Dear Antreas,

1. ABC, A"B"C" are perspective on OI line. Perspector: X(36)

2. Antipedal triangle of I (excentral triangle),A"B"C" are perspective. The first barycentric coordinate of the perspector is:

a(3a^6 - 4(b + c)a^5 + (8b*c - 3(b^2 + c^2))a^4 +

(10(b^3 + c^3) - 4b*c(b + c))a^3 +

(9b^2c^2 - 10b*c(b^2 + c^2) - b^4 - c^4)a^2 +

(8b*c(b^3 + c^3) - 6(b^5 + c^5))a +

b^6 + 2b^5c - b^4c^2 - 4b^3c^3 - b^2c^4 + 2b*c^5 + c^6)

a. HaHbHc, A"B"C" are perspective (on OI line of orthic=Euler line of ABC). Perspector: Infinity Point of the Euler line

b. ABC,A"B"C" are perspective.

The first barycentric coordinate of the perspector is (S, SA, SB, SC usual Conway notation):

SA/( (S^2-3SA^2) (SA(c^2-b^2)^2 + a^2(SB*SC-SA^2)) )

This point is the barycentric product X(265)X(2986) or X(69)X(1989))X(2986)

Best regards,

Angel Montesdeoca

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:

>

> Let ABC be a triangle and A'B'C' the circumcevian triangle of I.

>

> The circle (I,IA') intersects again the circumcircle at A",

> the (I,IB') at B" and the (I,IC') at C".

>

> 1. ABC, A"B"C" are perspective on OI line

>

> 2. Antipedal triangle of I (excentral triangle), A"B"C" are perspective.

>

> Orthic triangle Version:

>

> Let ABC be a triangle, HaHbHc the orthic triangle and

> A'B'C' the circumcevian triangle of H wrt NPC.

>

> The circle (H,HA') intersects again the NPC at A",

> the (H,HB') at B" and the (H, HC') at C".

>

> a. HaHbHc, A"B"C" are perspective (on OI line of orthic = Euler line of ABC)

>

> b. ABC, A"B"C" are perspective.

>

> Perspectors?

>

> APH

>