# Euler line

Euler's line (red) is a straight line through the centroid (orange), orthocenter (blue), circumcenter (green) and center of the nine-point circle (red).

In geometry, the Euler line, named after Leonhard Euler (US , OY-lər or UK , OY-lə), is a line determined from any triangle that is not equilateral. 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.[1]

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

## 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

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

## Parametric representation

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

$\cos A+t\cos B\cos C:\cos B+t\cos C\cos A:\cos C+t\cos A\cos B\,$

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

For example:

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

## Slope

In a Cartesian coordinate system, denote the slopes of the sides of a triangle as $m_{1},$ $m_{2},$ and $m_{3},$ and denote the slope of its Euler line as $m_{E}$. Then these slopes are related according to[4]:Lemma 1

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

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

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

Moreover, the Euler line is parallel to an acute triangle's side BC if and only if[4]:p.173 $\tan B\tan C=3.$

## Lengths of segments

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:[5]:p.102

$GH=2GO;$
$OH=3GO.$

The center of the nine-point circle lies along the Euler line midway between the orthocenter and the circumcenter.[1] 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 circumradius R2 by an amount equal to one-ninth the sum of the squares of the side lengths a, b, and c:[5]:p.71

$GO^{2}=R^{2}-{\tfrac {1}{9}}(a^{2}+b^{2}+c^{2}).$

## Right triangle

In a right triangle, the Euler line contains the median on the hypotenuse—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 altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse.