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Maximum number of intersections of all perpendiculars to 286

Source: AustriaMO1970 problem 2

June 29, 2011
combinatorics unsolvedcombinatorics

Problem Statement

In the plane seven different points P1,P2,P3,P4,Q1,Q2,Q3P_1, P_2, P_3, P_4, Q_1, Q_2, Q_3 are given. The points P1,P2,P3,P4P_1, P_2, P_3, P_4 are on the straight line pp, the points Q1,Q2,Q3Q_1, Q_2, Q_3 are not on pp. By each of the three points Q1,Q2,Q3Q_1, Q_2, Q_3 the perpendiculars are drawn on the straight lines connecting points different of them. Prove that the maximum's number of the possibles intersections of all perpendiculars is to 286, if the points Q1,Q2,Q3Q_1, Q_2, Q_3 are taken in account as intersections.
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