Thursday, March 10, 2011

Problem 593: Triangle, Circumcircle, Collinear Points

Geometry Problem
Click the figure below to see the complete problem 593.

Problem 593: Triangle, Circumcircle, Collinear Points.
Zoom at: Geometry Problem 593

Posted by Antonio Gutierrez at 8:41 PM
Labels: , , ,

2 comments:

  1. Partial Solution:

    AB":B"C
    = [AEB’]:[CEB”]
    = {AE.EB”.Sin<AEB”}:{CE.EB”.Sin<CEB”}
    = {AE.Sin<ABB’}:{CE Sin<CBB’}
    = (AE:CE).{Sin<ABB’:Sin<CBB’}
    = (AE:CE).{Sin<ABD:Sin<CBD}
    = (AE:CE)(AB.BD.Sin<ABD: BC.BD.Sin<CBD).(BC: AB)
    = (AE: CE)(BC: AB){[ABD]:[BCD]}

    Similarly
    CA":A"B = - (BE:AE)(AC: BC){[BCD]:[ACD]} with due regard to sign and

    BC":C"A = (CF:BF)(AB:AC){[ACD]:[ABD]}

    Multiplying out,
    (AB": B"C)(CA": A"B)(BC": C"A) = -1

    Hence by Converse of Menelau,
    A", B", C" are collinear

    ReplyDelete

  2. Consider 6 points A, B’, C, E, B and C’
    These points are co-cyclic.
    B” is the intersecting point of AC and B’E
    D is the intersecting point of BB’ and CC’
    C” is the intersecting point of AB and EC’
    Per Pascal’s theorem B”, D and C” are collinear
    Per Pravin’s solution above, A”, B” and C” are collinear
    So 4 points A”, B”, C” and D are collinear
    Peter Tran

    ReplyDelete

Subscribe to: Post Comments (Atom)