- Writing correct geometric property of the quintic Q068 in the message #21755:

Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the

circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A', B', C' w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.

A particular situation:

If P=X(3) the orthology centers of ABC y A"B"C" (on NPC) are Q=X(113) and R, nine-point-circle-antipode of X(3258).

R= (2*a^8 - a^4*(b^4-4*b^2*c^2+c^4)- 2*a^6*(b^2+c^2) + (b^2-c^2)^4 )*

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

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

(b^2 - c^2)^2*(b^4 + 3*b^2*c^2 + c^4)) : .... : ....

Angel M.

--- In Hyacinthos@yahoogroups.com, "Angel" <amontes1949@...> wrote:

>

>

>

> --- 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.

>

> Dear Antreas

>

> The locus of real point P such that the triangles ABC, A"B"C" are orthologic is Q068 (a circular equilateral quintic) in Higher Degree Related Curves.

> http://bernard.gibert.pagesperso-orange.fr/relatedcurves.html

>

>

> It is a result does not appear between the propiedades of Q068 in http://bernard.gibert.pagesperso-orange.fr/curves/q068.html

>

>

> We can express as:

>

> Let ABC be a triangle, A'B'C' the pedal triangle of point P and O' the circumcenter of A'B'C'. Let A", B", C" be the symmetrical points of A, B, C w/r to the line PO'. The triangles ABC, A"B"C" are orthologic if and only if P lies on Q068.

>

>

> (http://groups.yahoo.com/group/Hyacinthos/files/Hyacinthos21754A.eps)

>

>

> Angel M.

> - [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.