User:Hans G. Oberlack/QH 1.7189820

Shows the largest semicircle within a square.

General case

Base is the square $[ACEJ]$ of side length s.

The line segment $[AC]$ has the side length $s$ of the square. So has the line segment $[CE]$. So the line segment $[AE]$ has the length ${\sqrt {2}}\cdot s$. So we get the equation:
(1) $|AE|=|AK|+|KE|={\sqrt {2}}\cdot s$

The line segments $[KG]$, $[KF]$, $[KH]$, $[KD]$ and $[KB]$ have the length of the radius $r$ of the semicircle $[KBDHFG]$.

Since $|KF|=|KD|=r$ the rectangle $[KDEF]$ is a square with side length $r_{1}$. This leads to the equation:
(2) $|KE|={\sqrt {2}}\cdot r_{1}$

The line segment $[BG]$ is the diameter of the semicircle and has the length: $|BG|=2\cdot r_{1}$. The line segment $[AB]$ has length $a$. For symmetry reasons the line segment $[AG]$ has the same length, so $|AB|=|AG|=a$. Using the Pythagorean theorem we get equation:
(3) $a^{2}+a^{2}=(2\cdot r_{1})^{2}$
$\Rightarrow 2\cdot a^{2}=4\cdot r_{1}^{2}$
$\Rightarrow a^{2}=2\cdot r_{1}^{2}$
$\Rightarrow a={\sqrt {2}}\cdot r_{1}$

Applying the Pythagorean theorem to the triangle $\triangle ABK$ we get the equation
(4) $|AK|^{2}+r_{1}^{2}=a^{2}$

Using equations (3) and (4) we arrive at:
$|AK|^{2}+r_{1}^{2}=2\cdot r_{1}^{2}$
$\Rightarrow |AK|^{2}=2\cdot r_{1}^{2}-r_{1}^{2}$
$\Rightarrow |AK|^{2}=r_{1}^{2}$
$\Rightarrow |AK|=r_{1}$

Now we use this result together with equations (1) and (2).
$|AK|+|KE|={\sqrt {2}}\cdot s$
$\Rightarrow r_{1}+|KE|={\sqrt {2}}\cdot s$
$\Rightarrow r_{1}+{\sqrt {2}}\cdot r_{1}={\sqrt {2}}\cdot s$
$\Rightarrow r_{1}\cdot (1+{\sqrt {2}})={\sqrt {2}}\cdot s$
$\Rightarrow r_{1}={\frac {\sqrt {2}}{\cdot (1+{\sqrt {2}})}}\cdot s$

Segments in the general case

0) The side length of the square $=s$
1) Radius of the semicircle $r_{1}={\frac {\sqrt {2}}{(1+{\sqrt {2}})}}\cdot s$

Perimeters in the general case

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

Areas in the general case

0) Area of the base square $A_{0}=s^{2}$
1) Area of the semicircle $A_{1}={\frac {\pi \cdot r_{1}^{2}}{2}}={\frac {\pi }{2}}\cdot ({\frac {\sqrt {2}}{(1+{\sqrt {2}})}}\cdot s)^{2}={\frac {\pi }{2}}\cdot {\frac {2}{(1+{\sqrt {2}})^{2}}}\cdot s^{2}={\frac {\pi \cdot s^{2}}{(1+{\sqrt {2}})^{2}}}={\frac {\pi }{1+2{\sqrt {2}}+2}}\cdot s^{2}={\frac {\pi }{3+2{\sqrt {2}}}}\cdot A_{0}$

Centroids in the general case

Centroid positions are measured from the lower left point of the square.
0) Centroid positions of the base square: $x_{0}=0\quad y_{0}=0$
1) Centroid positions of the semicircle: If the center of the radius of the semicircle were positioned on $x_{0}=0\land y_{0}=0$ and the diameter were parallel to the y-axis then the centroid position would be $x_{c}={\frac {4\cdot r}{3\cdot \pi }}\quad y_{c}=0$. Since the center point is shift by distance $|AK|=r$ and rotated by 45 degrees the centroids are $x_{1}=(r_{1}+{\frac {4\cdot r_{1}}{3\cdot \pi }})\cdot \cos {45}-{\frac {s}{2}}$
$\Leftrightarrow x_{1}=r_{1}\cdot (1+{\frac {4}{3\cdot \pi }})\cdot \cos {45}-{\frac {s}{2}}$
$\Leftrightarrow x_{1}=r_{1}\cdot (1+{\frac {4}{3\cdot \pi }})\cdot {\frac {1}{\sqrt {2}}}-{\frac {s}{2}}$, since $\cos {45}={\frac {1}{\sqrt {2}}}$
$y_{c}=r\cdot (1+{\frac {4}{3\cdot \pi }})\cdot \cos {45}$

Normalised case

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

Segments in the normalised case

0) Segment of the base square $s=1$
1) Segment of the semicircle $r={\frac {{\sqrt {2}}\cdot 1}{(1+{\sqrt {2}})}}=0.5857864..$

Perimeters in the normalised case

0) Perimeter of base square$=4$
1) Perimeter of the semicircle $=(2+\pi ){\frac {\sqrt {2}}{(1+{\sqrt {2}})}}=3.0118752..$

Areas in the normalised case

0) Area of the base square $=1$
1) Area of the semicircle $={\frac {\pi }{(1+{\sqrt {2}})^{2}}}=0.539012...$

Centroids in the normalised case

Centroid positions are measured from the lower left point of the square.
0) Centroid positions of the base square: $x_{c}={\frac {1}{2}}\quad y_{c}={\frac {1}{2}}$
1) Centroid positions of the semicircle:
$x_{c}=r\cdot (1+{\frac {4}{3\cdot \pi }})\cdot \cos {45}={\frac {\sqrt {2}}{(1+{\sqrt {2}})}}\cdot (1+{\frac {4}{3\cdot \pi }})\cdot \cos {45}=0,59...$
$y_{c}={\frac {\sqrt {2}}{(1+{\sqrt {2}})}}\cdot (1+{\frac {4}{3\cdot \pi }})\cdot \cos {45}=0.59..$