Problem B. 3988. (March 2007)
B. 3988. The midpoints of the sides of a convex pentagon are F1, F2, F3, F4, F5, in this order. (The vertex A is between F5 and F1.) Let P be the point in the plane for which the quadrilateral PF2F3F4 is a parallelogram. Prove that the quadrilateral PF5AF1 is also a parallelogram.
(3 pont)
Deadline expired on April 16, 2007.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás: Legyenek az ötszög csúcsai az Fi pontoknak megfelelő körüljárási sorrendben A,B,C, D,E. Legyen továbbá , , , és . Az a,b,c,d,e vektorok összege 0, hiszen
Mivel a PF2F3F4 négyszög paralelogramma,
Ezért
tehát a PF5AF1 négyszög is paralelogramma.
Statistics:
160 students sent a solution. 3 points: 128 students. 2 points: 15 students. 1 point: 1 student. 0 point: 4 students. Unfair, not evaluated: 12 solutionss.
Problems in Mathematics of KöMaL, March 2007