Euler Lines Concurrent on the Nine-Point Circle: 10796

Abstract

Solution by W Weston Meyer, General Motors Corporation, Warren MI. Let H be the orthocenter of The nine-point circle circumscribes A'B'C' and bisects the segments AH, BH, and CH in points A, B, and C, respectively (Theorem 1.81, page 20 in H. S. M. Coxeter & S. L. Greitzer, Geometry Revisited, Random House, 1967), so and ABC are homothetic. Quadrilateral AB'HC' is cyclic, since it has right angles at B' and C'. Thus its circumcenter, and the circumcenter of AB'C', is A. Now AB'B and AC'C are similar, since two corresponding angles are congruent. This implies that AB'/AC' = A B/AC, and therefore AB'C' and ABC are similar. By the same reasoning, AB'C', A'BC', and A'B'C are all similar (Coxeter & Greitzer, Exercise 1, page 17) and have A, B, and C for their circumcenters, respectively. Let e(A), e(B), and e(C) denote the Euler lines of AB'C', A'BC', and A'B'C, respectively. If the Euler line of ABC is parallel to a side of then e(A), e(B), and e(C) concur in a vertex of ABC. Otherwise, say that e(A) and e(B) meet in a point P. Because of the similarity of AB'C' and A'B C', the angle that e(A) makes with AB' (or with AC) is the same as the angle e(B) makes with A'B (or with CB). So in APB the interior angle at P equals either ZACB or its supplement. Hence P lies on the circumcircle of ABC, the nine-point circle of By the same reasoning, the other two pairs of Euler lines also meet at points of the nine-point circle. If the three Euler lines were not concurrent, then one of them would meet the nine-point circle in three distinct points, which is impossible.