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I am trying to track down the first published solution to the following problem:

What curves within the unit disc in the plane and endpoints on the unit circle, minimize their length (within the ball) while dividing the area of the ball into two regions of a given ratio?

The solution is, of course, circular arcs orthogonal to the unit circle.

The only complete solution and proof that I've been able to find is embedded in a much stronger theorem within "Stability for Hypersurfaces of Constant Mean Curvature with Free Boundary" by A. Ros and E. Vergasta, Geometriae Dedicata, volume 56, 1995. But I suspect that there is an earlier proof for this specific case.

If anyone can point me in the right direction, I would very much appreciate it.

Edited for clarity of problem.

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Pólya posed an area-bisecting version in 1958, which might lead to earlier proofs. – Joseph O'Rourke Jan 16 '12 at 13:36
perhaps this post can also be tagged [calculus-of-variations] or [geometric-measure-theory]? Also, I don't quite understand the problem statement. Is "unit ball" the two dimensional disc? – Willie Wong Jan 16 '12 at 14:04
Edited as per Willie's comments – Brendan Foreman Jan 16 '12 at 15:19
I know that a related problem was studied in "on growth and form" by d'Arcy Thompson first published in 1917. I do not think there is a proof in the book, but there might be references. The problem studied is slightly different, since only the ratio $1/2$ is considered, but for more general shapes. More precisely, the question is to determine what can happen starting with a disk and dividing at each step each piece at hand in two equally large parts, using minimal curves to cut them. – Benoît Kloeckner Jan 16 '12 at 16:13
Check out these notes of Ros: In particular, Theorem 5 (the isoperimetric inequality for the ball) is attributed to Almgren '76 and Bukowski-Sperner '79. – Ian Agol Jan 16 '12 at 17:12

I think that a straight variational approach has a good chance of yielding the desired conclusion. Fix two points $z_0,z_1$ on the boundary of the unit disk. Denote by $C$ the positively oriented arc of the circle that runs from $z_0$ to $z_1$.

Let $\mathcal{P}_{z_0,z_1}$ the set of paths $[0,1]\to \mathbb{C}$ with endpoints $z_0$ $z_1$. $\mathcal{P}_{z_0,z_1}$ is an affine space modeled on the vector space of maps $[0,1]\to\mathbb{C}$ that vanish at endpoints.

If $\gamma\in \mathcal{P}_{z_0,z_1}$ is an embedded path inside the disk, then the area between $\gamma$ and $C$ is given by the integral

$$\int_C xdy -\int_\gamma xdy $$

Your constraint can now be given a simpler form

$$\int_\gamma xdy =const. $$

Now solve the constrained variational problem

$$ \min\left\lbrace \int_0^1 |\dot{\gamma}(t)| dt;\;\;\gamma\in \mathcal{P}_{z_0,z_1},\;\;\int_\gamma xdy =const\right\rbrace. $$

If we write $\gamma(t)= x(t) + i y(t)$ then the constraint equation can be rewritten as

$$ \int_0^1 x(t) \dot{y}(t) dt =const. $$

This defines a quadratic hypersurface in the affine space $\mathcal{P}_{z_0,z_1}$.

If we define

$$L, F: \mathcal{P}_{z_0,z_1}\to \mathbb{R}, $$ $$ L(\gamma)= \int_0^1 |\dot{\gamma}(t)| dt,\;\; F(\gamma)= \int_\gamma xdy$$

then the Euler-Lagrange equations for the above variational problem have the form

$$ dL +\lambda dG=0 \tag{$EL_\lambda$}$$

where $\lambda\in\mathbb{R}$ is a Lagrange multiplier and $dL$ and $dF$ are the differentials of $L$ resp. $F$. The differential of $F$ is a linear function so ($EL_\lambda$) can be viewed as a nonlinear eigenvalue problem. It can be written very explicitly and I believe that playing with it will yield the desired conclusion.

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Hi, Liviu, It is an easily solved problem. I was just curious if anyone knew of a published version of the solution in order to give credit where credit is due. – Brendan Foreman Jan 16 '12 at 17:27
Hi, Brendan. I never heard of this problem before and I thought it is interesting. I don't know any reference. – Liviu Nicolaescu Jan 16 '12 at 20:03
Thanks, Liviu. Your solution and interest are both gratifying. The problem is even more interesting given the context in which it was given to me -- namely, cellular anticlinal division. This is a case where Nature is doing the optimization for us. – Brendan Foreman Jan 16 '12 at 23:11

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