# Topics List

### G,I

(4)

Let ABC be a triangle and P a point. The AP line intersects the circle with diameter BC at A' on the negative side of BC (ie not on the side the vertex A is).
Antreas Hatzipolakis

Apr 23

### A NPC center on the OI line ?

(4)

... circumcenter of triangles BPC,CPA,APB. >Let A',B',C' be midpoints of AOa,BOb,COc then NPC center of A'B'C' lies on Euler line of ABC. Denote Np = the NPC
Antreas Hatzipolakis

Apr 20

### I -- Concurrent Circumcircles

(4)

Let ABC be a triangle and I1,I2,I3 the reflections of I in BC,CA,AB, resp. Denote: I12, I13 = the orthogonal projections of I1 on AC,AB, resp. I23, I21 = the
Antreas Hatzipolakis

Apr 19

Fetching Sponsored Content...

### Antipedal triangle -- Loci

(3)

Let ABC be a triangle, P a point and PaPbPc the antipedal triangle of P. Denote: AaAbAc, BaBbBc, CaCbCc = the pedal triangles of Pa, Pb, Pc, resp. A'B'C.' =
Antreas Hatzipolakis

Apr 19

### Orthologic triangles - Locus

(14)

Orthologic centers and homography Angel Montesdeoca, Centros ortológicos y homografía ...
Antreas Hatzipolakis

Apr 15

### Cyclologic triangles (Re: Orthologic triangles - Locus)

(5)

[A variation of a configuration by APH] Let ABC be a triangle, A'B'C' a variable triangle w/r to ABC and A1B1C1, A2B2C2 the medial triangles of ABC, A'B'C',
Antreas Hatzipolakis

Apr 13

### Orthologic triangles -- Locus

(5)

[APH]: Let ABC be a triangle and P a point. Denote: Pa = (Reflection of BC in PB) /\ (Reflection of BC in PC) Pb = (Reflection of CA in PC) /\ (Reflection of
Antreas Hatzipolakis

Mar 7

### I - Concurrent NPCs

(6)

On Wed, Mar 4, 2015 at 7:11 PM, Antreas Hatzipolakis wrote: Let ABC be a triangle. ... ************************ On Wed, Mar 4, 2015 at 8:51 PM, Moses, Peter J.
Antreas Hatzipolakis

Mar 6

### Reflections - Loci

(3)

1. La, Lb, Lc are concurrent? (the entire plane, with point of concurrence the isogonal conjugate of P) Confirmed 2. ABC, A'B'C' are perspective? (the entire
Antreas Hatzipolakis

Feb 26

View First Topic
Go to
View Last Topic