## Loci

Expand Messages
• 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
Message 1 of 26 , May 2, 2005
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
--
• ... I think that the more interesting is #3. I found that the locus is Darboux cubic + (Three Altitudes) Antreas
Message 2 of 26 , May 3, 2005
--- In Hyacinthos@yahoogroups.com, "Antreas P. Hatzipolakis" <xpolakis@o...> wrote:
> 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 ?

I think that the more interesting is #3.

I found that the locus is

Darboux cubic + (Three Altitudes)

Antreas
• An application to triangle geometry of a problem of Kostas Vittas (*) Let ABC be a triangle, P a point and A B C the cevian or pedal triangle of P. Denote:
Message 3 of 26 , May 28, 2014
• 159 KB

An application to triangle geometry of a problem of Kostas Vittas (*)

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

Denote:

Mab, Mac = the midpoints of BA', CA', resp.
M1 = the midpoint of MabMac

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 positive
side 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 points
of B**CA, C**AB, resp., which is the locus of P such that
ABC, KaKbKc are perspective?
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
Message 4 of 26 , Jan 17, 2015
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
• [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
Message 5 of 26 , Jan 18, 2015

[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

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