What is the geometrical importance of the Euler Line (ie, the line through the centroid, orthocenter, and circumcenter (and other points) of a non-equilateral triangle)?

What is meant by importance is that ---

  • What role it plays in mathematics...
  • Why is the Euler Line worthy of attention
  • Why is it valuable in geometry
  • Why should some more special points also lie on this magical Euler line – the Exeter point, the nine-point centre, and the Schiffler point

For example: $(a+b)^2$

Though it is very small compared to the vast field of mathematics, we all know how important it is and it would be a nightmare without its existence.

I hope that I was able to make clear what I wanted to ask. If there is any confusion please ask😊.
Anyone who sees this question, feel free to share what you think about this.
And upvote it if you are curious about it too but don't know the answer.

  • $\begingroup$ Personally I like your question but cannot really answer, I am curious to see how the guys here will receive it :D $\endgroup$
    – Thomas
    Jul 23 at 10:59
  • $\begingroup$ I will use a comparison. Imagine you are cycling in the country without map ; you are at a T junction and don't know if you have better go to the left or the right, you randomly take one of the orientations and after a while cross a fellow cyclist and ask him : is it interesting going there ? He answers you : "well if you are heading in this direction, it will be pleasant, its a nice flat road with tree shade, nice vista, etc... but don't go too far because finally it's a dead end... [to be continued] $\endgroup$
    – Jean Marie
    Jul 23 at 12:59
  • $\begingroup$ [Ctd] Indeed, Euler line which is present for example in Olympiad problems has no connection with "great mathematics" but it has pleasant properties to which some hours can be devoted for the sake of mathematical training... $\endgroup$
    – Jean Marie
    Jul 23 at 13:04
  • 1
    $\begingroup$ @JeanMarie: My comment was to address "Why should some more special points also lie on this black-magical Euler line?" The trilinear/barycentric coordinate relations neatly characterize points on the line. Any "deep" meaning in that characterization happens to elude me, but that could be a failure of imagination on my part. ... Anyway, as for "importance", I have no contribution; I can't think of a situation where "because $P$ is on the Euler line!" was the linch-pin revelation in a proof of some grand (or even not-so-grand) theorem, but, then again, my knowledge is far from encyclopedic. $\endgroup$
    – Blue
    Jul 24 at 9:08
  • 1
    $\begingroup$ @Blue: thanks for your comment. I am on the same line (intended pun). $\endgroup$
    – Jean Marie
    Jul 25 at 8:49

This answer was started as a comment, but soon there was no place in that margin...

A good place to go to is: ETC

ETC is a short cut for the longer "Encyclopedia of Triangle Centers". Points that are "important" for a general triangle are listed in some humanly chosen order.

In this long list, there is a first comment on the Euler line, that i am copy+pasting here for the convenience of the reader:

Introduced on March 21, 2015: Shinagawa coefficients for triangle centers on the Euler line

Suppose that $X$ is a triangle center given by barycentric coordinates $f(a,b,c)$ : $f(b,c,a)$ : $f(c,a,b)$. The Shinagawa coefficients of $X$ are the functions $G(a,b,c)$ and $H(a,b,c)$ such that $$ f(a,b,c) = G(a,b,c)\;S^2 + H(a,b,c)\;S_B\; S_C\ . $$ (Notations are as in loc. cit., for instance $S_B$, $S_C$ are the "Conway symbols", $S_B=\frac 12(a^2+c^2-b^2)=ac\cos B$, $S_C=\frac 12(a^2+b^2-c^2)=ab\cos C$. And $S$ is twice the area of $\Delta ABC$.)

For many choices of $X$, $G(a,b,c)$ and $H(a,b,c)$ are conveniently expressed in terms of the following:

$$ \begin{aligned} E &= (S_B + S_C)(S_C + S_A)(S_A + S_B)/S^2\ , &&\text{ so that} \\ E &= (abc/S)^2 = 4R^2\ , \\[4mm] F &= S_A S_B S_C/S^2\ , &&\text{ so that} \\ F &= \frac 12(a^2 + b^2 + c^2) - 4R^2 = S_\omega - 4R^2\ . \end{aligned} $$ Examples:

  • $X(2)$ has Shinagawa coefficients $(1, 0)$; i.e., $X(2) = 1\cdot S^2 + 0\cdot S_BS_C$.
  • $X(3)$ has Shinagawa coefficients $(1, -1)$.
  • $X(4)$ has Shinagawa coefficients $(0, 1)$.
  • $X(5)$ has Shinagawa coefficients $(1, 1)$.
  • $X(23)$ has Shinagawa coefficients $(E + 4F, \ -4E - 4F)$.
  • $X(1113)$ has Shinagawa coefficients $(R - |OH|, \ -3R + |OH|)$.

Just start a search for "Euler line" in the ETC-pages, to see how often it shows. For instance:

  • The line $X(13)X(15)$ is parallel to the Euler line.
  • The line $X(14)X(16)$ is parallel to the Euler line.
  • Centers $2-5$, $20-30$ (a lot of important points), $140$, $186$, $199$, $235$, $237$, $297$, $376-379$, $381-384$, $401-475$ (a lot of points), $546-550$, $631-632$ (and others) lie on the Euler line.
  • and there are many other points defined using the Euler line, or satisfying properties related to the Euler line.

See also §14 in Pamfilos' book on Barycentrics for some related thoughts in the world of trilinear and barycentric coordinates...

LATER EDIT: The comments to this answer in the form containing only the previous lines, show that the question wants an answer touching as much as possible a geometrical, rather basic level, and that results will be collected and inserted in some work. This is a new perspective, and initially i considered that the above is useless seen from this less geometrical, but more algebraic point of view, where "coincidences" (in projective geometry like colinearity, points being on a circle, on an ellipse, etc.) are shown based on computations. However, there may be a good idea to combine such properties, sometimes show some of them using purely geometric arguments, sometimes pass to the algebraic world when this gives the better insight.

An excellent introduction to barycentric coordinates is provided by:

Barycentric coordinates for the impatient, by Max Schindler and Evan Chen

Here is a quick introduction after a long motivating story. The story comes first, but one may want to skip it...

Consider a triangle $\Delta ABC$ with sides (of lengths) $a,b,c$. Many "centers" of the triangle (like mass center, incenter, circumcenter, orthocenter...) are defined in geometric terms. In the school, i first knew four such centers, and ignored some other ones (like the symmedian point), considering them to be "too artificial". A lot of beautiful interesting properties were connecting these points. However, while preparing for Olympiads, i saw there is a good idea to also "know" "the other" (very few) points mentioned in the one or the other article. At the age of $18$ i thought i knew everything about points related to a general triangle. After years, i discovered the "ETC", and saw how many points are still interesting and share beautiful, unexpected properties. One can stop after the first $1000$ points listed in the ETC as $X(1)$, $X(2)$, ... , $X(1000)$, but from time to time, some other point may cross the interest...

To see what kind of properties we may get and prove, here is an older problem (with a solution of mine, but please, please read now only the problem, since it is destroyed by the given solution):

The points $I=X(1)$, $H=X(500)$, and $P=X(501)$ satisfy $HI=HP$.

As given, it involves the three points denoted at loc. cit. by $I,H,P$, and the problem is stated in only three lines. How to solve it? The fine geometer who was posting it, communnites, says:

I have been thinking about this geometry test question for a long time, but I haven't made it yet.

Usually, trying to solve such a problem, one wants really to "know more" about the involved points. Of course, $I=X(1)$ is a well known protagonist, it has its own Hollywood star in ETC, everybody steps in and sees it. What about the two other points? As they are defined, in loc. cit., there is a symmetry in their construction w.r.t. permutations of $A,B,C$, so they are in this sense "centers" of the given $\Delta ABC$. Which "centers"? Do they occur in some other context? Do we know some other properties? (So that we may have the chance to use them and get a proof...) To "identify" them, the following procedure of "human ignorance" may be applied. We build the triangle with sides $6,9,13$, (for $a,b,c$) and we compute numerically for this one special triangle the points $H,P$. (Yes, numerically, sometimes the geometer is close to black despair...) Then we lookup for a match in the table

Search 6, 9, 13

and this is the way i could identify the two points from the problem as $H=X(500)$ and $P=X(501)$. Then i had immediately a list of all their known properties! For instance:


  • $X(500)$ lies on these lines: $\color{red}{1,30}$ and $3,6$ and $651,943$
  • $X(500)$ = inverse-in-Brocard-circle of $X(582)$
  • $X(500)$ = crosspoint of $X(1)$ and $X(35)$
  • $X(500)$ = crosssum of $X(1)$ and $X(79)$
  • $X(500)$ = $X(1)$-Ceva conjugate of $X(942)$

I know, again involved properties are listed, but one property was striking for me at the time is saw it, and it is also related to the present question. Which is the line $\color{red}{X(1)X(30)}$? It is the parallel through $X(1)$ to the Euler line! (Since $X(30)$ is the point at infinity of the Euler line, in other words the Euler line and the line $IH=X(1)X(500)$ intersect at infinity, so they are parallel.)

This was the long introduction, thought as a short motivation. I provided in my solution to the problem "$HI=HP$" many pictures, and the "less interesting" point $2070$ captured my attention. All the above shows why it may be interesting to study the "geometry in a triangle" from this point of view.

Now regarding the calculus with barycentric coordinates i will say some words, provide a quick introduction, and make explicit simple computations to get (not so simple) properties of the main points in a triangle.

What are barycentric coordinates related to a triangle $\Delta ABC$? Let us fix once for all times this triangles. It has sides $a,b,c$, say. Then the angles of $\Delta ABC$ have some values, when applying some special explicit expression involving trigonometric functions on them, and we will compute proportions, for instance:

$$\sin A:\sin B:\sin C = a:b:c\ . $$ (The sine theorem in $\Delta ABC$.) Or: $$\cos A:\cos B:\cos C = \frac{b^2+c^2-a^2}{2bc}\ :\ \frac{c^2+a^2-b^2}{2ca}\ :\ \frac{a^2+b^2-c^2}{2ab}\ . $$ (The cosine theorem in the triangle $\Delta ABC$.)

Many points have than formulas that involve such expressions in a symmetric way. How? Let us consider the "simplest center" when it comes to linearly destruct the geometry and use vectors, the mass center, the centroid $G=X(2)$. Let $X$ be some point (well "any" point, but we fix it) in the plane of the triangle, then in a vectorial notation: $$ XG = \frac 13(XA+XB+XC)\ . $$ (Here, $XG$ is a vector, as there are also $XA$, $XB$, $XC$.) Because replacing $X$ by some other point $Y$ leads to the "same" relation, $$ YG = \frac 13(YA+YB+YC)\ , $$ (because $YG=YX+XG$, and the sum of the "weights" on the R.H.S. is also one,) we can and do write instead $$ G=\frac 13(A+B+C)=\frac 13A+\frac 13B+\frac 13 C\ . $$ (If vectors are too complicated, an alternative way of seeing this is maybe simpler, write $A=(a_1,a_2)$ in cartesian coordinates, and the same for the other points. Then we have $(g_1,g_2)=\frac 13(a_1,a_2)+\frac 13(b_1,b_2)+\frac13(c_1,c_2)$, with "operations on components".) And from this writing we isolate the barycentric coordinates of $G$, they are $\left(\frac 13,\frac 13,\frac 13\right)$. But this is too complicated to write down, it is enough to mention the "proportionalities", and we write $$G=X(2)=[1:1:1]\ .$$

Let us do "the same" and find the coordinates for the incenter $I=X(1)$. Let $A',B',C'$ be on opposite sides w.r.t. $A,B,C$, so that $AA'$, $BB'$, $CC'$ are the angle bisectors in the given triangle. Then $BA':A'C=BA:AC=c:b$ becomes vectorially $$ AA' =\frac b{b+c} AB+\frac c{b+c}AC\ , $$ so "formally" we can and do write: $$ \tag{$*$} A'=\frac b{b+c} B+\frac c{b+c}C\ . $$ Let $I=AA'\cap BB'\cap CC'$ be now the incenter. Then we have some relation of the shape $$ I = xA+yB+zC\ ,\qquad x+y+z=1\ , $$ and we need and want to get the components $x,y,z$. We "test" the above in $A'$, and get $$ A'I=xA'A+yA'B+zA'C\ . $$ But the vector $A'I-xA'A$ is a vector in direction supported on the line $A'IA$, and $yA'B+zA'C$ is a vector supported on the line $BA'C$, so both vectors vanisch, so we have to take $y,z$ according to $(*)$ above i.e. $y:z = b/(b+c):c/(b+c)=b:c$. By symmetry, $x:y:z=a:b:c$, so we obtain $$ \begin{aligned} I=X(1) &=[a:b:c] \\ &=\left[\frac a{a+b+c}:\frac b{a+b+c}:\frac c{a+b+c}\right] \\ &=\frac a{a+b+c}A+\frac b{a+b+c}B+\frac c{a+b+c}C\ . \end{aligned} $$ The coefficients above are the (normed) barycentric coordinates of $I=X(1)$.

As seen, for $I$ it is enough to know where are the cevians $AI$, $BI$, $CI$ hitting the opposite sides, then patch together the three proportions to one fitting them together.

In a similar way, one has for the circumcenter $O=X(3)$ the barycentric coordinates $$ O=[\sin 2A:\sin 2B:\sin 2C]\ . $$ Why? We draw $\Delta ABC$, the circle $(ABC)$ centered in $O$ with radius $R$, take the intersection $A'=AO\cap BC$, and consider the triangles $\Delta OA'B$ and $\Delta OA'C$. Then: $$ \frac {BA'}{A'C} = \frac {BA'}{R} \cdot \frac {R}{A'C} = \frac {BA'}{BO} \cdot \frac {CO}{A'C} = \frac {\sin\widehat{BOA'}}{\sin\widehat{BA'O}} \cdot \frac {\sin\widehat{CA'O}}{\sin\widehat{COA'}} = \frac {\sin\widehat{BOA'}} {\sin\widehat{COA'}} = \frac {\sin(\pi-2C)} {\sin(\pi-2B)} = \frac {\sin(2C)} {\sin(2B)} \ . $$ Because of this, in a writing $O=X(3)=[x:y:z]$ we have $y:z=\sin 2B:\sin 2C$, so by symmetry $x:y:z=\sin 2A:\sin 2B:\sin 2C$.

Let us also get the barycentric coordinates of the orthocenter $H=X(4)$. Let $AA'$, $BB'$, $CC'$ be the heights in $\Delta ABC$, again we ask where is $A'$ positioned on $BC$, and of course $$ \frac{BA'}{A'C} = \frac{BA'}{AA'} \cdot \frac{AA'}{A'C} = \frac{\cot B}{\cot C}= \frac{\tan C}{\tan B}\ , $$ so in a writing $H=[x:y:z]$ we have $y:z=\tan B:\tan C$, so by symmetry $$ H=[\tan A:\tan B:\tan C]\ . $$

Let us now algebraically show that the points $G,O,H$ are on a line. Namely on the line of all $[x:y:z]$ with equation: $$ 0 = \begin{vmatrix} x & y& z\\ 1 & 1 & 1\\ \tan A &\tan B&\tan C\\ \end{vmatrix} := x(\tan C-\tan B)+y(\tan A-\tan C)+z(\tan B-\tan A) \ . $$ (You do not need for this the full theory of determinants, only the definition of $3\times 3$ determinants.)

This relation is immediately true for (the coordinates of) $G,H$ because two rows in the determinant coincide in each case. To see that also $G$ is on that line amounts to showing the relation between the coordinates of $G,O,H$: $$0 =\begin{vmatrix} 1 & 1 & 1\\ \sin 2A &\sin 2B &\sin 2C\\ \tan A &\tan B &\tan 2C\\ \end{vmatrix} \ . $$ (Left as an exercise.)

I know, the above "analytic" method to proceed is not illuminating from a geometric point of view, but it makes for instance easy to check if a specific point (with some given geometrical property) is on the Euler line or not. I will stop here, since the story goes too far (away from the framework of the OP) already.

  • $\begingroup$ I appreciate your effort and work as it has given me some idea about what I was seeking. But actually, I study in a lower grade and what you have written is going over my head. $\endgroup$ Jul 30 at 6:50
  • $\begingroup$ @dan_fulea-- Also, these maths are not at my level. So I am having difficulty understanding the calculations. So, can you please write some statements or concluding lines that I can comprehend. The main motive of my work is to explore the topic and not just copy something I don't understand. I will be grateful to you if you can write it without using the proofs and coordinate geometry so that I can understand and use it to write something meaningful. $\endgroup$ Jul 30 at 6:55
  • $\begingroup$ Also, can you explain what you mean by barycentric coordinates. Hope to see a positive reply. $\endgroup$ Jul 30 at 7:01

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