In the plane seven different points P1,P2,P3,P4,Q1,Q2,Q3 are given. The points P1,P2,P3,P4 are on the straight line p, the points Q1,Q2,Q3 are not on p. By each of the three points Q1,Q2,Q3 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,Q3 are taken in account as intersections. combinatorics unsolvedcombinatorics