## Σάββατο, 14 Ιανουαρίου 2012

### SEGOVIA POINT

Let ABC be a triangle, A1B1C1 the pedal triangle of H, A2B2C2 the antipodal triangle of A1B1C1 with respect the circumcircle of A1B1C1 (= NPC of ABC) and A'B'C' the antipodal triangle of ABC (= circumcevian triangle of O)

The triangles A2B2C2, A'B'C' are perspective.

Perspector?

Generalization:

A1B1C1 = the pedal triangle of a point P (instead of H).

APH
14 January 2012

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For H the perspector is

{(a^2 - b^2 - c^2) (3 a^8 - 8 a^4 b^4 + 4 a^2 b^6 + b^8 +
4 a^4 b^2 c^2 - 4 a^2 b^4 c^2 - 4 b^6 c^2 - 8 a^4 c^4 -
4 a^2 b^2 c^4 + 6 b^4 c^4 + 4 a^2 c^6 - 4 b^2 c^6 +
c^8), -(a^2 - b^2 + c^2) (a^8 + 4 a^6 b^2 - 8 a^4 b^4 + 3 b^8 -
4 a^6 c^2 - 4 a^4 b^2 c^2 + 4 a^2 b^4 c^2 + 6 a^4 c^4 -
4 a^2 b^2 c^4 - 8 b^4 c^4 - 4 a^2 c^6 + 4 b^2 c^6 +
c^8), -(a^2 + b^2 - c^2) (a^8 - 4 a^6 b^2 + 6 a^4 b^4 -
4 a^2 b^6 + b^8 + 4 a^6 c^2 - 4 a^4 b^2 c^2 - 4 a^2 b^4 c^2 +
4 b^6 c^2 - 8 a^4 c^4 + 4 a^2 b^2 c^4 - 8 b^4 c^4 + 3 c^8)}

not in ETC

Locus:

circumcircle + Linf + cubic (I think not cathalogued in B. Gibert's Cubics in Triangle Plane)

The cubic is

-2 a^6 b^2 c^4 x^3 y + 4 a^4 b^4 c^4 x^3 y -
2 a^2 b^6 c^4 x^3 y + 2 a^2 b^2 c^8 x^3 y - a^8 c^4 x^2 y^2 +
2 a^6 b^2 c^4 x^2 y^2 - 2 a^2 b^6 c^4 x^2 y^2 + b^8 c^4 x^2 y^2 +
3 a^6 c^6 x^2 y^2 - a^4 b^2 c^6 x^2 y^2 + a^2 b^4 c^6 x^2 y^2 -
3 b^6 c^6 x^2 y^2 - 3 a^4 c^8 x^2 y^2 + 3 b^4 c^8 x^2 y^2 +
a^2 c^10 x^2 y^2 - b^2 c^10 x^2 y^2 + 2 a^6 b^2 c^4 x y^3 -
4 a^4 b^4 c^4 x y^3 + 2 a^2 b^6 c^4 x y^3 - 2 a^2 b^2 c^8 x y^3 +
2 a^6 b^4 c^2 x^3 z - 2 a^2 b^8 c^2 x^3 z - 4 a^4 b^4 c^4 x^3 z +
2 a^2 b^4 c^6 x^3 z + 9 a^6 b^4 c^2 x^2 y z -
5 a^4 b^6 c^2 x^2 y z - 5 a^2 b^8 c^2 x^2 y z + b^10 c^2 x^2 y z -
9 a^6 b^2 c^4 x^2 y z + 7 a^2 b^6 c^4 x^2 y z -
2 b^8 c^4 x^2 y z + 5 a^4 b^2 c^6 x^2 y z -
7 a^2 b^4 c^6 x^2 y z + 5 a^2 b^2 c^8 x^2 y z + 2 b^4 c^8 x^2 y z -
b^2 c^10 x^2 y z - a^10 c^2 x y^2 z + 5 a^8 b^2 c^2 x y^2 z +
5 a^6 b^4 c^2 x y^2 z - 9 a^4 b^6 c^2 x y^2 z +
2 a^8 c^4 x y^2 z - 7 a^6 b^2 c^4 x y^2 z +
9 a^2 b^6 c^4 x y^2 z + 7 a^4 b^2 c^6 x y^2 z -
5 a^2 b^4 c^6 x y^2 z - 2 a^4 c^8 x y^2 z - 5 a^2 b^2 c^8 x y^2 z +
a^2 c^10 x y^2 z + 2 a^8 b^2 c^2 y^3 z - 2 a^4 b^6 c^2 y^3 z +
4 a^4 b^4 c^4 y^3 z - 2 a^4 b^2 c^6 y^3 z + a^8 b^4 x^2 z^2 -
3 a^6 b^6 x^2 z^2 + 3 a^4 b^8 x^2 z^2 - a^2 b^10 x^2 z^2 -
2 a^6 b^4 c^2 x^2 z^2 + a^4 b^6 c^2 x^2 z^2 + b^10 c^2 x^2 z^2 -
a^2 b^6 c^4 x^2 z^2 - 3 b^8 c^4 x^2 z^2 + 2 a^2 b^4 c^6 x^2 z^2 +
3 b^6 c^6 x^2 z^2 - b^4 c^8 x^2 z^2 + a^10 b^2 x y z^2 -
2 a^8 b^4 x y z^2 + 2 a^4 b^8 x y z^2 - a^2 b^10 x y z^2 -
5 a^8 b^2 c^2 x y z^2 + 7 a^6 b^4 c^2 x y z^2 -
7 a^4 b^6 c^2 x y z^2 + 5 a^2 b^8 c^2 x y z^2 -
5 a^6 b^2 c^4 x y z^2 + 5 a^2 b^6 c^4 x y z^2 +
9 a^4 b^2 c^6 x y z^2 - 9 a^2 b^4 c^6 x y z^2 + a^10 b^2 y^2 z^2 -
3 a^8 b^4 y^2 z^2 + 3 a^6 b^6 y^2 z^2 - a^4 b^8 y^2 z^2 -
a^10 c^2 y^2 z^2 - a^6 b^4 c^2 y^2 z^2 + 2 a^4 b^6 c^2 y^2 z^2 +
3 a^8 c^4 y^2 z^2 + a^6 b^2 c^4 y^2 z^2 - 3 a^6 c^6 y^2 z^2 -
2 a^4 b^2 c^6 y^2 z^2 + a^4 c^8 y^2 z^2 - 2 a^6 b^4 c^2 x z^3 +
2 a^2 b^8 c^2 x z^3 + 4 a^4 b^4 c^4 x z^3 - 2 a^2 b^4 c^6 x z^3 -
2 a^8 b^2 c^2 y z^3 + 2 a^4 b^6 c^2 y z^3 - 4 a^4 b^4 c^4 y z^3 +
2 a^4 b^2 c^6 y z^3;

Francisco Javier García Capitán
14 January 2012

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I name the perspector for P = H as SEGOVIA Point for three reasons:

1. Segovia is Spanish as Francisco

2. His name is Andrés as mine

3. He was a great guitarist, and I am wishing to my daughter, who started recently lessons in guitar, to be like him!

Antreas

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This point S lies on line H-X1209 and satisfies the ratio:

HX1209:X1209S = (p^2 - r^2 - 4*r*R)/(3*p^2 - 3*r^2 - 12*r*R - 10*R^2).

Francisco, Hyacinthos 20678

#### 1 σχόλιο:

1. For a synthetic proof of Segovia Point see here:-
http://www.artofproblemsolving.com/blog/65683