1. Two lemmas
Lemma 1. Any four points of the cube span a tetrahedron of volume at most 1/3.
V = |det(B − A, C − A, D − A)| / 6
With B, C, D fixed the determinant is affine in A, so its absolute value is convex and peaks at a vertex of the cube. Replacing A, B, C, D by vertices in turn never decreases the volume (two points moved to the same vertex give volume 0, which does not affect the bound). Of the 70 choices of 4 vertices out of 8, the largest volume is 1/3, attained by (0,0,0), (1,1,0), (1,0,1), (0,1,1).
Lemma 2. Any five points of the cube have a convex hull of volume at most 1/2.
Fix four points S; then f(p) = vol(conv(S ∪ {p})) is convex in p. If conv S is solid, f(p) = vol(conv S) + Σ_F max(0, ℓ_F(p)), where ℓ_F(p) is the signed volume of the cone over the face F of conv S with apex p (positive when p is beyond F), so each term is the positive part of an affine function. If S is planar, f(p) is the area of its hull times the distance from p to that plane, over 3, also convex. So again the five points can be moved to vertices one at a time without shrinking the hull. Of the 56 choices of 5 vertices out of 8, the largest hull has volume 1/2, for example (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,1): a corner tetrahedron of volume 1/6 plus the regular tetrahedron of volume 1/3.
2. The bound
If four points are coplanar, that tetrahedron has volume 0 and m = 0. Assume from here on that no four are coplanar. By Radon’s theorem, five points in space split into two groups whose convex hulls meet; in general position there are only two ways: one point lies inside the tetrahedron of the other four, or the segment joining two points crosses the interior of the triangle of the other three.
Case 1: E lies inside the tetrahedron ABCD. E splits ABCD into the four tetrahedra EBCD, AECD, ABED and ABCE, each of volume at least m, so by Lemma 1
4m ≤ vol(ABCD) ≤ 1/3, m ≤ 1/12 < 1/6.
Case 2: the segment DE crosses the interior of triangle ABC at a point P. Then D and E lie on opposite sides of the plane ABC, and the hull of the five points is the bipyramid DABC ∪ EABC. P divides triangle ABC into PAB, PBC and PCA, and since P lies on the segment DE, the tetrahedron ABDE is exactly DPAB ∪ EPAB, and likewise for the other two. So the three crossing tetrahedra ABDE, BCDE and CADE have disjoint interiors and together fill the whole bipyramid. By Lemma 2:
3m ≤ vol(ABDE) + vol(BCDE) + vol(CADE) = vol(conv) ≤ 1/2, m ≤ 1/6.
In both cases m ≤ 1/6. ∎
3. Attainment
Take A = (0,0,0), B = (1,1,1), C = (0,1,0), D = (1,0,0), E = (0,0,1). Without A the tetrahedron BCDE is regular, with volume 1/3; without any one of B, C, D, E the volume is 1/6. The minimum is 1/6. The coordinates are integers, so the site’s nine-decimal grid attains it exactly, a score of 10²⁷ / (6·10²⁷). The current record (#72, 1 September 2026) already has this value, and its holder stays the same.
Credit and verification
NUE_13 #31 emailed this proof on 5 October 2026, stating that DeepSeek-V4-Pro generated it and that they had not checked every step. MinMax Arena checked it step by step and it holds. We filled in two steps: in Case 2, that D and E lie on opposite sides of plane ABC comes from Radon’s theorem, and Lemma 2’s convexity needs the positive-part decomposition above. The original said the three crossing tetrahedra miss a region near the base; in fact they fill the bipyramid exactly, which leaves the inequality unchanged. We enumerated the vertex cases of both lemmas ourselves. This adopted contribution earns one permanent +2 proof award.
This is the first P72 row with a settled optimum; every other n of the problem in space remains open.