Monday, May 19, 2008

Elearn Geometry Problem 61



See complete Problem 61
Triangle, Trisection of Sides. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

3 comments:

  1. hi good day.uhm you had a great project. i just wnt to ask where did u get the values 3/20? thank you

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  2. It can be easily demonstrated that line AMF meeting AC in F1 is a median of the triangle ABC. Likewise AGD and CEH extended are also medians of Tr. ABC.
    Now we apply Menelaus's theorem repetitively to various triangles and their transversals. First we can see that A'M/MA*AF1/F1C*3/1=1 or A'M/MA*1/1*3/1=1 or A'M/MA=1/3 or A'M/AA'=1/4.
    AE/EA"*A"B/BC*CB"/B"A=1 or AE/EA"*2/3*1/2=1 or AE/EA"=3/1. Thus, A'H/HA*1/2*3/1=1 or A'H/HA=2/3 or A'H/AA'=2/5 or AH/AA'=3/5
    Now, HM = AA' - AH - MA' = AA'(1-3/5-1/4)=3/20*AA'(This the genesis of the magic fraction "3/20"). Similarly, it is possible to prove that: MD = 3/20 * CC", DE = 3/20 * BB", EF = 3/20*AA", FG = 3/20*CC', GH = 3/20*BB'.
    Hence, P=(3/20)(AA'+AA"+BB'+BB"+CC'+CC")
    Ajit: ajitathle@gmail.com

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  3. FA" = 4/10 AA"
    EA" = 1/4 AA" ( see P122, P123 )

    EF = 4/10 - 1/4

    EF = 3/20 AA"

    in the same way others

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