Abstract. The Caccetta–Häggkvist conjecture (denoted CHC) states that the directed girth (the smallest length of a directed cycle) [Formula: see text] of a directed graph [Formula: see text] on [Formula: see text] vertices is at most [Formula: see text], where [Formula: see text] is the minimum outdegree of [Formula: see text]. We consider a version involving all outdegrees, not merely the minimum one, and prove that if [Formula: see text] does not contain a sink, then [Formula: see text]. In the spirit of a generalization of the CHC to rainbow cycles in [ 1 ], this suggests the conjecture that given nonempty sets [Formula: see text] of edges of [Formula: see text], there exists a rainbow cycle of length at most [Formula: see text]. We prove a bit stronger result when [Formula: see text], thereby strengthening a result of DeVos et al. [ J. Graph Theory, 96 (2021), pp. 192–202]. We prove a logarithmic bound on the rainbow girth in the case that the sets [Formula: see text] are triangles.

We prove that if a family of compact connected sets in the plane has the property that every three members of it are intersected by a line, then there are three lines intersecting all the sets in the family. This answers a question of Eckhoff [ Discrete Comput. Geom., 9 (1993), pp. 203--214], who proved that, under the same condition, there are four lines intersecting all the sets. In fact, we prove a colorful version of this result under weakened conditions on the sets. Three sets $A,B,C$ form a tight triple if $\textrm{conv}(A\cup B)\cap \textrm{conv}(A\cup C)\cap \textrm{conv}(B\cap C)\neq \emptyset.$ This notion was first introduced by Holmsen, who showed that if $\mathcal{F}$ is a family of compact convex sets in the plane in which every three sets form a tight triple, then there is a line intersecting at least $\frac{1}{8}|\mathcal{F}|$ members of $\mathcal{F}$. Here we prove that if $\mathcal{F}_1,\dots,\mathcal{F}_6$ are families of compact connected sets in the plane such that every three sets, chosen from three distinct families $\mathcal{F}_i$, form a tight triple, then there exists $1\le j\le 6$ and three lines intersecting every member of $\mathcal{F}_j$. In particular, this improves $\frac{1}{8}$ to $\frac{1}{3}$ in Holmsen's result.

A family of sets satisfies the $(p,q)$ property if among every $p$ members of it some $q$ intersect. Given a number $0

Let [Formula: see text] be a set of [Formula: see text] lines in the real projective plane in general position. We show that there exists a vertex [Formula: see text] such that [Formula: see text] is positioned in a face of size at most 5 in the arrangement obtained by removing the two lines passing through [Formula: see text].