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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

Re: Point on the IH line

(4)
[APH]: The perpendiculars to HIa, HIb, HIc from A,B,C, resp. bound triangle A"B"C". Which point is the circumenter of A"B"C" ? Is it lying on the IH line? ...
Antreas Hatzipolakis
4
posts
Feb 24

Collinear Points

(10)
... *********************************************** [C├ęsar Lozada] It is X(1141). [Peter Moses] Hi Antreas, X(1141). Best regards Peter. [APH]: [APH]: Let ABC
Antreas Hatzipolakis
10
posts
Feb 19

Interesting circle

(4)
... Extraversions: Following are "theoretic" ie written without drawings. Hope they are true..... :-) We have two incircles in the configuration, so there
Antreas Hatzipolakis
4
posts
Feb 11

Cyclology of hexagon

(6)
From: Angel Montesdeoca [Attachment(s) <#14b5bc8ac33fefb1_TopText> from Angel Montesdeoca included below] Let's rewrite the Anopolis #2404 and Anopolis #2406
Antreas Hatzipolakis
6
posts
Feb 5

A property of X(597) and a locus

(6)
Dear Seiichi, The Eulerline through the centroids of quadrangles made from 4 centroids, 4 circumcenters, and 4 orthocenters of its component triangles is the
Antreas Hatzipolakis
6
posts
Jan 29

Reflections - NPCs - Cyclologic triangles

(4)
... Discussion here: https://www.facebook.com/photo.php?fbid=773944689348296&set=a.237986886277415.57448.100001983178784&type=1&theater APH
Antreas Hatzipolakis
4
posts
Jan 28
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