Monday, December 5, 2011

Problem 700: Equilateral Triangle, Circle, Circular Segment, Midpoint, Metric Relations

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

Click the figure below to see the complete problem 700.

Online Geometry Problem 700: Equilateral Triangle, Circle, Circular Segment, Midpoint, Metric Relations.
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2 comments:

  1. Peter TranDecember 6, 2011 at 3:19 PM

    http://img37.imageshack.us/img37/9728/problem700.png

    FD cut AB and AO at G and H ( see picture)
    Note that AD=AG=DG=6
    OA= AG*2/√3 =12/√3
    In triangle OHF , applying Pythagoras Theorem we get HF=3. √5
    So x=HF-HD=3(√5-1)
    Peter Tran

    ReplyDelete
  2. PravinDecember 8, 2011 at 8:36 AM

    Extend ED to cut AC at X and circle at Y
    EDXY ∥ AB
    So DE = 6 ; and XY = x (by symmetry)
    ED.DY = AD.DC
    x(6 + x) = 36
    x²+ 6x = 36
    (x + 3)² = 45
    x = 3(√5 - 1)

    ReplyDelete

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