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Piercing intersecting convex sets

Imre Bárány, Travis Dillon, Dömötör Pálvölgyi, Dániel Varga

TL;DR

This work addresses a Helly-type transversal problem for two finite convex families $\mathcal{A}$ and $\mathcal{B}$ in $\mathbb{R}^3$ with $A\cap B\neq\emptyset$ for all pairs; it proves a line transversal result in the special case where $\mathcal{A}$ and $\mathcal{B}$ consist of vertical polygons lying in parallel planes, guaranteeing a line that intersects all sets of one family. It then strengthens the result by showing a location-restricted fractional transversal: under the same setup, there is a line lying in the plane of some $A_i$ hitting $\frac{1}{6}|\mathcal{B}|$ sets in $\mathcal{B}$ (or symmetrically in a $B_j$-plane hitting $\frac{1}{6}|\mathcal{A}|$). The main technique combines affine normalization, a bilinear feasibility framework via Farkas' lemma, dualizing to a non-constructive existence proof, and a secondary fractional Helly argument to obtain a plane-restricted transversal; a higher-dimensional partial extension is outlined. These results advance understanding of Helly-type line transversals for intersecting convex families and provide a framework for constructive computation in fixed dimension.

Abstract

Assume two finite families $\mathcal A$ and $\mathcal B$ of convex sets in $\mathbb{R}^3$ have the property that $A\cap B\ne \emptyset$ for every $A \in \mathcal A$ and $B\in \mathcal B$. Is there a constant $γ>0$ (independent of $\mathcal A$ and $\mathcal B$) such that there is a line intersecting $γ|\mathcal A|$ sets in $\mathcal A$ or $γ|\mathcal B|$ sets in $\mathcal B$? This is an intriguing Helly-type question from a paper by Martínez, Roldan and Rubin. We confirm this in the special case when all sets in $\mathcal A$ lie in parallel planes and all sets in $\mathcal B$ lie in parallel planes; in fact, all sets from one of the two families has a line transversal.

Piercing intersecting convex sets

TL;DR

This work addresses a Helly-type transversal problem for two finite convex families and in with for all pairs; it proves a line transversal result in the special case where and consist of vertical polygons lying in parallel planes, guaranteeing a line that intersects all sets of one family. It then strengthens the result by showing a location-restricted fractional transversal: under the same setup, there is a line lying in the plane of some hitting sets in (or symmetrically in a -plane hitting ). The main technique combines affine normalization, a bilinear feasibility framework via Farkas' lemma, dualizing to a non-constructive existence proof, and a secondary fractional Helly argument to obtain a plane-restricted transversal; a higher-dimensional partial extension is outlined. These results advance understanding of Helly-type line transversals for intersecting convex families and provide a framework for constructive computation in fixed dimension.

Abstract

Assume two finite families and of convex sets in have the property that for every and . Is there a constant (independent of and ) such that there is a line intersecting sets in or sets in ? This is an intriguing Helly-type question from a paper by Martínez, Roldan and Rubin. We confirm this in the special case when all sets in lie in parallel planes and all sets in lie in parallel planes; in fact, all sets from one of the two families has a line transversal.
Paper Structure (5 sections, 6 theorems, 25 equations, 3 figures)

This paper contains 5 sections, 6 theorems, 25 equations, 3 figures.

Key Result

Theorem 1

With the setup of the previous paragraph, either

Figures (3)

  • Figure 1: An illustration of the $n=m=3$ case. The red sets form the family $\mathcal{A}$ and the blue sets form the family $\mathcal{B}$.
  • Figure 2: The quadrangle $Q$, the lines $\ell_A$ and $\ell_B$, and their intersection at $R$.
  • Figure :

Theorems & Definitions (13)

  • Theorem 1
  • proof
  • Theorem 2
  • Remark
  • Lemma 3
  • proof
  • Lemma 4: Fractional Helly for vertical line segments
  • proof : Proof of Theorem \ref{['th:fracH']}
  • Proposition 5
  • proof
  • ...and 3 more