Wednesday, April 16, 2014

Geometry Problem 1005: Triangle, Circle, Circumcenter, Interior and Exterior Angle Bisector, 90 Degree, Perpendicular

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1005.

Online Math: Geometry Problem 1005: Triangle, Circle, Circumcenter, Interior and Exterior Angle Bisector, 90 Degree, Perpendicular

5 comments:

  1. Let BD cut circle O at F
    Draw altitude BH of triangle ABC
    F is the midpoint of arc AC => OF ⊥AC
    ∠(OFB)= ∠(FBH)= ∠(BEH)
    Since ∆(OBF) and ∆(BQE) are isosceles => ∠(OBF)= ∠(QBE)
    So ∠(OBQ)=90 degrees

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  2. <DBE=90, and <EBQ=90-(<A+<B/2)=B/2+90-(<A+<B)=<ABD+<OBA=<OBD, so <OBQ=<DBE

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  3. Let be B,F, intersection points of circles.
    FQB isoceles, QO altitude => QO bisector => QB tg to OB

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  4. Since QD = QB, < QDB = < QBD so < QBC = < BAC

    Hence QB is a tangent to circle ABC at B and so OB has to be perpendicular to QB

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Problem 1005
    Solution 1
    In triangle ABC by the theorem of the internal and external dichotomy we have
    AB/BC=AD/DC=AE/EC. Is <DBE=90 so the Q is midpoint DE. From theorem
    Newton apply QD^2=QC.QA=QB^2 so QB is tangent meets the circle with center O.
    Solution 2
    <CBQ=<CBE-<QBE=(<BAC+<ACB)/2-<QEB=(90-<ABC/2)-(90-<QDB)=<QDB-<ABC/2=<BAC+<ABC/2-<ABC/2=<BAC. So QB is tangent meets the circle with center O.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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