Difference between revisions of "Euler line"

Revision as of 03:51, 20 October 2022

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

Another Proof

Let $M$ be the midpoint of $BC$ . Extend $CG$ past $G$ to point $H'$ such that $CG = \frac{1}{2} GH$ . We will show $H'$ is the orthocenter. Consider triangles $MGO$ and $AGH'$ . Since $\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}$ , and they both share a vertical angle, they are similar by SAS similarity. Thus, $AH' \parallel OM \perp BC$ , so $H'$ lies on the $A$ altitude of $\triangle ABC$ . We can analogously show that $H'$ also lies on the $B$ and $C$ altitudes, so $H'$ is the orthocenter.

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

Euler line for a triangle with an angle of 120 $^\circ.$

Let the $\angle C$ in triangle $ABC$ be $120^\circ.$ Then the Euler line of the $\triangle ABC$ is parallel to the bisector of $\angle C.$

Proof

Let $\omega$ be circumcircle of $\triangle ABC.$

Let $O$ be circumcenter of $\triangle ABC.$

Let $\omega'$ be the circle symmetric to $\omega$ with respect to $AB.$

Let $E$ be the point symmetric to $O$ with respect to $AB.$

The $\angle C = 120^\circ \implies O$ lies on $\omega', E$ lies on $\omega.$

$EO$ is the radius of $\omega$ and $\omega' \implies$ translation vector $\omega'$ to $\omega$ is $\vec {EO}.$

Let $H'$ be the point symmetric to $H$ with respect to $AB.$ Well known that $H'$ lies on $\omega.$ Therefore point $H$ lies on $\omega'.$

Point $C$ lies on $\omega, CH || OE \implies CH = OE.$

Let $CD$ be the bisector of $\angle C \implies E,O,D$ are concurrent. $OD = HC, OD||HC \implies CD || HO \implies$

Euler line $HO$ of the $\triangle ABC$ is parallel to the bisector $CD$ of $\angle C$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

Concurrent Euler lines and Fermat points

Consider a triangle $ABC$ with Fermat–Torricelli points $F_1$ and $F_2.$ The Euler lines of the $10$ triangles with vertices chosen from $A, B, C, F_1,$ and $F_2$ are concurrent at the centroid of triangle $ABC.$

Case 1 Let $F$ be Fermat point $F_1$ of $\triangle ABC$ maximum angle of which smaller then $120^\circ.$ Then the centroid of triangle $ABC$ lies on Euler line of the $\triangle ABF.$

@@ Line 56: / Line 56: @@
 '''vladimir.shelomovskii@gmail.com, vvsss'''
+==Concurrent Euler lines and Fermat points==
+[[File:Fermat point lines.png|400px|right]]
+Consider a triangle <math>ABC</math> with Fermat–Torricelli points <math>F_1</math> and <math>F_2.</math> The Euler lines of the <math>10</math> triangles with vertices chosen from <math>A, B, C, F_1,</math> and <math>F_2</math> are concurrent at the centroid of triangle <math>ABC.</math>
+<i><b>Case 1</b></i>
+Let <math>F</math> be Fermat point <math>F_1</math> of <math>\triangle ABC</math> maximum angle of which smaller then <math>120^\circ.</math> Then the centroid of triangle <math>ABC</math> lies on Euler line of the <math>\triangle ABF.</math>
 ==See also==

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