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Geometry Problem

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A, B, C be three points on a plane and O be the origin point on this plane. Put a = OA and b = OB, and c = OC, (a,b and c are vectors). P is a point inside the triangle ABC. Suppose that the ratio of the areas of the triangles PAB, PBC and PCA is 2:3:5
    (i) The straight line BP intersects the side AC at the point Q.
    Find AQ:QC

    (ii) Express OP in terms of a, b and c.

    The problem is from the 2009 Math paper (B) for the Japanese Government Scholarship Qualifying Exam. They have a solution and here's the link.
    http://www.studyjapan.go.jp/pdf/questions/09/ga-answers.pdf
    They are in Japanese by the way.


    2. Relevant equations



    3. The attempt at a solution
    For (i) The answer is 2:3, but I don't get it at all. What is the basis on saying that the line BQ drawn by extending BP will divide the areas of the triangle ABQ and BQC in the same ratio as that of ABP and APC? And why is it necessary that if ABQ and BQC are in the same ratio then AQ and QC are in the same ratio as well?

    Using the result from (i) I have the following:
    OP = OB + BP
    From the problem OB = b
    BP = kBQ
    BQ = BA + AQ
    BA = a-b
    AQ = (2/5)AC
    AC = c-a
    So BQ = (a-b)+(2/5)(c-a) = (3/5)a - b + (2/5)C = (1/5)(3a-5b+2c)
    The problem is I don't know what proportion of BQ, BP is. How can I determine the value of k?

    Thanks.
     
  2. jcsd
  3. Mar 5, 2012 #2
    Consider △ABP and △CBP,
    ∵they have the same base BP, and Area of △ABP : Area of △CBP = 2 : 3,
    ∴Height of △ABP : Height of △CBP = 2 : 3
    As △APQ and △CPQ also have the same base PQ,
    AQ : QC = 2 : 3

    Then, Area of △APQ : Area of △CPQ = 2 : 3,
    Consider △ABP and △APQ,
    Area of △ABP : Area of △APQ = 1 : 1
    ∵They have the same height,
    ∴Base of △ABP : Base of △APQ = BP : PQ = 1 : 1
    ∴[itex]\vec{BP}[/itex] = [itex]\frac{1}{2}[/itex][itex]\vec{BQ}[/itex]
     
  4. Mar 5, 2012 #3
    Thanks a lot. I get the first part now. But I still don't get the second part. We only know the two triangles share the same height so how can we deduce that their areas are in the same proportion without knowing the relationship between the heights.

    Sorry, but I really suck in geometry.
     
  5. Mar 6, 2012 #4
    Let Area of △X be AX, Base of △X be bX, Height of △X be hX
    AX = (0.5)(bX)(hX)
    Similarly for △Y, AY = (0.5)(bY)(hY)

    If AX = AY and hX = hY
    bX = bY
     
  6. Mar 6, 2012 #5
    Thanks, I get it now.
     
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