Summary edit

Shows the largest quarter circle within a right isosceles triangle.

Elements edit

Base is the right isosceles triangle of side length ${\displaystyle a_{0}}$ and centroid ${\displaystyle S_{0}}$

Inscribed is the largest possible quarter circle with radius ${\displaystyle r_{1}}$ and centroid ${\displaystyle S_{1}}$

General case edit

Segments in the general case edit

0) The side length of the base right triangle ${\displaystyle |AB|=|AC|=a_{0}}$
1) Radius of the quarter circle ${\displaystyle r_{1}={\frac {a_{0}}{\sqrt {2}}}\quad }$ (See calculation 1).

Perimeters in the general case edit

0) Perimeter of base triangle ${\displaystyle P_{0}=|AB|+|BC|+|AC|=a_{0}+{\sqrt {2}}\cdot a_{0}+a_{0}=(2+{\sqrt {2}})\cdot a_{0}}$
1) Perimeter of the quarter circle ${\displaystyle P_{1}=\left({\frac {4+\pi }{2{\sqrt {2}}}}\right)\cdot a_{0}\quad }$ (See calculation 2 )

Areas in the general case edit

0) Area of the base triangle ${\displaystyle A_{0}={\frac {1}{2}}\cdot a_{0}^{2}}$
1) Area of the inscribed circle ${\displaystyle A_{1}={\frac {\pi }{8}}\cdot a_{0}^{2}\quad }$ (See calculation 3)

Centroids in the general case edit

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 quarter circle: ${\displaystyle x_{1}=y_{1}=\left({\frac {{\sqrt {8}}-\pi }{3\pi }}\right)\cdot a_{0}}$, (see calculation 4)

Normalised case edit

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

Segments in the normalised case edit

0) Side length of the base triangle ${\displaystyle a_{0}={\sqrt {2}}\quad \approx 1.414}$
1) Radius of the inscribed quarter circle ${\displaystyle r_{1}={\frac {\sqrt {2}}{\sqrt {2}}}=1\quad }$

Perimeters in the normalised case edit

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 quarter circle ${\displaystyle P_{1}=\left({\frac {4+\pi }{2{\sqrt {2}}}}\right)\cdot {\sqrt {2}}=\left({\frac {4+\pi }{2}}\right)\quad \approx 3.5707963}$
S) Sum of perimeters ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 8.3992235}$

Areas in the normalised case edit

0) Area of the base triangle ${\displaystyle A_{0}=1}$
1) Area of the inscribed quarter circle ${\displaystyle A_{1}={\frac {\pi }{8}}\cdot ({\sqrt {2}})^{2}={\frac {\pi }{8}}\cdot 2={\frac {\pi }{4}}\quad \approx 0.785}$

Centroids in the normalised case edit

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: ${\displaystyle S_{1}:x_{1}=y_{1}=\left({\frac {{\sqrt {8}}-\pi }{3\pi }}\right)\cdot {\sqrt {2}}\quad \approx -0.047\quad }$

Distances of centroids edit

The distance between the centroid of the base element and the centroid of the circle is:
${\displaystyle d={\sqrt {(x_{0}-x_{1})^{2}+(y_{0}-y_{1})^{2}}}\approx 0.0664558}$

Identifying number edit

Apart of the base element there is only one shape allocated. Therefore the integer part of the identifying number is 1.
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.

${\displaystyle decimalpart(8.3992235+0.0664558)=decimalpart(8.4656792)=.4656792}$

So the identifying number is: ${\displaystyle 1.4656792}$

Calculations edit

Calculation 1 edit

Because FS_R is a right isosceles triangle the following equations hold:
(1) ${\displaystyle \quad |AB|=|AC|=a_{0}}$
(2) ${\displaystyle \quad |BC|={\sqrt {2}}\cdot a_{0}}$
(3) ${\displaystyle \quad |BD|=|CD|={\frac {{\sqrt {2}}\cdot a_{0}}{2}}={\frac {a_{0}}{\sqrt {2}}}}$
(4) ${\displaystyle \quad r_{1}=|AD|}$

Since ${\displaystyle {\overline {AB}}}$ is tangent to the quarter circle in point ${\displaystyle D}$ the triangle ${\displaystyle \triangle ABD}$ has an right angle in ${\displaystyle D}$. This means:
(5) ${\displaystyle \quad |AD|^{2}+|BD|^{2}=|AB|^{2}}$
${\displaystyle \quad \Leftrightarrow r_{1}^{2}+\left({\frac {a_{0}}{\sqrt {2}}}\right)^{2}=a_{0}^{2}\quad }$, applying equations (1),(3) and (4)
${\displaystyle \quad \Leftrightarrow r_{1}^{2}=a_{0}^{2}-{\frac {a_{0}^{2}}{2}}={\frac {a_{0}^{2}}{2}}\quad }$
${\displaystyle \quad \Leftrightarrow r_{1}={\frac {a_{0}}{\sqrt {2}}}\quad }$

Calculation 2 edit

Perimeter of the quarter circle:
${\displaystyle \quad P_{1}=2\cdot r_{1}+{\frac {2\pi r_{1}}{4}}\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=2\cdot r_{1}\cdot \left(1+{\frac {\pi }{4}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=2\cdot r_{1}\cdot \left({\frac {4+\pi }{4}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=r_{1}\cdot \left({\frac {4+\pi }{2}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}={\frac {a_{0}}{\sqrt {2}}}\cdot \left({\frac {4+\pi }{2}}\right)\quad }$
${\displaystyle \quad \Leftrightarrow P_{1}=\left({\frac {4+\pi }{2{\sqrt {2}}}}\right)\cdot a_{0}\quad }$

Calculation 3 edit

Area of the inscribed quarter circle:
${\displaystyle \quad A_{1}={\frac {\pi \cdot r_{1}^{2}}{4}}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{4}}\cdot r_{1}^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{4}}\cdot \left({\frac {a_{0}}{\sqrt {2}}}\right)^{2}}$
${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{4}}\cdot {\frac {a_{0}^{2}}{2}}}$

${\displaystyle \quad \Leftrightarrow A_{1}={\frac {\pi }{8}}\cdot a_{0}^{2}}$

Calculation 4 edit

Centroid of the inscribed circle measured from the centroid of the base triangle:
${\displaystyle \quad x_{1}=|Ax_{1}|-|Ax_{0}|}$
${\displaystyle \quad \Leftrightarrow x_{1}=|Ax_{1}|-{\frac {1}{3}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {4}{3\pi }}\cdot r_{1}-{\frac {1}{3}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}={\frac {4}{3\pi }}\cdot {\frac {a_{0}}{\sqrt {2}}}-{\frac {1}{3}}\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {4}{3\pi }}\cdot {\frac {1}{\sqrt {2}}}-{\frac {1}{3}}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {1}{3\pi }}\cdot {\frac {\sqrt {16}}{\sqrt {2}}}-{\frac {1}{3}}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {\sqrt {8}}{3\pi }}-{\frac {\pi }{3\pi }}\right)\cdot a_{0}}$
${\displaystyle \quad \Leftrightarrow x_{1}=\left({\frac {{\sqrt {8}}-\pi }{3\pi }}\right)\cdot a_{0}}$
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