# Topics List

### 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

7:56 AM

### NPC center -- Locus

(1)

Let ABC be a triangle, P a point and PaPbPc the circumcevian triangle of P. Let A',B',C' be the midpoints of APa, BPb, CPc, resp, Which is the locus of the NPC
Antreas Hatzipolakis

4:42 AM

### 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

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### 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

### Re: [EGML] Center on Euler line

(2)

Which is the locus of P such that A'B'C', A"B"C" are perspective? Not a simple locus: Locus = { a circum-sixtic (q6) } U { a circum-nonic (q9) } No
Antreas Hatzipolakis

Apr 16

### Some kind of reciprocal perspectivity

(1)

Dear friends: Let P=X(316) and Q=X(671) of a triangle ΔABC Let Pa,Pb,Pc and Qa,Qb,Qc be the reflections of P and Q in (AH), (BH), (CH), resp. (H is the
Antreas Hatzipolakis

Apr 16

### Center on Euler line

(2)

Very nice, César! Also, A"B"C" is the 2nd extouch triangle. Best regards, Randy Hutson Very nice, César! Also, A"B"C" is the 2nd extouch triangle. Best
Antreas Hatzipolakis

Apr 16

### 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

### Re: [EGML] Re: Cyclologic triangles (Re: Orthologic triangles - Locu

(1)

... *"Reflection triangle version*" is the version we get if we replace the excentral triangle with the reflection triangle. In details: Let ABC be a triangle
Antreas Hatzipolakis

Apr 13

### proof angle equality

(1)

Dear friends, Given 2 lines L1 and L2 with intersection point S. Points A1 and B1 lie on L1. Points A2 and B2 lie on L2. T = A1A2 ^ B1B2. Let La1 be the
Antreas Hatzipolakis

Apr 13

### Restricted triangle constructions

(1)

Old problems: Let ABC be a triangle and A'B'C', A1B1C1 the circumcevian triangles of G,K, resp. To constuct ABC if is given 1. A'B'C'. 2. A1B1C1 (*) Now, to
Antreas Hatzipolakis

Apr 13

### Locus

(235)

Let ABC be a triangle and P a point Which is the locus of P such that PA + PB + PC = 0 (signed segments) ? APH PS see special case
xpolakis@...

Apr 10

### X(110) generalization

(1)

Let ABC be a triangle and A'B'C' the orthic triangle Denote: Ba, Ca = the reflections of B, C in A', resp. Cb, Ab = the reflections of C, A in B', resp. Ac, Bc
Antreas Hatzipolakis

Mar 31

### Pedal triangle - orthocenters - loci

(2)

On Sat, Mar 28, 2015 at 9:01 PM, Antreas Hatzipolakis wrote: Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P. ...
Antreas Hatzipolakis

Mar 29

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