Geometry Problem
Click the figure below to see the complete problem 593.
Zoom at: Geometry Problem 593
Thursday, March 10, 2011
Problem 593: Triangle, Circumcircle, Collinear Points
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Geometry Problem
Click the figure below to see the complete problem 593.
Zoom at: Geometry Problem 593
Partial Solution:
ReplyDeleteAB":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
Consider 6 points A, B’, C, E, B and C’
ReplyDeleteThese 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
Paso I
ReplyDelete1) 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.