## Κυριακή, 30 Ιουνίου 2013

### ANOPOLIS CIRCLE (2)

Let ABC be a triangle and L1,L2,L3 the Euler lines of IBC,ICA,IAB, resp. (concurrent at Schiffler point).
Denote:
E1 = the point of concurrence of the reflections of L1 in the sidelines of IBC (on the circumcircle of IBC)
E2 = the point of concurrence of the reflections of L2 in the sidelines of ICA (on the circumcircle of ICA)
E3 = the point of concurrence of the reflections of L3 in the sidelines of IAB (on the circumcircle of IAB)

The points E1,E2,E3 lie on the circle with diameter IO.
Antreas P. Hatzipolakis, 30 June 2013

### Triangle Construction h_a, m_a, |b-c|/a

To construct triabgle ABC if are given:

altitude h_a, median m_a and the ratio |b-c|/a = m/n

Reference:

EUCLID, June 1969, p. 420-1

## Παρασκευή, 7 Ιουνίου 2013

### CONCURRENT EULER LINES

Let ABC be a triangle, IAb, IAc the trisectors of BIC (with Ab, Ac near B, C, resp.), IBc, IBa the trisectors of CIA (with Bc, Ba near C, A, resp.) and ICa, ICb the trisectors of AIB (with Ca, Cb near A, B, resp.)

Conjecture:

The Euler Lines of IBcCb, ICaAc, IAbBa are concurrent.

Antreas P. Hatzipolakis, 7 June 2013

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The Euler lines do not concur.

However, your configuration does provide a construction for a new ellipse through Ab, Ac, Bc, Ba, Ca, Cb, with some new (and one existing) centers:

Center of ellipse = (non-ETC search 1.883071694412417), on line 1,3604.

Let A' be the intersection of tangents to ellipse at Ab and Ac, and define B', C' cyclically.

Let A" be the intersection of tangents to ellipse at Ba and Ca, and define A", B" cyclically.

Let A* be the intersection of tangents to ellipse at Bc and Cb, and define B*, C* cyclically. The lines AA', BB', CC' concur in P1 = (non-ETC search 1.178825206518684), the perspector of the ellipse.

The lines AA"A*, BB"B*, CC"C* concur in X(3604).

The lines A'A", B'B", C'C" concur in P2 = (non-ETC search 1.249947393611434).

The lines A'A*, B'B*, C'C* concur in P3 = (non-ETC search 1.043484806530543).

P2 and P3 are collinear with X(3604).

Some variations to explore:

1) other starting points besides I

2) substitute external angle trisectors for internal

Randy Hutson Anopolis #377

## Κυριακή, 2 Ιουνίου 2013

### N1N2N3 - PERSPECTIVE

Let ABC be a triangle, A'B'C' the antipedal triangle of I (excentral tr.), N1,N2,N3 the NPC centers of IBC, ICA, IAB, resp. and Na,Nb,Nc the NPC centers of A'BC, B'CA, C'AB, resp.

1. The triangles A'B'C', N1N2N3 are perspective.

Perspector:

(aabc+a(a+b+c)(bb+4bc+cc-aa) , ... , ...)

Barry Wolk, Anopolis #347

ETC X(5506)

2. The triangles N1N2N3, NaNbNc are perspective (?).

Antreas P. Hatzipolakis, 2 June 2013

### CONCURRENT CIRCUMCIRCLES - N1N2N3, NaNbNc

Let ABC be a triangle, A'B'C' the antipedal triangle of I (excentral triangle) N1,N2,N3 the NPC centers of IBC, ICA, IAB, resp. and Na, Nb, Nc the NPC centers of A'BC, B'CA, C'AB, resp.

1. The circumcircles of ABC, AN2N3, BN3N1, CN1N2 are concurrent.

********************

It is now the center X(5606) in ETC

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2. The circumcircles of ABC, ANbNc, BNcNa, CNaNb are concurent.

Note: The circumcenter of NaNbNc is the O.

Antreas P. Hatzipolakis, 2 June 2013

Addendum (26 - 11 - 2013)

1'. The circumcircles of N1BC, N2CA, N3AB are concurrent.

2'. The circumcircles of NaBC, NbCA, NcAB are concurrent.

APH, Anopolis #1119

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*** 1'. X(502)

*** 2'. ( (b+c) (a^6- a^4(b^2+c^2) - a^2(b^4-3b^2c^2+c^4) - 2a b c(b-c)^2(b+c) + (b-c)^4(b+c)^2 ) : ... : ... )

with (6-9-13)-search number 2.5719845710987353841258271936

Angel Montesdeoca, Anopolis #1120

*******************************

( (b+c) (a^6- a^4(b^2+c^2) - a^2(b^4-3b^2c^2+c^4) - 2a b c(b-c)^2(b+c) + (b-c)^4(b+c)^2 ) : ... : ... ) = R X[65]-(2r+R) X[1365], is on lines {{1,149},{10,1109},{36,759},{37,115},{65,1365},{162,1838},{267,3336},{897,1738},{1054,1247},{1737,2166},{2218,2915}}.

isogonal conjugate X(5127)

X(2245) cross-conjugate of X(226)

trilinear pole of line X(661) X(2294)

X(3) isoconjugate of X(2074)

X(21) isoconjugate of X(5172)

trilinear product of X(523) & X(1290)

barycentric product of X(1290) & X(1577)

antigonal of X(502)

Peter J.C. Moses, 2 Dec 2013

***************

It is now X(5620) = ISOGONAL CONJUGATE OF X(5127) in ETC