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.
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3 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

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  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

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  3. Paso I
    1) Trazamos el triangulo ABC y su circuncirculo.
    2) Sobre el circuncirculo ubicamos 3 puntos arbitrarios A', B', E.
    3) Aplicamos el T.Pascal a las ternas (E,B,A) y (C,A',B',) entonces
    AA' , BB' intersectan en D
    EB' , CA intersectan en B"
    EA' , CB intersectan en A"
    Luego A", D y B" son colineales (recta de Pascal).

    Paso II
    1) Definimos C" y C' a las intersecciones de la recta de Pascal con AB, y de EC" con la circunferencia, respectivamente.
    2) Aplicamos el T.Pascal a las ternas (B',A,C') y (C,E,B) entonces como la recta de Pascal ya esta definida (B"C") y el punto D tambien (interseccion BB' y recta de Pascal) entonces BB' intersecta a CC' en D.
    Finalmente se puede ver que AA', BB' y CC' intersectan en D.

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