Monday, May 19, 2008

Geometry Problem 96



See complete Problem 96
Similar Triangles, Incenters, Parallelogram. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. join B to O1 & O extend to P ( P on AC )(same bisector)
    extend GO3 to Q ( Q on AC )

    ang APB = 1/2 B + C
    ang MQG = 1/2 G + C ( AB//MF => ang B = ang G )
    =>
    ang APB = ang MQG
    =>
    PB // QO2

    ang DEA = ang C ( DE // AC )
    =>1/2 DEA = 1/2 C
    =>
    O2O1 // CO

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  2. To Antonio:
    The written enunciate of problem 96 is a repetition of problem 94. The figure is all right.

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    Replies
    1. Thanks Nilton. The written enunciate of problem 96 has been updated.

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  3. Alternative solution to problem 96.
    Points O, O1 and B are collinear, on the bisector of ang(ABC). Points O2, G and O3 are collinear, on the bisector of ang(MGC). As triangles ABC, DBE, FGE and MGC are similar,
    then OB/BC = O1B/BE = O2G/EG = O3G/GC = (O1B+O2G+O3G)/(BE+EG+GC).
    But BE + EG + GC = BC, so OB = O1B + O2G + O3G, which is the same as OB – O1B = O2G + O3G. Besides OO1 = OB – O1B = O2G + O3G = O2O3.
    Being ABC and BGC similar, then
    ang(ABO) = ang(MGO3), and, as AB // MR, then
    OO1 // O2O3.
    We can see now that OO1O2O3 is a parallelogram because it has two opposite sides equal and parallel.

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