Monday, June 23, 2008

Elearn Geometry Problem 127



See complete Problem 127
Triangle, Centroid, Incenter, Parallel, Proportions. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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3 comments:

  1. AnonymousDecember 15, 2009 at 6:05 AM

    1)[AGC]=[ABC]/3=sr/3=br/2 hence b=(a+c)/2
    2)M midpoint of AC; bisector BI and AC meet at D
    AM=b/2; DC= ab/(a+c)=a/2
    triangles BGI and BMD are similar with the ratio
    2/3
    GI=(2/3)MD= (c-a)/6

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  2. Nilton LapaJune 10, 2012 at 4:22 AM

    Solution to problem 127.
    Let be BS the angle bisector and BM the median. From problem 126 we know that BI/IS = (a+c)/AC. Since GI and MS are parallel, then
    BI/IS = BG/GM = 2,
    so (a+c)/AC = 2 and AC = (a+c)/2.
    As BS is the angle bisector, we have
    c/AS = a/SC = (c-a)/(AS-SC) = (c+a)/(AS+SC) = (c+a)/AC = 2.
    Thus (c-a)/(AS-SC) = 2 (1).
    Besides, AS – SC = AM + MS – SC = MC + MS – SC = MS + SC + MS – SC = 2MS.
    As BGI and BMS are similar, MS = (3GI)/2, then AS – SC = 2.(3GI)/2 = 3GI.
    From (1) we get (c-a)/(3GI) = 2 and GI = (c-1)/6.

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  3. PravinJune 10, 2012 at 10:08 AM

    (GAC) = (IAC)
    ∆/3 = (1/2)(br)= b∆/2s
    3b = 2s
    3b = a + b + c
    b = (a + c)/2
    Extend GI either way to meet BA at D and BC at E.
    DG = GE = DE/2 = 2AC/6 = b/3 and BI/IE = sin C / sin (B/2)
    GI = GE - IE = b/3 - BI sin(B/2)/sin C = b/3 - r/sin C
    = b/3 - ra/h where h is altitude from B
    But r = h/3 since GI∥AC
    So GI = b/3 - a/3 = (b- a)/3 = (2b - 2a)/6 = (c - a)/6

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