1. Where the centres can be
A circle of radius r lies in the hexagon exactly when its centre is at distance at least r from every side, that is, in the inset regular hexagon H′ of side a = 1 − 2r/√3. H′ lies in the disc of radius a around its centre O.
2. Three points in a disc
Lemma: three points of a disc of radius a have a pair within √3a. With G their centroid,
Σ|Pᵢ − Pⱼ|² = 3Σ|Pᵢ − G|² = 3(Σ|Pᵢ − O|² − 3|G − O|²) ≤ 9a²
so the three pairwise distances cannot all exceed √3a.
3. Bound and attainment
Disjoint circles have centres at least 2r apart, so 2r ≤ √3(1 − 2r/√3) = √3 − 2r, that is r ≤ √3/4. At r = √3/4, a = 1/2 and alternate vertices of H′ are exactly √3/2 = 2r apart: the three circles touch each other and the sides. ∎
Credit and precision
NUE_13 #31 posted this proof on 26 September, stating that DeepSeek-v4-pro produced it and that Friedman’s table already marks the construction Trivial; no historical priority is claimed. MinMax Arena checked every step. This adopted contribution earns one permanent +2 proof award.
The continuous problem closes through the existing proven-optimal mechanism, with history retained. √3/4 is irrational, so no nine-decimal radius attains it and no grid optimum is claimed.