Monday, December 19, 2011

Problem 706: Triangle, Cevian, Circumcenters, Three Circumcircles, Concyclic Points

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

Click the figure below to see the complete problem 706.

Online Geometry Problem 706: Triangle, Cevian, Three Circumcenters, Circumcircles, Concyclic Points<br />

Posted by Antonio Gutierrez at 3:43 PM
Labels: , , , , , ,

5 comments:

  1. PravinDecember 19, 2011 at 9:15 PM

    ∠EBO =∠OBA-∠EBA =(90°-∠BCA)-(90°-∠BDA)
    =∠BDA-∠BCA=∠DBC=∠EFO (since EF⊥DB and OF⊥CB)
    ∴ E,O,F,B are concyclic

    ReplyDelete
  2. Peter TranDecember 20, 2011 at 7:44 AM

    Note that OE and OF are perpendicular bisectors of AB and AC.
    So angle(EOF) supplement to angle(EOF) and E,O,F,B are concyclic

    ReplyDelete
  3. Sumith PeirisSeptember 1, 2015 at 2:21 AM

    Sorry Peter cannot understand

    ReplyDelete
    Replies
    1. Peter TranFebruary 18, 2016 at 8:50 PM

      To Sumith

      You are right. " OE and OF are perpendicular bisectors of AB and AC." is not enough to conclude that E,O,F,B are cocyclic.
      addition work to show that angle(ABE)= angle(CBF) is required.
      Thanks Sumith

      Delete
  4. c.t.e.oFebruary 19, 2016 at 9:30 AM

    EO perpend AB, OF perp BC =>BE'E + BF'O = 180
    (E' on AB, F' on BC)

    ReplyDelete

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