# The incenter and Euler line.

It seems well known that the incenter of a triangle lies on the the Euler line if and only if the triangle is isosceles (or equilateral, but that is trivial). Searching the internet, I could not find a simple geometric proof of this fact. Can anyone provide such a proof? Also, when the incenter lies on the Euler line, does it do so in a set location? (For example, we know the centroid is a third of the way from the circumcenter to the orthocenter on the Euler line, does the incenter satisfy any nice ratios like that?)

• Have you tried various extreme forms of triangle to see what the possibilities are. It depends which relationships you are studying, but what would you have to do, do you think, to get an extreme position of the incentre in relation to the other centres? I think there are plenty of resources out there which would enable you to conduct some investigations (one extreme is the equilateral case, where almost any obvious centre is the same). Note that even in this "trivial" case there are centres outside the triangle which give circles tangent to the three sides. Jan 8 '12 at 21:58

Here's a nice proof by contradiction.

Let the incenter $I$ lie on the Euler line of $ABC$.

It is known that orthocenter $H$ and circumcenter $O$ are isogonal conjugates, i.e. $AI$ is the bisector of angle $HAO$.

So, (if the point $A$ does not lie on Euler line) $HA/AO=HI/IO$ (angle bisector theorem). Also $HB/BO=HC/CO=HI/IO=HA/AO$. And we know that all points $X$, such that $YX/ZX=const$, lie on a circle with center on the line $YZ$ (Appolonius circle)

So, $A, B, C$ and $I$ lies on the same circle, and that cannot be true. We have assumed that all of points $A, B, C$ don't lie on the Euler line, so, one of them lies on Euler line and that means $ABC$ is isosceles.

• Wow, it's been a while since I've thought about this problem. Thank you for supplying this wonderful proof, it is exactly what I was looking for.
– EuYu
Sep 6 '13 at 17:15

One approach could be to use trilinear coordinates and show that the incentre at $1:1:1$ is usually not collinear with for example the circumcentre at $\cos A :\cos B :\cos C$ and the orthocentre at $\sec A :\sec B :\sec C$ by looking at the determinate

$$\begin{vmatrix}1&1&1\\ \cos A &\cos B &\cos C\\ \sec A &\sec B &\sec C\end{vmatrix}$$

which is non-zero unless at least two of $A$, $B$ and $C$ are equal.

• Thank you for the answer. I am primarily interested in a geometrical proof without appealing to coordinates or analytic geometry.
– EuYu
Jan 9 '12 at 2:40

The Incenter of the ABC triangle of sides a, b and c is at a distance d from the Euler line given by the formula:

$$d=\frac{1}{2}\frac{|(a-b)(a-c)(b-c)|}{\sqrt{(abc)^2-(-a^2+b^2+c^2)(a^2-b^2+c^2)((a^2+b^2-c^2)}}$$

If the distance is equal to zero the Incenter is on the Euler line. The formula results in zero in the case of the isosceles triangle (a-b = 0 or a-c = 0 or b-c = 0) and an equilateral triangle (a = b = c).

• I just wrote a proof which is very close to yours but starts from the barycentric equation of the Euler line. Sep 19 '20 at 15:47

I would like to present the result of W. Massaro in a slightly different manner.

It is known (see for example p. 82 of his document](http://www.journal-1.eu/2017/Grozdev-Okumura-Dekov-Euler-Line-pp.81-85.pdf)) that the Euler line of a triangle with sidelengths $$a,b,c$$ has barycentric equation:

$$(b^2-c^2)(b^2+c^2-a^2)x+(c^2-a^2)(c^2+a^2-b^2)y+(a^2-b^2)(a^2+b^2-c^2)z=0\tag{1}$$

As the barycentric coordinates of the incenter are $$(x,y,z)=(a,b,c)$$, plugging them into (1) results in an expression which, factorized (using a Computer Algebra System !), becomes:

$$(a-b)(b-c)(c-a)(a+b+c)^2$$

Nothing astonishing that we find back the expression in the numerator of the result of W. Massaro.