Complexity of computing kissing numbers of triangles with given side lengths

Question

Question: Given three positive integers $a$, $b$ and $c$ such that the sum of any two of them is bigger than the third, how difficult is it algorithmically to determine the kissing number of triangles with side lengths $a$, $b$ and $c$ -- that is, the number of such triangles in the plane which can simultaneously touch one such triangle in at least one point each?

Remark: On the one hand, it seems quite possible that this can even be done with a bounded number of steps (arithmetic operations, case distinctions etc.) for any choice of $a$, $b$ and $c$, but on the other -- is there an obvious reason for this problem to be algorithmically solvable at all?

Answer 1

Too long for a comment:

I think there is a reason, why this is algorithmically solvable.

Given three side lengths $a,b$ and $c$, and a fixed number $k\in\mathbb{N}$, writing down all the (in)equalities, that prescribe $k$ non-intersecting triangles with side-length $(a,b,c)$ kissing a (fixed) triangle with the same side lengths yields a semi-algebraic set $C_k$; the configuration space of $k$ kissing triangles. For each $k$ it is decidable whether or not $C_k$ is empty (existential theory of the reals). Therefore an algorithm would start with $k=0$, increasing the $k$ until the first $k$ is found such that $C_k$ is empty and then returning $k-1$ as the kissing number. (The fact that this algorithm terminates comes from the fact that there is an upper bound on the kissing number.)