Difference between revisions of "Euler line"
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'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
− | ==Euler line as radical | + | ==Euler line as radical axis== |
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Let <math>\triangle ABC</math> with altitudes <math>AA_1, BB_1,</math> and <math>CC_1</math> be given. | Let <math>\triangle ABC</math> with altitudes <math>AA_1, BB_1,</math> and <math>CC_1</math> be given. |
Revision as of 15:46, 15 November 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
- 1 Proof Centroid Lies on Euler Line
- 2 Another Proof
- 3 Proof Nine-Point Center Lies on Euler Line
- 4 Analytic Proof of Existence
- 5 Triangles with angles of or
- 6 Euler lines of cyclic quadrilateral (Vittas’s theorem)
- 7 Concurrent Euler lines and Fermat points
- 8 Euler line of Gergonne triangle
- 9 Thebault point
- 10 Schiffler point
- 11 Euler line as radical axis
- 12 De Longchamps point of Euler line
- 13 See also
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
Triangles with angles of
or ![$120^\circ$](//latex.artofproblemsolving.com/f/3/f/f3fd424e953ac76b50bf6103875acc049d21d3c2.png)
Claim 1
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.
Claim 2
Let the in triangle
be
Then the Euler line of the
is perpendicular to the bisector of
Proof
Let be circumcircle, circumcenter, orthocenter and incenter of the
points
are concyclic.
The circle centered at midpoint of small arc
is rhomb.
Therefore the Euler line is perpendicular to
as desired.
Claim 3
Let be a quadrilateral whose diagonals
and
intersect at
and form an angle of
If the triangles PAB, PBC, PCD, PDA are all not equilateral, then their Euler lines are pairwise parallel or coincident.
Proof
Let and
be internal and external bisectors of the angle
.
Then Euler lines of and
are parallel to
and Euler lines of
and
are perpendicular to
as desired.
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Euler lines of cyclic quadrilateral (Vittas’s theorem)
Let be a cyclic quadrilateral with diagonals intersecting at
Prove that the Euler lines of triangles
are concurrent.
Proof
Let be the circumcenters (orthocenters) of triangles
Let
be the common bisector of
and
Therefore
and
are parallelograms with parallel sides.
bisect these angles.
So points
are collinear and lies on one straight line which is side of the pare vertical angles
and
Similarly, points
are collinear and lies on another side of these angles.
Similarly obtuse
so points
and
are collinear and lies on one side and points
and
are collinear and lies on another side of the same vertical angles.
We use Claim and get that lines are concurrent (or parallel if
or
).
Claim (Property of vertex of two parallelograms)
Let and
be parallelograms,
Let lines
and
be concurrent at point
Then points
and
are collinear and lines
and
are concurrent.
Proof
We consider only the case Shift transformation allows to generalize the obtained results.
We use the coordinate system with the origin at the point and axes
We use and get
points
and
are colinear.
We calculate point of crossing and
and
and
and get the same result:
as desired (if
then point
moves to infinity and lines are parallel, angles
or
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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 will 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 will 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
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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 as desired.
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Thebault point
Let and
be the altitudes of the
where
a) Prove that the Euler lines of triangles are concurrent on the nine-point circle at a point T (Thebault point of
)
b) Prove that if then
else
Proof
Case 1 Acute triangle
a) It is known, that Euler line of acute triangle cross AB and BC (shortest and longest sides) in inner points.
Let be circumcenters of
Let and
be centroids of
Denote is the circle
(the nine-points circle).
is the midpoint
where
is the orthocenter of
Similarly
is the midline of
Let cross
at point
different from
spiral similarity centered at
maps
onto
This similarity has the rotation angle acute angle between Euler lines of these triangles is
Let these lines crossed at point
Therefore
points
and
are concyclic
Similarly, as desired.
b)
Point
lies on median of
and divide it in ratio 2 : 1.
Point lies on Euler line of
According the Claim,
Similarly
Case 2 Obtuse triangle
a) It is known, that Euler line of obtuse cross AC and BC (middle and longest sides) in inner points.
Let be circumcenters of
Let and
be centroids of
Denote
is the circle
(the nine-points circle).
is the midpoint
where
is the orthocenter of
Similarly
is the midline of
Let cross
at point
different from
spiral similarity centered at
maps
onto
This similarity has the rotation angle acute angle between Euler lines of these triangles is
Let these lines crossed at point
Therefore
points
and
are concyclic
Similarly, as desired.
b)
Point lies on median of
and divide it in ratio
Point lies on Euler line of
According the Claim,
Similarly
Claim (Segment crossing the median)
Let be the midpoint of side
of the
Then
Proof
Let be
(We use sign
to denote the area of
Denote
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Schiffler point
Let and
be the incenter, circumcenter, centroid, circumradius,
and inradius of
respectively. Then the Euler lines of the four triangles
and
are concurrent at Schiffler point
.
Proof
We will prove that the Euler line of
cross the Euler line
of
at such point
that
.
Let and
be the circumcenter and centroid of
respectively.
It is known that lies on circumcircle of
Denote
It is known that is midpoint
point
lies on median
points
belong the bisector of
Easy to find that ,
We use sigh [t] for area of t. We get
Using Claim we get
Therefore each Euler line of triangles
cross Euler line of
in the same point, as desired.
Claim (Segments crossing inside triangle)
Given triangle GOY. Point lies on
Point lies on
Point lies on
Point lies on
Then
Proof
Let be
(We use sigh
for area of
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Euler line as radical axis
Let with altitudes
and
be given.
Let and
be circumcircle, circumcenter, orthocenter and circumradius of
respectively.
Circle centered at
passes through
and is tangent to the radius AO. Similarly define circles
and
Then Euler line of is the radical axis of these circles.
If is acute, then these three circles intersect at two points located on the Euler line of the
Proof
The power of point with respect to
and
is
The power of point with respect to
is
The power of point with respect to
is
The power of point with respect to
is
It is known that
Therefore points and
lies on radical axis of these three circles as desired.
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De Longchamps point of Euler line
Definition 1
The De Longchamps’ point of a triangle is the radical center of the power circles of the triangle. Prove that De Longchamps point lies on Euler line.
We call A-power circle of a the circle centered at the midpoint
point
with radius
The other two circles are defined symmetrically.
Proof
Let and
be orthocenter, circumcenter, and De Longchamps point, respectively.
Denote power circle by
power circle by
WLOG,
Denote the projection of point
on
We will prove that radical axes of power and
power cicles is symmetric to altitude
with respect
Further, we will conclude that the point of intersection of the radical axes, symmetrical to the heights with respect to O, is symmetrical to the point of intersection of the heights
with respect to
Point is the crosspoint of the center line of the
power and
power circles and there radical axis.
We use claim and get:
and
are the medians, so
We use Claim some times and get:
radical axes of
power and
power cicles is symmetric to altitude
with respect
Similarly radical axes of power and
power cicles is symmetric to altitude
radical axes of
power and
power cicles is symmetric to altitude
with respect
Therefore the point
of intersection of the radical axes, symmetrical to the heights with respect to
is symmetrical to the point
of intersection of the heights with respect to
lies on Euler line of
Claim (Distance between projections)
Definition 2
We call circle of a
the circle centered at
with radius
The other two circles are defined symmetrically. The De Longchamps point of a triangle is the radical center of
circle,
circle, and
circle of the triangle (Casey – 1886). Prove that De Longchamps point under this definition is the same as point under Definition 1.
Proof
Let and
be orthocenter, centroid, and De Longchamps point, respectively. Let
cross
at points
and
The other points
are defined symmetrically.
Similarly
is diameter
Therefore is anticomplementary triangle of
is orthic triangle of
So
is orthocenter of
as desired.
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See also
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