Consider a closed simple (i.e. with no self-intersection) curve of fixed length $L$, on a two-dimensional surface. What is the shape of this curve so that it encloses the largest possible area?
If the curve is drawn on a plane, the answer is easy and well-known: it is just the circle, saturating the iso perimetric inequality on the plane.
If the curve is drawn on a two-sphere of radius one and $L\leq 2\pi$, then the answer is also well-known and easy: the curve is the intersection of the sphere with a plane (curve of constant extrinsic curvature on the sphere), saturating the isoperimetric inequality on the sphere.
The question is: what happens if $L>2\pi$? It seems that the curve would have to be made of several pieces each having constant extrinsic curvature. What is the general solution of this problem, for all $L$?
A generalisation of this question is as follows. What are the metrics on a disk, with fixed boundary length $L$ and fixed positive Gaussian curvature $+1$, that maximise the area of the disk? The question is not obvious for $L>2\pi$; is the answer provided by the previous problem (the disk being represented as an embedding on the two-sphere with boundary a simple closed curve) or more general possibilities arise (with immersed disk on the two-sphere that are not embeddings)?