Saturday, March 16, 2013

Problem 863: Circumscribed trapezoid, Inscribed Circle, Tangent, Metric Relations

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

Click the figure below to see the complete problem 863.

Online Geometry Problem 863: Circumscribed trapezoid, Inscribed Circle, Tangent, Metric Relations

4 comments:

  1. EC=CF,GD=FD
    EC//GD ,so EK/KD=EC/GD , or EK/KD=CF/FD therefore KE//EC
    x/EC=DF/DC so, x/2=5/7 ,x=10/7

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  2. CF/CD = x/5
    FD/CD = x/2

    CF/CD + FD/CD = 1
    x/5 + x/2 = 1
    x = 10/7

    ReplyDelete
  3. http://img43.imageshack.us/img43/6426/problem863.png

    Note that Triangle ECK similar to ∆DGK ( case AA)
    We have CF/FD=CE/DG=CK/KG
    So KG//DG and ∆CKF similar to ∆CGD ( case SAS)
    From ∆CKF similar to ∆CGD we have
    DG/FK=5/x=CG/CK=(CK+KG)/CK= 1+KG/CK= 7/2
    So x=10/7

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  4. prof Radu Ion,Sc Bozioru,BuzauMarch 18, 2013 at 12:18 PM

    Fie ABCD un trapez cu bazele AB=a,CD=b in care notam cu O intersectia diagonalelor.Daca notam cu M si N intersectia paralelei prin O la baze cu laturile neparalele avem ca MN este media armonica a bazelor ,MN=2ab/a+b si ON=OM=MN/2 in general,In cazul particular al problemei 863 In trapezulECDG a=2,b=5 si KF=2.2.5/2(2+5)=10/7

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