We study the following problem: Given a set $S$ of $n$ points in the plane, how many edge-disjoint plane straight-line spanning paths of $S$ can one draw? A well known result is that when the $n$ points are in convex position, $\lfloor n/2\rfloor$ such paths always exist, but when the points of $S$ are in general position the only known construction gives rise to two edge-disjoint plane straight-line spanning paths. In this paper, we show that for any set $S$ of at least ten points, no three of which are collinear, one can draw at least three edge-disjoint plane straight-line spanning paths of~$S$. Our proof is based on a structural theorem on halving lines of point configurations and a strengthening of the theorem about two spanning paths, which we find interesting in its own right: if $S$ has at least six points, and we prescribe any two points on the boundary of its convex hull, then the set contains two edge-disjoint plane spanning paths starting at the prescribed points.

翻译：我们研究以下问题：给定平面上的 $n$ 个点构成的集合 $S$，其中可以画出多少条边不相交的平面直线生成路径？一个众所周知的结果是，当 $n$ 个点处于凸位置时，总是存在 $\lfloor n/2\rfloor$ 条这样的路径；但当 $S$ 中的点处于一般位置时，已知的构造仅能给出两条边不相交的平面直线生成路径。在本文中，我们证明：对于任意至少包含十个点且无三点共线的点集 $S$，至少可以画出三条边不相交的平面直线生成路径。我们的证明基于关于点配置平分线的结构定理，以及对两条生成路径定理的加强结果，后者本身也值得关注：如果 $S$ 至少包含六个点，并且我们任意指定其凸包边界上的两个点，则该点集包含两条边不相交的平面生成路径，且这两条路径分别以指定点为起点。