Thursday, January 5, 2017

Geometry Problem 1303: Triangle, 60 Degrees, Angle Bisector, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Geometry Problem 1303: Triangle, 60 Degrees, Angle Bisector, Measurement.

Posted by Antonio Gutierrez at 4:48 AM
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6 comments:

  1. APOSTOLIS MANOLOUDISJanuary 5, 2017 at 7:17 AM

    Problem 1303
    If AD and CE intersect at I then <AIC=90+<ABC/2=120. So <DIC=60=<EBD, then the EBDI is cyclic, or <IDC=<IEB. Let point F on the side AC (no in extension) such that AE=AF.
    Is triangle AEI=triangle AFI so <AEI=<AFI or <IEB=<IFC=<IDC so <FIC=<DIC and <DCI=<FCI.
    Therefore triangle IDC=triangle IFC so CD=CF.Then AE+CD=AF+FC=AC.
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

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  2. c.t.e.o.January 5, 2017 at 8:55 AM

    O, CE meet AD. Draw OK, OL OM, Perpendicular to tr sides
    Tr EOK = tr DOL,=> EK=DL => But AK=AM, CM=CL, =>
    => AK+CL=AM+MC => (AK-KE)+(CL+DL)=AM+MC
    =>AE+CD=AM+MC => AE+CD=AC

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  3. Sumith PeirisJanuary 5, 2017 at 10:58 AM

    Let AD, CE meet at X. Mark F on AC such that AE = AF.

    Tr.s AEX and AFX are congruent SAS and since < AXC = 120 (since < A + < C = 120),
    < EXA = < FXA = < FXC = <DXC = 60.

    Hence Tr.s DXC and FXC are congruent ASA.
    So CD = CF

    So AC = AF + FC = AE + CD

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Stan FulgerJanuary 5, 2017 at 11:00 AM

    Let I the incenter, clearly DI=EI ( from cyclic BEID, with <EBI=<DBI=30) and as constructed, AF=AE.
    If F is reflection of E about AD, it lies onto AC and <AIF=<AIE=60, that is, it remains that <CIF=60=<CID; with EF=IE=ID we infer that triangles IFC and IDC are congruent (s.a.s.), thus FC=DC and we are done.

    Best regards.

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  5. Sumith PeirisJanuary 5, 2017 at 11:20 AM

    Solution 2

    b^2 = a^2 + c^2 - ac ......(1) since < B = 60.

    Now AE + CD = cb/(a+b) + ab/(b+c)
    = b(bc+c^2+a^2+ab)/{(a+b)(b+c)}
    = b(bc+b^2+ac+ab)/{(a+b)(b+c)}
    = b

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Sumith PeirisJanuary 6, 2017 at 1:17 AM

    Solution 3

    If I is the incentre and r the inradius,
    BI = 2r. Let ID = IE =p so DE=√3p.

    Also a2 + c2 –ac = b2 since <B= 60

    Applying Ptolemy to cyclic quadrilateral BDIE,

    p.BE + p.BD = √3p.BI
    So BE + BD = 2√3.r
    c-AE + a-CD = 4√3.r = 4√3∆/s with the usual notation for area and semi-perimeter.

    So AE +CD = a+c - 4√3.(1/2.ac.√3/2)/{(a+b+c)/2}
    = a+c - 3ac/(a+b+c)
    = (a2 + c2 –ac +ab+bc)/(a+b+c)
    = (b2 + ab+bc)/(a+b+c)
    = b

    Sumith Peiris
    Moratuwa
    Sri Lanka

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