Monday, December 5, 2011

Problem 699: 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 699.

Online Geometry Problem 699: Equilateral Triangle, Circle, Circular Segment, Midpoint, Metric Relations.
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4 comments:

  1. Peter TranDecember 5, 2011 at 5:43 PM

    http://img821.imageshack.us/img821/6871/problem699.png
    BD cut EF at G ( see picture)
    Note that DG=OG (angle ABC=60)
    Triangle ABG similar to tri. EDG with ratio =BG/DG=3
    So x= AB/3= 4
    Peter Tran

    ReplyDelete
  2. PravinDecember 7, 2011 at 9:41 AM

    DAC = 30, AED = 120 imply ADE = 30
    So AE = ED = x
    FC = DF = x (by symmetry)
    So x = AE = EF = FC = AC/3 = 4

    ReplyDelete
  3. MarcoJanuary 6, 2024 at 11:51 PM

    Let radius of the circle be r,
    r^2+r^2-2r^2cos120=144
    r=4sqrt3
    BD perpendicular to AC and intersects at H
    BH=sqrt(12^2-6^2)=6sqrt3
    DH=DB-BH-2r-6sqrt3=2sqrt3
    x^2=EH^2+DH^2
    x^2=4+12
    x=4

    ReplyDelete
  4. Sumith PeirisJanuary 12, 2024 at 6:53 AM

    < ADC = 120
    Since D is the mid point of arc AC, < CAD = < ACD = 30

    Since < DEF = 60, AE = DE = x
    Similarly CF = x

    So 3x = 12 and x = 4

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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