Quadratrix property

[Function]

Hello everyone

 

In a dissertation for maths, I've mentioned once "quadratrix"... Can't really explain what it is, but I think I know what it is, and I've discovered what seems to be a pretty beautiful property of the quadratrix of a circle (commonly known as the quadratrix of Dinostratus)

 

So I thought: what if I 'completed' the quadratrix? (i.e. completing the circle and the quadratrix, resulting in a horizontal flip of the quadratrix over x=0, a vertical one over y=0, and a horizontal one of the vertically flipped one over x=0)

The result: an eye-shaped geometrical figure.

 

If you're eager to see this 'completed' quadratrix, you can ask me and I will try to upload it somewhere (.ggb, to be opened with GeoGebra).

In this file, I've also drawed what I like to call 'sub-quadratrices' (the blue lines).

My conclusion of doing this: if the number of equal parts (now there are 10, for more information, seek information on the quadratrix of Dinostratus) are going to infinity, you become this eye-shaped figure; the less equal parts you have, the more you will get like 2 strange flower petals above and under y=0, with the eye-shaped 'fulfilled' quadratrix around it (touching it at the extreme values ( (0,r) and (0,-r) with r the radius of the circle).

 

And then I thought: this figure I 'made' is so special, so unique per radius of the circle, that it must have some special properties. First thing I thought of, was the area.

So then I started playing with the numbers I had:

 

Let A be the area of this fulfilled quadratrix. In the file (i.e. for r = 10), A = 175.8804176

Note that this won't be the exact area of the fulfilled quadratrix, for it's only measured with 10 equal parts the quadratrix is based on...

 

The area of the circle (i.e. for r = 10) = 314.1592654

Dividing the latter area by the former results in 1.786209458 = B

Dividing [math]\pi[/math] by this number results in 1.758804176 = C

The square root of [math]\pi[/math] is 1.772453851

 

Then I thought that this square root is about the mean of B and C:

(B+C)/2 = 1.772506817

 

Not such a big difference, right?

 

And then I thought: if this is right, I should be able to find a nice formula to express the area of the fulfilled quadratrix in function of r:

 

[math]\frac{\left(\frac{r^2\pi}{A}+\frac{A\pi}{r^2\pi}\right)}{2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow \frac{r^4\pi+A^2}{2Ar^2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow 2Ar^2\sqrt{\pi}=r^4\pi+A^2[/math]

[math]\Leftrightarrow A^2-2Ar^2\sqrt{\pi}+r^4\pi=0[/math]

 

Giving a discriminant of 0 and a final result of [math]A=\sqrt{\pi}r^2[/math].

 

Nice enough, in my opinion. It's always possible that I'm not right, but, in my opinion, it's very beautiful that this discriminant is 0, so that must mean something!

 

Can someone tell me whether I'm right or not? Can't find any appropriate information on the internet..

Are there other special properties the (fulfilled) quadratrix has?

 

Thanks!

 

Function

Edited April 30, 2014 by Function

[Function]

Hello everyone

 

In a dissertation for maths, I've mentioned once "quadratrix"... Can't really explain what it is, but I think I know what it is, and I've discovered what seems to be a pretty beautiful property of the quadratrix of a circle (commonly known as the quadratrix of Dinostratus)

 

So I thought: what if I 'completed' the quadratrix? (i.e. completing the circle and the quadratrix, resulting in a horizontal flip of the quadratrix over x=0, a vertical one over y=0, and a horizontal one of the vertically flipped one over x=0)

The result: an eye-shaped geometrical figure.

 

If you're eager to see this 'completed' quadratrix, you can ask me and I will try to upload it somewhere (.ggb, to be opened with GeoGebra).

In this file, I've also drawed what I like to call 'sub-quadratrices' (the blue lines). (There are also lines which I'd like to call 'super-quadratrices', those are the lines outside of the quadratrix)

My conclusion of doing this: if the number of equal parts (now there are 10, for more information, seek information on the quadratrix of Dinostratus) are going to infinity, you become this eye-shaped figure; the less equal parts you have, the more you will get like 2 strange flower petals above and under y=0, with the eye-shaped 'fulfilled' quadratrix around it (touching it at the extreme values ( (0,r) and (0,-r) with r the radius of the circle).

 

And then I thought: this figure I 'made' is so special, so unique per radius of the circle, that it must have some special properties. First thing I thought of, was the area.

So then I started playing with the numbers I had:

 

Let A be the area of this fulfilled quadratrix. In the file (i.e. for r = 10), A = 175.8804176

Note that this won't be the exact area of the fulfilled quadratrix, for it's only measured with 10 equal parts the quadratrix is based on...

 

The area of the circle (i.e. for r = 10) = 314.1592654

Dividing the latter area by the former results in 1.786209458 = B

Dividing [math]\pi[/math] by this number results in 1.758804176 = C

The square root of [math]\pi[/math] is 1.772453851

 

Then I thought that this square root is about the mean of B and C:

(B+C)/2 = 1.772506817

 

Not such a big difference, right?

 

And then I thought: if this is right, I should be able to find a nice formula to express the area of the fulfilled quadratrix in function of r:

 

[math]\frac{\left(\frac{r^2\pi}{A}+\frac{A\pi}{r^2\pi}\right)}{2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow \frac{r^4\pi+A^2}{2Ar^2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow 2Ar^2\sqrt{\pi}=r^4\pi+A^2[/math]

[math]\Leftrightarrow A^2-2Ar^2\sqrt{\pi}+r^4\pi=0[/math]

 

Giving a discriminant of 0 and a final result of [math]A=\sqrt{\pi}r^2[/math].

 

Nice enough, in my opinion. It's always possible that I'm not right, but, in my opinion, it's very beautiful that this discriminant is 0, so that must mean something!

 

Can someone tell me whether I'm right or not? Can't find any appropriate information on the internet..

Are there other special properties the (fulfilled) quadratrix has?

 

Thanks!

 

Function

 

200th post

 

Dang.. Press edit and a new post pops up... 200th post gone to waste..

 

So the A(quadratrix)/A(circle) = 1/sqrt(pi). Conclusion: the area of a circle is sqrt(pi) times as large as that of its quadratrix.

Edited May 3, 2014 by Function