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.

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