Van Lamoen circle



In Euclidean plane geometry, the van Lamoen circle is a special circle associated with any given triangle $$T$$. It contains the circumcenters of the six triangles that are defined inside $$T$$ by its three medians.

Specifically, let $$A$$, $$B$$, $$C$$ be the vertices of $$T$$, and let $$G$$ be its centroid (the intersection of its three medians). Let $$M_a$$, $$M_b$$, and $$M_c$$ be the midpoints of the sidelines $$BC$$, $$CA$$, and $$AB$$, respectively. It turns out that the circumcenters of the six triangles $$AGM_c$$, $$BGM_c$$, $$ BGM_a$$, $$CGM_a$$, $$CGM_b$$, and $$AGM_b$$ lie on a common circle, which is the van Lamoen circle of $$T$$.

History
The van Lamoen circle is named after the mathematician Floor van Lamoen who posed it as a problem in 2000. A proof was provided by Kin Y. Li in 2001, and the editors of the Amer. Math. Monthly in 2002.

Properties
The center of the van Lamoen circle is point $$X(1153)$$ in Clark Kimberling's comprehensive list of triangle centers.

In 2003, Alexey Myakishev and Peter Y. Woo proved that the converse of the theorem is nearly true, in the following sense: let $$P$$ be any point in the triangle's interior, and $$AA'$$, $$BB'$$, and $$CC'$$ be its cevians, that is, the line segments that connect each vertex to $$P$$ and are extended until each meets the opposite side. Then the circumcenters of the six triangles $$APB'$$, $$APC'$$, $$BPC'$$, $$ BPA'$$, $$CPA'$$, and $$CPB'$$ lie on the same circle if and only if $$P$$ is the centroid of $$T$$ or its orthocenter (the intersection of its three altitudes). A simpler proof of this result was given by Nguyen Minh Ha in 2005.