Τετάρτη, 30 Ιανουαρίου 2013

St. Gregory the Theologian

.... μηδὲ μία χελιδὼν ἔαρ ποιεῖ, μηδὲ γραμμὴ μία τὸν γεωμέτρην, ἢ πλοῦς εἷς τὸν θαλάττιον.
.... one swallow does not make spring nor the geometer a line nor the sailor a sea trip.
Ἁγίου Γρηγορίου τοῦ Θεολόγου, Εἰς τὰ ἅγια Φῶτα.

COLLINEAR NPC CENTERS ?


Let ABC be a triangle, L a line passing through H (orthocenter), intersecting the sidelines BC,CA,AB at A',B',C', resp., La,Lb,Lc the reflections of L in the sidelines BC,CA,AB, resp. (concurrent at a point S on the circumcircle) and Ab,Ac the orthogonal projections of A' on Lb,Lc, resp., Bc,Ba the orthogonal projections of B' on Lc,La, resp. and Ca,Cb the orthogonal projections of C' on La,Lb, resp. The NPC centers Na,Nb,Nc of of A'AbAc, B'BcBa, C'CaCb resp. are collinear. (??)

Antreas P. Hatzipolakis, 30 Jan. 2013

Τρίτη, 29 Ιανουαρίου 2013

CONICS CENTERED AT O

Let ABC be a triangle and r1,r2,r3 three not equal line segments.

Denote:

a1 = the circle centered at A with radius r1. Similarly ....

a1b2 = the radical axis of the circles a1 and b2. Similarly .....

Six Radical centers:

(a1,b2,c3), (a1,b3,c2), (a2,b3,c1), (a2,b1,c3), (a3,b1,c2), (a3,b2,c1)

Six other points of concurrent radical axes:

(a1b2,b3c1,c2a3), (a1b3,b2c1,c3a2), (a2b3,b1c2,c3a1), (a2b1,b3c2,c1a3), (a3b1,b2c3,c1a2),(a3b2,b1c3,c2a1)

The 12gon has opposite sides parallel and equal. It is inscribed on a conic centered at the circumcenter O = radical center of (a1,b1,c1) and (a2,b2,c2) and (a3,b3,c3)

Antreas P. Hatzipolakis. 29 Jan. 2013

Σάββατο, 19 Ιανουαρίου 2013

A MALFATTI - LIKE PROBLEM

Let ABC be a triangle. To draw three circles, each of which is tangent to the other two and to one side of ABC and to the circumcircle of ABC.

Perspective triangles.

Let (Ka), (Kb), (Kc) be the three circles.

(Ka) is tangent to the circumcircle at A', to the BC at A" and to the other two circles (Kb), (Kc) at C*, B*, resp.

(Kb) is tangent to the circumcircle at B', to the CA at B" and to the other two circles (Kc), (Ka) at A*, C*, resp.

(Kc) is tangent to the circumcircle at C', to the AB at C" and to the other two circles (Ka), (Kb) at B*, A*, resp.

The triangles A'B'C', A*B*C* are perspective (??)


Antreas P. Hatzipolakis, 19 Jan. 2013

Τρίτη, 15 Ιανουαρίου 2013

ΕΝΑ ΠΡΟΒΛΗΜΑ ΤΟΥ ALEX MYAKISHEV


Έστω ΑΒC ένα οξυγώνιο τρίγωνο και (Ma), (Mb), (Mc) οι τρεις κύκλοι που εφάπτονται εσωτερικά του Κύκλου των Εννέα Σημείων και καθένας δύο πλευρών του τριγώνου (διαφορετικοί απο τον εγγεγραμμένο κύκλο). Τα κέντρα των κύκλων αυτών είναι συγγραμμικά.
Hyacinthos #21396
Να αποδειχθεί συνθετικά, δηλαδή με την Ευκλείδεια Γεωμετρία.

Ο Έλληνας λύτης του προβλήματος θα λάβει ως έπαθλο το τετράτομο έργο ΑΣΚΗΣΕΙΣ ΓΕΩΜΕΤΡΙΑΣ (γνωστό ως "Γεωμετρία των Ιησουιτών").


Παρασκευή, 11 Ιανουαρίου 2013

ORTH. PROJECTIONS OF MORLEY TRIANGLE VERTICES -2


Let A'B'C' be the Morley 1st Triangle.

Denote

A'b, A'c = the orthogonal projections of A' on CB', BC', resp.

B'c, B'a = the orthogonal projections of B' on AC', CA', resp.

C'a, C'b = the orthogonal projections of C' on BA', AB', resp.

1. A'b,A'c, B'c, B'a, C'a, C'b are con-conic (??).

2. The Euler Lines of A'AbAc, B'B'cB'a, C'C'aC'b are concurrent. (??)

Antreas P. Hatzipolakis, 11 Jan. 2013

ORTH. PROJECTIONS OF MORLEY TRIANGLE VERTICES -1

Let A'B'C' be the Morley 1st Triangle.

Denote

A'b, A'c = the orthogonal projections of A' on AB', AC', resp.

B'c, B'a = the orthogonal projections of B' on BC', BA', resp.

C'a, C'b = the orthogonal projections of C' on CA', CB', resp.

1. A'b,A'c, B'c, B'a, C'a, C'b are concyclic (??).

2. The Euler Lines of A'AbAc, B'B'cB'a, C'C'aC'b are concurrent (??).

Antreas P. Hatzipolakis, 11 jan. 2012


Κυριακή, 6 Ιανουαρίου 2013

HOMOTHETIC EQUILATERAL TRIANGLES

Let A'B'C', A"B"C" be two homothetic equilateral triangles.

Conjecture: The Euler lines of the triangles A'B"C", B'C"A", C'A"B" are concurrent, and also the Euler lines of the triangles A"B'C', B"C'A', C"A'B', if no one of the 6 triangles is degenerated.


A. P. Hatzipolakis, Hyacithos #21357

Πέμπτη, 3 Ιανουαρίου 2013

MORLEY TRIANGLES Conjecture

See the discussion in Hyacinthos #21341

Conjecture:

Let A'B'C' be the internal Morley triangle of triangle ABC and A"B"C" the (homothetic) Roussel equilateral triangle(*). The Euler Lines of A'B"C", B'C"A", C'A"B" are concurrent.

(*)The Roussel triangle Here and with figure Here

Antreas P. Hatzipolakis
 

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