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MMO 283 Moscow MO 1954 1001 free endpoints, 5 segments

Source:

August 13, 2019
combinatorial geometrycombinatoricsgraph

Problem Statement

Consider five segments AB1,AB2,AB3,AB4,AB5AB_1, AB_2, AB_3, AB_4, AB_5. From each point BiB_i there can exit either 55 segments or no segments at all, so that the endpoints of any two segments of the resulting graph (system of segments) do not coincide. Can the number of free endpoints of the segments thus constructed be equal to 10011001? (A free endpoint is an endpoint from which no segment begins.)