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 one of the Fermat points and a pair of vertices A, B, C. There are six such triangles. The other three 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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

Suffice it to consider one of the six triangles. We will actually consider two at once, viz., ΔBCF, where F is one of the Fermat points.

In any triangle, the Euler line passes through the centroid, the circumcenter, the orthocenter and the nine-point center. The circumcircle of ΔBCF, coincides with the circumcircle of the equilateral ΔBCA'', implying that A', the center of ΔBCA'', is the circumcenter of ΔBCF. What we are going to show is that the line GA' passes through the centroid, say, G' of ΔBCAF.


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?

Let Ma be the midpoint of side BC. Observe that GMa:CG = 1:2 and also A'Ma:A'A'' = 1:2 so that lines GA' and AA'' are parallel. Point G' lies on FMa and divides it in the same ratio: G'MA:FG' = 1:2. On the other hand, FMa is cut by GA' in the ratio 1:2. It follows that G' lie at the intersection of GA' and FMa and, in particular, on GA' making it the Euler line of ΔBCF.

References

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

Napoleon's Theorem

Related material
Read more...

  • The Euler Line and the 9-Point Circle
  • The altitudes and the Euler Line
  • Existence of the Euler Line: An Elementary Proof
  • Euler Line Cuts Off Equilateral Triangle
  • Fermat Points and Concurrent Euler Lines
  • Remarkable Line in Cyclic Quadrilateral
  • Nine Point Circle: an Elementary Proof
  • Three Euler Lines That Are Four
  • |Activities| |Contact| |Front page| |Contents| |Geometry|

    Copyright © 1996-2018 Alexander Bogomolny

    71471238