A metric space *(V,d)* will be called distance regular if for every distances *a>0, b, c* a nonnegative integer *p(a,b,c)* is defined, so that whenever *d(B,C)=a*, there are precisely *p(a,b,c)* points *A* such that *d(A,B)=c, d(A,C)=b*. 


The Euclidean plane is an example: *p(a,b,c)=0,1,* or *2* when the triangle inequality for *a,b,c*, correspondingly, fails, turns into equality, or is strict.


If we also require that *p(a,b,c)>0* whenever the triangle inequality does not fail, then I conjecture that this is the only possibility for the parameters *p(a,b,c)*. That is, there may be many non-isomorphic examples, but the parameters will be the same for all of them. (Thanks to Heather for this clarification.)


Has anybody formulated/proved/refuted this conjecture before? It looks very natural.


UPD. I should have mentioned this: "for every distances *a>0, b, c*" means all nonnegative reals, and the same for "whenever the triangle inequality does not fail". In particular this means that all positive real distances actually occur.


UPD2. After a week trial, the question seems to be new, open, and interesting. Anton suugested a line of attack, and I believe I can write down the proof of the first step: that *p(a,b,c)=1* when the triangle inequality turns into equality. Fedja produced examples showing that this first step is indeed essential.
I'm adding the "open problem" tag to the question.



distance regular metric spaces
Note that any metric with unique infinite geodesics on $\mathbb R^2$ has this property. 
In particular, hyperbolic plane as noted by Heather Macbeth.
It also includes Minkowski plane with smooth ball and all complete negatively curved Riemannian metrics on $\mathbb R^2$. 


So it is better to ask: 


* Is it true that the function $p$ is the only possible? ($p(a,b,c)=0,1,$ or $2$ when the triangle inequality for a,b,c, correspondingly, fails, turns into equality, or is strict.)
* Is it true that space has to be homeomorphic to $\mathbb R^2$?


**Sketch for locally compact case**
You condition implies existence of line (i.e. infinite minimizing geodesic) through any pair of points. 
Such line must be unique; otherwise for some values $p(a,b,a+b)=\infty$ for some values $a,b$. 


If space is locally compact then line depends continuousely on the points. 
For some points $a\_1, a\_2, a\_3, a\_4$ take geodesic $a\_1a\_2$ connect each point to $a\_3$ and connect each point of this trianle to $a\_4$. 
This way we get a cone-map of 3-simplex in your space. 
If it is *nondegenerate* (i.e. the image does not coinsides with image of its boundary) then it would be easy to get a triple of small numbers $a,b,c$ such that $p(a,b,c)=\infty$ thus we arrive to a contradiction. 


Thus the space must be 2-dimensional and with unique infinite geodesics --- I bet it should be homeomorphic to $\mathbb R^2$. Then it follows that $p(a,b,c)=0,1$ or $2$ as defined above.


