# The Dido problem with an arclength constraint

It is well known that the solution to the classical Dido problem is a semicircle, and that the solution to the classical isoperimetric problem is a circle. It's also reasonably obvious that the solution to the following variant is a circular arc:

Let $A$ and $B$ be fixed points on a plane, and let $l$ be a length greater than $\overline{AB}$. Which (smooth) curve through $A$ and $B$, of length $l$, maximises the area between itself and the line $AB$?

It's a straightforward exercise to verify extremality using the calculus of variations, but are there alternative proofs that do not invoke e.g. the Euler-Lagrange equations? This was originally a homework problem with the isoperimetric inequality given as a hint, and I'm just wondering what the intended solution was...

This diagram might help:

For any length $l > \overline{AB}$, we can find a circle passing through $AB$ such that the length of a circular arc between $A$ and $B$ is equal to $l$. In the diagram above, suppose that the red, dotted arc $ADB$ and the upper part of the circle both have length $l$. Note that the region bounded by $ADBC$ (i.e.: the pink area) has the same perimeter length as the circle.

• Ah. The diagram makes things a lot clearer. I think this must have been the intended solution, thanks! Commented Feb 21, 2011 at 16:38
• Nice! The 'use the other side of the circle' was the bit I was missing. Commented Feb 21, 2011 at 23:44

While I'm not sure that this will perfectly solve the matter for arbitrary $l$, it's easy to get this in special cases, in a way that may point the way to a general solution (in particular, by continuity) : build an isosceles triangle on $AB$ with the other two sides both equal to the radius of the appropriate circle, and consider the 'ice-cream-cone' with your curve and that the two radii, and piece together however many of these it takes to form a circle (note that this only works for arcs that are integral divisions of the appropriate circle!). The isoperimetric inequality then says that the total area-maximizing curve is a circle, which implies that the section of curve between $AB$ that maximizes the wedge area is a circular arc, and since the area of the triangle is constant you can just subtract it out...

• I had thought about this, but I wasn't really convinced because it wasn't obvious to me why the solution to the unconstrained problem should be related to the solution for the constrained problem. In effect, the assertion seems to be that, look, we can always draw $AB$ as a chord in the circle, so that must be the answer... but I'm not convinced. Commented Feb 18, 2011 at 20:42

A proof that doesn't necessarily invoke the Euler-Lagrange equations was given by E. Schmidt. He uses two algebraic inequalities, the Green's theorem and property of parameterised curves as the only pre-requisites for the proof.

Another simple proof was given was Peter D. Lax who again doesn't use Euler Lagrange equations to prove the Isoperimetric Inequality.

A paper which discusses the Isoperimetric Inequality extensively including the history of a lot of proofs was presented by Robert Osserman.

One of the first proofs was presented by Jakob Steiner, but his proof was deemed to be incomplete by Weierstrass, who by a new formalized mathematical system showed the proof was not rigorous, in the sense that Steiner assumed the existence of a curve of the maximum area without actually proving it. Weierstrass himself then provided the first rigorous proof as a corollary to his Theory of Calculus of Several Variables in 1870.

References:

[1] Hehl, A., The Isoperimetric Inequality, presented at Eberhard-Karls Universitaet Tue- Bingen, 9th February 2013.

[2] Lax, P. D., A Short Path to the Shortest Path, The American Mathematical Monthly, 102:2(1995), 158{159.

[3] Siegel, A., A Historical Review of the Isoperimetric Theorem in 2-D, and its place in Elementary Plane Geometry, Retrieved April 1, 2012, from http://www.cs.nyu.edu/faculty/siegel/SCIAM.pd