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Difference between revisions of "Gossard perspector"

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==Gossard perspector X(402) and Gossard triangle==
 
==Gossard perspector X(402) and Gossard triangle==
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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.
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Leonhard Euler proved in 1765 that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line as Euler line. Soon he 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.
  
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Gossard proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle is perspective with the given triangle and having the same Euler line. The center of perspective is known as Gossard perspector or Kimberling point <math>X(402).</math>
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Professor Harry Clinton Gossard in 1916 proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle is perspective with the given triangle and having the same Euler line. The center of perspective now is known as Gossard perspector or Kimberling point <math>X(402).</math>
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Let triangle <math>ABC</math> be given. Let Euler line cross <math>AB, BC,</math> and <math>AC</math> in points <math>D, E,</math> and <math>F. 13/01/2023</math> was found thas Gossard perspector is the centroid of the points <math>A, B, C, D, E, F.</math>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==Gossard perspector of right triangle==
 
==Gossard perspector of right triangle==

Revision as of 17:09, 14 January 2023

Gossard perspector X(402) and Gossard triangle

Leonhard Euler proved in 1765 that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line as Euler line. Soon he 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.

Professor Harry Clinton Gossard in 1916 proved that the three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle is perspective with the given triangle and having the same Euler line. The center of perspective now is known as Gossard perspector or Kimberling point $X(402).$

Let triangle $ABC$ be given. Let Euler line cross $AB, BC,$ and $AC$ in points $D, E,$ and $F. 13/01/2023$ was found thas Gossard perspector is the centroid of the points $A, B, C, D, E, F.$

vladimir.shelomovskii@gmail.com, vvsss

Gossard perspector of right triangle

Gossard 90.png

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

Gossard perspector and Gossard triangle for isosceles triangle

Gossard equilateral.png

It is clear that the Euler line of isosceles $\triangle ABC (AB = AC)$ 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 isosceles 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 $\triangle ABC.$ Let $H$ be the orthocenter of $\triangle ABC, O$ be the circumcenter of $\triangle ABC.$

It is clear that $Go$ is the midpoint of $AA'.$ $M = A'C' \cap AB$ 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 isosceles triangle and its Gossard triangle are congruent.

Any isosceles triangle and its Gossard triangle have the same Euler line.

The Gossard triangle of the isosceles $\triangle ABC$ is the reflection of $\triangle ABC$ in the Gossard perspector. Denote $\angle BAC = \alpha, \angle ABC = \angle ACB = \beta, AO = BO = R \implies$ \[\sin \beta = \cos \frac {\alpha}{2}, GoO = AO – AGo = R \cdot \frac{1 – \cos {\alpha}}{2},\] \[OH = AH – AO = R(2 \cos \alpha – 1) \implies \vec {GoO} = \vec {OH} \cdot \frac {1 – \cos \alpha}{2(2 \cos \alpha – 1)}.\]

vladimir.shelomovskii@gmail.com, vvsss

Euler line of the triangle formed by the Euler line and the sides of a given triangle

Euler Euler line.png

Let the Euler line of $\triangle ABC$ meet the sidelines $AB, AC,$ and $BC$ of $\triangle ABC$ at $D, E,$ and $F,$ respectively.

Euler line of the $\triangle ADE$ is parallel to $BC.$ Similarly, Euler line of the $\triangle BDF$ is parallel to $AC,$ Euler line of the $\triangle CEF$ is parallel to $AB.$

Proof

Denote $\angle A = \alpha, \angle B = \beta, \angle C = \gamma,$ smaller angles between the Euler line and lines $BC, AC,$ and $AB$ as $\theta_A, \theta_B,$ and $\theta_C,$ respectively. WLOG, $AC > BC > AB.$ It is known that $\tan \theta_A = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma}, \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}, \tan \theta_C = \frac{3 – \tan \beta \cdot \tan \alpha}{\tan \beta – \tan \alpha}.$

Let $O'$ be circumcenter of $\triangle ADE, KO'$ be Euler line of $\triangle ADE, K \in DE$ (line).

Similarly, $\tan \angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.$ \[3(\tan\alpha – \tan \gamma) (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \gamma) (3 – \tan \alpha \cdot \tan \beta) = (\tan^2 \alpha – 3) \cdot (3 –\tan \beta \cdot \tan \gamma),\] \[(3 – \tan \alpha \cdot \tan \gamma) \cdot (\tan\alpha – \tan \beta) – (3 – \tan \alpha \cdot \tan \beta) \cdot (\tan\alpha – \tan \gamma) = (\tan^2 \alpha – 3) \cdot (\tan \beta – \tan \gamma).\] Suppose, $\tan^2 \alpha \ne 3$ which means $\alpha \ne 60^\circ$ and $\alpha \ne 120^\circ.$ In this case \[\tan \angle O'KF = \frac{3 – \tan \beta \cdot \tan \gamma}{\tan \beta – \tan \gamma} = \tan \theta_A \implies \angle O'KF = \theta_A \implies O'K||BC.\]

Similarly one can prove claim in the other cases.

vladimir.shelomovskii@gmail.com, vvsss

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