User:Gaal Sandor/sandbox/geometry

Area and circumference of a circle
The idea of the mathematical constant $$\pi$$ is based on the assumption, that the circumference of a circle can be calculated from the difference of the perimeters of an inscribed and a circumscribed polygon. [] To test this theory, I start with squares.



r1=radius of the small circle; r3=radius of the larger circle; a=side of the small polygon; A1/P1=area/perimeter of the small circle; A2/P2=area/perimeter of the small polygon; A3/P3=area/perimeter of the larger circle; A4/P4=area/perimeter of the larger polygon; c=coefficient of the area/perimeter of the circle $$ r1=\tfrac{a}2;  r3=\tfrac{\sqrt{2}}2*a;   P1=2c*\tfrac{a}2=c*a;   P2=4*a;   P3=2c*\tfrac{\sqrt{2}}2*a=c*\sqrt{2}*a;   P4=4*\sqrt{2}*a $$

I continue with hexagons.



$$ r1=\tfrac{\sqrt{3}}2*a;  r3=a;   P1=2c*\tfrac{\sqrt{3}}2*a=c*\sqrt{3}*a;   P2=6*a;   P3=2c*a; P4=6*\tfrac{2}\sqrt{3}*a=\tfrac{12}\sqrt{3}*a $$

The number of the polygons' sides can be increased to infinite.

Problem #1: The mathematical constant $$\pi$$ is based on calculating with polygons with a certain number of sides. Such determination is a rough guess between 4 and infinite.

Problem #2: Despite of the difference decreases between the polygons' perimeter, as the number of their sides increases, the actual value of their perimeter can only be calculated with endless fractions. (See $$\sqrt{3}$$ above for instance.) That means decrease of accuracy.

Take the areas of the squares and the inscribed circles instead:

$$ A1=c*\tfrac{a^2}4; A2=a^2; A3=c*\tfrac{a^2}2; A4=2*a^2; $$

This proportion enables to exactly determine the area of the circle between the squares and vice versa: the square between the inscribed and the circumscribed circles. Defining the area of a circle means comparing the area of the circle to the area of a square. A possible way to compare the area of a circle to the area of a square, is to draw the square first and place four equal size quarter circles on it, with their origo on the corners of the square.

In this layout, if the radius of the quarter circles equals $$a/2$$, their arcs don't cross each other, but touch at the midpoint of the sides. Rearranging them into one, gives a circle inscribed in the square.

If their radius equals $$\sqrt{2}/2*a$$, their arcs cross at the center of the square and rearranging them into one, gives a circumscribed circle.

The area of the square equals the combined area of the four quarter circles, if their arcs cross at half way between the center of the square, and the midpoint of its side, because the square is in between the inscribed and the circumscribed circles.

The distance between the midpoint and the closest corner of the square equals the radius of the circle, and its ratio with the side of the square can be calculated using Pythagoras' theorem. A=area of the square / circle

a=side of the square

r=radius of the circle

$$ r^2=(a/4)^2+(a/2)^2 $$ $$ r=\sqrt{(a/4)^2+(a/2)^2}=\sqrt{(a/4)^2+(2a/4)^2}=\sqrt{1+4}*(a/4) $$ $$ a/4=\tfrac{r}\sqrt{5} $$ $$ a=1.789r $$ $$ A=8*\tfrac{r}\sqrt{5}*2*\tfrac{r} \sqrt{5}=16/5r^2=3.2r^2 $$

In contrast to Archimedes' method, when the circle is compared to other polygons, this approach is more direct and precise. The result of the equation is approximately 2% higher, than with the commonnly used $$\pi$$ constant.

The circumference (perimeter) of a circle can be derived from the area of the circle.

$$ C=\tfrac{3.2r^2-3.2(r-x)^2}x= \tfrac{3.2r^2-3.2(r^2-2rx+x^2)}x=\tfrac{6.4rx-3.2x^2}x=6.4r-3.2x $$ As x is just theoretical, with a value of almost 0, $$ C=6.4r $$

proof
$$ \tfrac{a}2<a*\tfrac{\sqrt{5}}4<\tfrac{a}2+\tfrac{{a*\tfrac{\sqrt{2}}2}-\tfrac{a}2}2=a*(\tfrac{1+\sqrt{2}}4) $$