## Fermat point and 9-point Centers: What Is This About? A Mathematical Droodle

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Explanation

### Explanation

The applet attempts to suggest an engaging result from C. Bradley's article, The Fermat Point Configuration, in the Math Gazette (2008, pp. 214-222.)

Let F be the Fermat point of ΔABC. Let Tc, Ta, Tb denote the 9-point centers of triangles ABF, BCF, CAF with centroids Gc, Ga, Gb, respectively. Then

1. the Euler lines of the three triangles are parallel to CF, AF, BF, respectively,
2. they meet at G - the centroid of ΔABC,
3. ΔTaTbTc is equilateral,
4. Points F, G, Ta, Tb, Tc are concyclic, with FG being a diameter of their circumcircle.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

First of all, the lines through centroids Ga, Gb, Gc parallel to AF, BF, and CF, respectively, meet at G. Also, because of the choice of F, the Euler line in ΔABF is parallel to FIc, where Ic is the incenter of ΔABF. But, the Fermat point may be constructed from the Napoleon triangle ABC', BCA', and CAB' as the intersection of AA', BB', and CC'. Each of these three lines bisects the angle formed by the other two. So it follows that the Euler line in ΔABF is parallel to CF. It also passes through Gc and, hence, through G. Triangles BCF and CAF can be treated similarly implying that the Euler lines ea, eb, ec of triangles BCF, CAF, and ABF meet in G where they are equally inclined to each other.

As we know, points Ta, Tb, Tc are the feet of perpendiculars from F to ea, eb, ec. This makes angles FTaG, FTbG, FTcG right so that the five points F, G, Ta, Tb, Tc are indeed concyclic. In addition, the perpendiculars from F to equally inclined lines are equally inclined meaning that ΔTaTbTc is equilateral.