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.

Blue lines are altitudes, orange medians, green perpendicular bisectors, and the red line is the Euler line

In the diagram above, the points
- $H$ represents the orthocenter,
- $N$ represents the nine-point center,
- $G$ represents the centroid,
- $O$ represents the circumcenter.

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.

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

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

Relative distances of triangle centers. The red line is twice as long as the green line, and the blue line is three times as long as the green line.

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,

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

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 $R$?

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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 $9\pi$. $_\square$

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.

• Circumcenter
• Orthocenter
• Centroid
• Nine point circle