An enlightening discussion about pi and tau.

The first point in the $$\tau$$ Manifesto is that it seems more natural to define the "Circle Constant" as
$$\frac{C}{r}$$
$$\frac{C}{d},$$
where $$C$$,$$r$$ and $$d$$ are respectively circle's circumference, radius and diameter. Indeed the circle is defined as the set of the points with fixed distance (radius) from a fixed center, so the radius is more "important" than the diameter.

My point is to forget that we are talking about circles, and consider a (planar) shape $$F$$, let's think of the points inside a closed simple curve. We can define its diameter as
$$d(F)=\sup \{ |x-y|:x,y\in F\}$$
and we can also define in a standard way (e.g. by integration) its area $$A(F)$$ and perimeter $$P(F)$$. The radius can be defined as half the diameter, but I don't see any geometrical meaning in it. If $$F$$ is convex then
$$P(F)\leq d(F)\pi$$
i.e. the perimeter of any convex shape is always smaller than the perimeter of the circle with the same diameter. The idea behind this inequality is that for any convex set, if you draw the circle with the same diameter centered in the barycenter of the set, the the circle contains the set. (I'm not entirely sure of this, if someone has a rigorous proof he's welcome!).

Then we have the following beautiful (at least for me) variational formulation of $$\pi$$:
$$\pi=\sup \{ \frac{P(F)}{d(F)}:F \text{convex}\}.$$
Moreover the supremum is a maximum (e.g. the circles are minimizers, but also the so called Reuleaux polygons). So $$\pi$$ is not only the "circle constant", but is the "every convex shape maximal constant". This because the diameter is an intrinsic value of ANY shape.

I hope you agree with my argument and that it is correct. I'm not a native English speaker, so I'm sorry if there are any kind of grammatical errors. Please do not kill me for whose.
Nyarly
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Interesting, I never thought to generalize a definition for 2d shapes other than circles in support of which constant is more natural.

Can you explain your first step, though? Where is the origin in your coordinate system containing $$F$$? For instance, if $$F$$ is a circle and the origin is at its (bary)center, wouldn't

$$sup\{|x-y|: x,y \in F\} = radius?$$

Or can the origin be anywhere in $$F$$? Forgive me if I've got it wrong, I'm more of an engineering/physics guy and not used to set notation.
Chris Park
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Chris Park wrote:Interesting, I never thought to generalize a definition for 2d shapes other than circles in support of which constant is more natural.

I'm a Ph.D. student in calculus of variation, so its the first thing that I thought

Chris Park wrote:Can you explain your first step, though? Where is the origin in your coordinate system containing $$F$$? For instance, if $$F$$ is a circle and the origin is at its (bary)center, wouldn't

$$sup\{|x-y|: x,y \in F\} = radius?$$

Or can the origin be anywhere in $$F$$? Forgive me if I've got it wrong, I'm more of an engineering/physics guy and not used to set notation.

it doesn't matter where the origin is, you can think of the diameter of a (closed) set as the length of the longest segment contained in it. In fact $$|x-y|$$ is the length of the segment with extremal points $$x$$ and $$y$$! So you take every couple of point $$x,y\in F,$$ construct the segment from $$x$$ to $$y$$ and take its length. Then you take the supremum/maximum of these lengths.

I hope it is clear.

By the way there is an error (I don't know if I can edit the post, so I write here) on the suggested proof of the inequality
$$P(F)\leq d(F)\pi$$
and I can't find a simple argument to justify it. (But I am 99.99% sure it is true).

Moreover I think that there is a similar "dual" point of view, considering the so-called "width" of a set, i.e. the width of the smallest stripe which contains $$F$$.
Nyarly
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Oooh, I see. I was thinking x and y were coordinates instead of vectors and the | | were absolute value instead of magnitude. I see then that the origin can be anywhere.

So if I understand the problem correctly, you want to prove $$P(F)\leq d(F)\pi$$ by integrating. I don't think you would integrate around the barycenter, but maybe around the midpoint of the diameter? (Bounds are zero to ... well let's just say all the way around the circle).

$$P1 = \int f(\theta) d\theta P2 = \int \frac{d}{2} d\theta = d(F)\pi$$

but usually

$$f(\theta) < \frac{d}{2}$$

so

$$(P1 = \int f(\theta) d\theta) < (P2 = \int \frac{d}{2} d\theta = d(F)\pi)$$

The problem is that "usually" above is not an "always". Hmmm.

Edit: By using a drawing, it is very easy to prove that a circle maximizes the perimeter of a concave shape with a given diameter. But this drawing would be very difficult to describe!
Chris Park
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Probably there is a proof in which you integrate something, but i think is more related to isoperimetric or Brunn-Minkowsky inequalities.

In your proof there is a more serious problem than the fact that $$f(\theta)$$ can be greater than $$\theta/2$$ (this is more or less related to the fact that my "hint" for the proof is not corrected). The problem is that there can be a lot of diameters (the circle has infty diameters!) and the midpoints of these diameters can be different (as in the Reuleaux triangle, http://en.wikipedia.org/wiki/Reuleaux_triangle).

Yes I think that there is an elementary proof "by drawing", but I can't visualize it. (for me it's 11:30 PM , and I stop thinking at 7pm)

By the way, my point on using diameter instead of radius, is that if you give me any convex shape I can always tell you what the diameter is, but I can tell you what the radius is only if I know that it is a circle, and I find it halving the diameter!

Edit: I think first of difficult thing, but the easiest shape with very distinct diameters is the equilateral triangle (the diameters are the edges).
Nyarly
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Yes, yes! My drawing leads me to a Reuleaux triangle! My mind is blown

Basically I started by drawing the diameter of a circle and calling it (the diameter) AB. Then I tried to determine the locus of points that would not create a chord that would exceed the length of AB. Obviously this area was the intersection of two bigger circles (each of diameter 2AB) drawn around points A and B. Since this football-shaped intersection was symmetric on either side of line AB, I just looked at the bottom half. Choosing the vertex of the new region as a third point and drawing yet another circle around it yields a Reuleaux triangle.

The triangle appears to be the 'worst case scenario' in terms of drawing a shape with diameter D whose $$f(\theta)$$ exceeds $$\frac{d}{2}$$ as much as possible without changing D. Even in this shape, however, the perimeter is still only $$\frac{\pi D}{\sqrt{3}}$$.

I agree that this description makes diameter seem like the more natural dimension. I've actually seen it before: in engineering, sometimes one characterizes a diameter to calculate properties of fluid flow in pipes. When a pipe isn't circular, this is done as $$D_{equiv} = \frac{4A}{P}$$, so that even unusual shapes can be characterized with a 'diameter'.
Chris Park
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Chris Park wrote:The triangle appears to be the 'worst case scenario' in terms of drawing a shape with diameter D whose $$f(\theta)$$ exceeds $$\frac{d}{2}$$ as much as possible without changing D. Even in this shape, however, the perimeter is still only $$\frac{\pi D}{\sqrt{3}}$$.

The perimeter of the Reuleaux triangle (and every Reuleaux polygon) is $$\pi D$$, in fact it is made of three circular arc of radius $$D$$ and subtended angle $$\pi/3$$. Then
$$P=3\left(\frac{\pi}{3} D\right).$$

The problem in the proof is that the circle is not the only shape for which $$P=\pi D$$, so, probably, the idea would be, given a shape, to find a "constant width set", i.e. a shape for which $$P=\pi D$$ that trivially has the perimeter bigger. An idea maybe is to take the convex hull of all the diameters of the set, and hope that the worst case scenarios are the Reuleaux polygons! I'm starting to think that is not an easy result...
Nyarly
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I don't have a rigorous proof but there might be stuff known already. One interesting result is Theorem B from "Some Inequalities for Convex and Star-Shaped Domains". This isn't exactly the same problem as their "r" is different than our "r".

We can also try looking at the radius of curvature and see if that gives anything interesting.
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The inequality
$$P<\pi d$$
is a known fact. e.g. see the book Convex figures by Jaglom, I. M.; Boltjanskiĭ and V. G. (MathSciNet http://www.ams.org/mathscinet-getitem?mr=123962, i don't know if it always work of if you need access to the MathSciNet database) is exercise 7.17 and the solution is in the appendix.

It is not a difficult proof, but it isn't elementary as I hoped...
Nyarly
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Nyarly wrote:The perimeter of the Reuleaux triangle (and every Reuleaux polygon) is $$\pi D$$, in fact it is made of three circular arc of radius $$D$$ and subtended angle $$\pi/3$$. Then
$$P=3\left(\frac{\pi}{3} D\right).$$

Sorry, my $$D$$ in that post is different from my $$d$$, which I failed to elaborate. $$D$$ is the diameter of the circle that circumscribes the Reuleaux, and that is (correct me if I'm wrong) $$\sqrt{3}d$$ So when written in terms of the original diameter, I also get $$\pi d$$

I think this picture tells the story pretty well. The red line is the diameter of some shape we want to draw. The black region is the locus of points that can be drawn without creating new chords longer than the diameter (excluding the original circle, which is left blue because MS paint wasn't cooperating). When you choose point C and maximize perimeter by tracing the outer edge of the remaining available points, you get a Reuleaux triangle. You can also see by symmetry that no chosen points between C and the original circle will ever increase the perimeter beyond that of the two situational extrema (which both gave pi D).

Interesting that we've just brought up the similarity of all Reuleaux n-gons in having perimeters $$\pi d$$. When you start looking at this subset of polygons, radius becomes a player again (since it's easy to define the radius of a Reuleaux n-gon).
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Chris Park
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Chris Park wrote:Interesting that we've just brought up the similarity of all Reuleaux n-gons in having perimeters $$\pi d$$. When you start looking at this subset of polygons, radius becomes a player again (since it's easy to define the radius of a Reuleaux n-gon).

Sorry, how do you define it? It's not so clear to me....
Nyarly
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(Nine months later) -- you can define it as the distance from the center to a vertex!
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Radius can also be generalized as the largest distance from the centroid of a figure to the outside of a figure. This actually wouldn't always work out to be half of the diameter, but it does in the case of a circle. This generalization of the radius is already used often for cyclic shapes and known as the circumradius. Since both radius and diameter can be generalized to other shapes, I think it is important to consider which is more important to the circle. The answer there is clear--the radius is the defining characteristic of a circle, or of any hypersphere in general. The circle is the only 2D figure with constant radius (or in general the n-sphere is the only n-dimensional figure with constant radius), but constant diameter is not sufficient to define the circle (or the n-sphere).

There are other reasons that the radius is more important than the diameter; it is used all the time in trigonometry, and the natural unit of angle measure is based on it. It's also used in calculus, for example, integrating with respect to radius. Diameter rarely shows up in mathematics, while radius is everywhere. Even formulas using π, the diameter-based constant, are always written in terms of radius.

Reulaux n-gons, show why diameter is not as fundamental as radius because they have a constant diameter yet are not circles. This is a win for the definition of tau over the definition of pi. They also provide a win for pi since the perimeter of a Reulaux n-gon is πd. However, it should be noted that a Reulaux n-gon is constructed from n intersecting circles, where the radius of the circles becomes the width of the n-gon, and each side of the n-gon is formed from 1/(2n) of one of these circles. That, of course means the perimeter of the n-gon will be τrn*1/(2n)=τ/2r=πr.
τ>π
1=0
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