Tuesday, February 24, 2009

Problem 256: Equilateral Triangle, Circumcircle, Point, Vertices, Distances

Proposed Problem
Problem 256. Equilateral Triangle, Circumcircle, Point, Vertices, Distances.

See complete Problem 256 at:
gogeometry.com/problem/p256_equilateral_triangle_circumcircle_distance_vertex.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

13 comments:

  1. Use Ptolemy's therem by which in this case we've:
    AD*BC = CA*BD + AB*CD or d*BC = e*CA + f*AB which leads us to d = e + f since AB = BC = CA
    Ajit: ajitathle@gmail.com

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  2. according to ptolemy's theorem, AB.CD + BD.AC = BC.AD. but triangle ABC is equilateral. so AB = BC = CA so we have AD = BD + CD that is d = e + f
    Q. E. D.

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  3. Obviously ∠BDA = ∠ADC = 60°.

    Rotate 60° anti-clockwise about D.
    Let B→B'. Then ΔBDB' is equilateral, and DB' = e.

    Rotate 60° anti-clockwise about B.
    Then B'→D, A→C, so B'A = DC = f.

    As a result, d = DA = DB' + B'A = e + f.

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  4. Problem 256 (solution 1)
    Following on the DB to point B take the point E such that DE=AD (<EDA=<BDA=<BCA=60).Then triangleADE is equilateral .But triangleAEb=triangleADC
    (AE=AD,AB=AC,<BAE=60-<BAD=<DAC ).Therefore AD=DE=BD+BE=BD+DC.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  5. Problem 256 (solution 2)
    Following on the BD to point D take the point F such that DF=DC (<BDA=<CDA=<CDF=60).Then triangleCDF is equilateral .But triangleBCF=triangleADC
    (BC=AC,CF=DC,<ACD=60+<BCD=<BCF ).Therefore AD=BF=BD+DF=BD+DC.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  6. Problem 256 (solution 3)
    On the side AB ( not to extend ) get the point K such that BD=BK.Then triangle BKD is equilateral(<BDK=60=
    <BKD ),so triangleABK=triangleCBD (AB=BC,BK=BD,<ABK=60-KBC=<CBD), so AK=DC. Therefore AD=AK+KD=DC+DB.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  7. Extend CD to F such that BF//AD.
    Then Tr. BDF is equilateral and
    Tr.s ABD ≡ CBF.

    Hence AD = CF = CD + DF so d = e + f

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  8. See the drawing

    CAD intercepts arc CD such as CBD => ∠CAD= ∠CBD
    Define D’ n AD such as AD’C is congruent to BDC
    => AD’=f and ∠ACD’= ∠BCD
    ∠ACD= ∠ACB+ ∠BCD=Π/3+ ∠BCD
    And ∠ACD= ∠ACD’+ ∠D’CD=∠BCD+ ∠D’CD => ∠D’CD = Π/3
    CDA intercepts arc CA such as ABC => ∠CDA= ∠ABC=Π/3
    ∠D’CD = Π/3 and ∠CDA= ∠CDD’= Π/3 => ΔD’CD is equilateral
    =>D’D=CD=e
    d=AD=AD’+D’D therefore d=e+f

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  9. Draw AE such that AE=e and forms a regular trapezium ADBE
    BDC and AEB are congruent => EB=f
    Drop Perpendiculars from B and E on AD and denote them as B' and E'
    =>B'E'=f
    Since m(BCA)=60=> m(BDA)=60=>Tr.BDB' is a 30-60-90 triangle
    =>DB'=E'A=e/2
    AD=d=DB'+B'E'+E'A=e+f

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  10. See diagram here.
    ∆ABE ~ ∆CDE ~ ∆ADB and ∆AEC ~ ∆BED
    Therefore a/AE = f/CE = d/a and a/AE = e/BE
    So a/AE = f/CE = e/BE = (f+e)/(BE+EC) = (f+e)/a = d/ad = e+f

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  11. Here is my solution...
    https://www.youtube.com/watch?v=tB1FLsVUoio

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  12. Let length of the triangle be x
    <ADC=<ACB=60 & <ADB=<ACB=60
    By cosine law, x^2=d^2+f^2-2dfcos60 & x^2=d^2+e^2-2decos60
    so, d^2+f^2-2dfcos60=d^2+e^2-2decos60
    f^2-df=e^2-de
    de-df=e^2-f^2
    d(e-f)=(e+f)(e-f)
    d=e+f

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