Wednesday, October 12, 2011

Problem 676: Circles, Diameter, Tangent, Chord, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 676.

Online Geometry Problem 676: Circles, Diameter, Tangent, Chord, Metric Relations

Posted by Antonio Gutierrez at 8:50 AM
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4 comments:

  1. Peter TranOctober 12, 2011 at 9:39 AM

    http://img18.imageshack.us/img18/5852/problem676.png

    Connect DB, DE and note that AB perpendicular to BC, DB per. To AC, DE per. To AB and EF per. To AC
    We have DB^2=AD.DC= a.b ( relations in right triangle)
    So DB=SQRT(a.b)
    Triangle AEF similar to tri. ABD and triangle AED similar to tri. ABC
    So we have x/DB=AE/AB=AD/AC=a/(a+b)
    And x=a.DB/(a+b)=a.SQRT(a.b)/(a+b)
    Peter Tran

    ReplyDelete
  2. HenkieOctober 13, 2011 at 4:39 AM

    Nice short solution!
    I needed more work :-(

    ReplyDelete
  3. PravinOctober 15, 2011 at 9:43 AM

    x /√(ab) = x / BD = AE / AB = AD / AC = a /(a+b)

    (∵ BD² = AD.DC; EF ∥ BD; DE ∥ CB)

    ReplyDelete
  4. AnonymousNovember 4, 2016 at 6:55 AM

    We have CB _|_ to AB => CB^2 = CD.CA
    => CB = Sqrt(b.(a+b))
    and AB = Sqrt(a.(a+b))

    Connect DE and consider the right triangle ADE. ADE is similar to ACB
    => AE/AB = AD/AC
    => AE = a.Sqrt(a.(a+b))/(a+b)
    => AE = a.Sqrt(a)/Sqrt(a+b)

    Now consider the right triangle AFE which is similar to ABC
    => EF/CB = AE/AC
    => EF = x = [a.Sqrt(a)/Sqrt(a+b)].Sqrt(b.(a+b))/(a+b)
    => x = a.Sqrt(ab)/(a+b)

    ReplyDelete

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