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 F_i 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 solutionss.

Problems in Mathematics of KöMaL, March 2007

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.

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