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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
4
posts
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
4
posts
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
4
posts
Apr 19

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
3
posts
Apr 19

Orthologic triangles - Locus

(14)
Orthologic centers and homography Angel Montesdeoca, Centros ortológicos y homografía ...
Antreas Hatzipolakis
14
posts
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
5
posts
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
5
posts
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
6
posts
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
3
posts
Feb 26
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