Difference between revisions of "Euler line"

Revision as of 01:45, 6 December 2018

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 $CG$ 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. $\square$

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

@@ Line 16: / Line 16: @@
 ==Another Proof==
 Let <math>M</math> be the midpoint of <math>BC</math>.
-Extend <math>CG</math> to point <math>H'</math> such that <math>CG = \frac{1}{2} GH</math>. We will show <math>H</math> is the orthocenter.
+Extend <math>CG</math> to point <math>H'</math> such that <math>CG = \frac{1}{2} GH</math>. We will show <math>H'</math> is the orthocenter.
-Consider triangles <math>MGO</math> and <math>AGH</math>. Since <math>\frac{MG}{GA}=\frac{HG}{GC} = \frac{1}{2}</math>, and they both share a vertical angle, they are similar by SAS similarity. Thus, <math>AH \parallel OM \perp BC</math>, so <math>H</math> lies on the <math>A</math> altitude of <math>\triangle ABC</math>. We can analogously show that <math>H</math> also lies on the <math>B</math> and <math>C</math> altitudes, so <math>H</math> is the orthocenter. <math>\square</math>
+Consider triangles <math>MGO</math> and <math>AGH'</math>. Since <math>\frac{MG}{GA}=\frac{H'G}{GC} = \frac{1}{2}</math>, and they both share a vertical angle, they are similar by SAS similarity. Thus, <math>AH' \parallel OM \perp BC</math>, so <math>H'</math> lies on the <math>A</math> altitude of <math>\triangle ABC</math>. We can analogously show that <math>H'</math> also lies on the <math>B</math> and <math>C</math> altitudes, so <math>H'</math> is the orthocenter. <math>\square</math>
 ==Proof Nine-Point Center Lies on Euler Line==

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