- 1. Let A' be a fixed point on the sideline BC of triangle ABC.To construct point A" on BC such that the NPC centers ofABA', ACA', ABA", ACA" are concyclic.The point A" is the trace on BC of the isogonal cevian of AA'(ie A" = BC /\ (reflection of AA' in the int. angle bisector of A)2. Let P,Q be two isogonal conjugate points and A'B'C', A"B"C"the cevian triangles of P,Q, resp.Denote:Oa = the center of the circle passing through theNPC centers of ABA', ACA', ABA", ACA"Similarly Ob, Oc.For P = O, Q = H, the triangles ABC, OaObOc are orthologic.[Cesar Lozada]:OrthologicCenters( P=O ) = { Po, X(546)=midpoint of (H,N) }

Po = a*((b^2+c^2)*a^2-(b^2-c^2)^2)/((b^2+c^2)*a^6-(b^4+c^4-4*b^2*c^2)*a^4-(b^2+c^2)* (b^2-c^2)^2*a^2+(b^2-c^2)^4) :: (trilinears)

ETC(__Po____) = 4.088637332117895, -1.05148019490158, 2.481548925092599__

__Po____not in any line through ETC centers__3. Let P be a point.

In the triangle PBC, let A',A" be the traces on BC of the (isogonal) ceviansthrough the circumcenter, orthocenter, resp. of PBC.[ie If O1,H1 the circumcenter, orthocenter, resp. of PBC, thenA' = BC /\ PO1, A" = BC /\ PH1].Let Oa be the center of the circle passing through the NPC centersof the triangles PBA', PCA', PBA", PCA"Similarly Ob, Oc.The triangles OaObOc, ABC are orthologic.The orthologic center P1 = (OaObPc, ABC) is lying on the line OP.The orthologic center P2 = (ABC, OaObOc) is lying on ?????4. Let P, Q be two isogonal conjugate points.Let Oa,Ob,Oc be the centers respective to P (above #3)and O'a,O'b,O'c the centers respective to Q (above #3)and let P1, P2 be the orthologic centers (OaObOc, ABC), (ABC, OaObOc), respand Q1,Q2 the orthologic centers (O'aO'bO'c, ABC), (ABC, O'aO'bO'c)Questions:i) Are the six centers Oa,Ob,Oc, O'a,O'b, O'c lying on a conic?ii) Are the four orthologic centers P1,P2,Q1,Q2 concyclic?Antreas

Let ABC be a triangle.

Denote:

Ab = AC /\ (perpendicular bisector of BC)

Ac = AB /\ (perpendicular bisector of BC)

Similarly Bc, Ba and Ca, Cb.

Ba, Ca, Cb, Bc are concyclic.

Reference (wording mine):

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=608711&p3635815

Let Oa be the center of the circle (Ba, Ca, Cb, Bc)

Simiarly Ob, Oc.

Are the triangles ABC, OaObOc perspective or orthologic ?

APH- [APH]:

Let ABC be a triangle.

Denote:

Ab = AC /\ (perpendicular bisector of BC)

Ac = AB /\ (perpendicular bisector of BC)

Similarly Bc, Ba and Ca, Cb.

Ba, Ca, Cb, Bc are concyclic.

Reference (wording mine):

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=608711&p3635815

Let Oa be the center of the circle (Ba, Ca, Cb, Bc)

Simiarly Ob, Oc.

Are the triangles ABC, OaObOc perspective or orthologic ?

APH[Peter Moses]:perspective at X(184)orthologic at X(4) and {{3,64},{4,54},{5,182},{6,1598},{20,110},{22,5562},{23,5889},{24,185},{25,389},{30,156},...}Best regards,Peter. - [APH]:

Let ABC be a triangle.

Denote:

Ab = AC /\ (perpendicular bisector of BC)

Ac = AB /\ (perpendicular bisector of BC)

Similarly Bc, Ba and Ca, Cb.

Ba, Ca, Cb, Bc are concyclic.

Reference (wording mine):

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=608711&p3635815

Let Oa be the center of the circle (Ba, Ca, Cb, Bc)

Simiarly Ob, Oc.

Are the triangles ABC, OaObOc perspective or orthologic ?

APH[Peter Moses]:perspective at X(184)orthologic at X(4) and {{3,64},{4,54},{5,182},{6,1598},{20,110},{22,5562},{23,5889},{24,185},{25,389},{30,156},...}Best regards,Peter.Synthetic proofs of the fact that the perspector is X(184) by Luis Gonzalez and Telv Cohl

www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=608711&p=3636331