Specifically, let , , be the vertices of , and let be its centroid (the intersection of its three medians). Let , , and be the midpoints of the sidelines , , and , respectively. It turns out that the circumcenters of the six triangles , , , , , and lie on a common circle, which is the van Lamoen circle of .[2]
History
The van Lamoen circle is named after the mathematician Floor van Lamoen[nl] who posed it as a problem in 2000.[3][4] A proof was provided by Kin Y. Li in 2001,[4] and the editors of the Amer. Math. Monthly in 2002.[1][5]
In 2003, Alexey Myakishev and Peter Y. Woo proved that the converse of the theorem is nearly true, in the following sense: let be any point in the triangle's interior, and , , and be its cevians, that is, the line segments that connect each vertex to and are extended until each meets the opposite side. Then the circumcenters of the six triangles , , , , , and lie on the same circle if and only if is the centroid of or its orthocenter (the intersection of its three altitudes), at which point the six circumcenters degenerate into the three Euler points of the nine-point circle.[6] A simpler proof of this result was given by Nguyen Minh Ha in 2005.[7]
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