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

  • 2
    $\begingroup$ Yes actually. You can also say that a sphere will have the greatest volume for a fixed surface area compared to any other shape. $\endgroup$ – user123 May 12 '13 at 11:37
  • 1
    $\begingroup$ See Isoperimetric Inequality $\endgroup$ – JavaMan May 12 '13 at 21:58
  • $\begingroup$ Strongly related. $\endgroup$ – Daniel Fischer Dec 10 '16 at 19:56

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 $(K_1,K_2)$


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).

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

I will try to add "Easier" solution yet it requires 2 assumptions:

  1. The shape is formed of edges.
  2. Given a shape formed of $ n $ edges and perimeter $ P $ the largest area will be if all edges are equal in their length.

Namely, if the question is limited to Polygon with equal edges the following will hold.

If we take those 2 assumptions something nice happens.
Since the shape is symmetric it has a center and the distance from its center to any vertex is equal.
Let's define the distance to each vertex as $ r $.
What is formed is Isosceles Triangle composed from two edges of the length $ r $ and a base of the length $ \frac{P}{n} $.
The area of this triangle is given by $ {A}_{n} = \frac{r}{2} \cos \left( \frac{\pi}{n} \right) \frac{P}{n} $.

Since there are $ n $ triangles just like this the area is given by $ A = \frac{r}{2} \cos \left( \frac{\pi}{n} \right) P $.

Now, to maximize the area one needs to maximize the term $ \cos \left( \frac{\pi}{n} \right) $ which is maximized for $ n \to \infty $ which suggests a shape of a circle.

Dedicated with love to Renana!


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.

  • $\begingroup$ That is not a proof $\endgroup$ – WetlabStudent Jun 9 '17 at 9:14

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