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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