TITLE: distance regular metric spaces
QUESTION [12 upvotes]: 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.

REPLY [4 votes]: It all would be nice if not for the following. Using the transfinite induction, you can easily construct a set in $\mathbb R^3$ that intersects each circle by any number of points greater than 2 you want depending on the radius of the circle only. Indeed, the cardinality of the set of all circles is continuum. Now take any complete ordering of this set so that each beginning interval has cardinality less than continuum and start adding points to the least circle that still has too few points on it. Clearly, the points that can overload other circles should lie on some circles with at least 3 points already, but those are fewer than continuum at each step of induction, so they intersect our circle by less than continuum points and there are continuum points on our circle to choose from. So, at each step we can make a good choice.
Needless to say, this can be generalized quite a lot, so I'm not sure if there are any restrictions on $p$ whatsoever as long as we talk about arbitrary metric spaces. Now, if we impose additional restrictions, the situation changes a lot. The minimal restriction to impose is that the space be complete. Can we make nontrivial examples now?