Monday, May 13, 2013

Problem 874: Isosceles Triangle, Circumcircle, Circumcenter, Congruence

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 874.

Online Geometry Problem 874: Isosceles Triangle, Circumcircle, Circumcenter, Congruence

Posted by Antonio Gutierrez at 2:47 PM
Labels: , , , , ,

5 comments:

  1. nakaMay 13, 2013 at 9:39 PM

    Let the mid point of O1O2 be M, M must lies on BD (the radical axis of two circles)

    Angle(BO1O2)
    =angle(BO1M)
    =angle(BAD)
    =angle(BCD)
    =angle(BO2M)
    =angle(BO2O1)

    So BO1O2 forms an isosceles triangle

    Q.E.D.

    ReplyDelete
  2. UnknownMay 14, 2013 at 1:43 AM

    build:
    O2D
    O2B
    O1B
    O1D
    A=C=a
    BO2D=2a=BO1D
    O2BD=O2DB=O1BD=O1DB=90-a
    O1BO2=O1DO2=180-2a
    BO1DO2 parallelogram
    O2D=O1B
    circumcircle O1 and O2 are congruent
    Q.E.D.

    ReplyDelete
  3. Jacob HA (EMK2000)May 14, 2013 at 8:48 AM

    2×BO1 = BD / sin∠BAC = BD / sin∠BCA = 2×BO2

    The result follows.

    ReplyDelete
  4. Sumith PeirisApril 27, 2016 at 11:06 PM

    < BO1D = 2A hence < ABO1 = 90-A - < ABD

    Similarly < CBO2,= 180 - 2A - < ABD - (90-A) = < ABO1

    So Tr.s ABO1 and BCO2 are congruent ASA and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Sumith PeirisApril 27, 2016 at 11:20 PM

    It also follows that BO1DO2 is a rhombus so BD and O1O2 bisect each other perpendicularly

    ReplyDelete

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