For P=X(5)-NPC center and P=X(6) symmedian point, X is non-ETC point. If P=X(6), X has first barycentric: (a^4 (b^2 -c^2)^2 (-3 a^6 + b^6 + 7 b^4 c^2 + 7 b^2 c^4 + c^6 + 7 a^4 (b^2 + c^2) - a^2 (5 b^4 + 14 b^2 c^2 + 5 c^4)):...:...)
Let ABC be a triangle with P, Q are two points on Euler line of ABC. A',B',C' are inverse images of A,B,C through circle diame...
For P=X(5)-NPC center and P=X(6) symmedian point, X is non-ETC point. If P=X(6), X has first barycentric: (a^4 (b^2 -c^2)^2 (-3 a^6 + b^6 + 7 b^4 c^2 + 7 b^2 c^4 + c^6 + 7 a^4 (b^2 + c^2) - a^2 (5 b^4 + 14 b^2 c^2 + 5 c^4)):...:...)
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