\sqrt {2}}{2}}\cdot i\cdot a_{2})\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left(a_{0}-a_{2}\right)\cdot i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left(a_{0}-{\frac {\sqrt {2\pi }}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot i)\quad }$, applying result from Calculation 1
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left(1-{\frac {\sqrt {2\pi }}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i\cdot a_{0})\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left(1-{\frac {\sqrt {2\pi }}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left({\frac {{\sqrt {2\pi }}+{\sqrt {2}}+1}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}-{\frac {\sqrt {2\pi }}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i)\cdot a_{0}\quad }$, extending the denominator
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left({\frac {{\sqrt {2\pi }}+{\sqrt {2}}+1-{\sqrt {2\pi }}}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {\sqrt {2}}{2}}\left({\frac {{\sqrt {2}}+1}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i)\cdot a_{0}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {1}{\sqrt {2}}}\left({\frac {{\sqrt {2}}+1}{{\sqrt {2\pi }}+{\sqrt {2}}+1}}\right)\cdot i)\cdot a_{0}\quad }$, since ${\displaystyle {\frac {\sqrt {a}}{a}}={\frac {1}{\sqrt {a}}}}$
${\displaystyle \quad \Leftrightarrow S_{2}=(0+{\frac {{\sqrt {2}}+1}{2{\sqrt {\pi }}+2+{\sqrt {2}}}}\cdot i)\cdot a_{0}\quad \approx (0+0.347i)\cdot a_{0}\quad }$, rearranging

Calculating the centroid of the inscribed circle as displayed:
${\displaystyle S_{1}=S_{0}+{\overrightarrow {S_{0}S_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=(0+0i)+{\overrightarrow {S_{0}C}}+{\overrightarrow {CS_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=\left(|S_{0}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }$, since ${\displaystyle {\overrightarrow {S_{0}C}}}$ is collinear to the diagonal of the square
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {{\sqrt {2}}\cdot a_{0}}{2}}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }$, since ${\displaystyle {\overrightarrow {S_{0}C}}}$ is half the diagonal of the square
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)+{\overrightarrow {CS_{1}}}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(|S_{1}C|\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left({\sqrt {2}}\cdot r_{1}\cdot {\frac {1}{\sqrt {2}}}\cdot (1+i)\right)\quad }$, since ${\displaystyle {\overrightarrow {S_{1}C}}}$ is the diagonal of the square ${\displaystyle \square {S_{1}FCG}}$ with side length ${\displaystyle r_{1}}$
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}\cdot (1+i)\right)-\left(r_{1}\cdot (1+i)\right)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-r_{1}\right)\cdot (1+i)\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {a_{0}}{2}}-{\frac {\sqrt {2}}{{\sqrt {\pi }}+{\sqrt {2}}+1}}\cdot a_{0}\right)\cdot (1+i)\quad }$, applying Calculation 1b