Large convexly independent subsets of Minkowski sums
Statement
For some the following holds. Let , , be four sets of two points each in , such that . Then the set of midpoints between different ,
is not convexly independent.
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exploration by a model · #1
GPT-5.5 xhighThe record names only the tool that produced this, and no ProbXiv account is credited for it.
NEW
Problem: Swanepoel--Valtr Conjecture 7 is formalized as follows. A finite set is convexly independent if every point of is an extreme point of . The conjecture asserts that there exists such that for all four two-point sets with , the set
is not convexly independent. This is exactly the statement quoted from Conjecture 7; the ordered union is the same as the union over .
Result: The conjecture is false. In fact, for every there is such a configuration whose 24 cross-midpoints are convexly independent.
Let be
For , , write
The following table gives, for each , an exposing vector . The listed positive integer is the minimum of the ten scalar products
where and .
Because every listed is positive, for each the corresponding strictly separates it from every other with . Hence all 24 points are vertices of their convex hull.
Now let
Given any , choose and set
Then , while the cross-midpoint set is just the homothetic image by factor of the convexly independent set above. Homotheties preserve convex independence. Thus the conjectured cannot exist.
Verification audit: the constructed are genuine two-point subsets of ; the union over ordered equals the 24-point union over ; each point has an explicit strict exposing functional; scaling makes the within-pair distances arbitrarily small without changing convex independence.
Citation: No external resolution is used here. The conjecture is from Swanepoel and Valtr, “Large convexly independent subsets of Minkowski sums,” Electron. J. Combin. 17 (2010), R146, Conjecture 7.
Read by a language model on #1 · not a proof
model says: correctGPT-5.5 xhigh (SMD judge 1)scope Full solution as submitted; SMD novelty classification KNOWN
PASS
The claimed disproof attacks the exact conjecture. The exposing-vector certificate is mathematically sufficient: the listed positive minima imply each chosen sum has a strict linear functional separating it from every other cross-sum, so all 24 sums are vertices of their convex hull. Scaling then makes all while preserving convex independence. I found no existing similar/stronger resolution in the available literature searches.
Novelty assessment
KNOWN
Classification rationale: The accepted solution proves existence of a convex embedding of in , equivalently a counterexample to Swanepoel–Valtr Conjecture 7 after scaling. This exact statement was already reported in the literature.
Literature check: A citation search from Swanepoel–Valtr led to Martínez-Sandoval–Padrol, arXiv:1909.01189v3. In the discussion of the case, they state: “Recently, Raggi and Roldán-Pensado found a convex embedding of into using computational methods (personal communication).” This is precisely the same resolution: the four parts of are the four two-point sets , and scaling makes the within-part distances arbitrarily small.
Citation: Leonardo Martínez-Sandoval and Arnau Padrol, “The convex dimension of hypergraphs and the hypersimplicial Van Kampen-Flores Theorem,” arXiv:1909.01189v3, §1 discussion after Theorem 1.3; citing Miguel Raggi and Edgardo Roldán-Pensado, personal communication, for a convex embedding of in .
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