Τετάρτη, 8 Φεβρουαρίου 2012

X3542


Let ABC be a triangle, A'B'C' the cevian triangle of H (orthic) and A"B"C" the circumcevian triangle of O with respect A'B'C'.


The line A"H intersects the circumcircle of ABC at A1,A2 with |HA1| < |HA2|. Similarly the points B1, B2 with |HB1| < |HB2| and C1, C2 with |HC1| < |HC2|.

Are the triangles ABC, A1B1C1 perspective ?

APH,  8 February 2012

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Yes, they're perspective with perspector Q=X3542, on the Euler line.
This point satisfies, HQ:QG = - 6 cosA cosB cosC

Francisco Javier García Capitán
9 February 2012

A2B2C2 and ABC perspective if and only P is  lies on the NPC or in the trilinear
polar of X2052.

 Francisco Javier García Capitán Hyacinthos #20815


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Generalization: Is it true if we replace O with a point P?
(The perspector on the HP line)
APH, Hyacinthos #20809

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Yes, it is true even if you replace O by P,
It follows from the following problem:-
It follows from the following problem which is a generalisation of a
problem you posted earlier:- Given a triangle ABC, A1B1C1 be the
circumcevian triangle of a point R and A2B2C2 be the circumcevian triangle
of a point Q. If A3B3C3 is the circumcevian triangle of R wrt A2B2C2, then
A1A3,B1B3,C1C3 concur on RQ.
Now take R= reflection of H on P and Q=H and the result follows.

CHANDAN, Hyacinthos #20810

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2 σχόλια:

  1. It follows from the following problem which is a generalisation of a problem you posted earlier:- Given a triangle ABC, A1B1C1 be the circumcevian triangle of a point P and A2B2C2 be the circumcevian triangle of a point Q. If A3B3C3 is the circumcevian triangle of P wrt A2B2C2, then A1A3,B1B3,C1C3 concurr on PQ.

    In this problem take P=De-Longchamps Point of ABC, Q=orthocenter of ABC and the result follows.

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  2. I think there is a mistake in the defintion of A1,A2,B1,B2 and so. Because if we take HA1<HA2 and so then it will not be always true. But if we take B" on the same side of B2 and B1 on the other side, then it will be always true.

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