File:FS QJ dia.png

The largest non-right isosceles triangle inscribed in a square of side length ${\displaystyle a_{0}}$

0) The side length of the base square: ${\displaystyle a_{0}}$
1) Side length of the inscribed isosceles triangle: ${\displaystyle a_{1}={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad \approx 1.118\cdot a_{0}\quad }$, see Calculation 1

0) Perimeter of base square: ${\displaystyle P_{0}=4\cdot a_{0}}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}=2\cdot a_{1}+a_{0}=2\cdot {\frac {\sqrt {5}}{2}}\cdot a_{0}+a_{0}=({\sqrt {5}}+1)\cdot a_{0}\quad \approx 3.236\cdot a_{0}}$
S) Sum of perimeters: ${\displaystyle P_{s}=4\cdot a_{0}+({\sqrt {5}}+1)\cdot a_{0}=({\sqrt {5}}+5)\cdot a_{0}\quad \approx 7.236\cdot a_{0}}$

0) Area of the base square: ${\displaystyle A_{0}=a_{0}^{2}}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {a_{0}}{2}}\cdot a_{0}={\frac {1}{2}}\cdot a_{0}^{2}={\frac {1}{2}}\cdot A_{0}}$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed \triangle measured from the centroid of the base shape: ${\displaystyle S_{1}=(0-{\frac {1}{6}}\cdot i)\cdot a_{0}\quad \approx (0-0.167i)\cdot a_{0}}$, see calculation 2

W) Weighted centroid: ${\displaystyle S_{w}=\left(0-{\frac {1}{18}}\cdot i\right)\cdot a_{0}\quad \approx (0-0.056i)\cdot a_{0}}$, see calculation 3

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

0) Side lenght of the base square ${\displaystyle a_{0}=1}$
1) Side length of the inscribed triangle: ${\displaystyle a_{1}={\frac {\sqrt {5}}{2}}\cdot 1\quad \approx 1.118\quad }$

0) Perimeter of base square${\displaystyle P_{0}=4\cdot a_{0}=4}$
1) Perimeter of the inscribed triangle: ${\displaystyle P_{1}={\sqrt {5}}+1\quad \approx 3.2360680}$
S) Sum of perimeters: ${\displaystyle P_{s}=P_{0}+P_{1}\quad \approx 7.2360680}$

0) Area of the base square is by definition ${\displaystyle A_{0}=1}$
1) Area of the inscribed triangle: ${\displaystyle A_{1}={\frac {1}{2}}\cdot A_{0}=0.500\quad }$

0) Centroid position of the base square: ${\displaystyle S_{0}=0+0i}$
1) Centroid position of the inscribed triangle: ${\displaystyle S_{1}=(0-{\frac {1}{6}}\cdot i)\quad \approx (0-0.167i)}$
W) Weighted centroid: ${\displaystyle S_{w}=0-{\frac {1}{18}}\cdot i\quad \approx (0-0.056i)}$

(1) ${\displaystyle |AB|=|BC|=|CD|=|AD|=|EF|=a_{0}}$, since ${\displaystyle AEBCFD}$ is a square
(2) ${\displaystyle |AE|=|BE|=|CF|=|DF|=|ES_{0}|=|FS_{0}|={\frac {1}{2}}\cdot a_{0}}$
(3) ${\displaystyle |AF|=|BF|=a_{1}\quad }$

${\displaystyle \quad |AE|^{2}+|EF|^{2}=|AF|^{2}\quad }$, applying the Pythagorean theorem on triangle ${\displaystyle \triangle {AEF}}$
${\displaystyle \quad \Leftrightarrow ({\frac {1}{2}}\cdot a_{0})^{2}+|EF|^{2}=|AF|^{2}\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow ({\frac {1}{2}}\cdot a_{0})^{2}+a_{0}^{2}=|AF|^{2}\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow ({\frac {1}{2}}\cdot a_{0})^{2}+a_{0}^{2}=a_{1}^{2}\quad }$, applying equation (3)
${\displaystyle \quad \Leftrightarrow {\frac {1}{4}}\cdot a_{0}^{2}+a_{0}^{2}=a_{1}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow {\frac {5}{4}}\cdot a_{0}^{2}=a_{1}^{2}\quad }$, multiplying
${\displaystyle \quad \Leftrightarrow a_{1}={\frac {\sqrt {5}}{2}}\cdot a_{0}\quad \approx 1.118\cdot a_{0}}$,

${\displaystyle \quad S_{1}={\overrightarrow {S_{0}E}}+{\overrightarrow {ES_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{1}=-|ES_{0}|\cdot i+|ES_{1}|\cdot i}$
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{2}}\cdot a_{0}\cdot i+|ES_{1}|\cdot i\quad }$, applying equation (2)
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{2}}\cdot a_{0}\cdot i+{\frac {1}{3}}\cdot |EF|\cdot i\quad }$, applying the formula for centroids of triangles
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{2}}\cdot a_{0}\cdot i+{\frac {1}{3}}\cdot a_{0}\cdot i\quad }$, applying equation (1)
${\displaystyle \quad \Leftrightarrow S_{1}=\left({\frac {1}{3}}-{\frac {1}{2}}\right)\cdot a_{0}\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=-{\frac {1}{6}}\cdot a_{0}\cdot i\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{1}=(0-{\frac {1}{6}}\cdot i)\cdot a_{0}\quad \approx (0-0.167i)\cdot a_{0}}$, rearranging

${\displaystyle S_{w}={\frac {S_{0}\cdot A_{0}+S_{1}\cdot A_{1}}{A_{0}+A_{1}}}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {S_{1}\cdot A_{1}}{A_{0}+A_{1}}}\quad }$,since ${\displaystyle S_{0}=0+0i=0}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {A_{1}}{A_{0}+A_{1}}}\cdot S_{1}\quad }$,rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {{\frac {1}{2}}\cdot A_{0}}{A_{0}+{\frac {1}{2}}\cdot A_{0}}}\cdot S_{1}\quad }$,since ${\displaystyle A_{1}={\frac {1}{2}}\cdot A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {\frac {1}{2}}{1+{\frac {1}{2}}}}\cdot S_{1}\quad }$, eliminating ${\displaystyle A_{0}}$
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {\frac {1}{2}}{\frac {3}{2}}}\cdot S_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {1}{3}}\cdot S_{1}\quad }$, rearranging
${\displaystyle \quad \Leftrightarrow S_{w}={\frac {1}{3}}\cdot (0-{\frac {1}{6}}\cdot i)\cdot a_{0}\quad }$, applying the result of Calculation (2)
${\displaystyle \quad \Leftrightarrow S_{w}=\left(0-{\frac {1}{18}}\cdot i\right)\cdot a_{0}\quad \approx (0-0.056i)\cdot a_{0}}$, rearranging

I, the copyright holder of this work, hereby publish it under the following license:

This file is licensed under the Creative Commons Attribution-Share Alike 4.0 International license.

You are free:

• to share – to copy, distribute and transmit the work
• to remix – to adapt the work

Under the following conditions:

• attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
• share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original.

https://creativecommons.org/licenses/by-sa/4.0CC BY-SA 4.0 Creative Commons Attribution-Share Alike 4.0 truetrue