Saturday, December 31, 2011

Problem 710: Parallelogram, Midpoint, Diagonal, Collinear Points, Metric Relations, Triangle

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 710 details.

Online Geometry Problem 710: Parallelogram, Midpoint, Diagonal, Collinear Points, Metric Relations, Triangle.
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2 comments:

  1. Peter TranDecember 31, 2011 at 2:13 PM

    1. AC bisect BD , CO is a median of triangle BCD
    So CD will pass through center of gravity N of triangle BCD and A,O,P,N, C are collinear and EF//BD
    2. Solution of problem 709 showed that AG/GE=AH/AF = 4 so GH//EF//BD
    In triangle BCD, since O is midpoint of BD so P is the midpoint of MQ
    In triangle BND, since O is midpoint of BD so P is the midpoint of GH
    So PG=PH=HQ=GM
    3. see item 2 above
    4. in quadrilateral MBDR, BD//MR and BM//DR so MQDR is a parallelogram and BD=MR

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  2. Peter TranJanuary 2, 2012 at 12:13 PM

    Additional clarifications for question 2 ( MG=GP=PH=HQ=QR )
    Let BD=x and BD cut AE and AF at S and T
    1. In tri. ABC , S is the centroid so BS=2/3 BO=x/3
    Similarly DT=x/3
    2. In tri. BND , since NO is a median so P is the midpoint of GH
    Similarly in tri. BCD , P is the midpoint of MQ
    3. Tri BGS~Tri. FGE so BG/GF=BS/EF=2/3  BG/BF=2/5  GM=x/5
    FG/GB=3/2  FG/FB=3/5 = GH/BT GH=3/5* 2x/3=2x/5  PQ=1/2GH=x/5=PH
    PQ=PM=GM+1/2GP=2x/5 HQ=PQ-1/GH=x/5
    4. BDRM is a parallelogram  MR=BD=x  QR=MR-MQ=x/5
    So MG=GP=PH=HQ=QR=x/5

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