Difference between revisions of "Euler line"

Revision as of 17:50, 5 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 $OG:GN:NH = 2:1:3$

Euler line is the central line $L_{647}$ .

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 $\frac{OG}{HG} = \frac{1}{2}$

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 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>
+Euler line is the central line <math>L_{647}</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 circle of <math>\triangle ABC</math>.

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Difference between revisions of "Euler line"

Revision as of 17:50, 5 August 2017

Contents

Proof Centroid Lies on Euler Line

Proof Nine-Point Center Lies on Euler Line

Analytic Proof of Existence

See also