## 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