Difference between revisions of "Euler line"

Revision as of 18:56, 4 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 $\frac{HN}{ON} = 1$ .

Analytic Proof of Existence

Let the circumcenter be represented by the vector $O = (0, 0)$ , and let vectors $A,B,C$ correspond to the vertices of the triangle. It is well known the that the orthocenter is $H = A+B+C$ and the centroid is $G = \frac{A+B+C}{3}$ . Thus, $O, G, H$ are collinear and $\frac{OG}{HG} = \frac{1}{2}$

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

@@ Line 13: / Line 13: @@
 Assuming that the [[nine point circle]] exists and that <math>N</math> is the center, note that a homothety centered at <math>H</math> with factor <math>2</math> brings the [[Euler point]]s <math>\{E_A, E_B, E_C\}</math> onto the circumcircle of <math>\triangle ABC</math>. Thus, it brings the nine-point circle to the circumcircle. Additionally, <math>N</math> should be sent to <math>O</math>, thus <math>N \in \overline{HO}</math> and <math>\frac{HN}{ON} = 1</math>.
-~always_correct
+==Analytic Proof of Existence==
+Let the circumcenter be represented by the vector <math>O = (0, 0)</math>, and let vectors <math>A,B,C</math> correspond to the vertices of the triangle. It is well known the that the orthocenter is <math>H = A+B+C</math> and the centroid is <math>G = \frac{A+B+C}{3}</math>. Thus, <math>O, G, H</math> are collinear and <math>\frac{OG}{HG} = \frac{1}{2}</math>
 [[Image:Euler Line.PNG||500px|frame|center]]

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Difference between revisions of "Euler line"

Revision as of 18:56, 4 August 2017

Proof Centroid Lies on Euler Line

Proof Nine-Point Center Lies on Euler Line

Analytic Proof of Existence