Difference between revisions of "Euler line"
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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> |
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+ | Euler line is the central line <math>L_{647}</math>. | ||
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+ | 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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+ | ==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]] <math>\triangle O_AO_BO_C</math>. It is similar to <math>\triangle ABC</math>. Specifically, a rotation of <math>180^\circ</math> about the midpoint of <math>O_BO_C</math> followed by a homothety with scale factor <math>2</math> centered at <math>A</math> brings <math>\triangle ABC \to \triangle O_AO_BO_C</math>. Let us examine what else this transformation, which we denote as <math>\mathcal{S}</math>, will do. | ||
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+ | It turns out <math>O</math> is the orthocenter, and <math>G</math> is the centroid of <math>\triangle O_AO_BO_C</math>. Thus, <math>\mathcal{S}(\{O_A, O, G\}) = \{A, H, G\}</math>. As a homothety preserves angles, it follows that <math>\measuredangle O_AOG = \measuredangle AHG</math>. Finally, as <math>\overline{AH} || \overline{O_AO}</math> it follows that | ||
+ | <cmath>\triangle AHG = \triangle O_AOG</cmath> | ||
+ | Thus, <math>O, G, H</math> are collinear, and <math>\frac{OG}{HG} = \frac{1}{2}</math>. | ||
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+ | ==Another Proof== | ||
+ | Let <math>M</math> be the midpoint of <math>BC</math>. | ||
+ | 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. | ||
+ | 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. | ||
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+ | ==Proof Nine-Point Center Lies on Euler Line== | ||
+ | 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>. | ||
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+ | ==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> | ||
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+ | [[Image:Euler Line.PNG||500px|frame|center]] | ||
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− | + | ==See also== | |
+ | *[[Kimberling center]] | ||
+ | *[[Central line]] | ||
+ | *[[De Longchamps point]] | ||
+ | *[[Gergonne line]] | ||
+ | *[[Gergonne point]] | ||
+ | *[[Evans point]] | ||
− | + | {{stub}} | |
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Revision as of 02:47, 6 December 2018
In any triangle , the Euler line is a line which passes through the orthocenter , centroid , circumcenter , nine-point center and de Longchamps point . 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, and
Euler line is the central line .
Given the orthic triangle of , the Euler lines of ,, and concur at , the nine-point circle of .
Contents
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 . It is similar to . Specifically, a rotation of about the midpoint of followed by a homothety with scale factor centered at brings . Let us examine what else this transformation, which we denote as , will do.
It turns out is the orthocenter, and is the centroid of . Thus, . As a homothety preserves angles, it follows that . Finally, as it follows that Thus, are collinear, and .
Another Proof
Let be the midpoint of . Extend past to point such that . We will show is the orthocenter. Consider triangles and . Since , and they both share a vertical angle, they are similar by SAS similarity. Thus, , so lies on the altitude of . We can analogously show that also lies on the and altitudes, so is the orthocenter.
Proof Nine-Point Center Lies on Euler Line
Assuming that the nine point circle exists and that is the center, note that a homothety centered at with factor brings the Euler points onto the circumcircle of . Thus, it brings the nine-point circle to the circumcircle. Additionally, should be sent to , thus and .
Analytic Proof of Existence
Let the circumcenter be represented by the vector , and let vectors correspond to the vertices of the triangle. It is well known the that the orthocenter is and the centroid is . Thus, are collinear and
See also
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