Erd\H{o}s originally conjectured this (in \cite{Er46b}) with no $3$ vertices equidistant, but Danzer found a {IMAGE=97-Danzer,convex polygon} on 9 points such that every vertex has three vertices equidistant from it (but this distance depends on the vertex). Danzer's construction is explained in \cite{Er87b}. Fishburn and Reeds \cite{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? In \cite{Er75f} Erd\H{o}s claimed that Danzer proved that this false for every constant - in fact, for any $k$ there is a convex polygon such that every vertex has $k$ vertices equidistant from it. Since this claim was not repeated in later papers, presumably Erd\H{o}s was mistaken here. Erd\H{o}s 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. References [Er46b] Erd\H{o}s, P., On sets of distances of {$n$} points. Amer. Math. Monthly (1946), 248--250. [Er75f] Erd\H{o}s, Paul, On some problems of elementary and combinatorial geometry. Ann. Mat. Pura Appl. (4) (1975), 99-108. [Er87b] Erd\H{o}s, P., Some combinatorial and metric problems in geometry. Intuitive geometry (Si\'{o}fok, 1985) (1987), 167-177. [FiRe92] Fishburn, P. C. and Reeds, J. A., Unit distances between vertices of a convex polygon. Comput. Geom. (1992), 81-91.