Wednesday, June 26, 2013

Problem 892: Triangle, 60 degrees, Orthocenter, Circumcenter

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demonstration of problem 892.

Online Geometry Problem 892: Triangle, 60 degrees, Orthocenter, Circumcenter. GeoGebra, HTML5 Animation for iPad and more tablets

3 comments:

  1. http://img15.imageshack.us/img15/6545/fyi.png
    let H’ and O’ are point of symmetry of H and O over AC.
    H’ and O’ will be on circle O ….( property of orthocenter and angle AOC= 120)
    Due to symmetric property, O and H will be on circle O’, radius=OO’ ( see sketch)
    We have Angle(AOO’)=angle(AO’O)=60
    And angle(AHO)= ½*angle(AO’O)=30

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  2. Extend EC, EC meet the circle at anoter point, T
    HCF=EBF
    HCF=TBE
    TBE=EBH && BE Orthogonal to HT => TE=EH
    => AH=AT
    ATH=60 => ATH Equilateral triangle
    let M midPoint of AT
    build HM, HM median and Altitude to AT=> O on HT =>
    OH an angle bisector => OHA=30

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  3. Problem 892
    Is <AOC=120=<AHC (<AHC=180-<ABC=180-60=120). Therefore AOHC is concyclic. So
    <AHO=<ACB=(180-120)/2=30.

    ReplyDelete