Proposed Problem
See complete Problem 21 at:
gogeometry.com/problem/p021_triangle_orthocenter_tangent.htm
Level: High School, SAT Prep, College geometry
Sunday, April 5, 2009
Problem 21: Acute triangle, Orthocenter, Diameter
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the circles of diameter BC and of diameter AO
ReplyDeleteintersect at the tangency points M,N.
the line MN is their radical line
AOE is a right triangle,E lies on the circle of
diameter AO
the circle power of H with respect to this circle
is P1(H)=-HA.HE
BDC is a rigth triangle,D lies on the circle of diameter BC
the circle power of H with respect of this circle
is P2(H)= -HB.HD
but HA.HE=HB.HD =4R²cosAcosBcosC
H lies on the radical line,
M,H,N are collinear
.-.
Let DF and BC intersect at P. Let the intersection of the tangents at F and B be X, and let the intersection of the tangents at D and C be Y. FC and BD intersect at H, so by Pascal's theorem on FFCBBD and DDFCCB, X,Y,H, and P are collinear. The polars of X and Y are BF and CD and intersect at A. Since X, Y, and H are collinear, the polar of H must concur with the polars of X and Y, that is, the polar of H must pass through A. It follows that H lies on the polar of A, which is line MN, so we are done.
ReplyDeletepascal's theorem is concerned with projective geometry ....... how can i prove it through elementary geometry ??????
ReplyDeleteMANEO is cyclic, AO its diameter. BFHE is cyclic, so AF.AB=AH.AE ( 1 ), AM is tangent to the circle (BMC), and AM^2=AF.AB ( 2 ); with (1) AM^2=AH.AE and AM is tangent to the circle (HEM), so <AMH=<MEA ( 3 ). In a similar way <ANH=<NEA ( 4 ). Since <MAN+<MEN = 180, H belongs to the line MN. Best regards, Stan Fulger
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