For $CAT(\kappa)$ spaces $X$ we have following rigidity result: if equality holds in any of the comparison distances between a triangle $\Delta$ in $X$ and the corresponding comparison triangle $\tilde\Delta$ in the model space $M\_\kappa$ of constant curvature $\kappa$, then the convex hull of $\Delta$ is isometric to the convex hull of $\tilde\Delta$.


In Alexandrov spaces the picture is different. We do not have such a rigidity anymore. A counterexample can be found by considering two copies of a spherical triangle glued by their boundaries. 
The rigidity result that we can obtain is following (I think at least, but I do not know any reference): under the assumption of equality in any of the comparison distances we can embed isometrically the convex hull of the comparison triangle $\tilde\Delta$ in $X$, such that two sides coincide with the corresponding two sides of $\Delta$, but it may happen that the third side does not coincide anymore. An example of this can be seen in the counterexample above.


So my question is following. In the example given above the space is singular. Does anybody know an non singular example (i.e. manifold of sectional curvature $\geq \kappa$), where we do not have the same rigidity as in the $CAT(\kappa)$ case?


Remark: We assume for our triangle $\Delta$ that its perimeter is $<2\pi/\kappa$ and each side has length $<\pi/\kappa$.


Edit: Ok, I will be more sprecific with my question:
Is there a Riemannian manifold $M$ with sectional curvature $\geq \kappa$ and a 
a triangle $(x,y,z)$ in $M$ and a point $p$ in the side $yz$
such that if $(\tilde x, \tilde y, \tilde z)$ is a comparison triangle in $M^2\_\kappa$, and $\tilde p$ the corresponding point in $\tilde y \tilde z$,
we have the equality $d(x,p)=d(\tilde x,\tilde p)$ but the triangle $(x,y,z)$ (that is, the 1-dimensional object) cannot be filled with a triangle of constant sectional curvature $\kappa$ (that is, the 2-dimensional object). Such an example is easy to construct if we admit singular Alexandrov spaces, but I do not know any manifold examples.



Rigidity of triangle comparison in Alexandrov spaces
The question is not stated precisely.
So I'm free to say anything :)


* If you are interested in "non-uniqueness" then the anser is "NO". In any such triangle $[x y z]$ there are at least two distinct geodesics between directions of $[x y]$ and $[x z]$;
thus the space of directions at $x$ can not be a sphere.
* If you are interested in "the third side is wrong", the answer is "YES". Take $x=(0,0,0)$, $y=(1,0,1)$ and $z=(0,1,-1)$ in $\mathbb R^2\times [-1,1]$.
Then glue $\mathbb R^2\times \{1\}$ to it-self along reflection $(u,v)\mapsto (-u,v)$ 
and
glue $\mathbb R^2\times \{-1\}$ to it-self along reflection $(u,v)\mapsto (u,-v)$.
You get a singular Alexandrov space with triangle $[x y z]$ where the side $[y z]$ might be wrong for some filling of the hinge at $x$. BUT this triangle admins a flat filling.
It is easy to smooth this spaces near both singular lines to obtain Riemannian manifold with the same property.
* I know one example of a triangle in singular Alexandrov 3-space such that all distances between points on sides are the same as the corresponding distances in the model triangle, but it can not be filled with a flat triangle. This is a bit tricky to construct. I can not make this example to be Riemannian (and I feel that it is impossible).


**P.S.** The answer to the "specific" question is "NO".


Take a 2-dimensional nonnegatively curved manifold $M$ with a pair of points $p,q$ such that there are two minimizing geodesics from $p$ to $q$. One can choose a triangle with vertexes $x=(0,p)$, $y=(1,q)$ and $z=(-1,q)$ in the product $\mathbb R\times M$ which does not admit flat filling.
On the other hand any such triangle satisfies your condition for midpoint of $[y z]$.


