Two recent questions were posed by Arjuba [1] [2] asking for counter-examples regarding whether two different figures could have the same perimeter and area. Responders quickly raised a number of such examples.

However, all were of the form of polygons rather than smooth curves. This doesn't pose any obvious conceptual obstacle, since one can certainly construct such a counter-example by a limiting process. But what I would be interested in seeing a construction for is a continuous family of smooth curves, all with identical area and perimeter. (A family of algebraic curves would be my preference). Are there well-known examples of such?

EDIT) To clarify my intent: What I really want is a continuous family of curves, either parameterized explicitly or given implicitly as an algebraic curve, which have the desired property.

EDIT 2) I've asked another question which suggests/ requests consultation on a possible variational strategy to finding examples with more smoothness.

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    $\begingroup$ Consider a not simply connected example formed by a smaller circle within a larger one. Displacing the smaller circle from a concentric position toward the larger circle, but not intersecting it, conserves the area between the circles and the perimeter of that region. $\endgroup$
    – hardmath
    Jul 25, 2014 at 1:13
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    $\begingroup$ I see your point. Nor does them being circles seem terribly important: enclose one Jordan curve inside another and 'wobble' it. $\endgroup$ Jul 25, 2014 at 1:17
  • $\begingroup$ Generically, any $n$-parameter family of curves (modulo rigid transformations) will give an $(n-2)$-parameter family of equal-area-and-perimeter curves, right? $\endgroup$
    – user856
    Aug 4, 2014 at 0:12
  • $\begingroup$ Yes---but can you construct one explicitly? That's the tricky part. So far, the examples are either smooth to finitely many derivatives, or the example isn't completely constructive. (Christian's answer below is so far the closest to an explicit example, though, and it represents a specific version of your suggestion.) I think that the trick is to convert this from a 'constraint' problem into some kind of differential equation problem (maybe a Hamiltonian system?). But I can't quite see it straight. @Rahul $\endgroup$ Aug 4, 2014 at 0:16

7 Answers 7


Still not explicitly paramaterised curves, but someone may paramaterise them for me: smooth curves from J to U animated smooth curves from J to U

  • $\begingroup$ Another nice one, though I should've stated in my bounty that I'd prefer to avoid piecewise parameterizations. (As compared to, say, the parameterization of something like a lemniscate.) $\endgroup$ Jul 29, 2014 at 20:51
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    $\begingroup$ Naturally you'd prefer something with the simplest description possible -- so would I. I was just so excited about my idea I had to put it up, bounty or not. :) $\endgroup$ Jul 29, 2014 at 21:09
  • $\begingroup$ Heh, fair enough. It's a bit interesting that both examples so far are $C^1$ smooth curves, i.e. I can only object if I want the curvature to be nice, too! $\endgroup$ Jul 29, 2014 at 21:13
  • $\begingroup$ Nice animation! :o) $\endgroup$
    – String
    Jul 30, 2014 at 23:24
  • $\begingroup$ Given the plethora of constructions you've given for this, I figured you earned being the accepted answer. $\endgroup$ Aug 6, 2014 at 16:27

(The following example avoids piecewise definitions; but there is nothing "special" about it.)

Consider the curve $\gamma$ with polar representation $$r=r_0(\phi):=2+\cos(3\phi)\ ,$$ which looks like a clover leaf, see the following figure. This curve has a length $L_0$ and encloses an area $A_0$.

enter image description here

We now set up a perturbation of $\gamma$ in the form $$r=r(\phi):=r_0(\phi)+ a + b\cos(3\phi)+c\cos\phi$$ with small parameters $a$, $b$, $c$. Define $$L(a,b,c):=\int_0^{2\pi}\sqrt{r^2(\phi)+r'^2(\phi)}\ d\phi, \qquad A(a,b,c):={1\over2}\int_0^{2\pi}r^2(\phi)\>d\phi\ .$$ The quantities $$a_{11}:={\partial L\over\partial a}\biggr|_{(0,0,0)},\quad a_{12}:={\partial L\over\partial b}\biggr|_{(0,0,0)},\quad a_{21}:={\partial A\over\partial a}\biggr|_{(0,0,0)},\quad a_{22}:={\partial A\over\partial b}\biggr|_{(0,0,0)}$$ are obtained by differentiation under the integral sign and putting $(a,b,c)=(0,0,0)$. Numerical integration then gives $$a_{11}=4.4197,\quad a_{12}=9.3105\ ,$$ whereas the other two can be computed explicitly: $$a_{21}=4\pi,\quad a_{22}=\pi\ .$$ At any rate, one has $\det[a_{ik}]\ne0$. From this we can draw the following conclusion:

The system of equations $$L(a,b,c)=L_0, \quad A(a,b,c)=A_0$$ has the solution $(0,0,0)$. Therefore it can be solved for $a$ and $b$ in the neighborhood of $(0,0,0)$, which means that there are two $C^1$-functions $c\mapsto \alpha(c)$, $c\mapsto\beta(c)$ with $\alpha(0)=\beta(0)=0$, such that $$L\bigl(\alpha(c),\beta(c),c\bigr)=L_0,\quad A\bigl(\alpha(c),\beta(c),c\bigr)=A_0$$ for all $c$ in a neighborhood of $0$.

EDIT from String. This is such a great answer, so I decided that it should have an animation of its own:

enter image description here

I approximated $a$ and $b$ by iterating linearly solving the system of equations defined by the Jacobian. Three such linear steps were taken for each $c\in[-0.7,0.7]$ (by steps of $0.01$). For $c$'s of numerical value greater than about $1$ it appears that the system is still solvable, but the pertubation begins to intersect itself via loops so then it has no relevance to the problem in question.

@Christian Blatter: I hope you don't mind this edit - otherwise we can just delete it again!

  • $\begingroup$ This is very nice. Moreover it seems quite generalizable to a generic polar curve expanded in a Fourier series: one looks for which 'flows' in the space of Fourier coefficients preserves area and length. (Which is very reminiscent of the notion of isospectral flows for eigenvalue problems.) So this answer definitely has the lead for the bounty. $\endgroup$ Aug 2, 2014 at 14:13
  • $\begingroup$ I'd also suspect this approach can be formulated in a way that provides a positive answer to this more recent question of mine, inspired by this one. $\endgroup$ Aug 2, 2014 at 14:23
  • $\begingroup$ Hey! Did you realise that the curve $r = 2 + \cos(n \phi)$ has the same area for every $n \in \mathbb{N}$? Sadly, the arc length increases with $n$. :( $\endgroup$ Aug 3, 2014 at 6:05
  • $\begingroup$ @DavidButlerUofA: It's the arc length that's the problem, yeah. I somewhat suspect that some other periodic decomposition (say, into elliptic functions) would be more tractable. But I don't rightly know. $\endgroup$ Aug 3, 2014 at 23:44
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    $\begingroup$ @String: That's fantastic! A thousand thanks. $\endgroup$ Aug 4, 2014 at 10:53

Though still not satisfying the OP's desire for smoothness of higher degree, the following construction turned out to be simpler than my initial analysis had predicted:

enter image description here

All the curves in this family have a perimeter of $L=10$ and an area of $A=4$. The $\color{blue}{\text{blue arc}}$ has radius $r$ variyng as a function of $t$. The parameter $t$ is not really important here - it just reparametrizes $r$ to make the animation look nicer. The $\color{red}{\text{red arc}}$ has radius $r+d$ so that $d$ is the width of the section between the $\color{blue}{\text{blue arc}}$ and the $\color{red}{\text{red arc}}$. The two black half circles both have $d$ as diameter. The angle $\theta$ denotes the common angle of the $\color{blue}{\text{blue arc}}$ and $\color{red}{\text{red arc}}$ measured in radians.

In my initial analysis, I fixed some $r$ and tried to set up equations involving $d$ and $\theta$ as variables in order to match $L$ and $A$. I quickly found that $$ \begin{align} 2\theta(d^2+2rd)+\pi d^2&=4A&&&(1)\\ \text{and}\\ \pi d+\theta(2r+d)&=L&&&(2) \end{align} $$ and solving $(2)$ for $\theta$, substituting that into $(1)$, and rearranging a bit then yields $$ (2 r+d)(\pi d^2-2L\cdot d+4K)=0 $$ where it turns out (btw. applying the zero product rule) that only the solution $$ d=\frac{L-\sqrt{L^2-4 \pi K}}{\pi} $$ has both $d$ and $\theta$ positive. Surprisingly (I find it so), the width $d$ does NOT depend on $r$ anymore in this solution. So in effect only $\theta$ varies with $r$. Though the above construction is NOT a family of $C^2$ curves (or higher), but just $C^1$, it still has the advantage of looking very nice in the way it bends and transforms, in my opinion.

BTW. I have chosen $r\in[1/3,2]$ in the animation above by defining $r=\dfrac{0.4}{1.2-s}$ and $s=\dfrac{1 + \sin\left(\pi (t-0.5)\right)}{2}$ and letting $t$ run through the interval $[0,1]$. The sine function was to make the change slow down near the extreme values of $t$ to make the bending speed change more smoothly.

  • $\begingroup$ This may be $C^1$, but it's a really nice example nonetheless. Though I think this indicates the limitations of such engineering of a piecewise examples. One has to work hard just to get the area and perimeter to stay fixed, so trying to get higher smoothness conditions on top of that seems quite a challenge. $\endgroup$ Jul 30, 2014 at 12:01
  • $\begingroup$ @Semiclassical: Glad you liked it! A family of $C^\infty$ curves would be nice, indeed, but obtaining such families piecewise seems impossible. On the other hand, finding nice and closed formulas for both perimeter and area for other curves than circles and straight lines may be very challenging too ... $\endgroup$
    – String
    Jul 30, 2014 at 12:34
  • $\begingroup$ indeed. I think the better approach would be to somehow derive a curve for which the fixed area and perimeter conditions are "built in" somehow. My thought was that some calc. of variations method might work (I.e. having the length/area conditions as constraints and then varying a given initial curve). Problem is, I'd think you'd need another functional to extremalize in that case, and I don't see a natural choice... $\endgroup$ Jul 30, 2014 at 12:47
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    $\begingroup$ In the spirit of my last comment, I've started a question proposing/requesting consultation on a variational approach to this problem. $\endgroup$ Jul 30, 2014 at 19:52
  • $\begingroup$ @Semiclassical: Exciting! Very nice, detailed, and specific introduction and requirements there! Perhaps you should link to it in the question text here too? $\endgroup$
    – String
    Jul 30, 2014 at 20:12

Circle with two bumps

Let $f$ be your standard bump function $$ f(x) = \begin{cases} e^{-\frac{1}{1-x^2}}, &|x| < 1\\ 0, &\text{otherwise} \end{cases} $$ Let $g(x) = f(5x)$ (this makes your bump narrower).

For $b \in [0, 1]$ (but not too close to 0), consider the curve: $$ r(t) = 1 + g(t-\frac{\pi}{2}) + g(t-\frac{\pi}{2} - b\pi) \quad \text{for } t \in [0, 2\pi] $$

All of these curves have the same area and perimeter and are also smooth (though they are technically piecewise-defined).

circles with bumps

Note: I'm not trying for the bounty, I'm just having trouble putting the problem down.

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    $\begingroup$ Why do I find so many of the animations of the curves for this question to be so hilariously amusing? $\endgroup$
    – heropup
    Aug 3, 2014 at 23:15
  • $\begingroup$ @heropup Because they actually are hilariously amusing! ;) $\endgroup$ Aug 3, 2014 at 23:23
  • $\begingroup$ They are pretty cool. This particular one gave me an idea, and I found a perfect example of it on Youtube. (I don't consider that an example per se, but I have a sneaking suspicion that something like this could give an ideal example. But I know only enough to suspect, not to establish it...) @heropup $\endgroup$ Aug 3, 2014 at 23:41
  • $\begingroup$ I actually tried with $f(x) = e^{-x^2}$ (and a bigger multiplier for $g$ and shifting one by $b\pi$ and the other by $-b\pi$), but I couldn't be absolutely certain the area and perimeter matched. $\endgroup$ Aug 3, 2014 at 23:52

Only two curves, rather than a family, and one of them is piecewise-defined, but I thought people might be able to use them as fuel for doing something better:

Let $$ f(x) = \sqrt{\frac{1-x^2}{1+(x-1)^2}} $$ Then we define two curves like this:

Curve 1: $$ \begin{align} y &= \phantom{-}f(x) \quad \text{for } -1 \leq x \leq 1\\ \text{and}\quad y &= -f(x) \quad \text{for } -1 \leq x \leq 1 \end{align} $$ Curve 1

Curve 2: $$ \begin{align} y &=f(x) \quad \text{for } -1 \leq x \leq 1\\ \text{and}\quad y &= -f(-x) \quad \text{for } -1 \leq x \leq 1 \end{align} $$ Curve 2

Note that Curve 1 has implicit form $y^2(1+(x-1)^2) + x^2 = 1$.

The idea was that if I had a curve with an implicit function, which was asymmetric, I could cut it in half and turn one half around to get a different shape with the same perimeter and area.

I realised that if the only x-term not attached to a y was x^2, and the constant was 1, then the curve would cut the x-axis at 1 and -1.

I also realised that in order to "cut and flip" I would need the tangents at those two points to be vertical. In order to do that, the y-derivative of the function at those points would need to be zero, which could be achieved if the only terms with $y$ had at least a power of 2.

Playing around with Geogebra I found this example. It gives me hope that there is actually a nice example with two implicit curves.

  • $\begingroup$ I suspect one could take this a bit farther by finding an example where the curvatures are equal at the two endpoints, and therefore are still matched after the 'cut and flip.' $\endgroup$ Aug 1, 2014 at 13:19
  • $\begingroup$ Could you simply achieve that by making the power on the y higher (but still even)? $\endgroup$ Aug 2, 2014 at 7:18
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    $\begingroup$ I think you're right about this giving a cheap way to increase the number of continuous derivatives, though sadly (to me) it seems limited to only a finite increase in such. $\endgroup$ Aug 4, 2014 at 11:31

Take your favorite family of smooth Jordan curves parametrized by $\lambda>1$ say, for example $x^\lambda+y^\lambda=1$. These all bound area between 2 and 4. Scale them to unit area by an appropriate factor depending on $\lambda$. The resulting curves don't have the same perimeter, but notice that each perimeter is less than 8, say. Now apply affine transformations of the form $\pmatrix{\mu & 0\\0 &\mu^{-1}}$. As $\mu$ tends to infinity the perimeter does also, so by the intermediate value theorem there is a $\mu=\mu(\lambda)$ making all perimeters equal to 8, yielding the desired family.

One can't do this with conic sections because there is not enough room, but cubic curves are enough; start for example with small perturbations of the circle: $x^2+y^2+\epsilon x^3=1$ with small $\epsilon$. These are algebraic and therefore analytic.

To make the examples more explicit, one needs to overcome the problem of the computation of arclength which typically leads to unmanageable integrals. Perhaps this is more tractable in polar coordinates where one can start with $r^2+\epsilon \sin\theta=1$ and hope for the best.

  • $\begingroup$ Can you construct this more explicitly? An elliptic curve construction seems quite natural (particularly since one may parametrize in terms of Weierstrass $\mathcal{P}$ functions) but I'm not versed enough to fill in the details. $\endgroup$ Aug 4, 2014 at 12:37
  • $\begingroup$ With regards to that last paragraph: I heartily agree. That's why I'd been hoping that elliptic functions (being naturally periodic) would be useful. But right now we're hunting for a useful 'guess' of a answer. It'd be better if we could proceed in a more principled way. (I suspect a connection to integrable systems a la the KdV equation, but I don't rightly know...) $\endgroup$ Aug 5, 2014 at 13:12

It will not be easy to come up with a "natural" explicit example, since you need a two-parameter family of curves whose lengths you can compute in an elementary way.

The following example is not very sophisticated, but does the job:

Take a square of side length $a>0$, and round off its corners using small circular arcs of radii $\rho_i>0$ $(1\leq i\leq 4)$, $\rho_3\ne\rho_4$. The circumference of the resulting shape computes to $$L=4a-\left(2-{\pi\over2}\right)\sum_i\rho_i\ ,$$ and its area to $$A=a^2-\left(1-{\pi\over4}\right)\sum_i\rho_i^2\ .$$ The system of equations $$\eqalign{f({\bf r})&:=\qquad r_1+r_2+r_3+r_4={4a-L\over 2-{\pi\over2}} \cr g({\bf r})&:=\qquad r_1^2+r_2^2+r_3^2+r_4^2={a^2-A\over 1-{\pi\over4}}\cr}\tag{1}$$ can be solved for $r_3$, $r_4$ using the implicit function theorem (or even explicitly): Since $$\det\left[\matrix{{\partial f\over\partial r_3}&{\partial f\over\partial r_4}\cr {\partial g\over\partial r_3}&{\partial g\over\partial r_4}\cr}\right]_{(\rho_1,\rho_2,\rho_3,\rho_4)}=2(\rho_4-\rho_3)\ne0\ ,$$ the system $(1)$ is in a neighborhood of $(\rho_1,\rho_2,\rho_3,\rho_4)$ equivalent to $$r_3=\phi(r_1,r_2),\quad r_4=\psi(r_1,r_2)$$ with $C^1$-functions $\phi$ and $\psi$. In this way we obtain a two-dimensional family of "rounded squares" having "corner radii" $r_i$ $(1\leq i\leq 4)$. All these "rounded squares" possess the same circumference $L$ and the same area $A$.

  • $\begingroup$ While I still hold out hope for an explicit example (maybe an Euler-Lagrane approach with constraints would work?) this is a very nice example; I'll probably make an animation out of it when I get a chance. $\endgroup$ Jul 29, 2014 at 16:04

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