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.

7 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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  3. Draw CI parallel to AB to intersect AG in I (see graph here :
    https://drive.google.com/file/d/1O5LQpv5h_kdsRi72etJfKxBlp8v3sJem/view?usp=sharing ).
    As O incenter of ΔABC and D, E and F tangency points of the incircle to ΔABC :
    - ΔOCF and ΔOCE are similar with common side OC : CE = CF = e
    - ΔOAD and ΔOAF are similar with common side OA : AD = AF = d
    - ΔOBD and ΔOBE are similar with common side OB : BD = BE and ΔDBE is isocele in B.
    Also : e.(d+e+x) = d.x
    => e.(d+e) = x.(d-e) => x = e.(d+e)/(d-e) QED

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  4. Missing part after "Also" : <CIG = <ADG by construction of I so <CIE = <BDE = <DEB = <CEI therefore ΔECI isocele in C and CI = CE = e.

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  5. Lastly : In ΔADG: CI//AD therefore CI/AD = CG/AG translates into e/d = x/(d+e+x) => e.(d+e+x) = d.x => e.(d+e) = x.(d-e) => x = e.(d+e)/(d-e) QED

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  6. Hello Antonio, what is the max number of characters allowed. My earlier messages were truncated I don't know why ?

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