- Let ABC be a triangle, P a point and PaPbPc the pedal

triangle of P.

Ab := AB /\ PPa, Ac := AC /\ PPa

A' := (perpendicular to AB at Ab) /\ (perpendicular to AC at Ac)

A" := the orthogonal projection of A' on BC

A# := BC /\ AA'

Similarly B',B",B# and C',C",C#

Which is the locus of P such that

1. ABC, A'B'C' are perspective

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

3. A#,B#, C# are collinear ?

Antreas

-- - --- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis" <xpolakis@o...> wrote:
> Let ABC be a triangle, P a point and PaPbPc the pedal

I think that the more interesting is #3.

> triangle of P.

>

> Ab := AB /\ PPa, Ac := AC /\ PPa

>

> A' := (perpendicular to AB at Ab) /\ (perpendicular to AC at Ac)

>

> A" := the orthogonal projection of A' on BC

>

> A# := BC /\ AA'

>

> Similarly B',B",B# and C',C",C#

>

> Which is the locus of P such that

>

> 1. ABC, A'B'C' are perspective

>

> 2. ABC, A"B"C" are perspective

>

> 3. A#,B#, C# are collinear ?

I found that the locus is

Darboux cubic + (Three Altitudes)

Antreas - M1 = the midpoint of MabMacMab, Mac = the midpoints of BA', CA', resp.Denote:Let ABC be a triangle, P a point and A'B'C' the cevian or pedal triangle of P.An application to triangle geometry of a problem of Kostas Vittas (*)

The circle (M1, M1Mab = M1, M1Mac) ie the circle with diameter MabMac,intersects the circles with diameters BA', CA' at M12, M13, resp. on the positiveside of BC (ie the side where A is located on).Denote:A* = MabM13 /\ MacM12

The line M12M13 intersects again the circles with diameters BA', CA' at M'12, M'13 resp.The lines A'A*, BM'12, CM'13 are concurrent at point A**.[see figure]Which is the locus of P such that:1. ABC, A*B*C*2, A'B'C', A*B*C*,3. ABC, A**B**C**are perspective ?Note: The line A**A' is the symmedian from A** of the triangle A**BC.If Ka is the symmedian point of A**BC, and Kb,Kc the symmedian pointsof B**CA, C**AB, resp., which is the locus of P such thatABC, KaKbKc are perspective?APH - APH1.2. Op les on the Euler line of ABC1.1. Op lies on the OP lineWhich is the locus of P such that:center of NaNbNc, resp.of PBC, PCA, PAB , resp and Op,Np the circumcenter, NPCProbably we have seen some of the following loci.Let ABC be a triangle P a point, Na,Nb,Nc the NPC centers

Anyways here are :-)

2.1. Np lies on the OP line

2.2. Np les on the Euler line of ABC

? [APH]:

Probably we have seen some of the following loci.

Anyways here are :-)Let ABC be a triangle P a point, Na,Nb,Nc the NPC centers

of PBC, PCA, PAB , resp and Op,Np the circumcenter, NPC

center of NaNbNc, resp.

Which is the locus of P such that:

1.1. Op lies on the OP line

1.2. Op les on the Euler line of ABC

2.1. Np lies on the OP line

2.2. Np les on the Euler line of ABC

?APH

[César Lozada]:

1.1) Circum-quintic:

Q11 = CyclicSum[(v*w*(-v*cos(2*B+C)+w*cos(B+2*C))*sin(A)+

u*((2*cos(A)*cos(B-C)+1)*v*w+u^2*cos(3*A))*sin(B-C))*v*w ] = 0

Through vertices of cevian of N and ETC´s: 3, 4, 5, 80, 1113, 1114

1.2) Excentral-circum-quintic:

Q12 = CyclicSum[(v*w*(cos(C)*(1+2*cos(2*B))*v-w*cos(B)*(1+2*cos(2*C)))*sin(A)+

u*(-3*u^2*cos(A)+v*w*(1+2*cos(2*A)))*sin(B-C))*v*w] = 0

Through vertices of cevian of X(265) and ETC´s: 1, 4, 5, 30, 1113, 1114, 1138

2.1) Circum-quintic:

Q21 = CyclicSum[(v*w*(-w*cos(B)+v*cos(C))*sin(A)+

u*((2*cos(A)*cos(B-C)+cos(2*A)+2)*v*w+(cos(3*A)-2*cos(A))*u^2)*sin(B-C))*v*w] = 0

Through vertices of cevian of N and ETC´s: 3, 4, 1113, 1114

2.2) Circum-quintic:

Q22=CyclicSum[(-v*w*((2*cos(C)+cos(2*B+C))*v+(-2*cos(B)-cos(B+2*C))*w)*sin(A)+

u*((4*cos(A)*cos(B-C)+cos(2*A))*v*w+(cos(A)+2*cos(3*A))*u^2)*sin(B-C))*v*w]

Through vertices of cevian of X(30) and ETC´s: 3, 4, 30, 1113, 1114, 5627

César Lozada