- Dear Antreas

2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,

A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the

second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with

the pedal circle of P, resp.

The locus of P such that the triangles ABC, A"B"C" are orthologic is the septic Qnnn:

-a^2 c^4 x^4 y^3 + b^2 c^4 x^4 y^3 + c^6 x^4 y^3 -

a^2 c^4 x^3 y^4 + b^2 c^4 x^3 y^4 - c^6 x^3 y^4 +

2 b^2 c^4 x^4 y^2 z - 2 a^2 c^4 x^2 y^4 z -

2 b^4 c^2 x^4 y z^2 - 2 b^4 c^2 x^3 y^2 z^2 +

2 b^2 c^4 x^3 y^2 z^2 + 2 a^4 c^2 x^2 y^3 z^2 -

2 a^2 c^4 x^2 y^3 z^2 + 2 a^4 c^2 x y^4 z^2 +

a^2 b^4 x^4 z^3 - b^6 x^4 z^3 - b^4 c^2 x^4 z^3 - 2 a^4 b^2 x^2

y^2 z^3 + 2 a^2 b^4 x^2 y^2 z^3 + a^6 y^4 z^3 - a^4 b^2 y^4 z^3 +

a^4 c^2 y^4 z^3 + a^2 b^4 x^3 z^4 + b^6 x^3 z^4 -

b^4 c^2 x^3 z^4 + 2 a^2 b^4 x^2 y z^4 - 2 a^4 b^2 x y^2 z^4 -

a^6 y^3 z^4 - a^4 b^2 y^3 z^4 + a^4 c^2 y^3 z^4=0,

(+ line at infinity (A'B'C' undefined triangle) + circumcircle (A', B' and C' points aligned) + a sextic ).

Points of the septic Qnnn:

A, B, C (triple points)

X(1) double , X(4)

excenters (which are double)

feet of the altitudes.

The sextic (without real points ???)is:

-b^2 c^4 x^4 y^2 - a^2 c^4 x^3 y^3 - b^2 c^4 x^3 y^3 +

c^6 x^3 y^3 - a^2 c^4 x^2 y^4 + a^2 b^2 c^2 x^4 y z -

b^4 c^2 x^4 y z - b^2 c^4 x^4 y z + a^2 b^2 c^2 x^3 y^2 z -

b^4 c^2 x^3 y^2 z + b^2 c^4 x^3 y^2 z - a^4 c^2 x^2 y^3 z +

a^2 b^2 c^2 x^2 y^3 z + a^2 c^4 x^2 y^3 z - a^4 c^2 x y^4 z +

a^2 b^2 c^2 x y^4 z - a^2 c^4 x y^4 z - b^4 c^2 x^4 z^2 +

a^2 b^2 c^2 x^3 y z^2 + b^4 c^2 x^3 y z^2 - b^2 c^4 x^3 y z^2 +

6 a^2 b^2 c^2 x^2 y^2 z^2 + a^4 c^2 x y^3 z^2 +

a^2 b^2 c^2 x y^3 z^2 - a^2 c^4 x y^3 z^2 - a^4 c^2 y^4 z^2 -

a^2 b^4 x^3 z^3 + b^6 x^3 z^3 - b^4 c^2 x^3 z^3 -

a^4 b^2 x^2 y z^3 + a^2 b^4 x^2 y z^3 + a^2 b^2 c^2 x^2 y z^3 +

a^4 b^2 x y^2 z^3 - a^2 b^4 x y^2 z^3 + a^2 b^2 c^2 x y^2 z^3 +

a^6 y^3 z^3 - a^4 b^2 y^3 z^3 - a^4 c^2 y^3 z^3 - a^2 b^4 x^2 z^4 - a^4 b^2 x y z^4 - a^2 b^4 x y z^4 + a^2 b^2 c^2 x y z^4 -

a^4 b^2 y^2 z^4=0.

Best regards

Angel Montesdeoca

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

>

> 1. Let ABC be a triangle, A'B'C' the pedal triangle of point P

> and A",B",C" the second intersections of the circles (P,PA'),

> (P,PB'), (P,PC') with the pedal circle of P, resp.

>

> Which is the locus of P such that the triangles ABC, A"B"C"

> are orthologic?

>

> O is on the locus.

>

> 2. Let ABC be a triangle, A'B'C' the pedal triangle of point P,

> A*B*C* the circumcevian triangle of P wrt A'B'C' and A",B",C" the

> second intersections of the circles (P,PA*), (P,PB*), (P,PC*) with

> the pedal circle of P, resp.

>

> Which is the locus of P such that the triangles ABC, A"B"C"

> are orthologic?

>

> H is on the locus.

>

> Figures:

> http://anthrakitis.blogspot.gr/2013/03/orthologic-triangles-locus.html

>

> APH

> - [Antreas]:Which point is this wrt tr. ABC and which wrt orthic tr. of ABC?Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C'are orthologic?Which is the locus of P such that A'B'C', A"B"C"A"B"C" = the pedal triangle of O'.O' = the circumcenter of A'B'C' ( = the center of the pedal circle of P)A'B'C' = the pedal triangle of PDenote:Let ABC be a triangle and P a point.On the locus:1. P = H (O' = N)

Orthologic center (A'B'C', A"B"C") = ?

2. P = Ο (O' = N)

Orthologic center (A"B"C", A'B'C') = the O of A'B'C' = the NPC center of ABC

Orthologic center (A'B'C', A"B"C") = ? on the Euler line of the orthic triangle[Peter Moses]:>Which is the locus of P such that A'B'C', A"B"C" are orthologic?

Circumcircle, infinity, K003. McCay cubic1)Orthologic center (A"B"C", A'B'C') = the NPC center of A'B'C' = X(143) of ABC

Orthologic center (A'B'C', A"B"C") = on ABCs lines {{4,93},{6,24},{25,195},...} = X(79) of Orthic?2)Orthologic center (A'B'C', A"B"C") = X(1209) of ABC, X(3651) of Orthic. (probably!)Best regardsPeter

**************************Thanks, Peter,What a surprise! A new point in such a simple construction !!!!I wouldn't bet a cent for it as a new point :-)Description of the point:Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of H,N, resp.A'B'C', A"B"C" are orthologic with orthologic center (A'B'C', A"B"C")on {4,93},{6,24},{25,195},.. lines of ABC.If you compute other "things" of the point (coordinates etc) please let me know(to inform Clark)Season's GreetingsAPH

[Peter Moses]:

A few extra properties of the point:Barys: a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^4-2 a^2 b^2+b^4-2 a^2 c^2-b^2 c^2+c^4) (a^4 b^2-2 a^2 b^4+b^6+a^4 c^2-2 a^2 b^2 c^2-b^4 c^2-2 a^2 c^4-b^2 c^4+c^6)::On lines {{4,93},{6,24},{25,195},{49,143},{52,539},{70,6145},{113,5446},{155,3060},{403,3574},{648,1179},{1112,2914},{1209,1216},{1843,5965},{1986,3575}}.Reflection of X(i) in X(j) for these {i,j}: {54,973},{1493,143},{2914,1112}.3 X[1209] - 2 X[1216].3 X[54] - 5 X[3567].6 X[973] - 5 X[3567].X(4)-ceva conjugate of X(1594).X(4)-crosspoint of X(3518).X(3)-crosssum of X(3519).X(2216)-isoconjugate of X(3519).Trilinear product X(1594) X(2964).Barycentric product X(1594) X(1994).X(79) Orthic triangle.{{1,21},{7,79},...} of Tangential triangle.Best regards,Peter.