Παρασκευή 3 Μαΐου 2013

MORLEY

Let ABC be a triangle and A'B'C' the internal Morley triangle. The trisectors of the angle BA'C intersect BC at Ab,Ac near to B,C, resp. Similarly Bc,Ba and Ca,Cb.

Denote:

A1 = AbBc /\ AcCb

A2 = BaCb /\ CaBc

Similarly B1,B2 and C1, C2.

Conjecture:

The lines A1A2,B1B2,C1C2 are concurrent.

Antreas P. Hatzipolakis, 3 May 2013

Conjecture 2: The same for A'B'C' = the adjunct triangle.

Antreas P. Hatzipolakis, 4 May 2013

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