Monday, May 19, 2008

Elearn Geometry Problem 50



See complete Problem 50
Triangle with Equilateral triangles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1 comment:

  1. 1) ▲DBC = ▲ABE ( BC=BE, AB=BD, DBC=EBA=60+B ) =>
    AE = DC
    2) BAE = BDC (from 1) => *ADBO cyclic*
    => DAO + DBO = 180 => 60+BAO+60+ABO=120+(BAO+ABO)
    =120+(BDO+ABO) => BDO+ABO=60
    ▲DBO => 60+BDO+ABO+BOD=180 =60+60+BOD=180 => BOD=60
    in the same way
    BOE = 60 =>
    =>
    DOE = 120
    3) FH = 1/2 DC, HG = 1/2 AE (as middle line)
    FH = HG ( DC = AE )
    4) FHG = 120 ( angle between parallel line)
    5) HFG = HGF = 60 : 2 = 30 (▲FHG isoceles)
    6) see 2)
    7) OBC = OEC ( have same arc, EBOC cyclic)
    OEC = HGC corresponding angles
    in the same way ADO = ABO = AFH
    8) see 2)
    9) AOB = 120 ( AOB + D = 180, cyclic)
    BOC = 120 in the same way
    10)CMG = HGF = 30 = HFG = HGF as corresponding angles
    11)OB perpendicular to FG
    from ▲MOT ( T , BO meet MN ) GMC = 30, MOB=60 =>
    MTO = 90
    12) DOA = 60 => POQ = 60
    LPN = 60 => OPQ = 60
    => OPQ equaliteral
    13) OPQ = 60, JHP = 60
    => JPH equilateral
    14) from 12) and 13)
    15) JH = HP = HQ + QP = HK + OP
    16) HL = 1/2 HG = 1/4 AE ( LGH = 30, & HG = 1/2 AE )
    17) FL = 1/2 FG
    ▲FLH => FL² = FH² - HL²
    FL² = 1/4 DC² - 1/16 DC² (LH=1/2 FH)
    FL² = 3/16 DC²
    FL = (DC/4)√3 (DC=AE, from 1)
    FL = (AE/4)√3

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