Thébault's Problem IV

Here is a problem proposed in 1949 by Victor Thébault in the American Mathematics Monthly, with the solution published a year later (American Mathematics Monthly, Vol. 58, No. 1 (Jan., 1951), p. 45).

Solution

The solution is a based on a general property of three directly similar figures.

The three triangles AB'C', A'BC', A'B'C are cut off from ΔABC by its orthic triangle formed by three antiparallels. The triangles are therefore directly similar, with centers of spiral similarity at the feet of the altitudes: A' for A'BC', A'B'C, B' for AB'C', A'B'C, and C' for AB'C', A'BC'. It follows that ΔA'B'C' is the triangle of similitude while the nine-point circle of ΔABC is the circle of similitude.

Under spiral similarities, circumcenters map on circumcenters, orthocenters map on orthocenters. Therefore, Euler lines map on Euler lines, i.e., Euler lines are homologous under spiral similarities. They thus meet in a point that lies on the circle of similarity, the nine-point circle in this case.

... to be continued ...