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Math Help - hard question!

  1. #1
    Junior Member
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    Exclamation really hard! need help plz!

    I am desperate to solve this question :

    Given 4 points A,B,C,D on a straight line, show how to construct a point P on the line such that PA.PD=PB.PC

    The theorems that can be applied concerns intersecting chords and tangents
    Last edited by calculus_jy; April 30th 2007 at 10:02 PM.
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  2. #2
    Super Member malaygoel's Avatar
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    I attempted to solve d problem and I think I got it for the case when AB and CD are not equal.

    Construction Steps:
    (1) Draw perpendicular bisector of AD. Mark any arbitrary point M on it.
    (2) Draw perpendicular bisector of BC. Mark N on it such that NB=AM. N and M are on same side of the line.
    (3) Draw perpendicular bisector of MN. The point of intersection of this bisector with the original line gives P.

    Someone please check.
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  3. #3
    Super Member
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    Quote Originally Posted by calculus_jy View Post
    I am desperate to solve this question :

    Given 4 points A,B,C,D on a straight line, show how to construct a point P on the line such that PA.PD=PB.PC

    The theorems that can be applied concerns intersecting chords and tangents
    Not sure what is intended by: PA.PD=PB.PC

    However,
    If the object is to find some point P,
    such that the distance PA times distance PD
    is equal to
    the distance PB times distance PC, then

    Assign x to the distance PA.
    Assign b to the distance AB
    assign c to the distance AC
    assign d to the distance AD

    Code:
     
    ..P....................A.......B............C.......D...
    ..|<--------x--------->|<--b-->|............|.......|...
    ..|<--------x--------->|<--------------c--->|.......|...
    ..|<--------x--------->|<-----------------------d-->|...
    x(x+d) = (x+b)(x+c)


    x^2 + dx = x^2 + bx + cx + bc


    which results in


    x = \frac{bc}{d-b-c}



    for some random points along the x-axis
    A = 37.97089586895891
    B = 41.26128066028468
    C = 46.5861084072385
    D = 54.18555800267495

    From the above
    x= 6.57854211961539

    P = A-x = 31.39235374934351

    The distances:
    PA = 6.57854211961539 = x
    PB = 9.86892691094117 = x+b
    PC = 15.1937546578949 = x+c
    PD = 22.7932042533314 = x+d

    PA*PD= 149.9460542215376
    PB*PC= 149.9460542215376

    There also exists a point P to the right of the A above.

    The problem has a solution.

    Are you asking how to construct a solution graphically?
    Is that the question?
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by aidan View Post
    x = \frac{bc}{d-b-c}

    Are you asking how to construct a solution graphically?
    Is that the question?
    I think the question is to find the point graphically(using geometry). Also, what if
    d=b+c

    PS: I solved for the case d=b+c in next post.
    Last edited by malaygoel; June 14th 2009 at 04:33 AM.
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  5. #5
    Super Member malaygoel's Avatar
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    the solution that I posted above was valid when AB is not equal to CD. This solution is valid when AB=CD.

    Construction Steps:
    (1)Draw perpendicular bisector of BC. Mark any arbitrary point "O" on it.

    Let OA=R
    and OB=r

    (2) Draw a right angled triangle LMN, right angled at N such that MN=R and LN=r.

    (3) Draw a isosceles right angled triangle LMK, right angled at K.

    let MK=t

    (4) Draw a circle with "O" as centre and radius=t. The points of intersection of the circle with the given line give two possible positions of P.
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