Editing Euler line

{{Short description|Line constructed from a triangle}}
[[Image:Triangle.EulerLine.svg|thumb|
{{legend-line|solid red|Euler's line, with [[Nine-point center|the center]] of the [[nine-point circle]]}}
{{legend-line|solid #D2691E|[[Median (geometry)|Medians]] (intersect at the [[centroid]])}}
{{legend-line|solid blue|[[Altitude (triangle)|Altitude]]s (intersect at the [[orthocenter]])}} 
{{legend-line|solid green|Perpendicular lines from the side midpoints (intersect at the [[circumcenter]])}}]]

In [[geometry]], the '''Euler line''', named after [[Leonhard Euler]] ({{IPAc-en|ˈ|ɔɪ|l|ər}} {{respell|OY|lər}}), is a [[line (mathematics)|line]] determined from any [[triangle]] that is not [[equilateral triangle|equilateral]]. It is a [[Central line (geometry)|central line]] of the triangle, and it passes through several important points determined from the triangle, including the [[orthocenter]], the [[circumcenter]], the [[centroid]], the [[Exeter point]] and the center of the [[nine-point circle]] of the triangle.<ref name="k">{{cite journal
 | author = Kimberling, Clark
 | title = Triangle centers and central triangles
 | journal = Congressus Numerantium
 | volume = 129
 | year = 1998
 | pages = i–xxv, 1–295}}</ref>

The concept of a triangle's Euler line extends to the Euler line of other shapes, such as the [[quadrilateral]] and the [[tetrahedron]].

==Triangle centers on the Euler line==

===Individual centers===
Euler showed in 1765 that in any triangle, the orthocenter, circumcenter and centroid are [[Line (geometry)|collinear]].<ref>{{cite journal
 | author = Euler, Leonhard
 | authorlink = Leonhard Euler
 | title = Solutio facilis problematum quorundam geometricorum difficillimorum
 |trans-title= Easy solution of some difficult geometric problems
 | journal = Novi Commentarii Academiae Scientarum Imperialis Petropolitanae
 | volume = 11
 | year = 1767
 | pages = 103–123
 | url = https://books.google.com/books?id=e1Y-AAAAcAAJ&pg=PA103 
 | id = <!--Enestrom number-->E325}} Reprinted in ''Opera Omnia'', ser. I, vol. XXVI, pp.&nbsp;139–157, Societas Scientiarum Naturalium Helveticae, Lausanne, 1953, {{MR|0061061}}.  Summarized at:  [http://math.dartmouth.edu/~euler/pages/E325.html Dartmouth College.]
</ref> This property is also true for another [[triangle center]], the [[nine-point center]], although it had not been defined in Euler's time. In equilateral triangles, these four points coincide, but in any other triangle they are all distinct from each other, and the Euler line is determined by any two of them.

Other notable points that lie on the Euler line include the [[de Longchamps point]], the [[Schiffler point]], the [[Exeter point]], and the [[Gossard perspector]].<ref name="k"/> However, the [[incenter]] generally does not lie on the Euler line;<ref>{{cite book | url=https://books.google.com/books?id=lR0SDnl2bPwC&pg=PA4 | title=Geometry Turned On: Dynamic Software in Learning, Teaching, and Research | publisher=The Mathematical Association of America |author1=Schattschneider, Doris |author2=King, James | year=1997 | pages=3–4 | isbn=978-0883850992}}</ref> it is on the Euler line only for [[isosceles triangle]]s,<ref>{{citation
 | last1 = Edmonds | first1 = Allan L.
 | last2 = Hajja | first2 = Mowaffaq
 | last3 = Martini | first3 = Horst
 | doi = 10.1007/s00025-008-0294-4
 | issue = 1–2
 | journal = [[Results in Mathematics]]
 | mr = 2430410
 | pages = 41–50
 | quote = It is well known that the incenter of a Euclidean triangle lies on its Euler line connecting the centroid and the circumcenter if and only if the triangle is isosceles
 | title = Orthocentric simplices and biregularity
 | volume = 52
 | year = 2008| s2cid = 121434528
 }}.</ref> for which the Euler line coincides with the symmetry axis of the triangle and contains all triangle centers.

The [[tangential triangle]] of a reference triangle is tangent to the latter's [[circumcircle]] at the reference triangle's vertices. The circumcenter of the tangential triangle lies on the Euler line of the reference triangle.<ref name=SL/>{{rp|p. 447}} <ref name="ac"/>{{rp|p.104,#211;p.242,#346}} The [[center of similitude]] of the [[orthic triangle|orthic]] and tangential triangles is also on the Euler line.<ref name=SL>{{citation
 | last1 = Leversha | first1 = Gerry
 | last2 = Smith | first2 = G. C.
 | date = November 2007
 | issue = 522
 | journal = [[Mathematical Gazette]]
 | jstor = 40378417
 | pages = 436–452
 | title = Euler and triangle geometry
 | volume = 91| doi = 10.1017/S0025557200182087
 | s2cid = 125341434
 }}.</ref>{{rp|p. 447}}<ref name="ac"/>{{rp|p. 102}}

== Proofs ==

===A vector proof===
Let <math>ABC</math> be a triangle. A proof of the fact that the [[circumscribed circle|circumcenter]] <math>O</math>, the [[centroid]] <math>G</math> and the [[altitude (triangle)#Orthocenter|orthocenter]] <math>H</math> are '''collinear''' relies on [[euclidean vector|free vectors]]. We start by stating the prerequisites. First, <math>G</math> satisfies the relation

:<math>\vec{GA}+\vec{GB}+\vec{GC}=0.</math>

This follows from the fact that the [[barycentric coordinate system|absolute barycentric coordinates]] of <math>G</math> are <math>\frac{1}{3}:\frac{1}{3}:\frac{1}{3}</math>. Further, the [[Sylvester's triangle problem|problem of Sylvester]]<ref name=Dorrie>Dörrie, Heinrich, "100 Great Problems of Elementary Mathematics. Their History and Solution". Dover Publications, Inc., New York, 1965, {{ISBN|0-486-61348-8}}, pages 141 (Euler's Straight Line) and 142 (Problem of Sylvester)</ref> reads as

:<math>\vec{OH}=\vec{OA}+\vec{OB}+\vec{OC}.</math>

Now, using the vector addition, we deduce that

:<math>\vec{GO}=\vec{GA}+\vec{AO}\,\mbox{(in triangle }AGO\mbox{)},\,\vec{GO}=\vec{GB}+\vec{BO}\,\mbox{(in triangle }BGO\mbox{)},\,\vec{GO}=\vec{GC}+\vec{CO}\,\mbox{(in triangle }CGO\mbox{)}.</math>

By adding these three relations, term by term, we obtain that

:<math>3\cdot\vec{GO}=\left(\sum\limits_{\scriptstyle\rm cyc}\vec{GA}\right)+\left(\sum\limits_{\scriptstyle\rm cyc}\vec{AO}\right)=0-\left(\sum\limits_{\scriptstyle\rm cyc}\vec{OA}\right)=-\vec{OH}.</math>

In conclusion, <math>3\cdot\vec{OG}=\vec{OH}</math>, and so the three points <math>O</math>, <math>G</math> and <math>H</math> (in this order) are collinear.

In Dörrie's book,<ref name=Dorrie /> the '''Euler line''' and the [[Sylvester's triangle problem|problem of Sylvester]] are put together into a single proof. However, most of the proofs of the problem of Sylvester rely on the fundamental properties of free vectors, independently of the Euler line.

== Properties ==

===Distances between centers===

On the Euler line the centroid ''G'' is between the circumcenter ''O'' and the orthocenter ''H'' and is twice as far from the orthocenter as it is from the circumcenter:<ref name="ac">Altshiller-Court, Nathan, ''College Geometry'', Dover Publications, 2007 (orig. Barnes & Noble 1952).</ref>{{rp|p.102}}

:<math>GH=2GO;</math>
:<math>OH=3GO.</math>

The segment ''GH'' is a diameter of the [[orthocentroidal circle]].

The center ''N'' of the nine-point circle lies along the Euler line midway between the orthocenter and the circumcenter:<ref name="k"/>

:<math>ON = NH, \quad OG =2\cdot GN, \quad NH=3GN.</math>

Thus the Euler line could be repositioned on a number line with the circumcenter ''O'' at the location 0, the centroid ''G'' at 2''t'', the nine-point center at 3''t'', and the orthocenter ''H'' at 6''t'' for some scale factor ''t''.

Furthermore, the squared distance between the centroid and the circumcenter along the Euler line is less than the squared [[circumradius]] ''R''<sup>2</sup> by an amount equal to one-ninth the sum of the squares of the side lengths ''a'', ''b'', and ''c'':<ref name="ac"/>{{rp|p.71}}

:<math>GO^2=R^2-\tfrac{1}{9}(a^2+b^2+c^2).</math>

In addition,<ref name="ac"/>{{rp|p.102}}

:<math>OH^2=9R^2-(a^2+b^2+c^2);</math>

:<math>GH^2=4R^2-\tfrac{4}{9}(a^2+b^2+c^2).</math>

==Representation==

===Equation===

Let ''A'', ''B'', ''C'' denote the vertex angles of the reference triangle, and let ''x'' : ''y'' : ''z'' be a variable point in [[trilinear coordinates]]; then an equation for the Euler line is

:<math>\sin (2A) \sin(B - C)x + \sin (2B) \sin(C - A)y + \sin (2C) \sin(A - B)z = 0.</math>

An equation for the Euler line in [[barycentric coordinates (mathematics)|barycentric coordinates]] <math>\alpha :\beta :\gamma</math> is<ref>Scott, J.A., "Some examples of the use of areal coordinates in triangle geometry", ''Mathematical Gazette'' 83, November 1999, 472-477.</ref>

:<math>(\tan C -\tan B)\alpha +(\tan A -\tan C)\beta + (\tan B -\tan A)\gamma =0.</math>

===Parametric representation===

Another way to represent the Euler line is in terms of a parameter ''t''.  Starting with the circumcenter (with trilinear coordinates <math>\cos A : \cos B : \cos C</math>) and the orthocenter (with trilinears <math>\sec A : \sec B : \sec C = \cos B \cos C : \cos C \cos A : \cos A \cos B),</math> every point on the Euler line, except the orthocenter, is given by the trilinear coordinates
:<math>\cos A + t \cos B \cos C : \cos B + t \cos C \cos A : \cos C + t \cos A \cos B</math>
formed as a [[linear combination]] of the trilinears of these two points, for some ''t''.

For example:
* The [[circumcenter]] has trilinears <math>\cos A:\cos B:\cos C,</math> corresponding to the parameter value <math>t=0.</math>
* The [[centroid]] has trilinears <math>\cos A + \cos B \cos C : \cos B + \cos C \cos A : \cos C + \cos A \cos B,</math> corresponding to the parameter value <math>t=1.</math>
* The [[nine-point center]] has trilinears <math>\cos A + 2 \cos B \cos C : \cos B + 2 \cos C \cos A : \cos C + 2 \cos A \cos B,</math> corresponding to the parameter value <math>t=2.</math>
* The [[de Longchamps point]] has trilinears <math>\cos A - \cos B \cos C : \cos B - \cos C \cos A : \cos C - \cos A \cos B,</math> corresponding to the parameter value <math>t=-1.</math>

===Slope===

In a [[Cartesian coordinate system]], denote the slopes of the sides of a triangle as <math>m_1,</math> <math>m_2,</math> and <math>m_3,</math> and denote the slope of its Euler line as <math>m_E</math>.  Then these slopes are related according to<ref name="BHS">Wladimir G. Boskoff, Laurent¸iu Homentcovschi, and Bogdan D. Suceava, "Gossard's Perspector and Projective Consequences", ''Forum Geometricorum'', Volume 13 (2013), 169–184. [http://forumgeom.fau.edu/FG2013volume13/FG201318.pdf]</ref>{{rp|Lemma 1}}

:<math>m_1m_2 + m_1m_3 + m_1m_E + m_2m_3 + m_2m_E + m_3m_E</math>
::<math> + 3m_1m_2m_3m_E + 3 = 0.</math>

Thus the slope of the Euler line (if finite) is expressible in terms of the slopes of the sides as

:<math>m_E=-\frac{m_1m_2 + m_1m_3 + m_2m_3 + 3}{m_1 + m_2 + m_3 + 3m_1m_2m_3}.</math>

Moreover, the Euler line is parallel to an acute triangle's side ''BC'' if and only if<ref name=BHS/>{{rp|p.173}} <math>\tan B \tan C = 3.</math>

==Relation to inscribed equilateral triangles==

The locus of the centroids of [[equilateral triangle]]s inscribed in a given triangle is formed by two lines perpendicular to the given triangle's Euler line.<ref name=Garcia>Francisco Javier Garc ́ıa Capita ́n, "Locus of Centroids of Similar Inscribed Triangles", ''[[Forum Geometricorum]]'' 16, 2016, 257–267 .http://forumgeom.fau.edu/FG2016volume16/FG201631.pdf</ref>{{rp|Coro. 4}}

==In special triangles==

===Right triangle===

In a [[right triangle]], the Euler line coincides with the [[median (triangle)|median]] to the [[hypotenuse]]&mdash;that is, it goes through both the right-angled vertex and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its [[Altitude (triangle)|altitudes]], falls on the right-angled vertex while its circumcenter, the intersection of its [[Bisection#Perpendicular bisectors|perpendicular bisectors]] of sides, falls on the midpoint of the hypotenuse.

===Isosceles triangle===

The Euler line of an [[isosceles triangle]] coincides with the [[axis of symmetry]]. In an isosceles triangle the [[incenter]] falls on the Euler line.

===Automedian triangle===

The Euler line of an [[automedian triangle]] (one whose [[median (geometry)|medians]] are in the same proportions, though in the opposite order, as the sides) is perpendicular to one of the medians.<ref name="parry">{{citation
 | last = Parry | first = C. F.
 | issue = 472
 | journal = The Mathematical Gazette
 | jstor = 3620241
 | pages = 151–154
 | title = Steiner–Lehmus and the automedian triangle
 | volume = 75
 | year = 1991| doi = 10.2307/3620241
 }}.</ref>

===Systems of triangles with concurrent Euler lines===

Consider a triangle ''ABC'' with [[Fermat–Torricelli point]]s ''F''<sub>1</sub> and ''F''<sub>2</sub>. The Euler lines of the 10 triangles with vertices chosen from ''A, B, C, F''<sub>1</sub> and ''F''<sub>2</sub> are [[concurrent lines|concurrent]] at the centroid of triangle ''ABC''.<ref>Beluhov, Nikolai Ivanov. "Ten concurrent Euler lines", ''Forum Geometricorum'' 9, 2009, pp. 271–274. http://forumgeom.fau.edu/FG2009volume9/FG200924index.html</ref>

The Euler lines of the four triangles formed by an [[orthocentric system]] (a set of four points such that each is the [[orthocenter]] of the triangle with vertices at the other three points) are concurrent at the [[nine-point center]] common to all of the triangles.<ref name=ac/>{{rp|p.111}}

==Generalizations==

===Quadrilateral===

In a [[Quadrilateral#Remarkable points and lines in a convex quadrilateral|convex quadrilateral]], the quasiorthocenter ''H'', the "area centroid" ''G'', and the [[quasicircumcenter]] ''O'' are [[collinear]] in this order on the Euler line, and ''HG'' = 2''GO''.<ref>{{citation
 | last = Myakishev | first = Alexei
 | journal = Forum Geometricorum
 | pages = 289–295
 | title = On Two Remarkable Lines Related to a Quadrilateral
 | url = http://forumgeom.fau.edu/FG2006volume6/FG200634.pdf
 | volume = 6
 | year = 2006}}.</ref>

===Tetrahedron===
{{Main article|Tetrahedron#Properties analogous to those of a triangle}}

A [[tetrahedron]] is a [[three-dimensional space|three-dimensional]] object bounded by four triangular [[face (geometry)|faces]].  Seven lines associated with a tetrahedron are concurrent at its centroid; its six midplanes intersect at its [[Monge point]]; and there is a circumsphere passing through all of the vertices, whose center is the circumcenter. These points define the "Euler line" of a tetrahedron analogous to that of a triangle. The centroid is the midpoint between its Monge point and circumcenter along this line. The center of the [[twelve-point sphere]] also lies on the Euler line.

===Simplicial polytope===

A [[simplicial polytope]] is a polytope whose facets are all [[simplex|simplices]] (plural of simplex). For example, every polygon is a simplicial polytope. The Euler line associated to such a polytope is the line determined by its centroid and [[circumcenter of mass]]. This definition of an Euler line generalizes the ones above.<ref>{{citation
 | last1 = Tabachnikov | first1 = Serge
 | last2 = Tsukerman | first2 = Emmanuel
 | date = May 2014
 | issue = 4
 | journal = [[Discrete and Computational Geometry]]
 | pages = 815–836
 | title = Circumcenter of Mass and Generalized Euler Line
 | doi=10.1007/s00454-014-9597-2
 | volume=51| arxiv = 1301.0496
 | s2cid = 12307207
 }}.</ref>

Suppose that <math>P</math> is a polygon. The Euler line <math>E</math> is sensitive to the symmetries of <math>P</math> in the following ways:

# If <math>P</math> has a line of reflection symmetry <math>L</math>, then <math>E</math> is either <math>L</math> or a point on <math>L</math>.
# If <math>P</math> has a center of rotational symmetry <math>C</math>, then <math>E=C</math>.

==Related constructions==
A triangle's Kiepert parabola is the unique parabola that is tangent to the sides (two of them [[extended side|extended]]) of the triangle and has the Euler line as its [[Directrix (conic section)|directrix]].<ref name=Scimemi>[http://forumgeom.fau.edu/FG2010volume10/FG201008.pdf Scimemi, Benedetto, "Simple Relations Regarding the Steiner Inellipse of a Triangle", ''Forum Geometricorum'' 10, 2010: 55–77.]</ref>{{rp|p. 63}}

== References ==
{{reflist}}

== External links ==
* [http://www.mathopenref.com/eulerline.html An interactive applet showing several triangle centers that lies on the Euler line].
* [http://demonstrations.wolfram.com/EulerLine/ "Euler Line"] and [http://demonstrations.wolfram.com/NonEuclideanTriangleContinuum/ "Non-Euclidean Triangle Continuum"] at the [[Wolfram Demonstrations Project]]
* [http://dynamicmathematicslearning.com/ninepointconic.html Nine-point conic and Euler line generalization], [http://dynamicmathematicslearning.com/furtherEuler.html A further Euler line generalization], and [http://dynamicmathematicslearning.com/quasi-euler-line-hexagon.html The quasi-Euler line of a quadrilateral and a hexagon] at [http://dynamicmathematicslearning.com/JavaGSPLinks.htm Dynamic Geometry Sketches]
* [[Alexander Bogomolny|Bogomolny, Alexander]], "[http://www.cut-the-knot.org/triangle/altEuler.shtml Altitudes and the Euler Line]" and "[http://www.cut-the-knot.org/triangle/EulerLine.shtml Euler Line and 9-Point Circle]", ''[[Cut-the-Knot]]''
* {{citation|url=http://faculty.evansville.edu/ck6/tcenters/class/eulerline.html|title=Triangle centers on the Euler line|first=Clark|last=Kimberling|authorlink=Clark Kimberling|work=Triangle Centers}}
* Archived at [https://ghostarchive.org/varchive/youtube/20211211/wVH4MS6v23U Ghostarchive]{{cbignore}} and the [https://web.archive.org/web/20160201162310/https://www.youtube.com/watch?v=wVH4MS6v23U&gl=US&hl=en Wayback Machine]{{cbignore}}: {{citation|url=https://www.youtube.com/watch?v=wVH4MS6v23U|first=Zvezdelina|last=Stankova|authorlink=Zvezdelina Stankova|title=Triangles have a Magic Highway|publisher=[[YouTube]]|work=[[Numberphile]]|date=February 1, 2016}}{{cbignore}}
* {{mathworld | title = Euler Line | urlname = EulerLine}}

[[Category:Straight lines defined for a triangle]]