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Level C · Reviewed Geometry P-dolnikov-piercing-conjecture

Dol'nikov's conjecture on piercing translates in the plane

Three finite families of translates of a planar convex body, with every two sets from different families intersecting: can one of the families always be pierced by 3 points?

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@misc{cairn-dolnikov-piercing-conjecture,
  title        = {Dol'nikov's conjecture on piercing translates in the plane},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/dolnikov-piercing-conjecture}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY-SA 4.0. Accessed 2026-10-04}
}

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The problem

The question

Let K be a compact convex set in R², and let F₁, F₂, F₃ be finite families of translates of K. Suppose that for all i ≠ j, every A ∈ F_i intersects every B ∈ F_j. Conjecture (Dol'nikov): some family F_j can be pierced by 3 points (3 points such that every member of F_j contains one of them).

What is known

Proved when K is centrally symmetric or a triangle (Jerónimo-Castro, Magazinov and Soberón, 2015), with stronger results for discs; Gómez-Navarro and Roldán-Pensado extended the class of bodies and proved a version with more piercing points.

What counts as progress

  • Proofs for further classes of convex bodies, or with 3 points in a more general setting.
  • A computer search for counterexamples among polygons K (piercing numbers of finite families are computable).

Source. Posed by Edgardo Roldán-Pensado in an extended abstract of the Oberwolfach workshop Discrete Geometry (2024), recorded in Oberwolfach Reports 3/2024, p. 172 (EMS Press, DOI 10.4171/OWR/2024/3), licensed under CC BY-SA 4.0. This page summarises the problem in our own words; as an adaptation it is shared under CC BY-SA 4.0 as well.