Let a triangle ABC, and three equilateral triangle A1A2A3,B1B2B3,C1C2C3. They have centroid is G. A2A3 is equal and parallel BC, B2B3 is equal and parallel AC, C2C3 equal and parallel AC.
Problem 1: Three triangle A1B3C2, B1C3A2, C1A3B2 is an equiletaral triangle and they equal.
Problem 2: D,E,F are centroid of three triangle A1B3C2, B1C3A2, C1A3B2 => DEF is equilateral triangle.
Problem 3: Nine lines are side of A1B3C2, B1C3A2, C1A3B2 intersection.
Three triangle (magenta) are equilateral triangle and equal.
Six triangle (red) are equilateral triangle and equal.
Three triangle (yellow) are equilateral triangle and equal.
Three triangle (cyan) are equilateral triangle and equal.
Problem 4: Ga,Gb,Gc are centroid of three triangle AA2A3, BB2B3, C1C2C3 => Ga,Gb,Gc is equilateral triangle
Problem 5: A4,A5,B4,B5,C4,C5 are midpoint AB,A2A3,CA,B2B3,AB,C2C3 => A4A5, B4B5, C4C5 are concurrent at Femat point of triangle A4B4C4
Problem 6: M,N,P are midpoint of C4C5,A4A5,B4B5 => MNP is the equilateral triangle.
Problem 8: Centroid of MNP at G
Problem 9: A1,B3,A,C2 are concyclic. B1,C3,B,A are concyclic. C1,A3,C,B2 are concyclic. And three circles are equal and AA1=BB1=CC1 and they are diamiter of circle
Problem 10: AA1,BB1,CC1 concurrent at Fermat point of ABC
