Saturday, February 28, 2015

Geometry Problem 1090. Circle, Diameters, Chord, Perpendicular, Metric Relations

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1090. Circle, Diameters, Chord, Perpendicular, Metric Relations.

Posted by Antonio Gutierrez at 8:32 AM
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11 comments:

  1. Peter TranFebruary 28, 2015 at 1:07 PM

    Let y= FB
    We have Triangle AOE similar to AFB => FA= 2.y
    We have y^2+4y^2= 16 => y= 4/sqrt(5)
    Applying Ptolemy's theorem in quad. ADBD we have
    AB.FD=DB.AF+FB.AD
    4.x= 2.sqrt(2).( 2y +y)
    So x= 3.y.sqrt(2)/2
    Replace value of y in above we have x= 12.sqrt(10)/5

    ReplyDelete
    Replies
    1. Peter TranFebruary 28, 2015 at 8:48 PM

      Thanks Antonio
      Yes, The correct answer is 6.sqrt(10)/5.
      the error is in the last step

      Peter

      Delete
  2. Jacob HA (EMK2000)February 28, 2015 at 6:07 PM

    cos∠FOD = -sin∠FOB
    = -2 sin∠FAB cos∠FAB
    = -2×1/√5×2/√5
    = -4/5

    In ΔFOD,
    x² = 2²+2²-2×2×2cos∠FOD = 72/5
    x = 6√10 / 5

    ReplyDelete
  3. ion raduMarch 1, 2015 at 9:18 AM

    Observam ca AE=√5 (TP in trAOE) apoi CE.ED=FE.EA<=>1.3=√5.FE,teorema medianei in tr CFO ne da relatiaFE^2=[2(FO^2+FC^2)-CO^2]/4 si in final TP in tr CFD =>FO^2=CD^2-CF^2 =>x=4√21/5

    ReplyDelete
    Replies
    1. ion raduMarch 2, 2015 at 11:06 AM

      Observam ca AE=√5 (TP in trAOE) apoi CE.ED=FE.EA<=>1.3=√5.FE,teorema medianei in tr CFO ne da relatiaFE^2=[2(FO^2+FC^2)-CO^2]/4 =>FE^2=8/5si in final TP in tr CFD =>FO^2=CD^2-CF^2 =>FO^2=16-8/5=>x=6√10 / 5

      Delete
    2. Antonio GutierrezMarch 2, 2015 at 12:45 PM

      Thanks Ion, the second is the right answer.

      Delete
  4. PravinMarch 1, 2015 at 4:32 PM

    Draw FG perp to CD.
    Triangles EFG, EAO are similar.
    So FG/AO = GE/EO =EF/AE.
    (i.e.) FG/2 = GE/1 = EF/√5.
    Note AE.EF = DE.EC = 3*1 = 3.
    (i.e.) √5*EF=3, EF = 3/√5.
    So FG=6/5, GE=3/5, DG = 3 + 3/5 =18/5
    So x^2 = DG^2 + FG^2
    = 324/25 + 36/25 = 360/25.
    Hence x = 6√10 / 5

    ReplyDelete
  5. PravinMarch 2, 2015 at 8:46 AM

    Choose O as origin, lines AOB, COD as coordinate axes.
    A=(-2,0),E=(0,1) imply equation of AE is x - 2y + 2=0.
    Eq of the circle is x^2 + y^2 = 4
    Solving these two equations, [x = 0, y = - 2] or [x = 6/5, y = 8/5]
    So F=(6/5 ,8/5). Already D = (0,-2)
    Hence DF^2 = 36/5 + 324/25 = 360/25, DF = 6√10 / 5

    ReplyDelete
  6. Sumith PeirisNovember 16, 2015 at 9:48 PM

    CF^2 = 16-x^2, AE • sqrt 5 and AD = sqrt 8

    Tr.s ECF and AED are similar so

    (16-x^2)/8 • 1/5 from which x • 6/5sqrt10

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Stan FulgerJanuary 1, 2016 at 12:24 AM

    Remark only: If G is the intersection of DF and AB, then S[DEG]=S[AEC].

    Happy New Year everybody!

    ReplyDelete
  8. MarcoMay 11, 2024 at 10:53 PM

    OE=1, OA=2, AE=sqrt5
    cos<OAE=2/sqrt5
    <OEF=180-(90-<OAE)=90+<OAE
    <AFD=45 & DE=3

    Consider triangle DEF
    sin<DEF/x=sin45/3
    sin(90+<OAE)/x=sin45/3
    cos<OAE/x=sin45/3
    x=3cos<OAE/sin45
    x=6sqrt10/5

    ReplyDelete

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