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Is it true that "Among all shapes with the same area, a circle has the shortest perimeter" ? and how to prove it ?

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Yes actually. You can also say that a sphere will have the greatest volume for a fixed surface area compared to any other shape. –  Mohammad Ali Baydoun May 12 '13 at 11:37
    
See Isoperimetric Inequality –  JavaMan May 12 '13 at 21:58

3 Answers 3

This is true and classical, and not quite as trivial as one might think. This also works in higher dimensions.

For a real-life argument - look at the shape water drops take! Or even more convincingly: look at soap bubbles (thanks to fgp for this example).

For an intuition why this should be true mathematically, try the following reasoning:

Let $F$ be the figure with the shortest perimeter from among figures with a given area. Try cutting the figure $F$ by a line $l$ in such a way that both parts have the same area. Call these parts $F_1 $ and $F_2$.

The illuminating observation is that you could now construct another figure $F'$ with same area as $F$. This is achieved by replacing $F_2$ by the reflection of $F_1$ with respect to $l$; i.e. take $F' := F_1 \cup R_l(F_1)$, where $R_l$ is the reflection with respect to $l$. Of course, we can also take $F'' := F_2 \cup R_l(F_2)$. Because the areas $\operatorname{Area}(F_1)$ and $\operatorname{Area}(F_2)$ are equal (and have the value $\frac{1}{2} \operatorname{Area}(F)$), the new figures have areas $\operatorname{Area}(F') = \operatorname{Area}(F'') = \operatorname{Area}(F) $. Likewise, we have $\operatorname{Perimeter}(F') = 2 \operatorname{Perimeter}(F_1)$ (here, we are not counting the part of the perimeter that lies on $l$), $\operatorname{Perimeter}(F'') = 2 \operatorname{Perimeter}(F_2)$ and of course: $$\operatorname{Perimeter}(F) = \operatorname{Perimeter}(F_1) + \operatorname{Perimeter}(F_2) = \frac{\operatorname{Perimeter}(F') + \operatorname{Perimeter}(F'')}{2} $$ If $F_1$ and $F_2$ had different perimeters, then one of $F'$, $F''$ would have a smaller perimeter than $F$, and the same area. But we assumed that $F$ was optimal, so $\operatorname{Perimeter}(F_1) = 2 \operatorname{Perimeter}(F_2)$.

Let us push this idea a little further. Note that $F'$ and $F''$ are now again optimal, in the sense that they have the same area as $F$. This means, they have to be convex (if a figure is not convex, you can make a "shortcut" at its concavity, making the perimeter shorter and area larger; but it is easy to get rid of the extra area without increasing the perimeter, for instance by scaling). This means, among other things, that the lines perpendicular to $l$ through the intersection points with $F$ are tangent to $F$ (so if $F$ has a well defined tangent at this points, the tangent is perpendicular to $l$). This doesn't yet show that $F$ has to be a circle, but we see that $F$ has a lot of symmetry that only the circle seems to have.

With a little more work, we can almost get from here to the solution. Note that instead of working with $F$, can work with $F'$, i.e. we can assume that the figure is symmetric with respect to $l$. But this is equivalent to just considering $F_1$. Let us denote the two points where $l$ meets $F$ by $A,B$, and let $C$ be a point at the boundary of $F_1$. The lines $AB,BC,CA$ cut $F_1$ into three regions: triangle $ABC$ and two irregular shapes touching lines $AC$ and $BC$. We can do two things now. The idea from Cut the knot is to transform $ABC$ by changing the length of $AB$ Because $\operatorname{Area}(ABC) = \frac{1}{2} |AC| \cdot |CB| \cdot \sin \angle C$, we can assume that $\angle C = \pi/2$. Another possibility is to reflect the area adjacent to $AC$ an conclude that the tangent to $F$ at $C$ meets $AC$ at the same angle as the tangent at $A$ meets $AC$. Doing the same for $BC$ an counting angles, we again conclude that $\angle C = \pi/2$. But $\angle C = \pi/2$ means that $C$ lies on the half-circle based on $F_1$. So $F_1$ is a half-circle, an we are done!

Of course, there is some cheating involve in assuming the best figure exists. Also, at a few steps I assumed existence of tangents. This can be explained away, but Im not sure if it's of interest here.

For a detailed discussion, this seems relevant: Cut the knot (which I used it here strongly, but most of the ideas are mathematical folklore).

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thank you all for the reflexion tracks you gave me :) –  mounaim May 12 '13 at 21:03

The question can be asked such as "Find the plane curve which encloses given area with shortest perimeter". Then by parametric formulation of coordinates $x(t),y(t)$ and assuming non-selfintersecting curve the area can be written as $$A=\frac 12 \int_{t_1}^{t_2}\big(xy'-x'y\big)dt$$ and the perimeter $$I=\int_{t_1}^{t_2}\sqrt{x'^2+y'^2}dt$$ Then the Euler-Lagrange becomes $$-\frac{d}{dt}\bigg(\frac{x'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(y'+\frac{d}{dt}\bigg(y\bigg)\bigg)=0$$ $$-\frac{d}{dt}\bigg(\frac{y'}{\sqrt{x'^2+y'^2}}\bigg)-\frac{\lambda}2\bigg(-x'-\frac{d}{dt}\bigg(x\bigg)\bigg)=0$$ If $ds$ is the element of arc the equations can be reduced to $$\frac{d^2x}{ds^2}+\lambda \frac{dy}{ds}=0$$ $$\frac{d^2y}{ds^2}-\lambda \frac{dx}{ds}=0$$ which has the solution $$x(t)=K_1+C_1\cos(\lambda s)+C_2\sin(\lambda s)$$ $$y(t)=K_2-C_2\cos(\lambda s)+C_2\sin(\lambda s)$$ It can be further simplified by "Angle sum and difference identities" $$x(t)=K_1+D\sin(\lambda s+\alpha)$$ $$y(t)=K_2-D\cos(\lambda s+\alpha)$$ where $D\sin(\alpha)=C_1$ and $D\cos(\alpha)=C_2$; and this solution represents a circle of radius $D$ and centre $(K_1,K_2)$

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If you have a rope with length $P$ the shape with the biggest area that you can create is the circle. If you have a shape, rather than circle, with the same area with the first circle that is because you have used a longer rope.

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