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Here is a nice proof of the isoperimetric inequality (equality part ommited):

Isoperimetric Inequality

If $\gamma$ is any simple closed piecewise $C^1$ curve of length $l$, with it's interior having area $A$, then $4\pi A \le l^2$. Furthermore, if equality holds then $\gamma$ is a circle.

Proof.

Take two parallel straight lines $L$ and $L'$ such that $\gamma$ is between them and move them together until they first touch the curve. See my nice picture below.

Figure 1

Let C be a circle as in the picture. Take $x$ and $y$ axes as shown. Let $\gamma = (x,y)$ be a parametrization of $\gamma$. Pick points $\gamma (s_0)$ and $\gamma (s_1)$ on both $L$ and $L'$ wherever the lines touch the curve, respectively.

Let $C$ be parametrized by $(x, \overline{y})$ where

$$ \overline{y}(s) = \begin{cases} + \sqrt{r^2 - x^2 (s)}, & \text{if } s_0 \le s \le s_1 \\ - \sqrt{r^2 - x^2 (s)}, & \text{if } s_1 \le s \le s_0 + l \end{cases} $$

Denote the derivative of $f$ with respect to $s$ as $f_s$. Using Green's Theorem, we write:

$$A + \pi r^2 = \int_{\gamma} x\,dy + \int_C -y\,dx = \int^l_0 x(s)y_s(s)\,ds - \int^l_0 \overline{y}(s)x_s(s)\,ds = $$

$$ = \int^l_0 ( x(s)y_s(s) - \overline{y}(s)x_s(s)) \,ds \le \int^l_0 \sqrt{ (x(s)y_s(s) - \overline{y}(s)x_s(s))^2} \,ds \stackrel{*}{\le}$$

$$ \stackrel{*}{\le} \int^l_0 \sqrt{ (x^2(s) + \overline{y}^2(s))} \,ds = lr$$

Where the starred inequality follows from the fact that:

$$(x y_s - \overline{y} x_s)^2 = [(x, - \overline{y}) \cdot (y_s, x_s)]^2 \le (x^2 + \overline{y}^2) \cdot (y^2_s + x^2_s) = x^2 + \overline{y}^2 $$

So we have that $A + \pi r^2 \le lr$. Next we employ the Geometric-Arithmetic Mean Inequality to find that:

$$\sqrt{A \pi r^2} \le \dfrac{A + \pi r^2}{2} \le \dfrac{lr}{2}$$

From which it directly follows that $4 \pi A \le l^2$, as needed. $\square$

I have seen other proofs of this Theorem, for example the one using Wirtinger's Inequality. This proof was presented to me by my professor, who said this it was rather mysterious, and I agree. I think this proof is rather beautiful and much simpler than the other proofs. Here are my questions:

How? How does it work? I do not mean to ask how to we get from one step to another. I mean to ask what makes this work intrinsically. In particular, I am bothered by this constructed circle, and its radius. The next picture is what I have in mind:

Funny picture.

For this curve $\gamma$, choosing two different pairs of lines $L$, $L'$ and $K$, $K'$ gives us two circles with different radii. Furthermore, note that it appears that the area of the smaller circle is less that the area traced out by $\gamma$, whereas it is not the case for the larger circle. I guess my question here can be rephrased as: why do the $r$'s magically fall out of the equation in the last step? I am not looking for an anwser of the type "because the math works out that way," rather some geometric insight/explanation.

One observation I have thought about is that in case of equality $4 \pi A = l^2$, i.e. when $\gamma$ is a circle, any circle given by our construction will always have equall radius, in fact the radius of the circle $\gamma$.

From that I was led to this question: Suppose we take $n$ pairs of parallel lines (where two distinct pairs of pairs are not mutually parallel), and construct circles for each. Now, as $n \rightarrow \infty$, what will happen to the average radius of these circles? What will a circle with this radius represent?

EDIT: I have found an example where this last question turns out to be rather uninteresting. However, what will happen if we assume the curve to be convex as well?

I do not know how to explore this last question with my knowledge whatsoever.

Finally, I want to ask if anyone knows how this proof came to be; what is the hidden motivation.

Thank you.

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    I think that all the magic is hidden in double integrals. Think about it this way: Green's theorem turns a line integral into a double integral over a region in the plane. There are lots of ways to evaluate a double integral, depending on the axes of the plane you choose. This freedom corresponds to choosing different projections from your region that give you circles of different sizes. Another way to think about it is: yes, the alternative projection gives you a smaller circle area, but it also hides a lot of the bounding curve whose length you are interested in. Not sure if that helps you.2011-02-02
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    Isn't this the proof presented in Manfredo Do Carmo's *Differential Geometry of Curves and Surfaces*?2011-02-02
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    Alex is right: all the work is being done here by Green's theorem. That said... Perhaps one thing which could be confusing is the apparent importance of the circle C. The circle is used only as a stepping stone to travel from an expression for the area to an expression for the perimeter. The reason the $r$ drops out at the end is exactly that your choice of projection doesn't *really* matter. If the circle is very small then the estimate from Green's theorem takes this into account already.2011-02-02
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    @Alex: what do you mean by "it also hides a lot of the bounding curve whose length you are interested in." This comment has the type of flavour I'm looking for - I would be happy if you could explain further.2011-02-03
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    @Didier Thanks! I'll check it out!2011-02-03
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    @Glen Can you please explain "circle is used only as a stepping stone to travel from an expression for the area to an expression for the perimeter" further. I do see that the math works out that way. Could you please elaborate on *why* a bit more. Thanks!2011-02-03
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    IMO there is a simpler proof strategy that relies on three assertions: (1) interior area is bounded, (2) any closed Jordan curve can be morphed into another with equal perimeter and convex interior with no smaller area, (3) a curve with maximal interior area would be symmetric in respect to any line that dissects it into two equiperimetric halves.2011-02-03
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    @Didier: Yes, but this particular proof is due to Schmidt (1938)2011-02-04
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    @milcak What I mean is this: the inequality bounds an area from above by the length of the bounding curve. There are two extreme ways of proving such an equality: either by showing that the length is large or that the area is small. In your first sketch, you are comparing with a large circle and saying that your actual area is smaller, while in the second sketch you are comparing with a small circle and saying that your actual curve is much longer, because it extends into the direction that you can't see due to your projection onto the y-axis.2011-02-04
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    @Bey: Would you know a reference for this?2011-02-05
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    Considering the second problem mentioned: There is an inequality by Bonnesen on the "isoperimetric defect", namely $$l^2-4\pi A \geq \pi^2(r_a -r_i)^2.$$2011-02-06
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    In the above comment $r_i$ denotes the radius of the largest circle contained in $A$ and $r_a$ the radius of the smallest circle containing $A$. – There is also an inequality by Bottema in the other direction, namely $$l^2 - 4\pi A \leq \pi^2 (\rho_\max -\rho_\min)^2$$ where $\rho$ stands for the radius of curvature of $\gamma$.2011-02-06
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    *Since I cannot post comments I'm forced to post this as an answer, though it certainly is not. Please excuse the misuse.* Related to this inequality you, or someone, might find the following papers interesting, the second being more recent and the one which pointed me to the first one. - [*The isoperimetric inequality*][1], - [*The Brunn-Minkowski inequality*][2] [1]: http://www.ams.org/journals/bull/1978-84-06/S0002-9904-1978-14553-4/ [2]: http://www.ams.org/journals/bull/2002-39-03/S0273-0979-02-00941-2/2011-02-06
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    @NonUser: there is a reason users without sufficient reputation cannot post comments. Please do not try to circumvent the system, and wait until you have enough reputation to comment correctly. (For now I have moved your answer into a comment.)2011-02-06
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    @milcak What I mean is the following: the key inequality is $A + \pi r^2 \le lr$. This is obviously equivalent to $(lr - \pi r^2) \ge A$. Now the length of a curve scales like $r$ so the "weighted length" is $lr$. The previous estimate then states in words "The weighted length of a curve minus error is greater than the area enclosed by the curve." In this sense we have used the circle (the "weight") to obtain a good estimate for the area, with a controllable (cancellable!) error, and it is in this sense that the circle is a stepping stone.2011-02-07
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    @milcak Wow. This is a very nice uncomplicated proof!2012-12-29
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    @joriki: Thanks for your insight. While I am reading the same proof in Carmo's book, the equality part is like misleading to the conclusion that $A=\pi r^2$ when it has a maximum area which is determined by the size of the circle. And further, the book shows that $ A $ must be a circle by showing $ x=\pm r y'$ and $ y=\pm r x'$ (that I don't understand). Can you help me out how to resolve this? Sorry, but I couldn't find "add comment" button.2013-03-21

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