Tuesday, June 14, 2011

Fox 342

Ahmet Arduç submitted this one. Geometric solutions will be appreciated.

http://www.8foxes.com/

4 comments:

  1. m(DOC)=2a, then m(FAE)=a, m(FEA)=90-a
    M is midpoint of DC, OM perpendicular DC. m(ODM)=90-a, m(DOM)=a
    then
    AFE=OMD

    AE=AO

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  2. [URL=http://imageshack.us/photo/my-images/535/foxgeometry.png/][IMG]http://img535.imageshack.us/img535/5163/foxgeometry.png[/IMG][/URL]

    The two shaded right-angled triangles are congruent.
    .: AE = OD = AO (radius of semicircle)

    jyu

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  3. http://imageshack.us/photo/my-images/535/foxgeometry.png/

    The two shaded right-angled triangles are congruent.
    .: AE = OD = AO (radius of semicircle)

    jyu

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  4. Draw DB intersecting AC at P.

    Inscribed angles PCD and PBA are equal, and vertical angles at P are equal making ∆ PCD ~ ∆ PBA.
    Then (PD/PA) = (DC/AB) = (2*EF/AB) = (EF/(1/2)AB) = (EF/AO).

    Also, since Angle ADB is a right angle, ∆ ADP ~ ∆ AFE.
    Then (PD/PA) = (EF/AE).

    Combining the two equations gives (EF/AO) = (EF/AE) or AO = AE.

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