Difference between revisions of "Euler line"
(→Concurrent Euler lines and Fermat points) |
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==Concurrent Euler lines and Fermat points== | ==Concurrent Euler lines and Fermat points== | ||
[[File:Fermat point lines.png|400px|right]] | [[File:Fermat point lines.png|400px|right]] | ||
− | Consider a triangle <math>ABC</math> with Fermat–Torricelli points <math>F</math> and <math>F'.</math> The Euler lines of the <math>10</math> triangles with vertices chosen from <math>A, B, C, F,</math> and <math>F'</math> are concurrent at the centroid of triangle <math>ABC.</math> | + | Consider a triangle <math>ABC</math> with Fermat–Torricelli points <math>F</math> and <math>F'.</math> The Euler lines of the <math>10</math> triangles with vertices chosen from <math>A, B, C, F,</math> and <math>F'</math> are concurrent at the centroid <math>G</math> of triangle <math>ABC.</math> |
<i><b>Case 1</b></i> | <i><b>Case 1</b></i> | ||
Line 79: | Line 79: | ||
<i><b>Case 2</b></i> | <i><b>Case 2</b></i> | ||
− | [[File:Fermat 1 | + | [[File:Fermat 1 150 lines.png|500px|right]] |
− | Let <math>F</math> be the first Fermat point of <math>\triangle ABC, \angle BAC > 120^\circ.</math> Then the centroid of triangle <math>ABC</math> lies on Euler line of the <math>\triangle ABF.</math> | + | Let <math>F</math> be the first Fermat point of <math>\triangle ABC, \angle BAC > 120^\circ.</math> Then the centroid <math>G</math> of triangle <math>ABC</math> lies on Euler line of the <math>\triangle ABF.</math> |
<i><b>Proof</b></i> | <i><b>Proof</b></i> | ||
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Let <math>\triangle ABD</math> be external for <math>\triangle ABC</math> equilateral triangle, <math>\omega</math> be circumcircle of <math>\triangle ABD \implies F = CD \cap \omega.</math> | Let <math>\triangle ABD</math> be external for <math>\triangle ABC</math> equilateral triangle, <math>\omega</math> be circumcircle of <math>\triangle ABD \implies F = CD \cap \omega.</math> | ||
− | Let <math> | + | Let <math>G_1</math> be the centroid of <math>\triangle ADF, \vec {G_1} = \frac {\vec A + \vec F + \vec D}{3}.</math> |
<math>\angle AFB = \angle BFD = 60^\circ,\angle AFD = 120^\circ \implies</math> | <math>\angle AFB = \angle BFD = 60^\circ,\angle AFD = 120^\circ \implies</math> | ||
− | Euler line <math> | + | Euler line <math>OG_1</math> of <math>\triangle ADF</math> is parallel to <math>BF.</math> |
− | Point <math> | + | Point <math>O</math> is the circumcenter of <math>\triangle ABF, \triangle ADF, \triangle ABD</math> and centroid of the equilateral <math>\triangle ABD</math> |
− | <cmath>\implies \vec | + | <cmath>\implies \vec O = \frac {\vec A + \vec B + \vec D}{3}.</cmath> |
− | Therefore < | + | Therefore |
+ | <cmath>\vec {OG_1} = \frac {\vec A + \vec F + \vec D}{3} - \frac {\vec A + \vec B + \vec D}{3} = \frac {\vec F - \vec B}{3},</cmath> | ||
− | <cmath>\vec { | + | <cmath>\vec {OG'} = \frac {\vec A + \vec F + \vec B}{3} - \frac {\vec A + \vec B + \vec D}{3} = \frac {\vec F - \vec D}{3} \implies \angle G'OG_1 = \angle BFD = 60^\circ.</cmath> |
− | <cmath> | + | <cmath>OG_1|| BF, \angle G'OG_1 = \angle BFD \implies OG' || CD.</cmath> |
− | <cmath>\vec | + | <cmath>\vec G = \frac {\vec A +\vec B + \vec C}{3},\vec G' = \frac {\vec A + \vec B + \vec F}{3} \implies \vec {G'G} = \frac {\vec {FC}}{3}.</cmath> |
− | Point <math> | + | Point <math>G'</math> lies on Euler line of <math>\triangle ABF</math>, this line is parallel to <math>FC, \implies M \in OG'</math> as desired. |
<i><b>Case 4</b></i> | <i><b>Case 4</b></i> |
Revision as of 05:00, 29 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
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
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
Proof
is bisector
As shown above, the Euler line of the
is parallel to
Let and
be orthocenter, centroid and circumcenter of
respectively.
Let
be centroid of
Point
lies on Euler line of
, this line is parallel to
Similarly, lies on Euler lines of
and
as desired.
Case 2
Let be the first Fermat point of
Then the centroid
of triangle
lies on Euler line of the
Proof
Let be external for
equilateral triangle,
be circumcircle of
Let be the centroid of
Euler line of
is parallel to
Point is the circumcenter of
and centroid of the equilateral
Therefore
Point
lies on Euler line of
, this line is parallel to
as desired.
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
See also
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