Difference between revisions of "Euler line"

m
Line 1: Line 1:
 
In any [[triangle]] <math>\triangle ABC</math>, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] <math>H</math>, [[centroid]] <math>G</math>, [[circumcenter]] <math>O</math>, [[nine-point center]] <math>N</math> and [[De Longchamps point | de Longchamps point]] <math>L</math>.  It is named after [[Leonhard Euler]].  Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points.  In particular, <math>\overline{OGNH}</math> and <math>OG:GN:NH = 2:1:3</math>
 
In any [[triangle]] <math>\triangle ABC</math>, the '''Euler line''' is a [[line]] which passes through the [[orthocenter]] <math>H</math>, [[centroid]] <math>G</math>, [[circumcenter]] <math>O</math>, [[nine-point center]] <math>N</math> and [[De Longchamps point | de Longchamps point]] <math>L</math>.  It is named after [[Leonhard Euler]].  Its existence is a non-trivial fact of Euclidean [[geometry]]. Certain fixed orders and distance [[ratio]]s hold among these points.  In particular, <math>\overline{OGNH}</math> and <math>OG:GN:NH = 2:1:3</math>
  
Given the [[orthic triangle]] <math>\triangle H_AH_BH_C</math> of <math>\triangle ABC</math>, the Euler lines of <math>\triangle AH_BH_C</math>,<math>\triangle BH_CH_A</math>, and <math>\triangle CH_AH_B</math> [[concurrence | concur]] at <math>N</math>, the nine-point center of <math>\triangle ABC</math>. <span style="color:#ff0000"> Is this true? </span>
+
Given the [[orthic triangle]] <math>\triangle H_AH_BH_C</math> of <math>\triangle ABC</math>, the Euler lines of <math>\triangle AH_BH_C</math>,<math>\triangle BH_CH_A</math>, and <math>\triangle CH_AH_B</math> [[concurrence | concur]] at <math>N</math>, the nine-point circle of <math>\triangle ABC</math>.
  
 
==Proof Centroid Lies on Euler Line==
 
==Proof Centroid Lies on Euler Line==

Revision as of 20:35, 3 August 2017

In any triangle $\triangle ABC$, the Euler line is a line which passes through the orthocenter $H$, centroid $G$, circumcenter $O$, nine-point center $N$ and de Longchamps point $L$. It is named after Leonhard Euler. Its existence is a non-trivial fact of Euclidean geometry. Certain fixed orders and distance ratios hold among these points. In particular, $\overline{OGNH}$ and $OG:GN:NH = 2:1:3$

Given the orthic triangle $\triangle H_AH_BH_C$ of $\triangle ABC$, the Euler lines of $\triangle AH_BH_C$,$\triangle BH_CH_A$, and $\triangle CH_AH_B$ concur at $N$, the nine-point circle of $\triangle ABC$.

Proof Centroid Lies on Euler Line

This proof utilizes the concept of spiral similarity, which in this case is a rotation followed homothety. Consider the medial triangle $\triangle O_AO_BO_C$. It is similar to $\triangle ABC$. Specifically, a rotation of $180^\circ$ about the midpoint of $O_BO_C$ followed by a homothety with scale factor $2$ centered at $A$ brings $\triangle ABC \to \triangle O_AO_BO_C$. Let us examine what else this transformation, which we denote as $\mathcal{S}$, will do.

It turns out $O$ is the orthocenter, and $G$ is the centroid of $\triangle O_AO_BO_C$. Thus, $\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}$. As a homothety preserves angles, it follows that $\measuredangle O_AOG = \measuredangle AHG$. Finally, as $\overline{AH} || \overline{O_AO}$ it follows that \[\triangle AHG = \triangle O_AOG\] Thus, $O, G, H$ are collinear, and $\frac{OG}{HG} = \frac{1}{2}$.

Proof Nine-Point Center Lies on Euler Line

Assuming that the nine point circle exists and that $N$ is the center, note that a homothety centered at $H$ with factor $2$ brings the Euler points $\{E_A, E_B, E_C\}$ onto the circumcircle of $\triangle ABC$. Thus, it brings the nine-point circle to the circumcircle. Additionally, $N$ should be sent to $O$, thus $N \in \overline{HO}$ and $\frac{HN}{ON} = 1$.

~always_correct


Euler Line.PNG



This article is a stub. Help us out by expanding it.

Invalid username
Login to AoPS