Let ABC be a triangle with de Longchamps point L. A'B'C' is pedal triangle of L. L' is de Longchamps point of triangle A'B'C'. L* is isogonal conjugate of L with respect to triangle A'B'C'. Prove that L, L* and L' are collinear.
[Angel Montesdeoca]:
L' =
= X(30)X(21651)∩X(51)X(5895)
a^2 ( a^12 (b^2+c^2)-4 a^10 (b^2-c^2)^2+5 a^8 (b^2-c^2)^2 (b^2+c^2) -40 a^6 b^2 c^2 (b^2-c^2)^2-5 a^4 (b^2-c^2)^2 (b^6-9 b^4 c^2-9 b^2 c^4+c^6) +4 a^2 (b^2-c^2)^2 (b^8-2 b^6 c^2-14 b^4 c^4-2 b^2 c^6+c^8) -(b^2-c^2)^4 (b^6+7 b^4 c^2+7 b^2 c^4+c^6)) : :
= lies on these lines: {30,21651}, {51,5895}, {185,1885}, {373,5893}, {1498,6090}, {1906,6247}, {2777,21649}, {3357,15030}, {3917,5894}, {4319,7355}, {4320,6285}, {5650,8567}, {5878,6816}, {6225,7386}, {10575,12362}, {10990,15105}, {12163,13093}, {12315,14641}, {13417,16879}, {14642,17807}
= the reflection of X(i) in X(j), for these {i, j}: {5562,20427}, {11381,64}, {12315,14641}
(6 - 9 - 13) - search numbers of L': (86.3230025239081, 92.0881467899617, -99.9540536914082).
L* = X(12111).
Triangle centers on the line (L, L', L*): X(i) for i= 20, 2979, 5562, 6000, 9833, 12058, 12111, 12250, 12279, 15318, 15606, 15644, 20213, 20427.
Angel Montesdeoca
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