Παρασκευή 16 Μαρτίου 2012

Perspective


Let ABC be a triangle, A'B'C' the cevian triangle of G and A"B"C" the circumcevian triangle of G with respect A'B'C'. The circles with diameters HA",HB",HC" intersect the NPC again at A1,B1,C1, resp.


The circles with diameters HA",HB",HC" intersect the NPC again at A1,B1,C1, resp.

The triangles ABC, A1B1C1 are perspective.

Perspector?

APH, 16 March 2012

--------------------------------------------

Not in ETC. It is the isotomic conjugate of

(SA (b^4 + c^4-a^4), SB (a^4 - b^4 + c^4), SC (a^4 + b^4 - c^4))

************************************

ADDENDUM (12/9/19)
Perspector: X(13854)
Isotomic conjugate: X(34254)
************************************

The locus of P, instead of H, is trilinear polar of X648 (containing H) + NPC + a cubic:

5 a^12 x^3 - 28 a^10 b^2 x^3 + 53 a^8 b^4 x^3 - 40 a^6 b^6 x^3 +
7 a^4 b^8 x^3 + 4 a^2 b^10 x^3 - b^12 x^3 - 28 a^10 c^2 x^3 +
110 a^8 b^2 c^2 x^3 - 120 a^6 b^4 c^2 x^3 + 36 a^4 b^6 c^2 x^3 +
4 a^2 b^8 c^2 x^3 - 2 b^10 c^2 x^3 + 53 a^8 c^4 x^3 -
120 a^6 b^2 c^4 x^3 + 42 a^4 b^4 c^4 x^3 - 8 a^2 b^6 c^4 x^3 +
b^8 c^4 x^3 - 40 a^6 c^6 x^3 + 36 a^4 b^2 c^6 x^3 -
8 a^2 b^4 c^6 x^3 + 4 b^6 c^6 x^3 + 7 a^4 c^8 x^3 +
4 a^2 b^2 c^8 x^3 + b^4 c^8 x^3 + 4 a^2 c^10 x^3 - 2 b^2 c^10 x^3 -
c^12 x^3 + a^12 x^2 y - 4 a^10 b^2 x^2 y - 7 a^8 b^4 x^2 y +
40 a^6 b^6 x^2 y - 53 a^4 b^8 x^2 y + 28 a^2 b^10 x^2 y -
5 b^12 x^2 y - 2 a^10 c^2 x^2 y + 24 a^8 b^2 c^2 x^2 y -
44 a^6 b^4 c^2 x^2 y + 48 a^4 b^6 c^2 x^2 y - 34 a^2 b^8 c^2 x^2 y +
8 b^10 c^2 x^2 y - 17 a^8 c^4 x^2 y + 46 a^4 b^4 c^4 x^2 y +
3 b^8 c^4 x^2 y + 36 a^6 c^6 x^2 y - 24 a^4 b^2 c^6 x^2 y +
4 a^2 b^4 c^6 x^2 y - 8 b^6 c^6 x^2 y - 17 a^4 c^8 x^2 y +
4 a^2 b^2 c^8 x^2 y + b^4 c^8 x^2 y - 2 a^2 c^10 x^2 y +
c^12 x^2 y - 5 a^12 x y^2 + 28 a^10 b^2 x y^2 - 53 a^8 b^4 x y^2 +
40 a^6 b^6 x y^2 - 7 a^4 b^8 x y^2 - 4 a^2 b^10 x y^2 + b^12 x y^2 +
8 a^10 c^2 x y^2 - 34 a^8 b^2 c^2 x y^2 + 48 a^6 b^4 c^2 x y^2 -
44 a^4 b^6 c^2 x y^2 + 24 a^2 b^8 c^2 x y^2 - 2 b^10 c^2 x y^2 +
3 a^8 c^4 x y^2 + 46 a^4 b^4 c^4 x y^2 - 17 b^8 c^4 x y^2 -
8 a^6 c^6 x y^2 + 4 a^4 b^2 c^6 x y^2 - 24 a^2 b^4 c^6 x y^2 +
36 b^6 c^6 x y^2 + a^4 c^8 x y^2 + 4 a^2 b^2 c^8 x y^2 -
17 b^4 c^8 x y^2 - 2 b^2 c^10 x y^2 + c^12 x y^2 - a^12 y^3 +
4 a^10 b^2 y^3 + 7 a^8 b^4 y^3 - 40 a^6 b^6 y^3 + 53 a^4 b^8 y^3 -
28 a^2 b^10 y^3 + 5 b^12 y^3 - 2 a^10 c^2 y^3 + 4 a^8 b^2 c^2 y^3 +
36 a^6 b^4 c^2 y^3 - 120 a^4 b^6 c^2 y^3 + 110 a^2 b^8 c^2 y^3 -
28 b^10 c^2 y^3 + a^8 c^4 y^3 - 8 a^6 b^2 c^4 y^3 +
42 a^4 b^4 c^4 y^3 - 120 a^2 b^6 c^4 y^3 + 53 b^8 c^4 y^3 +
4 a^6 c^6 y^3 - 8 a^4 b^2 c^6 y^3 + 36 a^2 b^4 c^6 y^3 -
40 b^6 c^6 y^3 + a^4 c^8 y^3 + 4 a^2 b^2 c^8 y^3 + 7 b^4 c^8 y^3 -
2 a^2 c^10 y^3 + 4 b^2 c^10 y^3 - c^12 y^3 + a^12 x^2 z -
2 a^10 b^2 x^2 z - 17 a^8 b^4 x^2 z + 36 a^6 b^6 x^2 z -
17 a^4 b^8 x^2 z - 2 a^2 b^10 x^2 z + b^12 x^2 z -
4 a^10 c^2 x^2 z + 24 a^8 b^2 c^2 x^2 z - 24 a^4 b^6 c^2 x^2 z +
4 a^2 b^8 c^2 x^2 z - 7 a^8 c^4 x^2 z - 44 a^6 b^2 c^4 x^2 z +
46 a^4 b^4 c^4 x^2 z + 4 a^2 b^6 c^4 x^2 z + b^8 c^4 x^2 z +
40 a^6 c^6 x^2 z + 48 a^4 b^2 c^6 x^2 z - 8 b^6 c^6 x^2 z -
53 a^4 c^8 x^2 z - 34 a^2 b^2 c^8 x^2 z + 3 b^4 c^8 x^2 z +
28 a^2 c^10 x^2 z + 8 b^2 c^10 x^2 z - 5 c^12 x^2 z +
10 a^12 x y z - 36 a^10 b^2 x y z + 22 a^8 b^4 x y z +
8 a^6 b^6 x y z + 22 a^4 b^8 x y z - 36 a^2 b^10 x y z +
10 b^12 x y z - 36 a^10 c^2 x y z + 124 a^8 b^2 c^2 x y z -
88 a^6 b^4 c^2 x y z - 88 a^4 b^6 c^2 x y z +
124 a^2 b^8 c^2 x y z - 36 b^10 c^2 x y z + 22 a^8 c^4 x y z -
88 a^6 b^2 c^4 x y z + 132 a^4 b^4 c^4 x y z -
88 a^2 b^6 c^4 x y z + 22 b^8 c^4 x y z + 8 a^6 c^6 x y z -
88 a^4 b^2 c^6 x y z - 88 a^2 b^4 c^6 x y z + 8 b^6 c^6 x y z +
22 a^4 c^8 x y z + 124 a^2 b^2 c^8 x y z + 22 b^4 c^8 x y z -
36 a^2 c^10 x y z - 36 b^2 c^10 x y z + 10 c^12 x y z + a^12 y^2 z -
2 a^10 b^2 y^2 z - 17 a^8 b^4 y^2 z + 36 a^6 b^6 y^2 z -
17 a^4 b^8 y^2 z - 2 a^2 b^10 y^2 z + b^12 y^2 z +
4 a^8 b^2 c^2 y^2 z - 24 a^6 b^4 c^2 y^2 z + 24 a^2 b^8 c^2 y^2 z -
4 b^10 c^2 y^2 z + a^8 c^4 y^2 z + 4 a^6 b^2 c^4 y^2 z +
46 a^4 b^4 c^4 y^2 z - 44 a^2 b^6 c^4 y^2 z - 7 b^8 c^4 y^2 z -
8 a^6 c^6 y^2 z + 48 a^2 b^4 c^6 y^2 z + 40 b^6 c^6 y^2 z +
3 a^4 c^8 y^2 z - 34 a^2 b^2 c^8 y^2 z - 53 b^4 c^8 y^2 z +
8 a^2 c^10 y^2 z + 28 b^2 c^10 y^2 z - 5 c^12 y^2 z - 5 a^12 x z^2 +
8 a^10 b^2 x z^2 + 3 a^8 b^4 x z^2 - 8 a^6 b^6 x z^2 +
a^4 b^8 x z^2 + b^12 x z^2 + 28 a^10 c^2 x z^2 -
34 a^8 b^2 c^2 x z^2 + 4 a^4 b^6 c^2 x z^2 + 4 a^2 b^8 c^2 x z^2 -
2 b^10 c^2 x z^2 - 53 a^8 c^4 x z^2 + 48 a^6 b^2 c^4 x z^2 +
46 a^4 b^4 c^4 x z^2 - 24 a^2 b^6 c^4 x z^2 - 17 b^8 c^4 x z^2 +
40 a^6 c^6 x z^2 - 44 a^4 b^2 c^6 x z^2 + 36 b^6 c^6 x z^2 -
7 a^4 c^8 x z^2 + 24 a^2 b^2 c^8 x z^2 - 17 b^4 c^8 x z^2 -
4 a^2 c^10 x z^2 - 2 b^2 c^10 x z^2 + c^12 x z^2 + a^12 y z^2 +
a^8 b^4 y z^2 - 8 a^6 b^6 y z^2 + 3 a^4 b^8 y z^2 +
8 a^2 b^10 y z^2 - 5 b^12 y z^2 - 2 a^10 c^2 y z^2 +
4 a^8 b^2 c^2 y z^2 + 4 a^6 b^4 c^2 y z^2 - 34 a^2 b^8 c^2 y z^2 +
28 b^10 c^2 y z^2 - 17 a^8 c^4 y z^2 - 24 a^6 b^2 c^4 y z^2 +
46 a^4 b^4 c^4 y z^2 + 48 a^2 b^6 c^4 y z^2 - 53 b^8 c^4 y z^2 +
36 a^6 c^6 y z^2 - 44 a^2 b^4 c^6 y z^2 + 40 b^6 c^6 y z^2 -
17 a^4 c^8 y z^2 + 24 a^2 b^2 c^8 y z^2 - 7 b^4 c^8 y z^2 -
2 a^2 c^10 y z^2 - 4 b^2 c^10 y z^2 + c^12 y z^2 - a^12 z^3 -
2 a^10 b^2 z^3 + a^8 b^4 z^3 + 4 a^6 b^6 z^3 + a^4 b^8 z^3 -
2 a^2 b^10 z^3 - b^12 z^3 + 4 a^10 c^2 z^3 + 4 a^8 b^2 c^2 z^3 -
8 a^6 b^4 c^2 z^3 - 8 a^4 b^6 c^2 z^3 + 4 a^2 b^8 c^2 z^3 +
4 b^10 c^2 z^3 + 7 a^8 c^4 z^3 + 36 a^6 b^2 c^4 z^3 +
42 a^4 b^4 c^4 z^3 + 36 a^2 b^6 c^4 z^3 + 7 b^8 c^4 z^3 -
40 a^6 c^6 z^3 - 120 a^4 b^2 c^6 z^3 - 120 a^2 b^4 c^6 z^3 -
40 b^6 c^6 z^3 + 53 a^4 c^8 z^3 + 110 a^2 b^2 c^8 z^3 +
53 b^4 c^8 z^3 - 28 a^2 c^10 z^3 - 28 b^2 c^10 z^3 + 5 c^12 z^3

Francisco Javier García Capitán
17 March 2012

1 σχόλιο:

  1. Suppose, O is the circumcenter of ABC.A2B2C2 be the circumcevian triangle of G wrt ABC. Clearly, A3B3C3 be the circumcevian triangle of H wrt A2B2C2. A1,B1,C1 be the midpoints of HA3,HB3,HC3.
    Using this problem of yours(http://anthrakitis.blogspot.in/2012/02/pp.html) we can prove that ABC and A1B1C1 are perspective.

    ΑπάντησηΔιαγραφή

REGULAR POLYGONS AND EULER LINES

Let A1A2A3 be an equilateral triangle and Pa point. Denote: 1, 2, 3 = the Euler lines of PA1A2,PA2A3, PA3A1, resp. 1,2,3 are concurrent. ...