# Fermat Points and Concurrent Euler Lines

Let F1 and F2 denote the (inner and outer) Fermat-Toricelli points of a given ΔABC. We prove that the Euler lines of the 10 triangles with vertices chosen from A, B, C, F1, F2 (three at a time) are concurrent at the centroid of ΔABC [Beluhov]. This is obviously true for ΔABC itself. The applet below illustrates the case where the triangles are formed by two Fermat's points and one of the vertices A, B, C. There are three such triangles. The other triangles are considered separately.

### 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?

Discussion In the applet,

• A'B'C' is the outer and a'b'c' the inner Napoleon's triangles,
• F and f are the corresponding Fermat's points,
• G and G' are the centroids of ΔABC and ΔFfC, respectively,
• O is the circumcenter of ΔFfC,
• P the intersection of B'C' with a'c',
• M is the midpoint of AB, m the midpoint of Ff,
• F' the reflection of f in M.

We only prove the claim for ΔFfC.

### 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?

As we know, the two Napoleon triangles share the centroid G. Both are invariant under the rotations through 120° with center G. One such rotation maps A'B' and a'b' onto B'C' and c'a', respectively. Therefore, it also maps O to P, implying that ∠OGP = 120°. Since ∠Oa'P = 120°, the four points O, G, a', P are concyclic. Their circumcircle also contains B' because ∠PB'O = 60°. Therefore,

(1)

∠OGB' = ∠Oa'B'.

The same rotation maps ∠Oa'B' onto ∠Pc'C', showing that the two are equal:

(2)

∠Oa'B' = ∠Pc'C'.

Since Pc'⊥Bf and C'c'⊥ BA,

(3)

∠Pc'C' = ∠fBA.

Further, since ∠BF'A = ∠AFB = 120° = 180° - ∠AfB, the quadrilateral AfBF' is cyclic, giving ∠fBA = ∠fF'A. It follows that

(4)

∠fF'A = ∠OGB'.

Now, from AF' || fB⊥ A'C' and B'G⊥ A'C', we see that AF' || B'G. This, together with (4), yields

(5)

F'f || GO.

Points G and G' divide the segments CM and Cm in ratio 2:1, causing GG' || Mm. By construction, FM = MF' and, since also Fm = mf, the same argument shows that Mm || F'f. Combining this with (5) shows that GG' || GO, implying the collinearity of G', G and O.

### References

1. N. I. Beluhov, Ten Concurrent Euler Lines, Forum Geometricorum, Volume 9 (2009) 271-274 ### Napoleon's Theorem 