Monday, October 13, 2008

Elearn Geometry Problem 190, Circle

Squares, Distances

See complete Problem 190 at:
www.gogeometry.com/problem/p190_tangent_circle_diameter_perpendicular.htm

Tangent circles, Tangent chord, Perpendicular, Distance. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. DB/2r=EO'/AO'=1/3, x/DB=1/3, so x=2r/9

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  2. angles FDB=DAB, (pependicular sides )=> right triangles AEO'and DFB are similar
    => x/r=DB/3r
    ( DB=4/3r because triangles AEO' and ADB are similar, EO'//DB)
    =>x/r=4r/3/3r => x=4r/9 or 2r/9 (r=1/2 R)

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  3. connects point D and B.
    BD^2 = FB . AB
    (2r/3)^2 = x . 2r
    4r^2/9 = x . 2r
    then x = 2r/9

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  4. Let r=1 for simplicity
    m(O'ED) and m(O'FD) are supplimentary angles => O'EFD are concyclic and m(EO'A)=m(EDF)
    hence triangles AEO',AFD and ADB are similar right triangles
    AE^2=AO*AB (Since AE is tangent to O')
    => AE=Sqrt(2)
    Since AEO' and ADB are similar, we have
    AD=4√2/3

    Similarly AEO' and AFD are similar,
    AF=AE*AD/AO'
    => AF=(√2)*(4√2/3)/(3/2)
    => AF=16/9

    Hence BF=AB-AF=2-16/9=2/9

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