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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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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

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Collinear Points
(10)
... *********************************************** [César Lozada] It is X(1141). [Peter Moses] Hi Antreas, X(1141). Best regards Peter. [APH]: [APH]: Let ABC

Antreas Hatzipolakis
10
posts
Feb 19

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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

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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

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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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H - Concurrent NPCs (Re: NPC - Orthologic triangles)
(5)
[APH] Are lying the 6 NPC centers N1 N2, N3, N_A, N_B, N_C on a conic (Circle?) ? They are concyclic on the circle with radius OH/2 and center X(6102)=19th

Antreas Hatzipolakis
5
posts
Jan 25

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Mapping of P on the pedal circle of P
(4)
[APH] 1. The circumcenter of IaIbIc is the isogonal cojugate of P. 2. The parallels through A', B', C' of the perpendicular bisectors of IbIc, IcIa, IaIb,

Antreas Hatzipolakis
4
posts
Jan 21

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Euler Line
(12)
From: César Lozada [APH] For which (m,n) the perspector lies on the Euler line? For all m,n. The perspector Z(m,n) has trilinears: Z(m,n) =

Antreas Hatzipolakis
12
posts
Jan 20

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A Miquel point
(4)
... Dear César and Randy Thanks !! Generalization. Let ABC be a triangle and A'B'C' the cevian triangle of a point P. The parallel to BC through the midpoint

Antreas Hatzipolakis
4
posts
Jan 16

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A Circumcenter
(3)
... Synthetic proof by Telv Cohl http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=620608&p=3708849#p3708849 We have now a locus problem: Let Q, Q'

Antreas Hatzipolakis
3
posts
Jan 11

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Pedal Triangle - Cyclologic Triangles
(3)
From: César Lozada Yes. Cyclologic centers for P=u:v:w (trilinears) Zp = f(a,b,c,u,v,w) : : Za = g(a,b,c,u,v,w) : : where f(a,b,c,u,v,w)=

Antreas Hatzipolakis
3
posts
Jan 11

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Collinear N's
(4)
Antreas Hatzipolakis: Let ABC be a triangle and N, Nn, Nh the NPC centers ... [Peter Moses]: Nn is on lines {{5,51},{195,5640},{5943,6153}} Midpoint X(143) &

Antreas Hatzipolakis
4
posts
Jan 8

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Orthologic triangles - Locus
(10)
From: César Lozada 5. OaObOc, O1O2O3 are orthologic? [César Lozada]: 5) Couln´t calculate ************************ I think H lies on the locus

Antreas Hatzipolakis
10
posts
Jan 7

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Pedal Triangle - Orthocenters - Collinear points
(8)
[APH] Equivalently: Let ABC be a triangle and P a point. Denote: Aa = (Parallel through B to PC) /\ (Parallel through C to PB) A1 = the isogonal conjugate of

Antreas Hatzipolakis
8
posts
Jan 3
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