Tuesday, June 10, 2008

Elearn Geometry Problem 119



See complete Problem 119
Area of Triangles, Incenter, Excircle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. name K point circle I meet BC, M meet AC, N meet AB
    name L point circle E meet BC

    SILH = SICL - SICH
    SILH = IK∙LC - SICH
    SILH = r∙LC - SICH (1)

    SHCF = SICF - SICH
    SHCF = IM∙CF - SICH
    SHCF = r∙CF - SICH
    SHCF = r∙LC - SICH (2) ( CF = LC ,tg from C )

    from (1) & (2)

    SILH = SHCF (3)

    in the same way
    SILG = IBL - SIBG
    SILG = r∙LB - SIBG (4)

    SBGD = IBD - SIBG
    SBGD = r∙BD - SIBG
    SBGD = r∙LB - SIBG (5) (BD = LB, tg from B )

    from (4) & (5)

    SILG = SBGD (6)
    add (3) & (6)

    SILH + SILG = SHCF + SGBD

    SIGH = SHCF + SBGD
    ------------------------------------------

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  2. Let s = semiperimeter of ABC.
    AD = AF = s
    Since area of ABC = rs and since
    (area of AID) = (area of AIF) = (area of ABC)/2
    (area of IDF) =(ADF) - [(ADI) + (AFI)]
    =(ADF) - (ABC)
    =(BCFD)
    Or (IDF) = (BCFD). Now subtract (GHFD) from both sides to complete the proof.

    ReplyDelete
  3. Alternative solution to problem 119.
    Let be p the semiperimeter of ABC and r the radius of its incircle. We have BD = p – c and CF = p – b.
    So S1 = S(DBI) – S(BGI) = (p-c)r/2 – S(BGI) and S2 = S(CFI) – S(CHI) = (p-b)r/2 - S(CHI).
    Besides S3 = S(BCI) – S(BGI) – S(CHI) =
    = ar/2 – S(BGI) – S(CHF).
    Thus
    S1 + S2 = (p-c)r/2 + (p-b)/2 - S(BGI) - S(CHI) =
    = (2p-b-c)r/2 - S(BGI) - S(CHI).
    But 2p – b – c = a, so
    S1 + S2 = ar/2 - S(BGI) - S(CHI) = S3.

    ReplyDelete
  4. Let r = inradius

    Easily BD + CF = BC

    So r.BD/2 + r.CF/2 = r.BC/2

    Hence S(IBD) + S(ICF) = S(IBC)

    So S1+S 2 = S3

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete