Friday, September 26, 2008

Elearn Geometry Problem 184

Triangle and angles

See complete Problem 184 at:
www.gogeometry.com/problem/p184_triangle_angles_problem.htm

Triangle and angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

12 comments:

  1. Evrenin, the answer x = 32 is not correct, try again. Good luck.

    ReplyDelete
  2. x=30. I have an algebraic proof. Antonio, do you have a geometric one?

    Orange Skid

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  3. Yes, I have a geometric solution.
    Thank you for your question.

    ReplyDelete
  4. Angle ABD=58 which makes angle ABC = 88 and angle ACB =44. AC/AD=sin(x+16)/sin(x) =AC/AB=sin(88)/sin(44) which leads us to a simple equation like: sin(x+16)/sin(x) = 2cos(44) which gives x=30 deg since sin(46)=cos(44). Thus, x=30 deg.
    I still would like to know how this done by plane geometry alone.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  5. Problem 184
    Forming an equilateral triangle ADE (E is above the AC),then AB=AE=AD=ED .But <EAB=4,
    <EBA=(180-4)/2=88=58+30, so<EBD=30. Then B, E,C are collinear.But <EDB=2,<DEC=32, so
    <EAC=44=<ECA ,so EC=AE.Therefore E is the center of the circle triangle ADC so
    <ACD=<AED/2=60/2=30.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  6. Let X be midpoint of BD. Let DZY be perpendicular to BC, Z on BC and Y on AX extended.

    BXZY is concyclic and hence Tr.s DXZ and BDY are both equilateral.
    So BX = DX = XZ = DZ = ZY and hence CD = CY.

    Now mark P on AY such that CP = CY (= CD). So < CYP = < CPY = < CDQ where Q is on AD extended.

    It follows that ADCY is concyclic and x = < AYD = 30

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Sumith: I do not understand "Let DZY be perpendicular to BC, Z on BC and Y on AX extended." Would you kindly elaborate or provide a diagram please.

      Delete
    2. Let me rephrase "Let DZY be perpendicular to BC, Z on BC and Y on AX extended."

      "Drop a perpendicular from D to BC to meet BC at Z. AX and DZ extended meet at Y"

      Trust this clarifies

      Delete
  7. It is special case of problem (t, 3t, 30) where t = 16.

    It is the same problem with following cevian problem (Circumcircle BCD meets AC at P)
    https://output.jsbin.com/fofecum#48,16,14,44

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  8. Sir Mr.Sumith, Please may I ask; kindly explain how are
    "BXZY is concyclic and hence Tr.s DXZ and BDY are both equilateral".
    Thanking you in advance.

    ReplyDelete
    Replies
    1. That BXZY is concyclic (with BY diameter) is clear because Tr.s ABX & ADX are congruent SSS, <BXZ = 90 = <BZD (by construction)

      Now BDZ is a 30-60-90 triangle so DXZ is equilateral

      Now Y lies on the perpendicular bisector of BD and hence BY = DY. But < BDY = 60 so Tr. BDY is equilateral

      Trust this clarifies

      Delete