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Quadrilateral Locus (Posted on 2012-05-05) Difficulty: 3 of 5


DEFINITION

Let [XYZ] denote the signed area of ΔXYZ ( positive
if the directed angle XYZ is counterclockwise ).

CONSIDER THE FOLLOWING

Let point P be in the plane of ABCD ( a convex quad-
rilateral that is not a parallelogram ). What is the
locus of of points P such that

    [APB] + [CPD] = [BPC] + [DPA] ?

Prove it.

  Submitted by Bractals    
Rating: 4.0000 (1 votes)
Solution: (Hide)


NOTATION:

  PQ denotes the vector from point P to point Q.

   denotes the vector cross product.

CLAIM:

Let E and F be the midpoints of diagonals AC and
BD respectively. Since ABCD is not a parallelogram,
points E and F are distinct. The locus of points P
is the line EF.

PROOF:

       [APB] + [CPD] = [BPC] + [DPA]

  <==> PA ⊗ PB + PC ⊗ PD = PB ⊗ PC + PD ⊗ PA

  <==> PA ⊗ PB + PA ⊗ PD + PC ⊗ PB + PC ⊗ PD = 0

  <==> PA ⊗ (PB + PD) + PC ⊗ (PB + PD) = 0

  <==> (PA + PC) ⊗ (PB + PD) = 0

  <==> (PE + EA + PE + EC) ⊗ (PF + FB + PF + FD) = 0

  <==> PE ⊗ PF = 0

  <==> P lies on line EF.
QED

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionHarry2012-05-08 13:00:49
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