Difference between revisions of "Euler line"
(→Analytic Proof of Existence) |
(→Euler line for a triangle with an angle of 120^\circ.) |
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<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
− | Let <math>\omega</math> be circumcircle of <math>\triangle ABC | + | Let <math>\omega</math> be circumcircle of <math>\triangle ABC.</math> |
− | The <math>\angle C = 120^\circ \implies O</math> lies on <math>\omega', E</math> lies on <math>\omega. EO</math> is the radius of <math>\omega</math> and <math>\omega' \implies </math> translation vector <math>\omega</math> to <math>\omega | + | Let <math>O</math> be circumcenter of <math>\triangle ABC.</math> |
− | Let <math>H'</math> be the point symmetric to <math>H</math> with respect to <math>AB.</math> Well known that <math>H'</math> lies on <math>\omega | + | |
− | Let <math>CD</math> be the bisector of <math>\angle C \implies E,O,D</math> are concurrent <math> | + | Let <math>\omega'</math> be the circle symmetric to <math>\omega</math> with respect to <math>AB.</math> |
+ | |||
+ | Let <math>E</math> be the point symmetric to <math>O</math> with respect to <math>AB.</math> | ||
+ | |||
+ | The <math>\angle C = 120^\circ \implies O</math> lies on <math>\omega', E</math> lies on <math>\omega.</math> | ||
+ | |||
+ | <math>EO</math> is the radius of <math>\omega</math> and <math>\omega' \implies </math> translation vector <math>\omega'</math> to <math>\omega</math> is <math>\vec {EO}.</math> | ||
+ | |||
+ | Let <math>H'</math> be the point symmetric to <math>H</math> with respect to <math>AB.</math> Well known that <math>H'</math> lies on <math>\omega.</math> | ||
+ | Therefore point <math>H</math> lies on <math>\omega'.</math> | ||
+ | |||
+ | Point <math>C</math> lies on <math>\omega, CH || OE \implies CH = OE.</math> | ||
+ | |||
+ | Let <math>CD</math> be the bisector of <math>\angle C \implies E,O,D</math> are concurrent. | ||
+ | <math>OD = HC, OD||HC \implies CD || HO \implies </math> | ||
+ | |||
+ | Euler line <math>HO</math> of the <math>\triangle ABC</math> is parallel to the bisector <math>CD</math> of <math>\angle C</math> as desired. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 23:37, 18 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
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
See also
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