Difference between revisions of "Euler line"

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<cmath>\triangle AHG = \triangle O_AOG</cmath>
 
<cmath>\triangle AHG = \triangle O_AOG</cmath>
 
Thus, <math>O, G, H</math> are collinear, and <math>\frac{OG}{HG} = \frac{1}{2}</math>.
 
Thus, <math>O, G, H</math> are collinear, and <math>\frac{OG}{HG} = \frac{1}{2}</math>.
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==Another Proof==
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Let <math>M</math> be the midpoint of <math>BC</math>.
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Extend <math>CG</math> past <math>G</math> to point <math>H'</math> such that <math>CG = \frac{1}{2} GH</math>. We will show <math>H'</math> is the orthocenter.
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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.
  
 
==Proof Nine-Point Center Lies on Euler Line==
 
==Proof Nine-Point Center Lies on Euler Line==
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===See also===
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==See also==
 
*[[Kimberling center]]
 
*[[Kimberling center]]
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*[[Central line]]
 
*[[De Longchamps point]]
 
*[[De Longchamps point]]
 
*[[Gergonne line]]
 
*[[Gergonne line]]

Latest revision as of 02:47, 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 $CG$ 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.PNG


See also

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