Τετάρτη, 25 Ιανουαρίου 2012

ZIG ZAG LOCUS


Let ABC be a triangle, P a point (not on the circumcircle), A1B1C1 the circumcevian triangle of P, A2B2C2 the circumcevian triangle of H with respect the triangle A1B1C1 and A3B3C3 the circumcevian triangle of O with respect the triangle A2B2C2 (antipodal triangle of A2B2C2).


Which is the locus of P such that the triangles ABC, A3B3C3 are perspective? And the locus of the perspectors?
Generalization: Replace H with a point Q.

APH, 25 January 2012

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The locus is the Euler line.
For an arbitrary Q, instead of H, the locus is the line OQ, and the locus of perspectors P' is the same line, so we have a map P -> P', although the formula is not simple.

I found the following formula showing the relationship between some P on line OQ such that OP:PQ = k, and the perspector P', also on line OQ such that OP':P'Q = k': k' = R^2 / (k OQ^2 - (k+2) R^2)

Francisco Javier García Capitán
25 January 2012

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Variation:

A2B2C2 = the circumcevian triangle of P* (= the isogonal or isotomic conjugate of P, instead of H) with respect the triangle A1B1C1

APH

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