\(s (10)\)$s(10)$ — solved

\(s(10) = 3 + \tfrac{1}{2}\sqrt{2} \approx 3.70710678\)$s(10)=3+\frac{1}{2}\sqrt{2}\approx 3.70710678$. Stromquist’s Memo II (October 15, 1984) gives the proof; his published Theorem 1 appeared in 2003. This is the immediate predecessor of the open case at \(n = 11\)$n=11$. The proof route summarized below is the published presentation.

The focused renderer gallery retains a shared-scale source-return diagnostic. It tests the refiner against the proved Göbel geometry; it is not the proof of \(s(10)\)$s(10)$.

The proof, and the shape it shares with Theorem 2

Stromquist defines ten points \(A, \ldots, J\)$A,\dots ,J$ in the container, with \(A = (1, 1)\)$A=(1,1)$, \(B = (.97, s/2)\)$B=(\mathrm{.97},s/2)$ and the rest placed symmetrically, and shows each region of the figure is covered by one of his nonavoidance Lemmas 1, 2 or 4. That makes the ten points genuinely unavoidable, so ten packed boxes must contain exactly one each — and the proof then names the boxes for their points and closes by case analysis. It does not finish by pigeonhole alone.

The 45° tilt, and why 11 is different

Like \(n = 5\)$n=5$, the optimal packing here uses a 45° tilt, and \(s(10)\)$s(10)$ is again degree 2. Gardner’s conjecture was precisely that \(n = 11\)$n=11$ is the first case where 45° is not enough — and Stromquist settled it in the same paper by bounding the 0°/45° class for \(n = 11\)$n=11$ from below at \(2 + (4/3)\sqrt{2} \approx 3.885618\)$2+(4/3)\sqrt{2}\approx 3.885618$, which Trump’s obliquely tilted packing beats at \(3.877084\)$3.877084$. Thus \(n = 11\)$n=11$ is the first failure of the \(0^\circ\)$0^{\circ }$/\(45 ^\circ\)${45}^{\circ }$ restriction.

Verification Code

The programs behind this case’s verified bounds, by their evidence. The code column says how the code that ran stands to the code its producer used. VERIFIERS.md says what each program is and whose it is.