1. Projections
Let θ ∈ [0, π/2] be the angle between the long side and the x-axis. The domino projects onto the axes with lengths 2cosθ + sinθ and 2sinθ + cosθ, and each projection must fit in the side L. The two expressions swap under θ ↦ π/2 − θ, so it suffices to take θ ≤ π/4, where the first is the larger.
2. The minimum over angles
2cosθ + sinθ = √5 · cos(θ − φ), tan φ = 1/2
For θ ∈ [0, π/4] the angle θ − φ stays inside (−π/2, π/2), where cosine is concave, so the minimum is at an endpoint: 2 at θ = 0, and 3/√2 > 2 at θ = π/4. Hence L ≥ 2, and the axis-aligned domino in the 2 × 2 square attains it. ∎
Credit and precision
LittleChasa #135 submitted the projections and the conclusion on 25 September; MinMax Arena supplied the concavity step that locates the minimum. This adopted contribution earns one permanent +2 proof award. The optimum 2 is attained exactly by the existing certificate; the instance closes through the existing proven-optimal mechanism, with history retained.