OPEN - $100 Does every convex polygon have a vertex with no other$4$vertices equidistant from it? Erdős originally conjectured this with no$3$vertices equidistant, but Danzer found a convex polygon on 9 points such that every vertex has three vertices equidistant from it (but this distance depends on the vertex), and Fishburn and Reeds [FiRe92] have found a convex polygon on 20 points such that every vertex has three vertices equidistant from it (and this distance is the same for all vertices). If this fails for$4$, perhaps there is some constant for which it holds? Erdős suggested this as an approach to solve [96]. Indeed, if this problem holds for$k+1$vertices then, by induction, this implies an upper bound of$kn$for [96]. The answer is no if we omit the requirement that the polygon is convex (I thank Boris Alexeev and Dustin Mixon for pointing this out), since for any$d$there are graphs with minimum degree$d$which can be embedded in the plane such that each edge has length one (for example one can take the$d$-dimensional hypercube graph on$2^d\$ vertices). One can then connect the vertices in a cyclic order so that there are no self-intersections and no three consecutive vertices on a line, thus forming a (non-convex) polygon.

Additional thanks to: Boris Alexeev and Dustin Mixon