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:  $$x_c= r \cdot (1+\frac {4}{3 \cdot \pi})\cdot \cos {45} = \frac{\sqrt 2 }{ (1+\sqrt 2)} \cdot (1+\frac {4}{3 \cdot \pi})\cdot \cos {45} = 0,59...$$ $$y_c=\frac{\sqrt 2 }{ (1+\sqrt 2)} \cdot (1+\frac {4}{3 \cdot \pi})\cdot \cos {45}= 0.59..$$