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1964. A Construction For X(18569)
Let A1B1C1 reflection triangle of X(3)-circumcenter on the side lines of ABC.
Tangents to circumcircle of A1B1C1 at A1, B1, C1 form a triangle A2B2C2.**Circumcenter of A2B2C2 lies on Euler line of ABC. It's X(18569).
1963. A Triangle Inscribed In Nine-Point Circle
Let A1B1C1 median triangle of ABC.
A2: Inverse of X(3)-Circumcenter of ABC in the circle with diameter B1C1.Define B2, C2 cyclically.
*A2B2C2 is inscribed in the Nine-Point Circle of ABC.
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1987. Circumcenter On Euler Line
Let H=X(4)-Orthocenter of ABC. A 1 B 1 C 1 cevian triangle of H. B1 A , C1 A are reflections of B1, C1 on A. L A , line through B1 A , C1...