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Difference between revisions of "Gossard perspector"

(Created page with "==Gossard perspector X(402) and Gossard triangle== Euler proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is par...")
 
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Gossard proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by twos, of a given triangle, form a triangle triply perspective with the given triangle and having the same Euler line. The orthocenters, circumcenters and centroids of these two triangles are symmetrically placed as to the center of perspective which known as Gossard perspector or Kimberling point <math>X(402).</math>
 
Gossard proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by twos, of a given triangle, form a triangle triply perspective with the given triangle and having the same Euler line. The orthocenters, circumcenters and centroids of these two triangles are symmetrically placed as to the center of perspective which known as Gossard perspector or Kimberling point <math>X(402).</math>
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==Gossard perspector for right triangle==
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It is clear that the Euler line of right triangle <math>ABC (\angle A = 90 ^\circ)</math> meet the sidelines <math>BC, CA</math> and <math>AB</math> of <math>\triangle ABC</math> at <math>A'</math> and <math>A,</math> where <math>A'</math> is the midpoint of <math>BC.</math>
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Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle AA'B, \triangle AA'C,</math> and the line <math>l</math> contains <math>A</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle AA'C</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle AA'B</math> and <math>l.</math> We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of <math>\triangle ABC.</math>
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Let <math>\triangle ABC</math> be any right triangle and let  <math>\triangle A'B'C'</math> be its Gossard triangle. Then the lines <math>AA', BB',</math> and <math>CC'</math> are concurrent. We call the point of concurrence <math>Go</math> as  the Gossard perspector of <math>\triangle ABC.</math> <math>Go</math> is the midpoint of <math>AA'.</math> <math>A</math> is orthocenter of <math>\triangle ABC, A'</math> is circumcenter of <math>\triangle ABC,</math> so <math>Go</math> is midpoint of <math>OH.</math>
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<math>M = A'C' \cap AB</math> is the midpoint <math>AB, N = A'B' \cap AC</math> is the midpoint <math>AC, \triangle A'BM = \triangle CNA' \sim \triangle CBA</math> with coefficient <math>k = \frac {1}{2}.</math>
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Any right triangle and its Gossard triangle are congruent.
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Any right triangle and its Gossard triangle have the same Euler line.
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The Gossard triangle of the right <math>\triangle ABC</math> is the reflection of <math>\triangle ABC</math> in the Gossard perspector.
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'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 12:06, 10 January 2023

Gossard perspector X(402) and Gossard triangle

Euler proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is parallel with the third side of the given triangle.

Gossard proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by twos, of a given triangle, form a triangle triply perspective with the given triangle and having the same Euler line. The orthocenters, circumcenters and centroids of these two triangles are symmetrically placed as to the center of perspective which known as Gossard perspector or Kimberling point $X(402).$

Gossard perspector for right triangle

It is clear that the Euler line of right triangle $ABC (\angle A = 90 ^\circ)$ meet the sidelines $BC, CA$ and $AB$ of $\triangle ABC$ at $A'$ and $A,$ where $A'$ is the midpoint of $BC.$

Let $\triangle A'B'C'$ be the triangle formed by the Euler lines of the $\triangle AA'B, \triangle AA'C,$ and the line $l$ contains $A$ and parallel to $BC,$ the vertex $B'$ being the intersection of the Euler line of the $\triangle AA'C$ and $l,$ the vertex $C'$ being the intersection of the Euler line of the $\triangle AA'B$ and $l.$ We call the triangle $\triangle A'B'C'$ as the Gossard triangle of $\triangle ABC.$ Let $\triangle ABC$ be any right triangle and let $\triangle A'B'C'$ be its Gossard triangle. Then the lines $AA', BB',$ and $CC'$ are concurrent. We call the point of concurrence $Go$ as the Gossard perspector of $\triangle ABC.$ $Go$ is the midpoint of $AA'.$ $A$ is orthocenter of $\triangle ABC, A'$ is circumcenter of $\triangle ABC,$ so $Go$ is midpoint of $OH.$ $M = A'C' \cap AB$ is the midpoint $AB, N = A'B' \cap AC$ is the midpoint $AC, \triangle A'BM = \triangle CNA' \sim \triangle CBA$ with coefficient $k = \frac {1}{2}.$ Any right triangle and its Gossard triangle are congruent. Any right triangle and its Gossard triangle have the same Euler line. The Gossard triangle of the right $\triangle ABC$ is the reflection of $\triangle ABC$ in the Gossard perspector.

vladimir.shelomovskii@gmail.com, vvsss

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