Παρασκευή, 7 Ιανουαρίου 2011

A CONFIGURATION

Let ABC be a triangle and P a point.

The circle (B, BP) intersects AB at Ac1 (between A,B) and Ac2 (on the extension of BA) AND BC at A1c (between B,C) and A2c (on the extension of BC). The circle (C, CP) intersects AC at Ab1 (between A,C) and Ab2 AND BC at A1b (between B,C) and A2b (on the extension of CB).


1. The lines Ab1Ac2 and Ac1Ab2 intersect at A1.


2. The lines Ab1A1c and Ac1A1b intersect at A2


3. The lines Ab1A1b and Ac1A1c intersect at A3.


4. The lines Ab1A2b and Ac1A2c intersect at A4.


5. The lines Ab1A2c and Ac1A2b intersect at A5.


6. The lines Ab2A1b and Ac2A1c intersect at A6.


7. The lines Ab2A1c and Ac2A1b intersect at A7.


8. The lines Ab2A2b and Ac2A2c intersect at A8.


9. The lines Ab2A2c and Ac2A2b intersect at A9.


10. The lines Ab2A2c and Ab1Ac1 intersect at A10.


Similarly Bi, Ci, i=1,2,...,10

Problem:

Which are the loci of P such that ABC, AiBiCi (i = 1,2,...10)
are perspective?

For i = 1, the locus is the Neuberg cubic (see Hyacinthos, Message 19680 by Francisco Javier García Capitán)

Addendum (12-1-2011)

For i = 2:
A2B2C2 and ABC are never perspective by calculation. In this case the calculations (that consist of a determinant of the three lines is always 1)
Francisco Javier García Capitán

For i = 3 See THIS

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