Mathematical and Physical Journal
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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 \ora{AF_1}=\ora{F_1B}=a, \ora{BF_2}=\ora{F_2C}=b, \ora{CF_3}=\ora{F_3D}=c, \ora{DF_4}= \ora{F_4E} = d és \ora{EF_5}=\ora{F_5A}=e. Az a,b,c,d,e vektorok összege 0, hiszen

2a+2b+2c+2d+2e=\ora{AB}+\ora{BC}+\ora{CD}+\ora{DE}+\ora{EA}=0.

Mivel a PF2F3F4 négyszög paralelogramma,

\ora{F_2P}=\ora{F_3F_4}=\ora{F_3D}+\ora{DF_4}=c+d.

Ezért

\ora{F_1P}=\ora{F_1B}+\ora{BF_2}+\ora{F_2P}=a+b+c+d=-e=\ora{AF_5},

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 solutions.

Problems in Mathematics of KöMaL, March 2007