Wednesday, June 1, 2011

Problem 611: Altitude of a Triangle, Perpendicular, Collinear Points, Parallel

Geometry Problem
Click the figure below to see the complete problem 611.

Problem 611: Altitude of a Triangle, Perpendicular, Collinear Points, Parallel

2 comments:

  1. Analytic Proof:
    Let D = (o,o), A = (a,0), C = (c,0), B = (0,b)
    The equations of AB, DEM are respectively
    bx + ay = ab and ax-by = 0
    Solving, we have
    E = [ab²/(a² + b²) , a²b/(a² + b²)]
    Similarly,
    F = [b²c/ (b² + c²), bc²/ (b² + c²)]
    It can be verified that
    the equation of EF is
    b (a + c)x – (b² - ac)y = abc
    Hence
    G = [0, abc/ (ac - b²)] and
    so the equation of CG is
    abx + (ac - b²)y =abc
    Solving it with the equation
    ax-by = 0 of DE
    we obtain (eliminating x)
    y-coord of M = b = y-coord of B
    M and B are at the same height over AC
    Hence BM is parallel to AC
    Similarly BH is parallel to AC
    Thus M, B, H are collinear

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  2. http://img840.imageshack.us/img840/7563/problem611.png

    Draw circle diameter BD ( see attached sketch)
    From B draw a line parallel to AC . This line cut DE and DF at M and H
    AH cut BD at G and MC cut BD at G’ .
    By the properties of right triangles we have BH.DC=BD^2= BM.AD
    So BM/DC=BH/AD
    ∆MBG similar to ∆ CDG => BG’/DG’= BM/DC
    ∆HBG similar to ∆ADG => BG/DG= BH/AD
    From above expressions we have BG’/DG’= BG/DG => G coincide to G’
    M, B and H are collinear and BH // AC

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