## Sunday, April 5, 2009

### Problem 21: Acute triangle, Orthocenter, Diameter

Proposed Problem See complete Problem 21 at:
gogeometry.com/problem/p021_triangle_orthocenter_tangent.htm

Level: High School, SAT Prep, College geometry

1. the circles of diameter BC and of diameter AO
intersect 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
.-.

2. 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.

3. pascal's theorem is concerned with projective geometry ....... how can i prove it through elementary geometry ??????

4. MANEO 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