Inequality in geodesic quadrupel Let $(M,g)$ be a compact complete Riemannian manifold. Consider the geodesic quadrupel $ABCD$ where $l:=d(A,B)=d(B,C)=d(C,D)=d(A,D) \geq \frac{1}{2}$ and $d(B,D),d(A,C) \geq 1$. Let $x \in CD$ and $y \in AB$. Is it true that then $d(x,y)\geq \frac{1}{4}$ ? I have tried a lot of things but non of them worked, mostly I tried to do it by the triangle inequaity. I even do not know if its true.
Greetings
Lena   
 A: No (unless you assume something like a lower curvature bound).
Consider two very long segments (say, of length 100) in $\mathbb R^3$ that meet orthogonally at their midpoints. Let $M$ be a smoothened boundary of an $\varepsilon$-neighborhood of the union of these segments. It is a Riemannian manifold invariant under a 90 degrees rotation. Let $A$ be a point of $M$ near one of the segments' endpoints and $B$, $C$, $D$ be its consecutive images under this rotation. Due to rotation invariance, one has $d(A,B)=d(B,C)=d(C,D)=d(D,A)$. And all distances between $A,B,C,D$ are greater than 50.
On the other hand, geodesic segments $AC$ and $BD$ go through the central part of the construction, namely an $O(\varepsilon)$ neighborhood of the center. Take $x$ and $y$ in that region and observe that $d(x,y)=O(\varepsilon)$.
A: Surface   $z = y^2 - x^2$,   with proper scaling (say, one scaling per   $\epsilon > 0$),   will provide counter-examples, with   $\frac 14$   replaced by arbitrary   $\epsilon > 0$.   (Perhaps something like this was suggested by  Anton Petrunin in his comment above).
By the way, all four mid-points of "edges" of   $ABCD$   will form a set of diameter   $\epsilon$   or less.
