Friday, June 28, 2019

Geometry Problem 1440 Intersecting Circles, Perpendicular Bisector, Collinear Points

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

Details: Click on the figure below.

Geometry Problem 1440 Intersecting Circles, Perpendicular Bisector, Collinear Points, iPad apps, Tutoring.
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4 comments:

  1. Sumith PeirisJune 30, 2019 at 2:50 PM

    If AC extended meets circle Q at E’, Tr.s OAC & E’CD are iscoceles and similar

    Then E’M is perpendicular to CD. But EM is also perpendicular to CD

    Hence E and E” coincide and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Gewürtztraminer68December 17, 2019 at 1:38 AM

      Bonus : EM intersects circle (Q) at N.
      Then, by the same method, N, C, B are collinear.

      Delete
  2. Sailendra TJuly 4, 2019 at 1:45 PM

    Extend OQ to meet cirlce Q at F and form the diameter OF.
    Let m(EDF)=x => ECD and OAC are two similar isosceles triangles with angles (2x,90-x,90-x)
    => AC/OC=CD/EC
    =>AC.EC=OC.CD
    Since AE and OD are two chords of Q => A,C,E must be collinear

    ReplyDelete
  3. Ludwig MerckJuly 9, 2019 at 2:54 PM

    If:

    OC=r , OD=a => CM=MD=(a-r)/2 , OM=(a+r)/2

    Define:
    ME=p

    So:
    tg(MEC)=(a-r)/2p => 90-arctg((a-r)/2p)=MCE

    tg(MEO)=(a+r)/2p => 90-arctg((a+r)/2p)=MOE

    We know:
    1.) OAC=OCA and AOC=alpha => OAC=90-alpha/2

    2.) Points A O E give a Tri., so AOE+OEA+OAE=180

    So:
    MEO-MEC+AOC+OAC+MOE=180 => alpha=2*arctg(a-r)/2p=2*MEO

    _____

    90-alpha/2=MCE => OCA=MCE

    So AC and CE are on the line.


    ReplyDelete

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