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Summary

DescriptionFS CQ dia.png	English: Largest square in a circle Deutsch: Größtes Quadrat in einem Kreis
Date	7 January 2022
Source	Own work
Author	Hans G. Oberlack

Shows the largest square that can be inscribed into a circle.

General case

Base is the circle around point $S$ of radius $r_{0}$ .

In order to find the side length $r_{1}$ of the square $[ABCD]$ the following calculations have to be done:

The line line segments $[AC],[BD]$ are the diameter of the circle. This leads to
(1) $\quad |AC|=|BD|=2\cdot r_{0}$

The line segments $[AD],[AB],[BC],[CB]$ have the side length $r_{1}$ of the square. Thus we have
(2) $\quad |AD|=|AB|=r_{1}$

Applying the Pythagorean theorem we get:
(3) $\quad |AD|^{2}+|AB|^{2}=|BD|^{2}$

Using the results of (1) and (2) gives:

$\quad |AD|^{2}+|AB|^{2}=|BD|^{2}$

$\quad \Leftrightarrow r_{1}^{2}+r_{1}^{2}=(2\cdot r_{0})^{2}$

$\quad \Leftrightarrow 2\cdot r_{1}^{2}=4\cdot r_{0}^{2}$

$\quad \Leftrightarrow r_{1}^{2}=2\cdot r_{0}^{2}$

$\quad \Leftrightarrow r_{1}={\sqrt {2}}\cdot r_{0}$

Segments in the general case

0) The radius of the circle $=r_{0}$
1) The side length of the inscribed square $r_{1}={\sqrt {2}}\cdot r_{0}$

Perimeters in the general case

0) Perimeter of base circle $P_{0}=2\pi \cdot r_{0}$
1) Perimeter of the square $P_{1}=4\cdot r_{1}=4\cdot {\sqrt {2}}\cdot r_{0}$

Areas in the general case

0) Area of the base circle $A_{0}=\pi r_{0}^{2}$
1) Area of the square $A_{1}=r_{1}^{2}=({\sqrt {2}}\cdot r_{0})^{2}=2\cdot r_{0}^{2}$

Covered surface of the base shape: $C_{b}={\frac {A_{1}}{A_{0}}}={\frac {2\cdot r_{0}^{2}}{\pi \cdot r_{0}^{2}}}={\frac {2}{\pi }}\quad \approx 63.7\%$

Centroids in the general case

Centroid positions are measured from centroid point of the base shape.
0) Centroid positions of the base circle: $S_{0}=S=0+0i$
1) Centroid positions of the square: $S_{1}=S=0+0i$

Normalised case

In the normalised case the area of the base is set to 1.
$||\circ ABCD||=1\Rightarrow \pi r_{0}^{2}=1\Rightarrow r_{0}={\sqrt {\frac {1}{\pi }}}={\frac {1}{\sqrt {\pi }}}$

Segments in the normalised case

0) Radius of the base circle $r_{0}={\frac {1}{\sqrt {\pi }}}=0.5641896...$
1) Side length of the square $r_{1}={\sqrt {2}}\cdot {\frac {1}{\sqrt {\pi }}}={\sqrt {\frac {2}{\pi }}}=0.7978846...$

Perimeters in the normalised case

0) Perimeter of base circle $P_{0}=2\pi \cdot {\frac {1}{\sqrt {\pi }}}=2{\sqrt {\pi }}=3.5449077...$
1) Perimeter of the square $P_{1}=4\cdot {\sqrt {\frac {2}{\pi }}}=3.1915382...$
S) Sum of perimeters $P_{s}=P_{0}+P_{1}=6.7364459...$

Areas in the normalised case

0) Area of the base circle $A_{0}=1$
1) Area of the square $A_{1}=\left({\sqrt {\frac {2}{\pi }}}\right)^{2}={\frac {2}{\pi }}=0.6366198...$

Centroids in the normalised case

Centroid positions are measured from the centroids of the base shape
0) Centroid positions of the base circle: $S_{0}=0+0i$
1) Centroid positions of the inscribed square: $S_{1}=S_{0}=0+0i$

Licensing

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Category:CC-BY-SA-4.0#FS%20CQ%20dia.png Category:Self-published work

Category:Construction diagrams of basic sangaku shapes Category:Circle geometry Category:Squares (shape) Category:Files by User Hans G. Oberlack/Basic sangaku shapes#CQ%20dia