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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
13
posts
9:27 AM

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
5
posts
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
5
posts
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
5
posts
Dec 6

Cevian, NPC, Orthologic

(3)
[APH]: Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P. Denote: Ma, Mb, Mc = the midpoints of AA', BB', CC', resp. Na, Nb, Nc = the NPC
Antreas Hatzipolakis
3
posts
Dec 5

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
4
posts
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
10
posts
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
5
posts
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
4
posts
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
4
posts
Nov 28

O, Concurrent Euler lines

(3)
[APH]: Let ABC be a triangle, P a point and A'B'C' the cevian triangle of Ο, Denote: Ma, Mb, Mc = the midpoints of AA', BB', CC', resp. Mab, Mac = the
Antreas Hatzipolakis
3
posts
Nov 27

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
3
posts
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
7
posts
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
5
posts
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
4
posts
Nov 13
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