Monday, June 16, 2008

Elearn Geometry Problem 122



See complete Problem 122
Marion Walter's Theorem, Area of triangle and hexagon. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. SAC'F = SFCA" (1)
    SAEB' = SEBA" (2) ( Both: S/3 - common area )

    from (1), and the fact h1 = h/3, h2 = h/2
    (h alt of ABA",h1 of AC'F, h2 of FCA")
    =>
    AF = 6/10 AA", FA" = 4/10 AA"
    SAFC' = SFCA" = 2/15 S

    from (2)
    =>
    AE = 3/4 AA"
    SAFC = 1/3 - 2/15
    SAFC = 1/5 S (3)

    SECA" = 1/4 ∙1/3
    SECA" = 1/12

    SFEC = 2/15 - 1/12
    SFEC = 1/20 (4)

    SAGFEC = 1/20 + 1/5 + 1/20
    SAGFEC = 3/10

    for three sides of ABC

    SAGFEC + SAGHMB + SBMDEC = 9/10

    SFGHMDE = 10/10 - 9/10

    SFGHMDE = 1/10
    -----------------------------------------

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  2. Solution to problem 122.
    Let K be the center of gravity, and AL, BN, CP the medians of triangle ABC. Let S be the area of ABC. Applying Ceva’s theorem to the cevians AA”, BN and CC’, we have
    (AN/NC).(CA”/A”B).(BC’/C’A)=(1/1).(1/2).(2/1)=1.
    Thus F lies on the median BN. Similarly D and H lie on the medians AL and CP, respectively.
    For cevians AA’, BN and CC”, we have
    (AN/NC).(CA’/A’B).(BC”/C”A)=(1/1).(2/1).(1/2)=1,
    so M lies on the median BN, and similarly G and E lie on the medians AL and CP, respectively.
    Now we have the same situation of problem 121, where it was proved that S(KEF) = S/60. The hexagonal central region is the union of six triangles KDE, KEF, KFG, KGH, KHM, and KMD, each one if them with area S/60.
    Hence S1 = 6.(S/60) = S/10.

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