Difference between revisions of "Euler line"

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Let <math>ABC</math> be a triangle, points <math>H, N, G, O, L</math> as <math>\triangle ABC</math>'s [[orthocenter]], [[nine-point center]], [[centroid]], [[circumcenter]], [[De Longchamps point]] respectively, then these points are collinear(regardless of <math>\triangle ABC</math>'s shape). And the line passes through points <math>H, N, G, O, L</math> is called as Euler line, which is named after [[Leonhard Euler]].
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Let <math>ABC</math> be a triangle, points <math>H, N, G, O, L</math> as <math>\triangle ABC</math>'s [[orthocenter]], [[nine-point center]], [[centroid]], [[circumcenter]], [[De Longchamps point]] respectively, then these points are [[collinear]](regardless of <math>\triangle ABC</math>'s shape). And the line passes through points <math>H, N, G, O, L</math> is called as Euler line, which is named after [[Leonhard Euler]].
  
 
An interesting property of distances between these points on the Euler line:
 
An interesting property of distances between these points on the Euler line:
 
* <math>OG:GN:NH\equiv2:1:3</math>
 
* <math>OG:GN:NH\equiv2:1:3</math>
  
Construct an [[orthic triangle]]<math>\triangle H_A,H_B,H_C</math>, then Euler lines of <math>\triangle AH_BH_C</math>,<math>\triangle BH_CH_A</math>,<math>\triangle CH_AH_B</math> concur at <math>\triangle ABC</math>'s [[nine-point center]].
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Construct an [[orthic triangle]]<math>\triangle H_AH_BH_C</math>, then Euler lines of <math>\triangle AH_BH_C</math>,<math>\triangle BH_CH_A</math>,<math>\triangle CH_AH_B</math> concur at <math>\triangle ABC</math>'s [[nine-point center]].

Revision as of 11:46, 5 November 2006

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Let $ABC$ be a triangle, points $H, N, G, O, L$ as $\triangle ABC$'s orthocenter, nine-point center, centroid, circumcenter, De Longchamps point respectively, then these points are collinear(regardless of $\triangle ABC$'s shape). And the line passes through points $H, N, G, O, L$ is called as Euler line, which is named after Leonhard Euler.

An interesting property of distances between these points on the Euler line:

  • $OG:GN:NH\equiv2:1:3$

Construct an orthic triangle$\triangle H_AH_BH_C$, then Euler lines of $\triangle AH_BH_C$,$\triangle BH_CH_A$,$\triangle CH_AH_B$ concur at $\triangle ABC$'s nine-point center.