Monday, May 19, 2008

Elearn Geometry Problem 64



See complete Problem 64
Triangle, Incircle, Transversal. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. By Menelaus's theorem, AG/GC*CE/EB*BD/DA = -1
    Remembering that CE=CF=e,DB=BE and FA=AD=d. we've (x+d+e)/x * e/EB *EB/d = 1. Hence,
    (x+d+e)/x * e/d = 1 or x = -e^2-ed/(e-d)
    = e(d+e)/(d-e)
    QED.
    Ajit: ajitathle@gmail.com

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  2. We see that AD = d and that CE =e.

    Also :

    ∠BDE = ∠BED = ∠GEC= t

    and

    ∠ADG=180-t
    ∠CGE = u

    than we use the sine rule on triangle ECG :

    sin(t)/x = sin(u)/e
    sin(t)/sin(u) = x/e (1)

    and using the sine rule on triangle ADG and the fact that sin(180-t)=sin (t)

    sin(180-t)/(x+d+e) = sin(u)/d
    sin(t)/sin(u) = (x+d+e)/d (2)

    From equation 1 and 2 we get :

    x/e = (x+d+e)/d
    xd = xe+e(d+e)
    x(d-e) = e(d+e)

    x = e(d+e)/(d-e)


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