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### Isogonal conjugates, coaxial

(4)

[APH]: Let ABC be a triangle, P, P* two isogonal conjugate points and A'B'C' the pedal triangle of P. Denote: A", B", C" = the antipodes of A', B', C' , resp.
Antreas Hatzipolakis

1:57 PM

### Circumhyperbola center

(4)

Generalization. Theorem by Kadir Altintas: The center of every circumconic of a triangle ABC passing through G lies on the Steiner inellipse.
Antreas Hatzipolakis

10:48 AM

### I, Coaxial

(13)

[APH]: Let ABC be a triangle and A'B'C' the pedal triangle of I. Denote: A", B", C" = the antipodes of A', B', C', resp. in the incircle. Na, Nb, Nc = the NPC
Antreas Hatzipolakis

Dec 10

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### Altintas hyperbola

(5)

*[Kadir Altintas]* *The second intersection, other than G, of the Nagel line IG and X(2) - Altintas hyperbola is the point with simple barycentrics* *(b +
Antreas Hatzipolakis

Dec 8

### In a given triangle (not isosceles and not equilateral) two distinct

(5)

Hi Peter and Antreas, Yes. I know there are some {6, 9, 13} “collisions”, but please observe the numeric coordinates for collisions remarked by Peter: X(
Antreas Hatzipolakis

Dec 8

### I,H, Euler lines, Parallelogic

(5)

[APH]: Let ABC be a triangle and A'B'C', A"B"C" the cevian, antipedal triangle of I, resp . Denote: Ma, Mb, Mc = the midpoints of AA', BB', CC', resp. Mab, Mac
Antreas Hatzipolakis

Dec 6

### An Incenter on the OI line

(4)

[Kadir Altintas]: Let ABC be a triangle and (Oi), i=1,2,3,4 four congruent circles such that: (O1) touches AB, AC and (O4) (O2) touches BC, BA and (O4) (O3)
Antreas Hatzipolakis

Dec 4

### H, perspective

(10)

[APH]: Let ABC be a triangle and P a point. Denote: PaPbPc = the pedal triangle of P P1, P2, P3 = the orthogonal projections of P on HA, HB, HC, resp. PaPbPc,
Antreas Hatzipolakis

Dec 4

### Isogonal conjugates 1

(5)

[APH]: Let Q, Q* be two isogonal conjugate points. Which is the locus of midpoint of QQ* as Q moves on a given line L? For L = Euler Line, Brocard Axis, OI
Antreas Hatzipolakis

Dec 2

### Euler lines, collinear

(4)

[APH]: Let ABC be a triangle, P a point and A'B'C' the cevian or pedal triangle of P. The Euler line of AB'C' intersects B'C' at A" The Euler line of BC'A'
Antreas Hatzipolakis

Nov 30

### H,O, radical axes

(4)

[APH]: Variation: Let ABC be a triangle, A'B'C' the cevian triangle of H and A"B"C" the cevian triangle of O. Denote: Ma, Mb, Mc = the midpoints of AA", BB",
Antreas Hatzipolakis

Nov 28

### O,N, NPC, collinear

(3)

[APH]: Let ABC be a triangle and A'B'C' the cevian triangle of O. The NPC centers Na, Nb, Nc of NAA', NBB', NCC', resp. are collinear. The line NaNbNc is
Antreas Hatzipolakis

Nov 23

### Midway

(7)

Let ABC be a triangle and P a point. Denote: Oa, Ob, Oc = the corcumcenters of PBC, PCA, PAB, resp. OaaOabOac = the midway triangle of Oa (ie Oaa, Oab, Oac =
Antreas Hatzipolakis

Nov 19

### N, Euler lines

(5)

In general : Let ABC be a triangle and L1, L2, L3 three arbitrary lines passing through O Denote: Aa, Ab, Ac = the orthogonal projections of A on L1, L2, L3,
Antreas Hatzipolakis

Nov 16

### O, NPC, circumcyclologic

(4)

[APH]: Let ABC be a triangle and A'B'C', A"B"C" the antimedial, medial triangle, resp. Denote: Na, Nb, Nc = the NPC centers of HB'C', HC'A', HA'B', resp. (H of
Antreas Hatzipolakis

Nov 13

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## Trending Topics

See All- Isogonal conjugates, coaxial 4 Posts
- Circumhyperbola center 4 Posts
- I, Coaxial 13 Posts
- Altintas hyperbola 5 Posts
- In a given triangle (not... 5 Posts

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