# Euler Line

The **Euler line** of a triangle is a line going through several important triangle centers, including the orthocenter, circumcenter, centroid, and center of the nine point circle. The fact that such a line exists for *all* non-equilateral triangles is quite unexpected, made more impressive by the fact that the relative distances between the triangle centers remain constant.

In the diagram above, the points

- $H$ represents the orthocenter,

- $N$ represents the nine-point center,

- $G$ represents the centroid,

- $O$ represents the circumcenter.

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## Proof of Existence

The simplest proof makes use of homothety, specifically the one with center at the centroid $G$ of the triangle, with scale factor $-\frac{1}{2}$. This sends vertices $A, B, C$ to the midpoints of the opposite sides, since the centroid divides the medians in a 2:1 ratio, meaning that triangle $ABC$ is sent to the medial triangle.

Therefore, the orthocenter of triangle $ABC$ is sent to the orthocenter of the medial triangle, which is the circumcenter of $ABC$. As a result, the orthocenter, circumcenter, and centroid are all collinear, as desired.

## Properties

The proof above shows more than collinearity: since $H$ is sent to $O$ through a scale factor of $-\frac{1}{2}$, we have $HG=2GO$. In fact, more is true: if $N$ is the nine-point center of the triangle, we have

$\begin{aligned}&HG = 2GO, &ON = NH, &OG = 2GN, &NH = 3GN.\end{aligned}$

The Euler line also contains dozens of other triangle centers, some of which have only been discovered recently. Of special note are the following points:

- The
**De Longchamps point**the reflection of the orthocenter over the circumcenter. - The
**Exeter point**defined as follows: if $DEF$ is the tangential triangle of $ABC$, and $A', B', C'$ are the intersections of the medians of $ABC$ with the circumcircle of $ABC$, then $DA', EB', FC'$ concur at the Exeter point. - The
**Schiffler point**: if $I$ is the incenter of $ABC$, then the Euler lines of $\triangle ABI, \triangle BCI, \triangle CAI$, and $\triangle ABC$ concur at the Schiffler point.

Additionally, if $A_1$ is the foot of the altitude from $A$ to $BC$ and $M_a$ is the midpoint of $BC$ (with $B_1$ and $M_b$ defined analogously), then

The intersection of $A_1M_b$ and $B_1M_a$ lies on the Euler line.

The slope of the Euler line relates to the slope of the sides in a nice way: If $m_1, m_2, m_3$ are the slopes of the three sides of a triangle $ABC$, and $m_e$ is the slope of the Euler line, then

$m_1m_2+m_2m_3+m_3m_1+m_1m_e+m_2m_e+m_3m_e+3m_1m_2m_3m_e+3=0,$

or equivalently,

$m_e=-\frac{m_1m_2+m_2m_3+m_3m_1+3}{m_1+m_2+m_3+3m_1m_2m_3}.$

## Applications

One important consequence of the Euler line is that information about any one of the centroid, orthocenter, and circumcenter can be derived from information about the other two. For example,

## Points $A, B,$ and $C$ are randomly chosen on a unit circle. The locus of the orthocenter of $\triangle ABC$ is a region $R$. What is the area of $R$?

Instead of focusing on the orthocenter, it helps to focus on the other two major triangle centers: the centroid and the circumcenter. The circumcenter is always the center of the unit circle, so it is only necessary to note that the centroid can lie anywhere within the unit circle, and nowhere else (why?). Since $HG=2GO$, this implies that the maximum possible value of $H$ is 3, and the region $R$ is a circle of radius 3. The area of $R$ is thus $9\pi$. $_\square$

Consider a triangle$\Delta ABC$ whose circumcenter is at the origin. If in $\Delta ABC$, the coordinates of the centroid $G$ are $\left(x_G,y_G\right)$ and the coordinates of the orthocenter $H$ are $\left(x_H,y_H\right)$.

Find the ratio $\dfrac{x_Hy_H}{x_Gy_G}$.

**Bonus**: Don't forget the lonely incenter!

Another result due to the properties of the Euler line:

If $ABC$ is a triangle with circumcenter $O$ and orthocenter $H$, then one of the areas of $\triangle AOH$, $\triangle BOH$, and $\triangle COH$ is the sum of the areas of the other two.

Mohit, influenced by Akshat and Anshuman, took a right $\bigtriangleup ABC$ (right angled at $B$), where $AB=5.5 \mathrm{ cm}$ and $BC=8 \mathrm{ cm}$, and started doing aimless constructions, the steps of which are given below:

($1$) He drew perpendicular bisector $X_1Y_1$ of $AC$ which intersects $AC$ at a point $O,$ and another perpendicular bisector $X_2Y_2$ of $AB$ which intersects $AB$ at a point $G$.

($2$) Then he constructed a circle taking center $O$ and radius of the circle as $OA$ . Then he constructed an $\angle ACE$ which is equal to $\angle ACB$ such that $E$ lies on the circle.

($3$) He then joined $BE$ which meets $AC$ at $F$.

($4$) Then he joined $CG$ which intersects $EB$ at $H$ and $X_1Y_1$ at $J$.

Mohit then wondered what $\frac{BH}{OJ}$ could possibly be equal to.