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

A Mathematical Droodle

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Copyright © 1996-2017 Alexander Bogomolny### 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 T_{c}, T_{a}, T_{b} denote the 9-point centers of triangles ABF, BCF, CAF with centroids G_{c}, G_{a}, G_{b}, respectively. Then

- the Euler lines of the three triangles are parallel to CF, AF, BF, respectively,
- they meet at G - the centroid of ΔABC,
- ΔT
_{a}T_{b}T_{c}is equilateral, - Points F, G, T
_{a}, T_{b}, T_{c}are concyclic, with FG being a diameter of their circumcircle.

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First of all, the lines through centroids G_{a}, G_{b}, G_{c} 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 FI_{c}, where I_{c} 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 G_{c} and, hence, through G. Triangles BCF and CAF can be treated similarly implying that the Euler lines e_{a}, e_{b}, e_{c} of triangles BCF, CAF, and ABF meet in G where they are equally inclined to each other.

As we know, points T_{a}, T_{b}, T_{c} are the feet of perpendiculars from F to e_{a}, e_{b}, e_{c}. This makes angles FT_{a}G, FT_{b}G, FT_{c}G right so that the five points F, G, T_{a}, T_{b}, T_{c} are indeed concyclic. In addition, the perpendiculars from F to equally inclined lines are equally inclined meaning that ΔT_{a}T_{b}T_{c} is equilateral.

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Copyright © 1996-2017 Alexander Bogomolny62383822 |