Difference between revisions of "Euler line"
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==Euler line of Gergonne triangle== | ==Euler line of Gergonne triangle== | ||
+ | [[File:Euler line of Gergonne triangle.png|500px|right]] | ||
Prove that the Euler line of Gergonne triangle of <math>\triangle ABC</math> passes through the circumcenter of triangle <math>ABC.</math> | Prove that the Euler line of Gergonne triangle of <math>\triangle ABC</math> passes through the circumcenter of triangle <math>ABC.</math> | ||
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Other wording: Tangents to circumcircle of <math>\triangle ABC</math> are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line of the original triangle. | Other wording: Tangents to circumcircle of <math>\triangle ABC</math> are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line of the original triangle. | ||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>H</math> and <math>I</math> be orthocenter and circumcenter of <math>\triangle DEF,</math> respectively. | ||
+ | Let <math>A'B'C'</math> be Orthic Triangle of <math>\triangle DEF.</math> | ||
+ | |||
+ | Then <math>IH</math> is Euler line of <math>\triangle DEF,</math> | ||
+ | <math>I</math> is the incenter of <math>\triangle ABC,</math> <math>H</math> is the incenter of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | <math>\angle DEF = \angle DB'C' = \angle BDF = \frac { \overset{\Large\frown} {DF}}{2} \implies B'C' || BC.</math> | ||
+ | |||
+ | Similarly, <math>A'C' || AC, A'B' || AB \implies A'B'C'\sim ABC \implies</math> | ||
+ | |||
+ | <math>AA' \cap BB' \cap CC' = P,</math> where <math>P</math> is the perspector of triangles <math>ABC</math> and <math>A'B'C'.</math> | ||
+ | |||
+ | Under homothety with center P and coefficient <math>\frac {B'C'}{BC}</math> the incenter <math>I</math> of <math>\triangle ABC</math> maps into incenter <math>H</math> of <math>\triangle A'B'C'</math>, circumcenter <math>O</math> of <math>\triangle ABC</math> maps into circumcenter <math>I</math> of <math>\triangle A'B'C' \implies P,H,I,O </math> are collinear. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 09:39, 30 October 2022
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
[hide]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
Euler line for a triangle with an angle of 120![$^\circ.$](//latex.artofproblemsolving.com/2/6/4/264848e4bfe378a141e723827c54edf11b23c0c8.png)
Let the in triangle
be
Then the Euler line of the
is parallel to the bisector of
Proof
Let be circumcircle of
Let be circumcenter of
Let be the circle symmetric to
with respect to
Let be the point symmetric to
with respect to
The lies on
lies on
is the radius of
and
translation vector
to
is
Let be the point symmetric to
with respect to
Well known that
lies on
Therefore point
lies on
Point lies on
Let be the bisector of
are concurrent.
Euler line of the
is parallel to the bisector
of
as desired.
vladimir.shelomovskii@gmail.com, vvsss
Concurrent Euler lines and Fermat points
Consider a triangle with Fermat–Torricelli points
and
The Euler lines of the
triangles with vertices chosen from
and
are concurrent at the centroid
of triangle
We denote centroids by
, circumcenters by
We use red color for points and lines of triangles
green color for triangles
and blue color for triangles
Case 1
Let be the first Fermat point of
maximum angle of which smaller then
Then the centroid of triangle
lies on Euler line of the
The pairwise angles between these Euler lines are equal
Proof
Let and
be centroid, circumcenter, and circumcircle of
respectevely.
Let
be external for
equilateral triangle
is cyclic.
Point is centroid of
Points
and
are colinear, so point
lies on Euler line
of
Case 2
Let be the first Fermat point of
Then the centroid
of triangle
lies on Euler lines of the triangles
and
The pairwise angles between these Euler lines are equal
Proof
Let be external for
equilateral triangle,
be circumcircle of
is cyclic.
Point is centroid of
Points and
are colinear, so point
lies on Euler line
of
as desired.
Case 3
Let be the second Fermat point of
Then the centroid
of triangle
lies on Euler lines of the triangles
and
The pairwise angles between these Euler lines are equal
Proof
Let be internal for
equilateral triangle,
be circumcircle of
Let and
be circumcenters of the triangles
and
Point
is centroid of the
is the Euler line of the
parallel to
is bisector of
is bisector of
is bisector of
is regular triangle.
is the inner Napoleon triangle of the
is centroid of this regular triangle.
points
and
are collinear as desired.
Similarly, points and
are collinear.
Case 4
Let and
be the Fermat points of
Then the centroid of
point
lies on Euler line
is circumcenter,
is centroid) of the
Proof
Step 1. We find line which is parallel to
Let be midpoint of
Let
be the midpoint of
Let be point symmetrical to
with respect to
as midline of
Step 2. We prove that line is parallel to
Let be the inner Napoleon triangle. Let
be the outer Napoleon triangle. These triangles are regular centered at
Points and
are collinear (they lies on bisector
Points and
are collinear (they lies on bisector
Points and
are collinear (they lies on bisector
angle between
and
is
Points and
are concyclic
Points
and
are concyclic
points
and
are concyclic
Therefore
and
are collinear or point
lies on Euler line
vladimir.shelomovskii@gmail.com, vvsss
Euler line of Gergonne triangle
Prove that the Euler line of Gergonne triangle of passes through the circumcenter of triangle
Gergonne triangle is also known as the contact triangle or intouch triangle. If the inscribed circle touches the sides of at points
and
then
is Gergonne triangle of
.
Other wording: Tangents to circumcircle of are drawn at the vertices of the triangle. Prove that the circumcenter of the triangle formed by these three tangents lies on the Euler line of the original triangle.
Proof
Let and
be orthocenter and circumcenter of
respectively.
Let
be Orthic Triangle of
Then is Euler line of
is the incenter of
is the incenter of
Similarly,
where
is the perspector of triangles
and
Under homothety with center P and coefficient the incenter
of
maps into incenter
of
, circumcenter
of
maps into circumcenter
of
are collinear.
vladimir.shelomovskii@gmail.com, vvsss
See also
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