Difference between revisions of "Gossard perspector"
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==Euler line of the triangle formed by the Euler line and the sides of a given triangle== | ==Euler line of the triangle formed by the Euler line and the sides of a given triangle== | ||
[[File:Euler Euler line.png|500px|right]] | [[File:Euler Euler line.png|500px|right]] | ||
− | Let the Euler line of <math>\triangle ABC</math> meet the sidelines <math>AB, AC,</math> and <math>BC</math> of <math>\triangle ABC</math> at <math>D, E,</math> and <math>F,</math> respectively. Euler line of the <math>\triangle ADE</math> is parallel to <math>BC.</math> Similarly, Euler line of the <math>\triangle BDF</math> is parallel to <math>AC,</math> Euler line of the <math>\triangle CEF</math> is parallel to <math>AB.</math> | + | Let the Euler line of <math>\triangle ABC</math> meet the sidelines <math>AB, AC,</math> and <math>BC</math> of <math>\triangle ABC</math> at <math>D, E,</math> and <math>F,</math> respectively. |
+ | |||
+ | Euler line of the <math>\triangle ADE</math> is parallel to <math>BC.</math> Similarly, Euler line of the <math>\triangle BDF</math> is parallel to <math>AC,</math> Euler line of the <math>\triangle CEF</math> is parallel to <math>AB.</math> | ||
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
+ | |||
+ | Denote <math>\angle A = \alpha, \angle B = \beta, \angle C = \gamma,</math> smaller angles between the Euler line and lines <math>BC, AC,</math> and <math>AB</math> as <math>\theta_A, \theta_B,</math> and <math>\theta_C,</math> respectively. WLOG, <math>AC > BC > AB.</math> | ||
+ | It is known that <math>\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}.</math> | ||
+ | *[[Euler line]] | ||
+ | Let <math>O'</math> be circumcenter of <math>\triangle ADE, KO'</math> be Euler line of <math>\triangle ADE, K \in DE</math> (line). | ||
+ | |||
+ | Similarly, <math>\tan \angle O'KF = \frac{3 – \tan \theta_B \cdot \tan \theta_C}{\tan \theta_C – \tan \theta_B}.</math> | ||
+ | <cmath>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),</cmath> | ||
+ | <cmath>(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) \implies</cmath> | ||
+ | <cmath>\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.</cmath> | ||
+ | |||
+ | Similarly one can prove claim in the other cases. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 15:36, 10 January 2023
Contents
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
Gossard perspector of right triangle
It is clear that the Euler line of right triangle meet the sidelines
and
of
at
and
where
is the midpoint of
Let be the triangle formed by the Euler lines of the
and the line
contains
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
We call the triangle as the Gossard triangle of
Let be any right triangle and let
be its Gossard triangle. Then the lines
and
are concurrent. We call the point of concurrence
as the Gossard perspector of
is the midpoint of
is orthocenter of
is circumcenter of
so
is midpoint of
is the midpoint
is the midpoint
with coefficient
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 is the reflection of
in the Gossard perspector.
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Gossard perspector and Gossard triangle for equilateral triangle
It is clear that the Euler line of equilateral meet the sidelines
and
of
at
and
where
is the midpoint of
Let be the triangle formed by the Euler lines of the
and the line
contains
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
We call the triangle as the Gossard triangle of
Let be any equilateral triangle and let
be its Gossard triangle. Then the lines
and
are concurrent. We call the point of concurrence
as the Gossard perspector of
Let
be the orthocenter of
be the circumcenter of
It is clear that is the midpoint of
is the midpoint
is the midpoint
with coefficient
Any equilateral triangle and its Gossard triangle are congruent.
Any equilateral triangle and its Gossard triangle have the same Euler line.
The Gossard triangle of the equilateral is the reflection of
in the Gossard perspector.
Denote
vladimir.shelomovskii@gmail.com, vvsss
Euler line of the triangle formed by the Euler line and the sides of a given triangle
Let the Euler line of meet the sidelines
and
of
at
and
respectively.
Euler line of the is parallel to
Similarly, Euler line of the
is parallel to
Euler line of the
is parallel to
Proof
Denote smaller angles between the Euler line and lines
and
as
and
respectively. WLOG,
It is known that
Let be circumcenter of
be Euler line of
(line).
Similarly,
Similarly one can prove claim in the other cases.
vladimir.shelomovskii@gmail.com, vvsss