Tuesday, January 28, 2014

Geometry Problem 970: Equilateral Triangle, Circumcircle, Secant, Metric Relations

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

Click the figure below to see the complete problem 970.

Online Geometry Problem 970: Equilateral Triangle, Circumcircle, Secant, Metric Relations

Posted by Antonio Gutierrez at 6:45 PM
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4 comments:

  1. Jacob HA (EMK2000)January 28, 2014 at 7:30 PM

    Let BD=y.

    DE×DA=DB×DC
    y(x+y)=12

    In ΔABD, using cosine law,
    36=x²+y²+xy=x²+y(x+y)=x²+12

    x²=24
    x=2√6

    ReplyDelete
  2. Sumith PeirisSeptember 22, 2015 at 12:46 AM

    Let BD = p

    From similar triangles p(p+x) = 2(2+4) = 12...(1)
    From Pythagoras AD^ 2 = 3x^2/4 + (p+x/2)^2 ....(2)

    (2) - (1) ; x^2 = AD^2 - 12 = 24

    so x = sqrt 24 = 2sqrt6

    ReplyDelete
  3. Benjamin LeisNovember 18, 2017 at 2:27 PM

    AEBC is a cyclic quad from that 3 similar triangles fall out: ABE, CDE and ABD. Note angles AEC = CEB = DEB = 60 degrees.

    Consider just ABE and ABD since they're similar from those we get 4/x = x/6
    and x^2 = 24 => x = 2 sqrt(6)

    ReplyDelete
  4. Sailendra TJanuary 27, 2021 at 2:13 PM

    Let BD=y
    As DE.DA=DB.DC
    => 12=y.(x+y)=> y^2+xy=12
    Let F be the foot of perpendicular from A to BC
    AF^2+DF^2=DA^2
    =>3/4x^2+x^2/4+y^2+xy=36
    =>x^2=24

    ReplyDelete

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