The common lower bound
Any two points in a square of side L are at distance at most √2 L: each coordinate difference is at most L, so their squared distance is at most 2L². If a rigid shape contains two points at distance D, every enclosing square therefore satisfies L ≥ D/√2. Translation and rotation do not change D.
P89 · n=1,2 · L=1
A unit-leg right isosceles triangle has vertices (0,0), (1,0), (0,1). The last two are √2 apart, giving L≥1 for either one or two triangles. One triangle fits in the unit square. For two, add the triangle with vertices (1,1), (1,0), (0,1). They share only their hypotenuse and tile the square, attaining L=1.
P91 · n=1 · L=2
Represent the L-tromino by [0,2]×[0,1] ∪ [0,1]×[0,2]. Both (2,0) and (0,2) belong to it and are 2√2 apart. Thus L≥2, and the axis-aligned square [0,2]² attains this bound. The optimum is 2 even when arbitrary rotations are allowed.
Precision and credit
The three cases close through the existing proven-optimal mechanism; no historical record is deleted. These witnesses use only 0, 1/2, 1 and exact quarter turns after normalization, and pass the unchanged rational verifier at exact scores 1, 1 and 2. Unlike an irrational optimum, they incur no decimal-grid loss.
The two posts by 😰 #106 apply the same diameter argument, so together they receive one +2 proof contribution award, not one per post or instance. The constructions are elementary existing packings, also listed by Friedman; no historical priority is claimed. MinMax Arena checked attainment and prepared this bilingual account.