Morley configuration has more miracles per square meter than any other configuration of triangle geometry.
(Antreas)
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Let ABC be a triangle, A'B'C' the internal Morley Triangle and A"B"C" the adjunt triangle.
1. TRISECTORS OF BA'C, CB'A, AC'B:
Trisectors of BA'C:
AbA'bA"b with Ab on BC near B, A'b on AC, A"b on AB
AcA'cA"c with Ac on BC near C, A'c on AB, A"c on AC
Similarly the trisectors of CB'A, AC'B.
1.1. Denote:
A1 = AbBc /\ AcCb
A2 = BaCb /\ CaBc
Similarly B1,B2 and C1, C2.
Conjecture 1:
The lines A1A2,B1B2,C1C2 are concurrent (at a point P1).
Antreas P. Hatzipolakis, Anopolis #213
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The lines are concurrent, at a non-ETC point (search 1.372808061836111). I could not find any collinearities with existing Morley-related centers.
Also, the points Ab, Ac, Bc, Ba, Ca, Cb all lie on a common ellipse, with center at non-ETC point (search 1.839251887562039) and perspector at non-ETC point (search 1.102316990783906). I also could find non collinearities with these centers.
Randy Hutson Anopolis #217
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1.2. Denote:
A'1 = A'bB'c /\ A'cC'b
A'2 = B'aC'b /\ C'aB'c
Similarly B'1,B'2 and C'1, C'2.
Conjecture 2:
The lines A'1A'2,B'1B'2,C'1C'2 are concurrent (at a point P2).
The points A'b, A'c, B'c, B'a, C'a, C'b lie on a conic.
1.3. Denote:
A"1 = A"bB"c /\ A"cC"b
A"2 = B"aC"b /\ C"aB"c
Similarly B"1,B"2 and C"1, C"2.
Conjecture 3:
The lines A"1A"2,B"1B"2,C"1C"2 are concurrent (at a point P3).
The points A"b, A"c, B"c, B"a, C"a, C"b lie on a conic.
Antreas P. Hatzipolakis, 4 May 2013
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