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I don't know whether I should ask this question or not, but I am asking this:

(Before anything, this is an analysis of closed 2D figures.)

I was analyzing $\pi$ . It seems to be a constant, But being a constant is my curiosity. Can we say that $\pi$ is a property of circle which makes circle a circle. Same as there must be constant for each and every figure possible. So I suggested a new theory of $\pi$ or a new definition of $\pi$, which says -

$Figure_\pi = {L \over D}$

where, L = Length of the curve,

D = Maximum distance possible between any two points

So, $Circle_\pi = {Circumference \over Diameter} \implies Circle_\pi = \pi$

Also, $Square_\pi = {Perimeter \over Diagonal } \implies Square_\pi = {4a \over a\sqrt2} = {2\sqrt2}$

And, $Rectangle_\pi = {2(a+b) \over \sqrt{a^2+b^2}}$

Now I got stuck, as for rectangle I am not getting a constant, but I realized that this is obvious. Two squares or circles are always similar but two rectangles are not necessarily similar, because for two rectangles to be similar, their ratios of lengths and breadths must be constant. So I suggested types of figure on the basis of their variables.

As a circle or a square can be defined by a single variable, I called them single variable figures. For a rectangle, a rhombus, or an ellipse, we must have two variables, so I called them 2 variable figures. For random figure to be similar we must know by some means the length of the curve and maximum distance between any two points, and we will get a constant calling it $Figure_\pi$

My question is this: Am I thinking in the right direction or not, because I am finding different constants for different 2D figures as different variable figures?

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  • $\begingroup$ On a tangential note, you might be interested in looking up the "shape factor" of various shapes. $\endgroup$ – abnry Aug 25 '12 at 15:18
  • $\begingroup$ Yes, exactly, and $\pi$ must be one of them. $\endgroup$ – Rahul Aug 25 '12 at 15:20
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    $\begingroup$ It might interest you to know that all parabolas are similar. $\endgroup$ – J. M. is a poor mathematician Aug 25 '12 at 15:31
  • $\begingroup$ Thnks J.M., I didn't know that $\endgroup$ – Rahul Aug 25 '12 at 15:35
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    $\begingroup$ What would be a more interesting way (for me, at least) is to define it as the maximum of $\frac{4*Area}{Perimeter^2}$ over every closed figure where this quantity makes sense. $\endgroup$ – ronno Aug 25 '12 at 18:09
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For any given shape you can define the ratio between the circumference and the longest chord, which will be a constant regardless of scale. This is because when you double the size of the shape you double the circumference and double the chord, so the ratio stays the same. You are correct that there is not a single ratio for all rectangles, but there is one for rectangles of a given proportion. Your $Rectangle_\pi$ can be expressed as $$\frac {2+2\frac ba}{\sqrt{1+(\frac ba)^2}}$$ where the dependence on the proportions of the rectangle is explicit. For a regular hexagon, the constant is 3. It will approach $\pi$ as the number of sides increases. You might see Wikipedia under Polygon approximation era for the story of Archimedes calculating $\pi$ in exactly this way.

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  • $\begingroup$ Thanks, I already figured it out, even for rhombus, parallelogram etc. Means I am right. $\endgroup$ – Rahul Aug 25 '12 at 15:33
  • $\begingroup$ like, for Rhombus, $Rhombus_\pi = {4a \over 2 cos(\theta/2)}$ Where, $\theta$ is smaller angle in rhombus. As per my assumption, this is a 2 variable figure. $\endgroup$ – Rahul Aug 25 '12 at 15:45
  • $\begingroup$ @RahulTaneja: it shouldn't depend upon the side length, only the angle. The long diagonal will scale with the side length. $\endgroup$ – Ross Millikan Aug 25 '12 at 16:14
  • $\begingroup$ I don't think so. You can say that the long diagonal will depend on the angle only but $Rhombus_\pi$ will depend on both, And you can see putting $\theta = 90 $degree $\endgroup$ – Rahul Aug 25 '12 at 16:25
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    $\begingroup$ @RahulTaneja: The units are wrong. a has units of length, while $\pi$ is dimensionless. The perimeter is $4a$ and the diagonal is $2a \cos \frac \theta 2$ giving $Rhombus_\pi=\frac 2{\cos \frac \theta 2}$ $\endgroup$ – Ross Millikan Aug 25 '12 at 21:06

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