Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 10 1 Solution Created 2026-10-03 Updated 2026-10-06
(A) For a finite point set and a family of distinct unit circles in , the Szemerédi–Trotter theorem for unit circles stateswith an absolute constant . Here counts point-circle incidences. Distinctness matters: repeated copies of the same circle are not separate members of this geometric family. A dilation gives the same bound for circles of any one fixed positive radius, with the same constant.
(B) Write and . The comparison is a bound on the cardinality of the distinct-distance set, rather than on the set itself. For each positive distance , take the circles of radius centred at points of . Their incidences between points and curves count exactly the ordered pairs at distance . After dilation by , part (A) bounds this number byEvery ordered pair of distinct points contributes to exactly one of these counts. Thus the unit-circle method for a distinct-distance lower bound givesFor , , soFor the distance set is and the conclusion holds after adjusting the absolute constant; the empty set causes no difficulty.
(C) A direct incidence bound from two-point multiplicity suffices. Put and . Count unordered pairs of distinct points on each curve. By double counting,Writing , this givesThe Cauchy-Schwarz inequality now yieldsIf with , then . ConsequentlyIn fact the argument proves the stronger bound. The two-point multiplicity hypothesis alone controls these incidences between points and curves; the algebraic degree bound is not needed for the requested estimate.
For every positive distance, draw equal-radius circles centred at the points of . Their incidences between points and curves count the ordered pairs at that distance. The Szemerédi–Trotter theorem for unit circles bounds each distance class by . Summing over all classes accounts for pairs, proving .