Euler line

In

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 collinear.^[2] 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.^[1] However, the incenter generally does not lie on the Euler line;^[3] it is on the Euler line only for isosceles triangles,^[4] for which the Euler line coincides with the symmetry axis of the triangle and contains all triangle centers.

The

Let ${\displaystyle ABC}$ be a triangle. A proof of the fact that the circumcenter ${\displaystyle O}$, the centroid ${\displaystyle G}$ and the orthocenter ${\displaystyle H}$ are collinear relies on free vectors. We start by stating the prerequisites. First, ${\displaystyle G}$ satisfies the relation

${\displaystyle {\vec {GA}}+{\vec {GB}}+{\vec {GC}}=0.}$

This follows from the fact that the absolute barycentric coordinates of ${\displaystyle G}$ are ${\displaystyle {\frac {1}{3}}:{\frac {1}{3}}:{\frac {1}{3}}}$. Further, the problem of Sylvester^[7] reads as

${\displaystyle {\vec {OH}}={\vec {OA}}+{\vec {OB}}+{\vec {OC}}.}$

Now, using the vector addition, we deduce that

${\displaystyle {\vec {GO}}={\vec {GA}}+{\vec {AO}}\,{\mbox{(in triangle }}AGO{\mbox{)}},\,{\vec {GO}}={\vec {GB}}+{\vec {BO}}\,{\mbox{(in triangle }}BGO{\mbox{)}},\,{\vec {GO}}={\vec {GC}}+{\vec {CO}}\,{\mbox{(in triangle }}CGO{\mbox{)}}.}$

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

${\displaystyle 3\cdot {\vec {GO}}=\left(\sum \limits _{\scriptstyle {\rm {cyc}}}{\vec {GA}}\right)+\left(\sum \limits _{\scriptstyle {\rm {cyc}}}{\vec {AO}}\right)=0-\left(\sum \limits _{\scriptstyle {\rm {cyc}}}{\vec {OA}}\right)=-{\vec {OH}}.}$

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

In Dörrie's book,^[7] the Euler line and the 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:^[6]: p.102

${\displaystyle GH=2GO;}$

${\displaystyle OH=3GO.}$

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:^[1]

${\displaystyle ON=NH,\quad OG=2\cdot GN,\quad NH=3GN.}$

Thus the Euler line could be repositioned on a number line with the circumcenter O at the location 0, the centroid G at 2t, the nine-point center at 3t, and the orthocenter H at 6t for some scale factor t.

Furthermore, the squared distance between the centroid and the circumcenter along the Euler line is less than the squared

In addition,[6]^: p.102

${\displaystyle OH^{2}=9R^{2}-(a^{2}+b^{2}+c^{2});}$

${\displaystyle GH^{2}=4R^{2}-{\tfrac {4}{9}}(a^{2}+b^{2}+c^{2}).}$

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

${\displaystyle \sin(2A)\sin(B-C)x+\sin(2B)\sin(C-A)y+\sin(2C)\sin(A-B)z=0.}$

An equation for the Euler line in

${\displaystyle \alpha :\beta :\gamma }$

${\displaystyle (\tan C-\tan B)\alpha +(\tan A-\tan C)\beta +(\tan B-\tan A)\gamma =0.}$

Parametric representation

Another way to represent the Euler line is in terms of a parameter t. Starting with the circumcenter (with trilinear coordinates ${\displaystyle \cos A:\cos B:\cos C}$) and the orthocenter (with trilinears ${\displaystyle \sec A:\sec B:\sec C=\cos B\cos C:\cos C\cos A:\cos A\cos B),}$ every point on the Euler line, except the orthocenter, is given by the trilinear coordinates

${\displaystyle \cos A+t\cos B\cos C:\cos B+t\cos C\cos A:\cos C+t\cos A\cos B}$

formed as a linear combination of the trilinears of these two points, for some t.

For example:

• The
  circumcenter
  has trilinears ${\displaystyle \cos A:\cos B:\cos C,}$ corresponding to the parameter value ${\displaystyle t=0.}$
• The centroid has trilinears ${\displaystyle \cos A+\cos B\cos C:\cos B+\cos C\cos A:\cos C+\cos A\cos B,}$ corresponding to the parameter value ${\displaystyle t=1.}$
• The nine-point center has trilinears ${\displaystyle \cos A+2\cos B\cos C:\cos B+2\cos C\cos A:\cos C+2\cos A\cos B,}$ corresponding to the parameter value ${\displaystyle t=2.}$
• The de Longchamps point has trilinears ${\displaystyle \cos A-\cos B\cos C:\cos B-\cos C\cos A:\cos C-\cos A\cos B,}$ corresponding to the parameter value ${\displaystyle t=-1.}$

Slope

In a Cartesian coordinate system, denote the slopes of the sides of a triangle as ${\displaystyle m_{1},}$ ${\displaystyle m_{2},}$ and ${\displaystyle m_{3},}$ and denote the slope of its Euler line as ${\displaystyle m_{E}}$. Then these slopes are related according to^{[9]: Lemma 1}

${\displaystyle m_{1}m_{2}+m_{1}m_{3}+m_{1}m_{E}+m_{2}m_{3}+m_{2}m_{E}+m_{3}m_{E}}$

${\displaystyle +3m_{1}m_{2}m_{3}m_{E}+3=0.}$

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

${\displaystyle m_{E}=-{\frac {m_{1}m_{2}+m_{1}m_{3}+m_{2}m_{3}+3}{m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}}}.}$

Moreover, the Euler line is parallel to an acute triangle's side BC if and only if^[9]: p.173 ${\displaystyle \tan B\tan C=3.}$

Relation to inscribed equilateral triangles

The locus of the centroids of equilateral triangles inscribed in a given triangle is formed by two lines perpendicular to the given triangle's Euler line.^{[10]: Coro. 4}

In special triangles

Right triangle

In a

The Euler line of an automedian triangle (one whose medians are in the same proportions, though in the opposite order, as the sides) is perpendicular to one of the medians.^[11]

Systems of triangles with concurrent Euler lines

Consider a triangle ABC with

A

Simplicial polytope

A simplicial polytope is a polytope whose facets are all 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.^[14]

Suppose that ${\displaystyle P}$ is a polygon. The Euler line ${\displaystyle E}$ is sensitive to the symmetries of ${\displaystyle P}$ in the following ways:

1. If ${\displaystyle P}$ has a line of reflection symmetry ${\displaystyle L}$, then ${\displaystyle E}$ is either ${\displaystyle L}$ or a point on ${\displaystyle L}$.

2. If ${\displaystyle P}$ has a center of rotational symmetry ${\displaystyle C}$, then ${\displaystyle E=C}$.

3. If all but one of the sides of ${\displaystyle P}$ have equal length, then ${\displaystyle E}$ is orthogonal to the last side.

Related constructions

A triangle's Kiepert parabola is the unique parabola that is tangent to the sides (two of them