Let $A$ and $B$ be two closed, 2-dimensional, non-positively-curved Riemannian disks (not necessarily with convex boundary). Suppose that their boundaries $\partial A$ and $\partial B$ have the same length, and identify $\partial A$ with $\partial B$ with a length-preserving diffeomorphism of circles. Suppose also that for every pair of points $p,q \in \partial A = \partial B$, there is a distance inequality $d\_A(p,q) \le d\_B(p,q)$. Does it follow that the area of $A$ is less than or equal to the area of $B$, with equality only when $A$ and $B$ are isometric?


This is a variation of a question that arose in a joint paper that I am trying to finish. Another version which is more relevant to the paper is about $\mathrm{CAT}(0)$ disks which are tiled by Euclidean unit equilateral triangles.



A comparison question for non-positively curved disks
The answer is yes. Moreover you don't need to assume that $B$ is nonpostively curved. (And, if you are not interested in the equality case or can afford a convex boundary, the nonpositive curvature of $A$ can be replaced by a weaker assumption that the geodesics in $A$ have no conjugate points).


This follows from a filling inequality that I have proved some time ago (and quite proud of it) and boundary rigidity results of Croke, Otal, or Pestov-Uhlmann.


**Inequality.**
There are two papers covering it:


1) S.Ivanov, *On two-dimensional minimal fillings*. Algebra i Analiz 13 (2001), no. 1, 26-38 (Russian); English translation in St. Petersburg Math. J., 13 (2002), no.1, 17-25. A preprint is [here](http://www.pdmi.ras.ru/~svivanov/papers/filling.pdf). In this paper the inequality $area(A)\le area(B)$ is proved under the assumptions that $A$ is free of conjugate points and has convex boundary (sorry;)) and for arbitrary $B$ such that $d\_A(p,q)\le d\_B(p,q)$ for all $p,q$ on the boundary.


2) S.Ivanov, *Filling minimality of Finslerian 2-discs*. [arXiv:0910.2257](http://arxiv.org/abs/0910.2257), to appear in Trudy Mat. Inst. Steklov (= Proc. Steklov Inst. Math). In this paper the same thing is proved for Finslerian metrics $A$ and $B$ and without convex boundary assumptins. This is the maximum generality I can imagine. But the paper is harder to digest if you don't like Finsler metrics. 


The equality of areas implies that the boundary distances of $A$ and $B$ coincide, and then you can apply boundary rigidity results mentioned below.


**Rigidity.**
The case you are asking for is covered by C.Croke, *Rigidity for surfaces of nonpositive curvature*, Comment. Math. Helv. 65 (1990), no. 1, 150–169. He proves that if $A$ is a nonpositively curved and $B$ is arbitrary Riemannian such that $d\_A(p,q)=d\_B(p,q)$ for all $p,q$ on the boundary, then $A$ and $B$ are isometric. He carefully works through the details of non-convex boundaries.


I have not read J.-P. Otal's paper mentioned by Igor Rivin, but from mathscinet review it seems that he assumes that both $A$ and $B$ are strictly negatively curved.


For the same result without curvature assumptions but with convex boundary, see L.Pestov and G.Uhlmann, *Two dimensional compact simple Riemannian manifolds are boundary distance rigid*. Ann. of Math. (2) 161 (2005), no. 2, 1093–1110. 


