Friday, January 31, 2014

Geometry Problem 972: Equilateral Triangle, Vertices, Three Parallel, Equal Circles, Perpendicular Bisector

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

Click the figure below to see the complete problem 972.

Online Geometry Problem 972: Equilateral Triangle, Vertices, Three Parallel, Equal Circles, Perpendicular Bisector

Posted by Antonio Gutierrez at 9:53 AM
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3 comments:

  1. Ivan BazarovFebruary 1, 2014 at 1:42 PM

    D is circumcenter of CAO, and QA=QO, so DQ is perpendicular to AO and parallel to CA. Then <ODQ=<OCA, but <OEQ=<ODQ from cyclic quadrilateral OQDE. If L2 meets AC at P, then PCEO is cyclic because <OCA=<OEQ. <AOQ=<CPE=<COE=60

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  2. Sumith PeirisMarch 23, 2016 at 3:10 AM

    DA = DO and Tr. AQO is equilateral so < OQM = 30 since Tr.s AQD and OQD are congruent (DQ, AO meet at M)

    Since OQDE is cyclic therefore < DEO = < OQM = 30 = < CED

    Hence CEO is equilateral

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. APOSTOLIS MANOLOUDISAugust 31, 2016 at 12:42 AM

    Problem 972
    Is the triangle AOQ equilateral.Let AC intersects the line L_2 and the circle Q in points F, C respectively.Then AFQO is cyclic(<FAO=90=<FQO, FA=FQ tangent parts), with <CFE=60=<QOA. Also the points G,Q, and O are collinear (<ΟΑΓ=90 ,ΓΟ is diameter) and
    OQ=OG.But OD=DC so QD//GC or <FCO=<QDO. Is <OQE=90=<ODE so OQDE is cyclic.
    Then <QDO=<QEO.So FCO=<FEO.Therefore OFCE is cyclic .So <COE=<CFE=60=<OCE.
    Therefore triangle OCE is equilateral.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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